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# Review of 'Student Evaluations of Teaching (Mostly) Do Not Measure Teaching Effectiveness'

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### Student Evaluations of Teaching (Mostly) Do Not Measure Teaching Effectiveness

(2016)
Student evaluations of teaching (SET) are widely used in academic personnel decisions as a measure of teaching effectiveness. We show: SET are biased against female instructors by an amount that is large and statistically significant the bias affects how students rate even putatively objective aspects of teaching, such as how promptly assignments are graded the bias varies by discipline and by student gender, among other things it is not possible to adjust for the bias, because it depends on so many factors SET are more sensitive to students' gender bias and grade expectations than they are to teaching effectiveness gender biases can be large enough to cause more effective instructors to get lower SET than less effective instructors. These findings are based on nonparametric statistical tests applied to two datasets: 23,001 SET of 379 instructors by 4,423 students in six mandatory first-year courses in a five-year natural experiment at a French university, and 43 SET for four sections of an online course in a randomized, controlled, blind experiment at a US university.
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### Review information

10.14293/S2199-1006.1.SOR-EDU.AETBZC.v1.RUIBVH

This work has been published open access under Creative Commons Attribution License CC BY 4.0, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Conditions, terms of use and publishing policy can be found at www.scienceopen.com.

 Keywords: nonparametric statistics, disparate impact, gender bias, permutation tests

### Review text

This article presents findings from two studies on the effectiveness of student evaluations of teaching as a measure of teaching effectiveness. First study is from a five-year natural experiment at a French university (23,011 evaluations of 379 instructors by 4,423 students in six mandatory first-year courses) and the second study is from a randomized, controlled, blind experiment at a US university (43 evaluations for four sections of an online course). The methodology uses nonparametric statistical tests, which is presented as an improvement over previous analyses of the French dataset.

Overall the designs of the two studies presented are interesting and the findings are eye-opening. Below are my comments on mostly the methodology used in the analysis and some of the assumptions made.

• Comparison to previous work

The methods section of the article references a previous analysis of the data where the tests assumed that SET of male and female instructors are independent random samples from normally distributed populations with equal variances and possibly different means". I assume the authors are referring to MacNell et al. [2014]. This article states the following:

We used Welch’s t-tests (an adaptation of the Student’s t-test that does not assume equal variance) to establish the statistical significance of each difference. We also ran two general linear multivariate analyses of variance (MANOVAs) on the set of 12 variables to test the effects of instructor gender (perceived and actual) on all of the questions considered as a group.

While some of the analyses used in the previous study required normally distribution of data and equal variances, others didn't. The text in this article makes it seem like all previous analyses used methods requiring these conditions. It would be best to clarify exactly where improvements to methods were introduced and present evidence that the data did not meet these conditions.

Another issue is that the authors do not compare their findings to those from MacNell et al. [2014]. It should be explicitly stated in the paper whether the findings agreed or disagreed.

• Assumptions on relationship between teaching effectiveness and student performance

Section 3.1 on page 9 states the following:

To the extent that final exams are designed well, scores on these exams reflect relevant learning outcomes for the course. Hence, in each course each semester, students of more effective instructors should do better on the final, on average, than students of less effective instructors.

This would be definitely true if the material covered in each section is exactly the same. It is not clear from the article whether this was the case  or how this was evaluated. It is possible that students of a very effective teacher score poorly on an exam if the exam assesses material that was not emphasized in the course.

• Controls for non-independence

In the French dataset there are 23,001 student evaluations of 379 instructors. We are told that instructors see the students in groups of 10-24 students. It seems like the same instructor must appear in the dataset multiple times. The analyses do not appear to control for this non-independence structure in the data.

It also seems like students might be repeated in the dataset as well since each student has to take these required courses. There does not appear to be any control for this either in the analysis.

• Multivariate analysis

All analyses presented in the article are bivariate, exploring the relationship between a given factor and the response variable of SET scores.

First, this can potentially result in problems associated with multiple testing. The authors address this issue in Section 6, however I would expect this discussion to appear earlier in the paper.

Second, it would potentially be more informative to analyze multivariate relationships instead of exclusively bivariate relationships.  A regression based approach, instead of a series of tests, would allow for this exploration, as well as conclusions that evaluate the relationship between SET and certain factors while controlling for others.

• Discussion of effect size and practical significance

The article does not discuss effect sizes and considerations around practical significance. This is especially important for the French dataset which has a large number of observations. Indeed in Section 6 the authors discuss the importance of finding small p-values with the small US dataset, however a similar discussion is not included for the French dataset.

We thank Professor Cetinkaya-Rundel for her thoughtful review.

Most of her concerns are already addressed in the paper; others we agree with in part but not entirely. For instance, Prof. Cetinkaya-Rundel asserts that

“while some of the analyses in the previous study required normally distribution (sic) of data and equal variances, others didn’t.”

She points to Welch’s two-sample t-test and MANOVA as examples. Both of these tests rely on the assumption that data are sampled independently at random from populations with normal distributions. That stochastic assumption contradicts the actual randomization in the MacNell et al. experiment and the as-if randomization in the SciencesPo natural experiment: the data arise from randomizing a fixed group of individuals into different treatments, not from sampling from larger populations at random. And the variables involved do not have normal distributions; indeed, they are dichotomous (e.g., gender), ordinal categorical (e.g., SET), or have compact support (e.g., grades). The null hypotheses for Welch’s t-test and MANOVA are simply not relevant to the problem. Asymptotically, those tests might have the correct level under the randomization actually performed, but for finite populations, the actual levels may differ considerably from the nominal levels, as Table 8 of the paper shows.

Prof. Cetinkaya-Rundel writes:

“Another issue is that the authors do not compare their findings to those from MacNell et al. [2014]. It should be explicitly stated in the paper whether the findings agreed or disagreed.”

Table 8 compares our results to those from MacNell et al. [2014] (the rightmost columns show the p-values from our permutation test side-by-side with the p-values from t-tests in MacNell et al.) and we discuss the differences near the end of page 8, in the section entitled “SET and perceived instructor gender.” Our omnibus test of the hypothesis that perceived instructor gender plays no role in SET matches the spirit of the MANOVA test (but for a more relevant null hypothesis); they do not report details of those tests, only that the p-values were less than 0.05. The permutation omnibus test we performed gives a p-value of essentially zero for the hypothesis that students rate instructors the same, regardless of the gender of the instructor’s name.

Prof. Cetinkaya-Rundel raises the concern that instructors and students appear in the data more than once, across years and across courses, producing correlation. In the Neyman model, the responses are fixed, not random: only the assignment of instructors to class sections is random. There is no assumption that responses random at all, much less independent. Moreover, instructors appear at most once per course per year—i.e., once per stratum. Because the natural experiment amounts to randomizing within each stratum, independently across strata, there is no dependence across strata, regardless of the identity of the instructors. Prof. Cetinkaya-Rundel continues:

“ A regression based approach, instead of a series of tests, would allow for this exploration, as well as conclusions that evaluate the relationship between SET and certain factors while controlling for others.”

A regression-based approach requires far stronger modeling assumptions (e.g., linearity of effects and IID additive errors), assumptions for which there is no basis here. Indeed, those assumptions seem rather far-fetched, for instance, because SET is an ordered categorical scale, not a linear scale. A model that posits that SET (or other variables) are observed with independent, identically distributed additive errors is also unjustified for these data: the randomness is in the assignment of students to instructors, not measurement noise. The Neyman model allows each individual (student or class section) to have its own response to each treatment condition, with no assumption about the relationship among responses of different individuals. Randomization in the experimental design mitigates confounding with no need to model the effect of covariates: that is the power of a randomized experiment. Moreover, the variables one would presumably want to control for are pretreatment variables, such as prior knowledge of the subject matter, while, the covariates available (e.g., final exam scores) are measured post-treatment.

Prof. Cetinkaya-Rundel also advocates reporting effect sizes. We reported empirical mean effect sizes in the tables (e.g., column 2 of Tables 3, 4, 5, 7, and 8). We intentionally did not report confidence intervals for mean effect sizes. Confidence intervals could be constructed by inverting the permutation tests, but that would require strong assumptions about the structure of the effect that have no basis here. For instance, it would suffice to assume that the effect of instructor gender on SET is the same for every student—but the data give evidence that the effect varies by student gender, at least. Indeed, one of our main conclusions is that there are such differences! Finally, from a policy perspective, effect size per se isn’t particularly interesting: the issue is disparate impact. The data show that the effect of gender on SET can overwhelm differences in actual instructor effectiveness, as measured by exam performance.

2016-03-17 11:27 UTC
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# Prove that a function is equal to a constant.

Assume that $f(x)$ is continuous on $[a,b]$. And for any continuous function $g$ if $\int_a^bg(x)dx=0$ then $\int_a^bf(x)g(x)dx=0$, show that $f(x)$ is a constant.

I tried to convert this question to show$f'(x)\equiv0$ but this seems impossible by using the mean value theorem or the Rolle theorem. Any ideas?

• What do you mean any integration? Do you mean that for any two real numbers $(a,b)$ the following holds or for any function $g(x)$ with $\int_{a}^{b}g(x)dx=0$? – Tucker Aug 11 '15 at 1:54
• @Tucker a and b are two real numbers and $g(x)$ is a random function meet $\int_a^bg(x)dx=0$. Thanks to point out – Rowan Aug 11 '15 at 2:02
• No, $g(x)$ is no "a random" function. It has to be true in general that if $g(x)$ is any function with this property, then $\int_a^b f(x)g(x)\,dx =0$ If $g(x)$ is one "random" function, whatever that is, you do not get this result. Language is really important in mathematics. – Thomas Andrews Aug 11 '15 at 2:08
• @ThomasAndrews Thanks a lot. I finished my problem by using the outline. And have you deleted your answer? I can't find it now. – Rowan Aug 11 '15 at 2:21
• @ThomasAndrews Now it appears again. It's my network error. – Rowan Aug 11 '15 at 2:28

Take $g(x)=f(x)-\dfrac{1}{b-a}\int_a^b f(x) dx$. Certainly for this choice of $g$, we have that $g$ is continuous and $\int_a^b g(x) dx=0$

We then observe that if $\int_a^b f(x)\,g(x)\,dx=0$, then

$$\int_a^b f^2(x)dx=\dfrac{1}{b-a}\left(\int_a^b f(x) dx\right)^2$$

But by the Cauchy-Scwartz Inequality

$$\left(\int_a^b f(x) dx\right)^2\le \int_a^b f^2(x)\, dx\,\,\int_a^b (1)^2\,dx\implies\,\int_a^b f^2(x)\, dx\ge \dfrac{1}{b-a}\left(\int_a^b f(x) dx\right)^2$$

with equality holding only when $f(x)$ is a constant. And that is that!

• You did it again. Nice!!! +1 ;) Do you have an account for chatting by the way? – johannesvalks Aug 11 '15 at 2:22
• @johannesvalksThanks!!! We can chat here somehow and I can give you my personal e-mail. – Mark Viola Aug 11 '15 at 2:27
• Using Cauchy-Schwarz was really innovative. +1 – Paramanand Singh Aug 11 '15 at 3:45
• @ParamanandSingh Wow! Hey, thank you!! As you might know, I had my forth surgery in the past 15 months just 6 weeks ago and am in recovery. You kind words means a lot to me!! – Mark Viola Aug 11 '15 at 3:48
• Would the cowardly down voter care to state his/her rationale? – Mark Viola Feb 22 '16 at 20:16

Outline, with some middle steps missing.

Show that if the above is true for $f$, the it is true for $f_1(x)=f(x)-C$ for $C$ any constant. Then show:

$$\int_a^b f_1(x)\,dx = 0$$

For a parrticular $C$.

So, let $g(x)=f_1(x)$ and we get tht:

$$\int_a^b f_1(x)^2 \,dx = 0$$

Therefore, show $f_1(x)=0$ for all $x$, and thus that $f(x)=C$ for all $x$.

Your idea of showing $f'(x)\equiv 0$ doesn't work because $f$ is not necessarily differentiable. However, it can be done if we assume that $f$ has a continuous derivative.

Let $G$ be a differentiable function on $[a,b]$ with $G(a)=G(b)=0$. Now $g=G'$ has $\int_a^b g(x) dx = 0$, so integrating by parts, we get $$\int_a^b f'(x) G(x) \,dx = \left[f(x) G(x) \right]_{x=a}^b - \int_a^b f(x)g(x)\,dx = 0.$$

If $f'(x)$ is nonzero at any point, then because $f'$ is continuous there is an open interval $(c,d) \subseteq [a,b]$ where $f'$ has only one sign. Without loss of generality, let's assume $f'$ is positive in $(c,d)$ (otherwise we may consider $-f'$ in the following). Let $G$ be a function which is zero outside $(c,d)$ and has positive values in $(c,d)$. (Such a function exists but giving a full definition here would be troublesome because $G$ needs to be continuous so I'll skip it.) Then $$\int_a^b f'(x) G(x) \,dx > 0,$$ which is a contradiction.

• Thank you. I spent a lot of time to understand your answer :) Something used to be vague now seems more clear. – Rowan Aug 11 '15 at 15:49

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# Raindrops are falling vertically at a speed of 120 km/h through air that is moving horizontally at a speed of 40 km/h(i.e. the wind is 40 km/h). At what angle do the drops strike the ground?

Oct 3, 2017

${18}^{\circ}$ from vertical or ${72}^{\circ}$ from horizontally.

#### Explanation:

Air is moving horizontally with 40 km/h speed
Horizontal velocity vector of Air is shown in figure.

Velocity Vector of Raindrops is vertically downward .
Let resulting Velocity of Raindrops makes angle $\theta$ from the
vertical.
Then angle
$\tan \theta = {v}_{\text{Horizontal"/v_"vericle}} = \frac{40}{120} = \frac{1}{3}$

$\theta = {\tan}^{-} 1 \left(\frac{1}{3}\right) \approx {18}^{\circ}$

Hence Raindrops makes angle ${18}^{\circ}$ from the vertical or make angle ${72}^{\circ}$ from Horizontal.


text:
## The First Problem

A car has to travel a distance $s$ on a straight road. The car has maximum acceleration $a$ and maximum deceleration $b$. It starts and ends at rest.

Show that if there is no speed limit, the time is given by

$\displaystyle \sqrt{2s\cdot \frac{(a+b)}{ab}}$.

Solution: We are going to use two pieces of information:

• the area under the velocity-time graph is the distance travelled,
• the slope of the velocity-time graph is the acceleration.

First, a rough sketch of the velocity-time graph:

The slope of the first line segment is $a$, while the slope of the second line segment is $b$. The area under the curve is given by $s$, the total distance. The base is $t_1+t_2=T$, the total time, and the height is the maximum speed.

Let $v$ be the maximum speed reached. Using the area of a triangle, where $T=t_1+t_2$ is the total time:

$\displaystyle s=\frac12 Tv \Rightarrow v=\frac{2s}{T}$.

Now using the fact that the slope of the line segments, rise/run, is equal to the accelerations, we derive:

$\displaystyle a=\frac{v}{t_1}\Rightarrow t_1=\frac{v}{a}$, and

$\displaystyle b=\frac{v}{t_2}\Rightarrow t_2=\frac{v}{b}$.

Thus

$\displaystyle T=t_1+t_2=\frac{v}{a}+\frac{v}{b}=\frac{va}{ab}+\frac{vb}{ab}=v\cdot \frac{(a+b)}{ab}$.

Now, using $v=\frac{2s}{T}$:

$\displaystyle T=\frac{2s}{T}\frac{(a+b)}{ab}$

$\displaystyle \underset{\times T}{\Rightarrow } T^2=2s\cdot \frac{(a+b)}{ab}$.

Taking square roots (and noting $T>0$) completes the proof.

## The Second Problem (1997)

A particle, moving in a straight line, accelerates uniformly from rest to a speed $v$. It continues at this constant speed for a time and then decelerates uniformly to rest, the magnitude of the deceleration being twice that of the acceleration. The distance traveled while accelerating is 6 m. The total distance traveled is 30 m and the total time taken is 6 s.

i. Draw a velocity-time graph and hence, or otherwise, find $v$.

ii. Calculate the distance travelled at constant velocity.

Solution:

i. First the velocity-time graph:

There are number of things we need to use:

• the area under the velocity-time graph is the distance traveled
• the slope of the velocity-time graph is the acceleration

The question gives the following:

• the total distance is 30 m
• the total time is 6 s
• the distance traveled from $t=0$ to $t=t_1$ is 6 m.

In the first triangle:

$\displaystyle a=\frac{v}{t_1}\Rightarrow t_1=\frac{v}{a}$.

In the second triangle:

$\displaystyle 2a=\frac{v}{t_2}\Rightarrow t_2=\frac{v}{2a}=\frac{1}{2}t_1$,

and also $t_1=2t_2$. Define Let $t:=t_2$ so that $t_1=2t$. Therefore, using the fact that the total time is 6 s:

$t_1+t_2+t_3=2t+t_2+t=6\Rightarrow t_2=6-3t$.

Now, using the fact that the area under the first triangle is 6:

$\frac{1}{2}t_1v=tv=6$,

and the total area is 30:

$6+vt_2+\frac12 tv=30$

$\Rightarrow 6+v(6-3t)+3=30$

$\Rightarrow 6+6v-3\underbrace{tv}_{=6}+3=30$

$\Rightarrow 6v=30-6+18-3=39\Rightarrow v=\frac{13}{2}$ m/s.

ii. Using $tv=6$ we have $t=\frac{12}{13}$ so that the time spent at constant velocity is:

$\displaystyle t_2=6-3t=6-\frac{36}{13}=\frac{42}{13}$,

and therefore the distance traveled is:

$s_2=\frac{13}{2}\times \frac{42}{13}=21 \text{ m}$.

Alternatively note that $s_3=\frac12 s_1=3$ and so $s_2=30-s_1-s_2=21$ m.


text:
# Doomsday differential equation

#### Trousers

1. Let c be a positive number. a differential equation of the form
dy/dx=ky^(1+c)
Determine the solution that satisfies the initial condition y(0)=y(subzero)

The solution is y(subzero)=(1)/((ckT)^(1/c))

but I can't understand how to find it. ok when i integrate

dy/y^(1+c)=kdt, I get y^(-c)/(-c)=kt+C and I don't know what do from there.

Related Calculus and Beyond Homework News on Phys.org

#### Defennder

Homework Helper
I googled the term "doomsday differential equation" and guess what? The top 2 entries were this thread itself and an older one which concerns the same DE:

EDIT: Does anyone know why it's called the doomsday equation?

#### Trousers

that doesn't help me.

#### foxjwill

Try leaving it in the form

$$y^c = f(x)$$

and then solve for C (the integration constant, not the one in the form I just mentioned).

#### arildno

Homework Helper
Gold Member
Dearly Missed
1. It is called the Doomsday equation because for any c>0, there exists a finite time T when y blows up (i.e, graphing y(t) with t as the argument axis, there would be a vertical asymptote to the graph at t=T, which the graph of y(t) will follow into infinity as t approaches T)

2. Now, for the solution:
We have that $$-\frac{1}{cy(t)^{c}}=kt+C in (')$$ where our initial condition y(0)=y_{0} now can be used to determine C:
$$-\frac{1}{cy_{0}^{c}}=C$$
Inserting in (*) and simplifying, we have:
$$\frac{1}{y(t)^{c}}=\frac{1}{y_{0}^{c}}-ckt=\frac{1-y_{0}^{c}ckt}{y_{0}^{c}}$$
Or, reciprocating, and taking the c-root on both sides:
$$y(t)=\frac{y_{0}}{(1-y_{0}^{c}ckt)^{c}}$$

Now, the blow-up instant T must fulfill the equation:
$$1-y_{0}^{c}ckT=0\to{T}=\frac{1}{cky_{0}^{c}}$$
Using the blow-up time T as our independent parameter, rather than y_0, we get from the same equation: $$y_{0}=\frac{1}{(ckT)^{\frac{1}{c}}}$$

The first result is simplest to use when you DO know the value of y_0, and wants to know WHEN the solution will blow up.

The second result is most convenient when you know when the solution is going to blow up, and wants to know what was the initial value of y, i.e, y_0.

thank you.

#### HallsofIvy

1. Let c be a positive number. a differential equation of the form
dy/dx=ky^(1+c)
Determine the solution that satisfies the initial condition y(0)=y(subzero)

The solution is y(subzero)=(1)/((ckT)^(1/c))

but I can't understand how to find it. ok when i integrate

dy/y^(1+c)=kdt, I get y^(-c)/(-c)=kt+C and I don't know what do from there.
I have no idea what you are saying. "The solution is y(subzero)=(1)/((ckT)^(1/c))" can't possibly be right- there is no y(t) in it so it can't be a solution to the differenital equation. Also, is T supposed to be t are is it a specific value of t?

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text:
# Gravity Renormalizable?

1. Jan 22, 2008

### BenTheMan

I remember marcus claiming somewhere that one of the QG researchers had claimed that Gravity was renormalizable. He had some graph of renormalization group flow or something, and it was some big result at the last loops conference.

The thing that has bugged me (and perhaps this has been asked already here) is that there are lots of effective field theories that have dimensionful coupling constants (eg Fermi theory), that are clearly effective field theories. What happens if the same treatment is applied to them? Why should gravity be special, in that respect?

Again, apologies if this has been asked before, but I know there are many here who can answer this question.

2. Jan 22, 2008

### marcus

Reuter's results have been discussed some here, but not enough. I'm glad you started a new Reuter thread.

I think Reuter's stuff (NONperturbative renormalizable) is extremely interesting and (in terms used in Percacci's review paper) could either lead to a theory that is effective or fundamental. At this point it is not clear which.

Percacci's review, an invited chapter for a book, is called Asymptotic Safety and would definitely be something to look at if you are interested. I'll get some links later.

Another thing to look at is the most recent paper by Frank Saueressig, which CHALLENGES some of Reuter's conclusions. This is especially important because Saueressig has been a junior co-author of several Reuter papers since 2001, and he now has this paper where he is the senior author where finds something different. Saueressig publishes both in string and non-string quantum gravity----that is, he does string papers and also asymptotic safety Reuter-type papers.

You should be very cautious about terminology. NONperturbative renormalizable may confuse you because you may be used to thinking only in the perturbation series context.

Reuter is NOT talking about taking the first few terms of a perturbation series.

Basically the concept was discovered by Steven Weinberg in around 1976-1979, if I remember right, and Weinberg gave up on it after he and others tried unsuccessfully, and it has almost not been heard of until Reuter's paper from around 1998.

You may also misunderstand what Percacci means by an EFFECTIVE theory, if you think of it in the perturbative context. You have to read his paper to understand. He doesnt mean what you might immediately think of, having heard the term "effective" in other QFT contexts.

Donoghue also treats gravity as an effective theory and gave an invited paper at Loops 07 which generated some interest. I am not familiar with his work. I recall either Reuter referred to Donoghue's work positively in his talk IIRC, or some of the blog commentary did. But I didn't follow up on that.

3. Jan 22, 2008

### marcus

First I will get a link to Saueressig's paper because it challenges Reuter's findings (making it of special interest). My impression is Saueressig is young and extremely smart. he is at Utrecht ('t Hooft's ITP same place as Renate Loll)
=====
Oh wait, I just ran across Percacci's review, so I will list that.
http://arxiv.org/abs/0709.3851
Asymptotic Safety
R. Percacci
To appear in "Approaches to Quantum Gravity: Towards a New Understanding of Space, Time and Matter", ed. D. Oriti, Cambridge University Press
(Submitted on 24 Sep 2007)

"Asymptotic safety is a set of conditions, based on the existence of a nontrivial fixed point for the renormalization group flow, which would make a quantum field theory consistent up to arbitrarily high energies. After introducing the basic ideas of this approach, I review the present evidence in favor of an asymptotically safe quantum field theory of gravity".
=========

I also found a recent Reuter-Saueressig collaboration

http://arxiv.org/abs/0708.1317
Functional Renormalization Group Equations, Asymptotic Safety, and Quantum Einstein Gravity
Martin Reuter, Frank Saueressig
Based on lectures given by M.R. at the 'First Quantum Geometry and Quantum Gravity School', Zakopane, Poland, March 2007, and the 'Summer School on Geometric and Topological Methods for Quantum Field Theory', Villa de Leyva, Colombia, July 2007, and by F.S. at NIKHEF, Amsterdam, The Netherlands, June 2006
(Submitted on 9 Aug 2007)

"These lecture notes provide a pedagogical introduction to a specific continuum implementation of the Wilsonian renormalization group, the effective average action. Its general properties and, in particular, its functional renormalization group equation are explained in a simple scalar setting. The approach is then applied to Quantum Einstein Gravity (QEG). The possibility of constructing a fundamental theory of quantum gravity in the framework of Asymptotic Safety is discussed and the supporting evidence is summarized."

As a sample, here are Saueressig papers from the past year
http://arxiv.org/find/grp_physics/1/au:+Saueressig/0/1/0/past/0/1
1. arXiv:0712.0445 [ps, pdf, other]
Title: On the renormalization group flow of f(R)-gravity
Authors: Pedro F. Machado, Frank Saueressig

2. arXiv:0710.4931 [ps, pdf, other]
Title: Recent results in four-dimensional non-perturbative string theory
Authors: Frank Saueressig
Comments: 7 pages, to appear in the proceedings of the European Physical Society HEP 2007 Conference, 19-25 July 2007, Manchester, England

3. arXiv:0708.1317 [ps, pdf, other]
Title: Functional Renormalization Group Equations, Asymptotic Safety, and Quantum Einstein Gravity
Authors: Martin Reuter, Frank Saueressig
Comments: Based on lectures given by M.R. at the First Quantum Geometry and Quantum Gravity School'', Zakopane, Poland, March 2007, and the Summer School on Geometric and Topological Methods for Quantum Field Theory'', Villa de Leyva, Colombia, July 2007, and by F.S. at NIKHEF, Amsterdam, The Netherlands, June 2006

4. arXiv:0707.0838 [ps, pdf, other]
Title: Membrane instantons from mirror symmetry
Authors: Daniel Robles-Llana, Frank Saueressig, Ulrich Theis, Stefan Vandoren

5. arXiv:0704.2229 [ps, pdf, other]
Title: Conifold singularities, resumming instantons and non-perturbative mirror symmetry
Authors: Frank Saueressig, Stefan Vandoren

You can see he has chosen to do both string and nonstring QG research. Interesting guy.

===========
HERE'S the paper that I think calls some finding of Reuter into question (but the issue is not yet resolved)
http://arxiv.org/abs/0712.0445
On the renormalization group flow of f(R)-gravity
55 pages, 7 figures
(Submitted on 4 Dec 2007)

"We use the functional renormalization group equation for quantum gravity to construct a non-perturbative flow equation for modified gravity theories of the form $$S = \int d^dx \sqrt{g} f(R)$$. Based on this equation we show that certain gravitational interactions monomials can be consistently decoupled from the renormalization group (RG) flow and reproduce recent results on the asymptotic safety conjecture. The non-perturbative RG flow of non-local extensions of the Einstein-Hilbert truncation including $$\int d^dx \sqrt{g} \ln(R)$$ and $$\int d^dx \sqrt{g} R^{-n}$$ interactions is investigated in detail. The inclusion of such interactions resolves the infrared singularities plaguing the RG trajectories with positive cosmological constant in previous truncations. In particular, in some $$R^{-n}$$-truncations all physical trajectories emanate from a Non-Gaussian (UV) fixed point and are well-defined on all RG scales. The RG flow of the $$\ln(R)$$-truncation contains an infrared attractor which drives a positive cosmological constant to zero, thereby providing a dynamical explanation of the tiny value of Lambda observed today."

Last edited: Jan 22, 2008
4. Jan 22, 2008

### marcus

This is an important paper of Percacci, which Reuter cites in his talks. It is probably the strongest evidence yet found for the existence of the UV fixed point of the RG flow

http://arxiv.org/abs/0705.1769
Ultraviolet properties of f(R)-Gravity
Alessandro Codello, Roberto Percacci, Christoph Rahmede
4 pages
(Submitted on 12 May 2007)

"We discuss the existence and properties of a nontrivial fixed point in f(R)-gravity, where f is a polynomial of order up to six. Within this seven-parameter class of theories, the fixed point has three ultraviolet-attractive and four ultraviolet-repulsive directions; this brings further support to the hypothesis that gravity is nonperturbatively renormalizable."

You can see why this is in a certain way a new field, even though Reuter has been working on it since 1998. It is only last year that this Percacci paper came out, and I expect quite a few people would only begin to notice based on this result.

5. Jan 22, 2008

### BenTheMan

Right, but none of this answers any questions as to why you can do this with gravity but not with, say, Fermi theory. Both have a coupling constant with the same mass dimensions, and both are (were) considered effective field theories. Except in the case of Fermi theory, the completion is clear (the SM), whereas in gravity, Reuter is telling me that there is no completion, and it is non-perturbatively renormalizable (whatever that means).

6. Jan 22, 2008

### marcus

You've put your finger on what you need to understand in order to grasp what is being talked about.

If your mind stays stuck in a perturbation theory context then you will not understand what any of these people (Saueressig, Percacci, Reuter etc) are saying, and you will necessarily be confused.

the other theories you are talking about are normally handled perturbatively.
I may be wrong of course---I may not understand what you are trying to say about these other theories. But it sounds to me like you may be trying to compare apples and oranges.

Maybe someone else can clarify for you.

7. Jan 22, 2008

### BenTheMan

I don't quite understand the difference. In one case, the Fermi constant goes like one over the W mass squared, and the Newton's constant goes like one over the Planck mass squared. Based on the effective field theory approach, I don't quite understand how you can claim the two are different---new physics comes in at the W mass and provides a natural UV completion for the theory.

8. Jan 22, 2008

### marcus

Have you looked at any of the papers I linked to?

9. Jan 22, 2008

### BenTheMan

I don't have time to read these things...I was hoping that someone could offer me a (semi-)technical explanation instead of linking me to a bunch of papers that I could have found on the arXiv myself.

10. Jan 23, 2008

### Haelfix

What Reuter originally did was to truncate gravity to first order (not surprisingly as divergences already appear at two loop), and then apply the 'exact renormalization group flow' to this quantity (see eg papers by Polchinksi which pioneer this method). Eg cosmological constant + (Einstein Hilbert terms + 1 loop contributions).

Distler explains the scheme here: http://golem.ph.utexas.edu/~distler/blog/archives/000648.html

Its a bit of a clever trick that you can indeed apply to say Fermi theory (though you won't find anything interesting), and it does give some control on the perturbative beta function, as this is now encoded at 1 loop, which he then uses to surmise a fixed point.

But... This is not a *proof* of a fixed point, but more of a hypothesis (that Reuter then goes on to test in a variety of circumstances, including strong coupling regimes). Indeed the effective action is not in the same universality class as the full shebang a priori, and there is absolutely no guarentee that it works at strong coupling (see Distlers objection) as you still need to consider the beta functions for the infinite set of couplings in the effective lagrangian and these are by no means guarenteed to vanish in the UV.

Last edited: Jan 23, 2008
11. Jan 23, 2008

### Demystifier

I think that the correct answer to your question is slightly philosophical. A priori, there is no reason to treat gravity in a different way than Fermi theory. However, many theoretical physicists find the classical Einstein theory of gravity so beautiful and elegant that they do not like the idea that such a beautiful theory is nothing but an approximate effective theory. For that reason, they desperately want to save that theory as a basis for quantum gravity as well. On the other hand, the Fermi theory is not that beautiful, so no one is too unhappy when it turns out that this theory is merely an effective one.
Is there a way to accept that the Einstein theory is merely an effective theory without sacrificing the elegance? Only if we replace it with something that seems even more elegant. And that is why string theory is believed by many physicists to be the right stuff, even without any observational evidence for it.

Of course, should the criteria of elegance and beauty be taken as valid scientific criteria, it's up to you. If you ask me, I think it should, but only as a guiding principle when no other scientific evidence is available. Quantum gravity is certainly one of such branches of physics.

Last edited: Jan 23, 2008
12. Jan 23, 2008

### Haelfix

The whole discussion hinges on the use of Polchinksi's form of the exact renormalization equation. Its primary purpose was to remove an approximation in the standard Wilson derivation (eg it takes into account irrelevant couplings), and it has a bit of a better bound on suppressing fluctuations above the cutoff scale (which is now smooth as opposed to sharp) as opposed to other exact renormalization group techniques. The issue is whether or not it sees nonperturbative UV effects and not just the IR ones since we have a regulator that is only perturbative in nature (eg its not on a lattice).

In the comment section of Distlers blog there was an interesting discussion arguing that it does not, and that make good sense, especially when we do not have any a priori knowledge about how much control we have in the UV. In general we expect (and this can be shown explicitly in exactly solvable models) quite a bit of extra nonperturbative corrections that are non analytic at the locus and that perturbation series will miss order by order.

13. Jan 23, 2008

### marcus

that was in 1998
as of 2007, the truncation is now at SIXTH order

the fixed point lies on a 3D hypersurface called the critical hypersurface
roughly speaking the fixed point is an attractor from three directions, and repells in four directions

Last edited: Jan 23, 2008
14. Jan 23, 2008

### marcus

Many people seem slow in registering this paper. I can understand this because it took me a while. My interest in the work of Reuter, Percacci, Saueressig etc was slight until I began to realize there was considerable suggestive, if not conclusive, evidence for the existence of the fixed point.
It could still prove NOT to exist, but there's reason to be interested in the approach. Another feature is that it looks like there might be only THREE attractive dimensions which means that there are only three parameters which need to be determined experimentally. If you get those numbers right then you are on track and get drawn in towards the fixed point. The physical significance of those three parameters is interesting to consider---one might be newton's G, one might be the cosmological constant. what is the third parameter?

15. Jan 23, 2008

### shalayka

If one keeps in mind the repulsive nature of the strong force at very close distances (http://arxiv.org/abs/nucl-th/0611096), then something like this may be a starting point:

Attractive:
1) Gravitation of spherically symmetric sources (stars)
2) Gravitation of axially symmetric sources (galaxies, dark matter)
3) Gravitation of minimally symmetric sources (filaments, dark matter)

Repulsive:
1) Weak nuclear force
2) Electromagnetic force
3) Strong force
4) Cosmological expansion

Last edited: Jan 23, 2008
16. Jan 23, 2008

### Coin

So... I am trying to understand this. As far as I get what you and Distler are saying, the idea is that this one-loop renormalization method is physically accurate in SOME models, but not ALL, and the models in which it is accurate are not models which we expect to find describing quantum gravity. The complaint however seems to be not that the model for quantum gravity CANNOT be one where the one-loop renormalization trick works, but that such a model would be contrived and there is no justification for adopting one.

Although this seems like a very good reason to expect the one-loop normalization approach wrong, it doesn't seem to rule it out completely since we don't really seem to know what the proper underlying model for quantum gravity is. So I'm curious-- what if, just for the sake of argument, we ASSUME the one-loop normalization approach is correct, and therefore spacetime and/or gravity must be described by some model to which the one-loop normalization hack can be applied? If this were true, what would it tell us? What if, for example, spacetime were "on a lattice"-- or rather, is there any choice of model in which the sentence "the regulator is on a lattice" can be made sensical in the qg arena? Alternately, what if we just assume some mechanism must exist which allows us to truncate to a fixed number of terms in the effective lagrangian, what would postulating the existence of that mechanism tell us about physics?

Please excuse me if this question is a little garbled :)

Last edited: Jan 23, 2008
17. Jan 23, 2008

### marcus

It sounds as if Haelfix and Distler are saying the same thing and that neither understands what has been happening.

In fact there IS reason to expect that only three constants are non-zero.
The conjecture is that at the UV fixed point you simply don't have an infinite list of coupling constants.

The conjecture is, once you get those three numbers right, the law is CORRECT.

There are various sorts of evidence for this, but the one that I've already pointed to here is the 2007 paper by Percacci Codello Rahmede where they tried out varying 7 parameters (a 6 degree polynomial in the curvature) and found that only the three lowest were nonzero. Reuter and the other already knew about those first three.

So you get the first three right and then you look at the next four and you find they vanish. OK what is your guess about the next? Very likely it is zero too. and it is just a matter of doing some more calculation to determine that.

It would seem there are people who are so used to looking at perturbation series that they have the illusion they are confronted with an infinite series of coupling constants even when they aren't!
What they've been doing is expanding around the wrong fixed point, so they get an endless bunch of coupling constants.

Start at the right fixed point and you may well avoid all that mess.

Last edited: Jan 23, 2008
18. Jan 23, 2008

### Coin

Well to the extent this is true I think Haelfix was basically paraphrasing Distler? The Distler quote is from the post Haelfix linked.

Hm, but that's only true if there is a UV fixed point, correct? I think Haelfix/Distler understand the argument here but are doubtful the UV fixed point exists.

I mean-- I don't know if I am understanding what you're saying right: Is the conjecture that given a fixed point, exactly those 3 terms will be non-negligable? Or is it known that if there is such a fixed point there will be exactly 3 relevant terms, and the conjecture is that the fixed point exists?

So are you saying there's a potentially soluble calculation that could show that terms > 3 in this effective lagrangian are zero?

If they did finish this calculation and proved, oh look at that, terms seven through infinity come out to zero-- would this prove the existence of the UV fixed point? Or would it prove that, if we assume the existence of the UV fixed point, gravity is renormalizable via this method?

19. Jan 23, 2008

### marcus

No Coin, what i am saying is something you may not recognize as kosher physics at all!
This is not an analytical calculation! I am not at all expert in these matters so I could be wrong, but what I think is that they did a NUMERICAL calculation. They ran the renormalization group flow using a computer.

Admittedly there are pitfalls when you use numerical calcuation, because you do not have infinite precision. But also it opens up new opportunities. I am talking about the Percacci Codello Rahmede paper. I will go back and check---perhaps I'm wrong and they were working analytically (as opposed to numerically).

Numerically they can never prove this! If they used a giant supercomputer and found that terms #4 thru #20 went to zero, it would still not PROVE anything. Because nature could be tricking us and term #21 might be significant!

I think what they are doing is, however, bold and exciting. It is using a computer to explore an infinite dimensional theory-space.

they explore the infinite dimensional space by projecting it down to----say, 6-degree polynomicals----a seven dimensional space and cruising around in that space.

and then there they find a 3D hypersurface on which the UV fix point is an attractor, so if you get on, it homes in on the fixed point.

It is fascinating to think that these structures, which one discovers empirically using the computer, might actually exist in nature. that this fixed point which for over 6 years keeps re-appearing, keeps coming up in study after study, might actually exist for real, and be the real theory. And always it is the computer sniffing the thing out for them, operating in effect on its own by an iterative process outside their direct control.

The aim is to have a predictive theory. To get the fixed point identified precisely enough so that one can use that action to calculate with and make predictions. We'll see.

To me the most worrying thing is not what Distler says (which I doubt matters much because of where he's coming from) but what Frank Saueressig says. His latest paper, with Machado, for me puts the UV fixed point in question. I am somewhat on pins and needles until the questions raised in that December 2007 paper are resolved.

20. Jan 25, 2008

### Haelfix

Coin. There are basically two seperate issues with what Reuter is doing.

One is that he truncates the series at some order (1, 7, 1000, it doesnt matter). This is *fine*, people do this all the time, however the theory that remains is an effective one, and not *a priori* in the same universality class as the full theory.

Now, the way that it can be *sometimes* in the same class, is if all those infinite counterterms that are being discarded have a certain pattern within them such that they cancel or at least are bounded to within some finite range. This happens when there is a fixed point within the theory. Basically, for a given truncated theory (gravity or other), there can be roughly three things that can happen. One is there is no fixed point (either in the effective truncated lagrangian or in the full renormalized lagrangian), two, there can be a fixed point in the effective truncated lagrangian but not the full renormalized lagrangian, and vice versa or both. The point is those last two options need not be the same theory, and absent some other way of solving the full path integral, its very hard to *know* the difference since we have no knowledge about what quantum gravity is. Reuters *hypothesis* is that they are coincident.

The other seperate issue that Distler mentions is I admit, a bit technical. It has to do with Polchinski's exact renormalization group method. Basically (very roughly) the technique starts by imposing a regularization cutoff that is a smooth function as opposed to the standard hardcutoff, the idea being eventually you send it to infinity after you have done the calculations and filtered out the high frequency modes. The whole point of using renormalization group flow is that it tends to give you some information about the nonperturbative behaviour of the theory in question. Now the issue is that this particular choice of regularization obviously lives in the perturbative sector, so consequently it will miss nonperturbative effects in the UV (but not the IR since its sensitive to these). You can contrast that with using a lattice regulator, which clearly is UV nonperturbative as you take infinite volume and zero lattice spacing limits. The point being that whatever fixed point you may find, could be spoiled by nonperturbative effects (even instanton like ones if you go to Euclidean space) and absent some degree of control from somewhere else, that remains a big assumption. (im glancing over a whole host of issues, like renormalon behaviour and so forth).

Anyway, you can see that its a hypothesis. And thats fine, but it should be clear there is good reason to be skeptical ipso facto and the burden of proof is high (we are after all talking about a full solution to QG if he were correct). I mean you can use the same technique for other nonrenormalizable field theories, as the OP originally asked. For instance certain quenched QCD models are probable with exact renormalization group flow methods.

Last edited: Jan 25, 2008


text:
### memset123's blog

By memset123, 4 years ago, ,

Recently, in order the celebrate the new year :), I was looking at some codeforces problems. I found one specific data structure problem (here) to be very interesting. I found a blog where much of this problem was explained (as there is no editorial..) but I was wondering if anyone could elaborate on how the blog writer used segment tree to solve this problem.. I could not figure that part out.

•
• +10
•

 » 4 years ago, # |   0 Any ideas?
 » 4 years ago, # |   +6 I've used a segment tree to handle following queries: 1. modify (l, r, x) : for each l <= i < r, a[i] = x 2. get (i) : return a[i] Look at my code : 7839071 Tell me, if you didn't understand some part of it!
•  » » 4 years ago, # ^ |   0 I see that, somehow, you are greedily pushing numbers into the answer vector, once you arrive at the indice i. Why is that optimal?
•  » » » 4 years ago, # ^ |   +5 You can prove this using induction on n. Consider any optimal sequence b1, b2, ..., b4k. If my output is m1, m2, ..., m4k', obviously if k ≥ 1, then k' ≥ 1 (Because I'll find b1, b2, b3, b4 greedily!) Induction step: My solution has guaranteed that positionm4 ≤ positionb4. By induction hypothesis, my solution will find an answer with length equal to 4k - 4 for the rest of the array (apositionm4 + 1, apositionm4 + 2, ..., an) and by adding them after m1, m2, m3, m4, we have an answer of length equal to 4k! (Sorry for my bad English ... Tell me if it's unclear!)
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   0 Ok, this is clear now. Can you explain how you are using the "modify" and "get" commands to match numbers between the indice i and the next occurrence of i. To elaborate on this question:Say you have numbers:1 2 1 3 2, and we are at indice 1, and obviously next[arr[i]] is indice 3. How do you match the middle 2, at indice 2, to the outer 2 at indice 5?
•  » » » » » 4 years ago, # ^ |   +5 We'll have another array s. When we're at indice 1, we won't do anything because it's the first occurrence of 1. (Also indice 2 is the same) When we're at the indice 3, we modify s[2] = 1 (We modify the segment between last occurrence of a[i] and i to a[i]) So when we're at the indice 5, we'll have a look at the last occurrence of a[i]! In this case, s[last[a[5]] = 1. So we know the last 2 is surrounded by two '1's and we found 1 2 1 2.
•  » » » » » » 4 years ago, # ^ |   0 Ahhh I finally got it :). Thank you so much. May I ask, what was the motivation behind that solution?
•  » » » » » » » 4 years ago, # ^ |   0 I'm glad that it helped. :) First of all I tried to find some subsequence of the input (Not the largest one) that satisfies the given conditions. During that process, the greedy solution comes in my mind and then the the only problem was "how to find the first abab fast?". After that I used some kind of dfs to find the idea and it just worked well!
•  » » » » 10 months ago, # ^ |   0 "My solution has guaranteed that positionm4 ≤ positionb4" could you explain this part more clearly? I mean, I understood the line before this, that is if k>= 1, then k' ≥ 1-but don't get how this leads to the conclusion that position of m4 <= position of b4


text:
# The infinity of primes

by kenewbie
Tags: euclid proof primes
 P: 236 Euclid's proof: 1) Assume there is a finite number of primes. 2) Let Pn be the largest prime. 3) Let X be the P1 * P2 ... * Pn + 1 At this point the statement is that "X cannot be divided by P1 through Pn", but why is that? This is not self-obvious to me. How can I know this? k
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,552 Because dividing by any of those primes gives a remainder of 1, not 0: $$\frac{P_1P_2...P_{i-1}P_iP_{i+1}...Pn+ 1}{Pi}= P_1P_2...P_{i-1}P_{i+1}...Pn+ \frac{1}{P_i}$$ The first term is an integer but the second is not.
 Emeritus Sci Advisor PF Gold P: 16,091 Minor aside: "the infinity of primes" is bad grammar. The noun form is wrong here -- you want the adjective "infinite", such as in "the infinite set of primes".
Emeritus
PF Gold
P: 16,091
The infinity of primes

 Quote by kenewbie How can I know this?
Try dividing.
P: 236
 Quote by HallsofIvy $$\frac{P_1P_2...P_{i-1}P_iP_{i+1}...Pn+ 1}{Pi}= P_1P_2...P_{i-1}P_{i+1}...Pn+ \frac{1}{P_i}$$ The first term is an integer but the second is not.
That made perfect sense, thank you!

k

 Related Discussions Linear & Abstract Algebra 3 Calculus & Beyond Homework 8 Calculus & Beyond Homework 3 Quantum Physics 31 Calculus & Beyond Homework 15


text:
## Cryptology ePrint Archive: Report 2020/654

Proximity Gaps for Reed-Solomon Codes

Eli Ben-Sasson and Dan Carmon and Yuval Ishai and Swastik Kopparty and Shubhangi Saraf

Abstract: A collection of sets displays a proximity gap with respect to some property if for every set in the collection, either (i) all members are $\delta$-close to the property in relative Hamming distance or (ii) only a tiny fraction of members are $\delta$-close to the property. In particular, no set in the collection has roughly half of its members $\delta$-close to the property and the others $\delta$-far from it.

We show that the collection of affine spaces displays a proximity gap with respect to Reed-Solomon (RS) codes, even over small fields, of size polynomial in the dimension of the code, and the gap applies to any $\delta$ smaller than the Johnson/Guruswami-Sudan list-decoding bound of the RS code. We also show near-optimal gap results, over fields of (at least) linear size in the RS code dimension, for $\delta$ smaller than the unique decoding radius. Finally, we discuss several applications of our proximity gap results to distributed storage, multi-party cryptographic protocols, and concretely efficient proof systems.

We prove the proximity gap results by analyzing the execution of classical algebraic decoding algorithms for Reed-Solomon codes (due to Berlekamp-Welch and Guruswami-Sudan) on a formal element of an affine space. This involves working with Reed-Solomon codes whose base field is an (infinite) rational function field. Our proofs are obtained by developing an extension (to function fields) of a strategy of Arora and Sudan for analyzing low-degree tests.

Category / Keywords: cryptographic protocols / Interactive Oracle Proofs, Reed Solomon codes, Verifiable Secret Sharing, Property Testing

Original Publication (in the same form): Electronic Colloquium on Computational Complexity


text:
My Math Forum improper integral

 Calculus Calculus Math Forum

 March 29th, 2014, 07:13 AM #1 Senior Member   Joined: Apr 2012 Posts: 135 Thanks: 1 improper integral i have the following integral: integral from -1 to pi/2 arcsinx/(x^2). The integral from -1 to 0 gives -oo and from 0 to pi/2 +oo... what is the conclusion here? Or integral from -1 to 1 1/x is this equal with 0? because we have the same amount of "infinity" on both sides. It is necessary to solve the other side of the improper integral if the first one is divergent? And the last question: It is possible to have -oo on one side , +oo on the other side and the integral to converge? Thank you! Thanks from zaidalyafey
 April 1st, 2014, 08:47 AM #2 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions $$\int^1_{-1} \frac{1}{x} \, dx \neq 0$$ because the integral actually diverges. The function has no antiderivative on the specified region. Thanks from alexmath

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text:
### Towards Post-Quantum Security for Cyber-Physical Systems: Integrating PQC into Industrial M2M Communication

Sebastian Paul, Patrik Scheible, and Friedrich Wiemer

##### Abstract

The threat of a cryptographically relevant quantum computer contributes to an increasing interest in the field of post-quantum cryptography (PQC). Compared to existing research efforts regarding the integration of PQC into the Transport Layer Security (TLS) protocol, industrial communication protocols have so far been neglected. Since industrial cyber-physical systems (CPS) are typically deployed for decades, protection against such long-term threats is needed. In this work, we propose two novel solutions for the integration of post-quantum (PQ) primitives (digital signatures and key establishment) into the industrial protocol Open Platform Communications Unified Architecture (OPC UA): a hybrid solution combining conventional cryptography with PQC and a solution solely based on PQC. Both approaches provide mutual authentication between client and server and are realized with certificates fully compliant to the X.509 standard. We implement the two solutions and measure and evaluate their performance across three different security levels. All selected algorithms (Kyber, Dilithium, and Falcon) are candidates for standardization by the National Institute of Standards and Technology (NIST). We show that Falcon is a suitable option - especially - when using floating-point hardware provided by our ARM-based evaluation platform. Our proposed hybrid solution provides PQ security for early adopters but comes with additional performance and communication requirements. Our solution solely based on PQC shows superior performance across all evaluated security levels in terms of handshake duration compared to conventional OPC UA but comes at the cost of increased handshake sizes. In addition to our performance evaluation, we provide a proof of security in the symbolic model for our two PQC-based variants of OPC UA. For this proof, we use the cryptographic protocol verifier ProVerif and formally verify confidentiality and authentication properties of our quantum-resistant variants.

Note: This is an extended version of "Towards Post-Quantum Security for Cyber-Physical Systems" that originally appeared in Computer Security - ESORICS 2020, Springer, pp. 295-316, DOI: 10.1007/978-3-030-59013-0_15.

Available format(s)
Category
Implementation
Publication info
Published elsewhere. Journal of Computer Security
DOI
10.3233/JCS-210037
Keywords
cyber-physical systemspost-quantum cryptographyformal security modelsOPC UAProVerif
Contact author(s)
sebastian paul2 @ de bosch com
History
2021-12-02: revised
See all versions
Short URL
https://ia.cr/2021/1563

CC BY

BibTeX

@misc{cryptoeprint:2021/1563,
author = {Sebastian Paul and Patrik Scheible and Friedrich Wiemer},
title = {Towards Post-Quantum Security for Cyber-Physical Systems: Integrating PQC into Industrial M2M Communication},
howpublished = {Cryptology ePrint Archive, Paper 2021/1563},
year = {2021},
doi = {10.3233/JCS-210037},
note = {\url{https://eprint.iacr.org/2021/1563}},
url = {https://eprint.iacr.org/2021/1563}
}

Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.


text:
# Test vectors for ristretto255

ristretto255 encodes group elements using 255 bits and provides a prime-order group of size 2^252. It can be implemented using Curve25519.

For ristretto255, the parameters are \begin{aligned} p &= 2^{255}-19 \\ a_2 = a &= -1 \\ d_2 = d &= -121665/121666. \end{aligned} Field elements are considered negative when their low bit is set, as in Ed25519. The Elligator map uses $$n = +\sqrt{-1}$$ as the quadratic nonresidue.

// The test vectors below also have code to test them against the
// curve25519_dalek implementation.
//
// These tests are run in CI against the curve25519-dalek implementation.

extern crate sha2;
extern crate hex;
extern crate curve25519_dalek;

# fn main() {
// The following are the byte encodings of small multiples
//     [0]B, [1]B, ..., [15]B
// of the basepoint, represented as hex strings.

let encodings_of_small_multiples = [
// This is the identity point
"0000000000000000000000000000000000000000000000000000000000000000",
// This is the basepoint
"e2f2ae0a6abc4e71a884a961c500515f58e30b6aa582dd8db6a65945e08d2d76",
// These are small multiples of the basepoint
"94741f5d5d52755ece4f23f044ee27d5d1ea1e2bd196b462166b16152a9d0259",
"e882b131016b52c1d3337080187cf768423efccbb517bb495ab812c4160ff44e",
"f64746d3c92b13050ed8d80236a7f0007c3b3f962f5ba793d19a601ebb1df403",
"44f53520926ec81fbd5a387845beb7df85a96a24ece18738bdcfa6a7822a176d",
"903293d8f2287ebe10e2374dc1a53e0bc887e592699f02d077d5263cdd55601c",
"02622ace8f7303a31cafc63f8fc48fdc16e1c8c8d234b2f0d6685282a9076031",
"bce83f8ba5dd2fa572864c24ba1810f9522bc6004afe95877ac73241cafdab42",
"46376b80f409b29dc2b5f6f0c52591990896e5716f41477cd30085ab7f10301e",
];

// Test the encodings of small multiples

use curve25519_dalek::constants;
use curve25519_dalek::traits::Identity;
use curve25519_dalek::ristretto::RistrettoPoint;

let B = &constants::RISTRETTO_BASEPOINT_POINT;
let mut P = RistrettoPoint::identity();
for i in 0..16 {
assert_eq!(
hex::encode(P.compress().as_bytes()),
encodings_of_small_multiples[i],
);
P = P + B;
}

// The following are invalid encodings, which should all be rejected.
// These are designed to test each of the checks during decoding.

// These are all bad because they're non-canonical field encodings.
"00ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff",
"ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff7f",
"f3ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff7f",
"edffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff7f",
// These are all bad because they're negative field elements.
"0100000000000000000000000000000000000000000000000000000000000000",
"01ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff7f",
"ed57ffd8c914fb201471d1c3d245ce3c746fcbe63a3679d51b6a516ebebe0e20",
"c34c4e1826e5d403b78e246e88aa051c36ccf0aafebffe137d148a2bf9104562",
"c940e5a4404157cfb1628b108db051a8d439e1a421394ec4ebccb9ec92a8ac78",
"47cfc5497c53dc8e61c91d17fd626ffb1c49e2bca94eed052281b510b1117a24",
"f1c6165d33367351b0da8f6e4511010c68174a03b6581212c71c0e1d026c3c72",
// These are all bad because they give a nonsquare x^2.
"26948d35ca62e643e26a83177332e6b6afeb9d08e4268b650f1f5bbd8d81d371",
"4eac077a713c57b4f4397629a4145982c661f48044dd3f96427d40b147d9742f",
"bcab477be20861e01e4a0e295284146a510150d9817763caf1a6f4b422d67042",
"2a292df7e32cababbd9de088d1d1abec9fc0440f637ed2fba145094dc14bea08",
"f4a9e534fc0d216c44b218fa0c42d99635a0127ee2e53c712f70609649fdff22",
"8268436f8c4126196cf64b3c7ddbda90746a378625f9813dd9b8457077256731",
"2810e5cbc2cc4d4eece54f61c6f69758e289aa7ab440b3cbeaa21995c2f4232b",
// These are all bad because they give a negative xy value.
"3eb858e78f5a7254d8c9731174a94f76755fd3941c0ac93735c07ba14579630e",
"8a2e1d30050198c65a54483123960ccc38aef6848e1ec8f5f780e8523769ba32",
"227142501b9d4355ccba290404bde41575b037693cef1f438c47f8fbf35d1165",
"5c37cc491da847cfeb9281d407efc41e15144c876e0170b499a96a22ed31e01e",
"445425117cb8c90edcbc7c1cc0e74f747f2c1efa5630a967c64f287792a48a4b",
// This is s = -1, which causes y = 0.
"ecffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff7f",
];

// Test that all of the bad encodings are rejected

use curve25519_dalek::ristretto::CompressedRistretto;

let mut bad_bytes = [0u8; 32];
}

// The following are a list of UTF-8 encoded strings, and the byte
// encodings of performing hash-to-point with SHA-512.

let labels = [
"Ristretto is traditionally a short shot of espresso coffee",
"made with the normal amount of ground coffee but extracted with",
"about half the amount of water in the same amount of time",
"by using a finer grind.",
"This produces a concentrated shot of coffee per volume.",
"Just pulling a normal shot short will produce a weaker shot",
"and is not a Ristretto as some believe.",
];

let encoded_hash_to_points = [
"3066f82a1a747d45120d1740f14358531a8f04bbffe6a819f86dfe50f44a0a46",
"f26e5b6f7d362d2d2a94c5d0e7602cb4773c95a2e5c31a64f133189fa76ed61b",
"006ccd2a9e6867e6a2c5cea83d3302cc9de128dd2a9a57dd8ee7b9d7ffe02826",
"f8f0c87cf237953c5890aec3998169005dae3eca1fbb04548c635953c817f92a",
"ae81e7dedf20a497e10c304a765c1767a42d6e06029758d2d7e8ef7cc4c41179",
"e2705652ff9f5e44d3e841bf1c251cf7dddb77d140870d1ab2ed64f1a9ce8628",
"80bd07262511cdde4863f8a7434cef696750681cb9510eea557088f76d9e5065",
];

// Test the encodings above.

use sha2::{Sha512, Digest};

for i in 0..7 {
let point = RistrettoPoint::hash_from_bytes::<Sha512>(labels[i].as_bytes());
assert_eq!(
hex::encode(point.compress().as_bytes()),
encoded_hash_to_points[i],
);
}
# }


text:
# Compute the following integral?

Posted 7 months ago
1030 Views
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|
1 Total Likes
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 I let Mathematica run overnight the following code Integrate[(r^2 Cos[A r Cos[x]] Exp[-r^2/2])/(a^2 Cos[x]^2 + b^2 Sin[x]^2), {r, 0, Infinity}, {x, 0, Pi/2}] but it was unable to compute the closed form in that time. However, Mathematica also didn’t “give up”. Do you think Mathematica can do it?
Answer
3 Replies
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Posted 7 months ago
 I suspect that the Cos inside a Cos makes it impossible. It looks like an integral in polar coordinates. Might be easier in Cartesian coordinates.
Answer
Posted 7 months ago
 Another thing to point out, is that there isn't always an answer to an integral, as my teacher best put it "99% of all integrals don't have antiderivatives." So try making those constant coefficients into variables and write the integral as a function of 3 variables (assuming those three constants are A, a, and b) so that you can work around that instead. You can also plot the newly defined function with respect to two variables and change the third one each time you want to run it. That's my best suggestion. If you want to know more, just ask.
Answer
Posted 7 months ago
 Doing the r under assumptions is the first step In[2]:= Integrate[(r^2 Cos[A r Cos[x]] Exp[-r^2/2])/(a^2 Cos[x]^2 + b^2 Sin[x]^2), {r, 0, Infinity},(* {x,0,Pi/2} *) Assumptions -> {A > 0, a > 0, b > 0}] Out[2]= (E^(-(1/2) A^2 Cos[x]^2) Sqrt[\[Pi]/2] (1 - A^2 Cos[x]^2))/( a^2 Cos[x]^2 + b^2 Sin[x]^2) now look for a meaningful substitution, you need $0 \leq x \leq \frac{\pi}{2}$. And as a preliminary step try $a = b$ to get rid of the denominator.
Answer
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Probability Topics

# Terminology

OpenStaxCollege

[latexpage]

Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin twice is an example of an experiment.

A result of an experiment is called an outcome. The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter S is used to denote the sample space. For example, if you flip one fair coin, S = {H, T} where H = heads and T = tails are the outcomes.

An event is any combination of outcomes. Upper case letters like A and B represent events. For example, if the experiment is to flip one fair coin, event A might be getting at most one head. The probability of an event A is written P(A).

The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, zero and one and all numbers between these values). P(A) = 0 means the event A can never happen. P(A) = 1 means the event A always happens. P(A) = 0.5 means the event A is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads).

Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head (H) and a Tail (T) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.

To calculate the probability of an event A when all outcomes in the sample space are equally likely, count the number of outcomes for event A and divide by the total number of outcomes in the sample space. For example, if you toss a fair dime and a fair nickel, the sample space is {HH, TH, HT, TT} where T = tails and H = heads. The sample space has four outcomes. A = getting one head. There are two outcomes that meet this condition {HT, TH}, so P(A) = $$\frac{2}{4}$$ = 0.5.

Suppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event E = rolling a number that is at least five. There are two outcomes {5, 6}. P(E) = $$\frac{2}{6}$$. If you were to roll the die only a few times, you would not be surprised if your observed results did not match the probability. If you were to roll the die a very large number of times, you would expect that, overall, $$\frac{2}{6}$$ of the rolls would result in an outcome of “at least five”. You would not expect exactly $$\frac{2}{6}$$. The long-term relative frequency of obtaining this result would approach the theoretical probability of $$\frac{2}{6}$$ as the number of repetitions grows larger and larger.

This important characteristic of probability experiments is known as the law of large numbers which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.)

It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair, or biased. Two math professors in Europe had their statistics students test the Belgian one Euro coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44% of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to draw a more accurate conclusion about such bias. Some dice may be biased. Look at the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely.

“OR” Event:An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B. For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}. A OR B = {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are NOT listed twice.

“AND” Event:An outcome is in the event A AND B if the outcome is in both A and B at the same time. For example, let A and B be {1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then A AND B = {4, 5}.

The complement of event A is denoted A′ (read “A prime”). A′ consists of all outcomes that are NOT in A. Notice that P(A) + P(A′) = 1. For example, let S = {1, 2, 3, 4, 5, 6} and let A = {1, 2, 3, 4}. Then, A′ = {5, 6}. P(A) = $$\frac{4}{6}$$, P(A′) = $$\frac{2}{6}$$, and P(A) + P(A′) = $$\frac{4}{6}+\frac{2}{6}$$ = 1

The conditional probability of A given B is written P(A|B). P(A|B) is the probability that event A will occur given that the event B has already occurred. A conditional reduces the sample space. We calculate the probability of A from the reduced sample space B. The formula to calculate P(A|B) is P(A|B) = $$\frac{P\left(A\text{AND}B\right)}{P\left(B\right)}$$ where P(B) is greater than zero.

For example, suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}. Let A = face is 2 or 3 and B = face is even (2, 4, 6). To calculate P(A|B), we count the number of outcomes 2 or 3 in the sample space B = {2, 4, 6}. Then we divide that by the number of outcomes B (rather than S).

We get the same result by using the formula. Remember that S has six outcomes.

P(A|B) = $$\frac{P\left(A\phantom{\rule{2pt}{0ex}}\text{AND}\phantom{\rule{2pt}{0ex}}B\right)}{P\left(B\right)}=\frac{\frac{\left(\text{the number of outcomes that are 2 or 3 and even in}\phantom{\rule{2pt}{0ex}}S\right)}{6}}{\frac{\left(\text{the number of outcomes that are even in}\phantom{\rule{2pt}{0ex}}S\right)}{6}}=\frac{\frac{1}{6}}{\frac{3}{6}}=\frac{1}{3}$$

Understanding Terminology and SymbolsIt is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem several times if necessary. Clearly identify the event of interest. Determine whether there is a condition stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any.

The sample space S is the whole numbers starting at one and less than 20.

1. S = _____________________________

Let event A = the even numbers and event B = numbers greater than 13.

2. A = _____________________, B = _____________________
3. P(A) = _____________, P(B) = ________________
4. A AND B = ____________________, A OR B = ________________
5. P(A AND B) = _________, P(A OR B) = _____________
6. A′ = _____________, P(A′) = _____________
7. P(A) + P(A′) = ____________
8. P(A|B) = ___________, P(B|A) = _____________; are the probabilities equal?
1. S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}
2. A = {2, 4, 6, 8, 10, 12, 14, 16, 18}, B = {14, 15, 16, 17, 18, 19}
3. P(A) = $$\frac{9}{19}$$, P(B) = $$\frac{6}{19}$$
4. A AND B = {14,16,18}, A OR B = {2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19}
5. P(A AND B) = $$\frac{3}{19}$$, P(A OR B) = $$\frac{12}{19}$$
6. A′ = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19; P(A′) = $$\frac{10}{19}$$
7. P(A) + P(A′) = 1 ($$\frac{9}{19}$$ + $$\frac{10}{19}$$ = 1)
8. P(A|B) = $$\frac{P\left(A\text{AND}B\right)}{P\left(B\right)}$$ = $$\frac{3}{6}$$, P(B|A) = $$\frac{P\left(A\text{AND}B\right)}{P\left(A\right)}$$ = $$\frac{3}{9}$$, No
Try It

The sample space S is the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)).

1. S = _____________________________

Let event A = the sum is even and event B = the first number is prime.

2. A = _____________________, B = _____________________
3. P(A) = _____________, P(B) = ________________
4. A AND B = ____________________, A OR B = ________________
5. P(A AND B) = _________, P(A OR B) = _____________
6. B′ = _____________, P(B′) = _____________
7. P(A) + P(A′) = ____________
8. P(A|B) = ___________, P(B|A) = _____________; are the probabilities equal?
1. S = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)}
2. A = {(1,1), (1,3), (2,2), (2,4), (3,1), (3,3)}

B = {(2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)}

3. P(A) = $$\frac{1}{2}$$, P(B) = $$\frac{2}{3}$$
4. A AND B = {(2,2), (2,4), (3,1), (3,3)}

A OR B = {(1,1), (1,3), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)}

5. P(A AND B) = $$\frac{1}{3}$$, P(A OR B) = $$\frac{5}{6}$$
6. B′ = {(1,1), (1,2), (1,3), (1,4)}, P(B′) = $$\frac{1}{3}$$
7. P(B) + P(B′) = 1
8. P(A|B) = $$\frac{P\text{(}A\text{AND}B\text{)}}{P\text{(}B\text{)}}\text{=}\frac{1}{2}$$, P(B|A) = $$\frac{P\text{(}A\text{AND}B\text{)}}{P\text{(}B\text{)}}\text{=}\frac{2}{3}$$, No.

A fair, six-sided die is rolled. Describe the sample space S, identify each of the following events with a subset of S and compute its probability (an outcome is the number of dots that show up).

1. Event T = the outcome is two.
2. Event A = the outcome is an even number.
3. Event B = the outcome is less than four.
4. The complement of A.
5. A GIVEN B
6. B GIVEN A
7. A AND B
8. A OR B
9. A OR B′
10. Event N = the outcome is a prime number.
11. Event I = the outcome is seven.
1. T = {2}, P(T) = $$\frac{1}{6}$$
2. A = {2, 4, 6}, P(A) = $$\frac{1}{2}$$
3. B = {1, 2, 3}, P(B) = $$\frac{1}{2}$$
4. A′ = {1, 3, 5}, P(A′) = $$\frac{1}{2}$$
5. A|B = {2}, P(A|B) = $$\frac{1}{3}$$
6. B|A = {2}, P(B|A) = $$\frac{1}{3}$$
7. A AND B = {2}, P(A AND B) = $$\frac{1}{6}$$
8. A OR B = {1, 2, 3, 4, 6}, P(A OR B) = $$\frac{5}{6}$$
9. A OR B′ = {2, 4, 5, 6}, P(A OR B′) = $$\frac{2}{3}$$
10. N = {2, 3, 5}, P(N) = $$\frac{1}{2}$$
11. A six-sided die does not have seven dots. P(7) = 0.

[link] describes the distribution of a random sample S of 100 individuals, organized by gender and whether they are right- or left-handed.

Right-handed Left-handed
Males 43 9
Females 44 4

Let’s denote the events M = the subject is male, F = the subject is female, R = the subject is right-handed, L = the subject is left-handed. Compute the following probabilities:

1. P(M)
2. P(F)
3. P(R)
4. P(L)
5. P(M AND R)
6. P(F AND L)
7. P(M OR F)
8. P(M OR R)
9. P(F OR L)
10. P(M’)
11. P(R|M)
12. P(F|L)
13. P(L|F)
1. P(M) = 0.52
2. P(F) = 0.48
3. P(R) = 0.87
4. P(L) = 0.13
5. P(M AND R) = 0.43
6. P(F AND L) = 0.04
7. P(M OR F) = 1
8. P(M OR R) = 0.96
9. P(F OR L) = 0.57
10. P(M’) = 0.48
11. P(R|M) = 0.8269 (rounded to four decimal places)
12. P(F|L) = 0.3077 (rounded to four decimal places)
13. P(L|F) = 0.0833

# References

“Countries List by Continent.” Worldatlas, 2013. Available online at http://www.worldatlas.com/cntycont.htm (accessed May 2, 2013).

# Chapter Review

In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive.

# Formula Review

A and B are events

P(S) = 1 where S is the sample space

0 ≤ P(A) ≤ 1

P(A|B) = $$\frac{P\text{(}A\text{AND}B\text{)}}{P\text{(}B\text{)}}$$

In a particular college class, there are male and female students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts a through j. (Note that you cannot find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.)

• Let F be the event that a student is female.
• Let M be the event that a student is male.
• Let S be the event that a student has short hair.
• Let L be the event that a student has long hair.
1. The probability that a student does not have long hair.
2. The probability that a student is male or has short hair.
3. The probability that a student is a female and has long hair.
4. The probability that a student is male, given that the student has long hair.
5. The probability that a student has long hair, given that the student is male.
6. Of all the female students, the probability that a student has short hair.
7. Of all students with long hair, the probability that a student is female.
8. The probability that a student is female or has long hair.
9. The probability that a randomly selected student is a male student with short hair.
10. The probability that a student is female.
1. P(L′) = P(S)
2. P(M OR S)
3. P(F AND L)
4. P(M|L)
5. P(L|M)
6. P(S|F)
7. P(F|L)
8. P(F OR L)
9. P(M AND S)
10. P(F)

Use the following information to answer the next four exercises. A box is filled with several party favors. It contains 12 hats, 15 noisemakers, ten finger traps, and five bags of confetti.

Let H = the event of getting a hat.

Let N = the event of getting a noisemaker.

Let F = the event of getting a finger trap.

Let C = the event of getting a bag of confetti.

Find P(H).

Find P(N).

P(N) = $$\frac{15}{42}$$ = $$\frac{5}{14}$$ = 0.36

Find P(F).

Find P(C).

P(C) = $$\frac{5}{42}$$ = 0.12

Use the following information to answer the next six exercises. A jar of 150 jelly beans contains 22 red jelly beans, 38 yellow, 20 green, 28 purple, 26 blue, and the rest are orange.

Let B = the event of getting a blue jelly bean

Let G = the event of getting a green jelly bean.

Let O = the event of getting an orange jelly bean.

Let P = the event of getting a purple jelly bean.

Let R = the event of getting a red jelly bean.

Let Y = the event of getting a yellow jelly bean.

Find P(B).

Find P(G).

P(G) = $$\frac{20}{150}$$ = $$\frac{2}{15}$$ = 0.13

Find P(P).

Find P(R).

P(R) = $$\frac{22}{150}$$ = $$\frac{11}{75}$$ = 0.15

Find P(Y).

Find P(O).

P(O) = $$\frac{150-22-38-20-28-26}{150}$$ = $$\frac{16}{150}$$ = $$\frac{8}{75}$$ = 0.11

Use the following information to answer the next six exercises. There are 23 countries in North America, 12 countries in South America, 47 countries in Europe, 44 countries in Asia, 54 countries in Africa, and 14 in Oceania (Pacific Ocean region).

Let A = the event that a country is in Asia.

Let E = the event that a country is in Europe.

Let F = the event that a country is in Africa.

Let N = the event that a country is in North America.

Let O = the event that a country is in Oceania.

Let S = the event that a country is in South America.

Find P(A).

Find P(E).

P(E) = $$\frac{47}{194}$$ = 0.24

Find P(F).

Find P(N).

P(N) = $$\frac{23}{194}$$ = 0.12

Find P(O).

Find P(S).

P(S) = $$\frac{12}{194}$$ = $$\frac{6}{97}$$ = 0.06

What is the probability of drawing a red card in a standard deck of 52 cards?

What is the probability of drawing a club in a standard deck of 52 cards?

$$\frac{13}{52}$$ = $$\frac{1}{4}$$ = 0.25

What is the probability of rolling an even number of dots with a fair, six-sided die numbered one through six?

What is the probability of rolling a prime number of dots with a fair, six-sided die numbered one through six?

$$\frac{3}{6}$$ = $$\frac{1}{2}$$ = 0.5

Use the following information to answer the next two exercises. You see a game at a local fair. You have to throw a dart at a color wheel. Each section on the color wheel is equal in area.

Let B = the event of landing on blue.

Let R = the event of landing on red.

Let G = the event of landing on green.

Let Y = the event of landing on yellow.

If you land on Y, you get the biggest prize. Find P(Y).

If you land on red, you don’t get a prize. What is P(R)?

$$P\left(R\right)=\frac{4}{8}=0.5$$

Use the following information to answer the next ten exercises. On a baseball team, there are infielders and outfielders. Some players are great hitters, and some players are not great hitters.

Let I = the event that a player in an infielder.

Let O = the event that a player is an outfielder.

Let H = the event that a player is a great hitter.

Let N = the event that a player is not a great hitter.

Write the symbols for the probability that a player is not an outfielder.

Write the symbols for the probability that a player is an outfielder or is a great hitter.

P(O OR H)

Write the symbols for the probability that a player is an infielder and is not a great hitter.

Write the symbols for the probability that a player is a great hitter, given that the player is an infielder.

P(H|I)

Write the symbols for the probability that a player is an infielder, given that the player is a great hitter.

Write the symbols for the probability that of all the outfielders, a player is not a great hitter.

P(N|O)

Write the symbols for the probability that of all the great hitters, a player is an outfielder.

Write the symbols for the probability that a player is an infielder or is not a great hitter.

P(I OR N)

Write the symbols for the probability that a player is an outfielder and is a great hitter.

Write the symbols for the probability that a player is an infielder.

P(I)

What is the word for the set of all possible outcomes?

What is conditional probability?

The likelihood that an event will occur given that another event has already occurred.

A shelf holds 12 books. Eight are fiction and the rest are nonfiction. Each is a different book with a unique title. The fiction books are numbered one to eight. The nonfiction books are numbered one to four. Randomly select one book

Let F = event that book is fiction

Let N = event that book is nonfiction

What is the sample space?

What is the sum of the probabilities of an event and its complement?

1

Use the following information to answer the next two exercises. You are rolling a fair, six-sided number cube. Let E = the event that it lands on an even number. Let M = the event that it lands on a multiple of three.

What does P(E|M) mean in words?

What does P(E OR M) mean in words?

the probability of landing on an even number or a multiple of three

# Homework

The graph in [link] displays the sample sizes and percentages of people in different age and gender groups who were polled concerning their approval of Mayor Ford’s actions in office. The total number in the sample of all the age groups is 1,045.

1. Define three events in the graph.
2. Describe in words what the entry 40 means.
3. Describe in words the complement of the entry in question 2.
4. Describe in words what the entry 30 means.
5. Out of the males and females, what percent are males?
6. Out of the females, what percent disapprove of Mayor Ford?
7. Out of all the age groups, what percent approve of Mayor Ford?
8. Find P(Approve|Male).
9. Out of the age groups, what percent are more than 44 years old?
10. Find P(Approve|Age < 35).

Explain what is wrong with the following statements. Use complete sentences.

1. If there is a 60% chance of rain on Saturday and a 70% chance of rain on Sunday, then there is a 130% chance of rain over the weekend.
2. The probability that a baseball player hits a home run is greater than the probability that he gets a successful hit.
1. You can’t calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100%
2. A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs.

## Glossary

Conditional Probability
the likelihood that an event will occur given that another event has already occurred
Equally Likely
Each outcome of an experiment has the same probability.
Event
a subset of the set of all outcomes of an experiment; the set of all outcomes of an experiment is called a sample space and is usually denoted by S. An event is an arbitrary subset in S. It can contain one outcome, two outcomes, no outcomes (empty subset), the entire sample space, and the like. Standard notations for events are capital letters such as A, B, C, and so on.
Experiment
a planned activity carried out under controlled conditions
Outcome
a particular result of an experiment
Probability
a number between zero and one, inclusive, that gives the likelihood that a specific event will occur; the foundation of statistics is given by the following 3 axioms (by A.N. Kolmogorov, 1930’s): Let S denote the sample space and A and B are two events in S. Then:

• 0 ≤ P(A) ≤ 1
• If A and B are any two mutually exclusive events, then P(A OR B) = P(A) + P(B).
• P(S) = 1
Sample Space
the set of all possible outcomes of an experiment
The AND Event
An outcome is in the event A AND B if the outcome is in both A AND B at the same time.
The Complement Event
The complement of event A consists of all outcomes that are NOT in A.
The Conditional Probability of A GIVEN B
P(A|B) is the probability that event A will occur given that the event B has already occurred.
The Or Event
An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B.


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# Tag Info

## Hot answers tagged supermassive-black-hole

Accepted

### Why not take a picture of a closer black hole?

I was surprised too when I first heard they were trying to image M87's black hole. The short answer is because it's really, really big. It is 1500 times bigger (diameter) than our Sagittarius A*, and ...
• 3,968

### If two black hole event horizons overlap (touch) can they ever separate again?

You have already got some good answers, but I'll just try to provide one more intuitive solution on why the event horizons will never separate again if overlapping each other: First, imagine a speck ...

### If two black hole event horizons overlap (touch) can they ever separate again?

If the event horizons ever touch and become one continuous surface, their fate is sealed - the two black holes will merge all the way in. They can never separate again, no matter what. There are ...
• 17.6k

### What keeps galaxies together?

The galaxy is kept together by the combined mass of the matter in the galaxy, of which the supermassive black hole is a negligible part. There are galaxies that don't have a central black hole (such ...
• 94.2k
Accepted

### Why didn't the Event Horizon Telescope team mention Sagittarius A*?

There was a mention of Sagittarius A* during the Q+A portion of the press conference; the team indicated that they hope to produce an image sometime in the future (although they were careful to make ...
• 34.2k

### Why do things float in space, though the gravity of our star is always present?

It is not true that "objects float around" in the solar system. Perhaps you have seen video from the space station, and you can see things floating. This is not because there is no gravity, but ...
• 94.2k

### Why not take a picture of a closer black hole?

There are a few criteria necessary to see a black hole with the Event Horizon Telescope. They are, in importance: Active Feeding: you need a thick accretion disk with lots of matter accreting onto ...
• 586

### Why do things float in space, though the gravity of our star is always present?

To help with James K's excellent answer, a visual representation might help. Let's look at a thought experiment - Newton's Cannonball. Let's say you have a cannon, high enough that it's being held ...
• 421

### What keeps galaxies together?

This was studied many years ago. Not only do galaxies have to hold together, but there also has to be enough matter to hold it tightly enough to spin at the speed it turns. (Imagine swinging an object ...
• 974
Accepted

### What will happen to the shape of a galaxy when a super massive black hole lying in its center dies(evaporates out)?

Presumbably we rotate beacuse of the BH. No. The galaxy is being held in one piece due to its own total gravity. The black hole is only a small fraction of that. Basically, the BH doesn't matter. ...
• 17.6k
Accepted

### What is it exactly about these flares of infrared light from Sgr A* that "confirms" it is a supermassive black hole?

We have reasonably good measurements of the mass of Sagittarius A*, thanks to measurements of the movements of stars like S0-2 over several decades. It's been well-established that the mass of the ...
• 34.2k

### Can an entire star pass through the event horizon of a black hole unharmed?

In order to survive, the star's self-gravitation must be larger than the tidal stretching forces provided by the black hole. If not, then the star will get spaghettified before it crosses the event ...
• 120k

### Why do some supermassive black holes have a * in their name?

How the name came to be chosen is discussed in this paper1 (§3, pp. 4–5): Eight years after the discovery, one of us (Brown) invented the name Sgr A∗ to distinguish the compact source from the other ...
• 1,179
Accepted

### Time according to the gravity of Sagittarius A*?

Not at all a dumb question. As you have heard, it is true that time is affected by gravity. The stronger the gravitational field, the slower time passes. If you're far from any gravitating matter, ...
• 34.2k
Accepted

### Why is there no color shift on the photo of the M87 black hole?

The picture isn't a "colour" picture - it is monochrome. i.e. It is obtained at a single microwave wavelength of 1.3 mm, and so not at any wavelength you could see (Akiyama et al. 2019). ...
• 120k
Accepted

### Could there be dark matter black holes?

The problem with trying to form a black hole with dark matter is that dark matter can only weakly interact (if at all) with normal matter and itself, other than by gravity. This poses a problem. To ...
• 120k
Accepted

### How can a supermassive black hole cause so much energy to enlighten its matter when its massive gravity prevents light to escape?

It is quite correct that a black hole has so much mass that light cannot escape from a region around the black hole. The edge of this region is called the event horizon. If you cross an event horizon ...
• 94.2k

### What will happen to the shape of a galaxy when a super massive black hole lying in its center dies(evaporates out)?

Answer: Not much The Milky Way's central black hole (BH) masses about 5 million suns, while the galaxy masses 100 billion to a trillion suns. Consequently, the central BH is pretty much irrelevant ...
• 7,390
Accepted

### Why can't supermassive black holes merge? (or can they?)

The main problem is angular momentum. In order for two gravitationally bound objects to merge (whether black holes, supermassive black holes, planets, stars, etc.), they must shed enough angular ...
• 2,563
Accepted

### How can a supermassive black hole be 13 billion years old?

The answer to this is unknown at the present time. The issue is that an accreting "seed black hole" can only accrete at a limited rate. The limitation is provided by radiation pressure from ...
• 120k

### Why do some supermassive black holes have a * in their name?

Sgr A is a radio source from near the centre of the galaxy. In early radio wavelength maps it wasn't clear if this was a single source or multiple. With higher resolution it is clear that Sgr A is ...
• 94.2k

### Why didn't the Event Horizon Telescope team mention Sagittarius A*?

I've found an explanation in Dutch here by Heino Falcke, one of the EHT founders. Translation: Hard to photograph It was easiest to take a picture of M87. "It is very difficult to photograph the ...
• 4,700

### Why not take a picture of a closer black hole?

As Ingolifs says, Sgr A* and M87* are the obvious candidates. At the press conference, Heino Falcke explained why they got a picture of M87* first: But it would take some more time because ...
• 301

### Why do things float in space, though the gravity of our star is always present?

Ok, gotta quote XKCD on this. This is not how space works: This is: Gravity in low Earth orbit is almost as strong as gravity on the surface. The Space Station hasn't escaped Earth's gravity at all; ...

### How do we know that supermassive black holes can gain mass by means other than merging with other supermassive black holes?

The idea behind the paper (Shannon et al. 2013) that article is based on is to measure the gravitational wave background (GWB) produced by mergers of supermassive black holes, and determine which ...
• 34.2k
Accepted

### Can magnetism escape a black hole?

Nothing "escapes" a BH - in the sense that a signal originating inside the event horizon remains forever inside. If something is observed moving away from the BH, then it was generated outside the ...
• 17.6k
Accepted

### Why black holes are extremely cold?

Under General Relativity (GR) alone, a Black Hole's (BH's) event horizon is a point of no return -- anything that passes through the event horizon is lost and gone forever, and nothing comes out. ...
• 7,390
Accepted

### Would Hubble Space Telescope improve black hole image observed by EHT if it joined array of telesopes?

No, it would not, because it operates in the visible spectrum and the EHT is an array of radio telescopes. For the "very long baseline interferometry" technique to work, all the telescopes ...
• 2,585
Part of the answer is easy. The strain measured in that event was about $0.25\times 10^{-21}$. That is an object $1m$ long would be squeezed by $0.25\times 10^{-21} m$ in one direction and stretched ...


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Bug Tracker

ID: 152 🔗 Closed (Fixed) glossaries-extra 1.42 Nicola Talbot 2020-02-28 15:27:06 Infinite loop caused by \glslabel in \glsxtrentryfmt

Report

If \glslabel is used as the first argument of \glsxtrentryfmt it causes an infinite loop.

MWE

\documentclass{article}

\usepackage[record=only]{glossaries-extra}

\newcommand*{\sampleformat}[1]{\glsxtrentryfmt{\glslabel}{#1}}

\glssetcategoryattribute{general}{textformat}{sampleformat}

\begin{document}
\gls{lorem}
\printunsrtglossaries
\end{document}

Evaluation

Fixed in v1.43 (2020-02-28).

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