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# Intuitively, why does Bayes' theorem work?

I'm not looking for a cryptic math demonstration. Rather, I'm interested in the intuition behind the theorem that reveals the a posteriori probability, given the prior probability $\times$ the likelihood.

• Would be helpful if you could be a little more "down to earth" here and give a concrete example rather than just "throwing questions in the air" (i.e., "Why does Bayes' theorem work in the following case..."). – barak manos Mar 28 '15 at 15:14

Draw a Venn diagram to help you understand $P(A|B)=P(A\cap B)/P(B)$. Then use this to relate the quantities $P(A|B)$ and $P(B|A)$ algebraically. Let's discuss the first point.

Suppose we have a finite sample space so we can count the number outcomes in each possible "event." To determine $P(A|B)$, we're essentially asking what the probability of getting an outcome in $A$ is if we uniformly at random (for simplicity) pick an outcome in event $B$.

For example, consider a collection of 100 objects. Say 64 are balls and 36 are blocks. Suppose further that among the balls, 24 are red balls and 40 are blue balls. If $A$ is the event of being a red object (we don't know how many red blocks there are, but it won't matter) and $B$ is the event of being a ball, then $P(A|B)$ is the probability of picking a red object given the fact that the object you picked was a ball, or equivalently of picking a red ball out of all of the balls, which will equal the number of red balls divided by the number of balls, or $P(A\cap B)/P(B)=\frac{24}{64}$.

Now, given $P(A|B)P(B)=P(A\cap B)=P(B|A)P(A)$ it shouldn't be hard to finish.

• Thank you, this is very useful! – NLK511 Mar 28 '15 at 15:23
• @NLK511: It also helps to draw the Venn diagram in a rectangular fashion, as one can see depicted in the answer of user2023861 below. – Lee Mosher Sep 16 '15 at 15:29

The prior distribution and the likelihood function (based on data) both contain information about a parameter. Bayes' theorem allows these two kinds and sources of information to put together into a posterior distribution. The combined information from the posterior distribution can be used make inferences about the parameter. A couple of examples illustrate this process.

Screening test for a disease. Suppose we wonder whether a particular person has a disease. The prevalence in the population to which the subject belongs is 2%, so this can be considered as our prior information about the subject. P(Subj has Disease) = .02. A quick and inexpensive,, but imperfect, screening test for the disease is available. Its characteristics are described in conditional probabilities: P(Pos test | Dis) = .99, P(Neg test | No Dis) = .97. Suppose our data is that the subject tests positive.

Then using the elementary form of Bayes' Theorem we can find the posterior probability P(Dis | Pos test) = 0.4024. Some people, focusing on P{Pos test | Dis} = .99) are surprised the posterior probability is so small.

However, the appropriate focus for our purposes is that the data (positive test result) has gone together with the prior probability of 0.02 to give us a posterior probability about 0.40. The screening test is imperfect, but data from it has made a considerable change in our assessment of the subject's probability of disease. A subject with a 40% chance of having a serious disease should be evaluated with further and perhaps more time consuming and expensive tests.

Public Opinion Poll. A newly hired consultant for a political campaign to elect Candidate A feels that the candidate will win, but not overwhelmingly. Suppose her prior distribution on the probability $\psi$ of winning is $Beta(330, 270)$, which has mean 0.55 and 95% of its probability in the interval $(0.51,0.59).$ Then a poll of 1000 randomly selected potential voters shows 620 of them in favor of Candidate A. This is our data and it is reflected in the binomial likelihood function with kernel $\psi^{620}(1-\psi)^{380}$.

Bayes' Theorem melds the prior distribution with the likelihood function encoding the data to give the posterior distribution $Beta(950, 650)$, where multiplying the prior by the likelihood gives the posterior beta parameters $330+620=950$ and $270+380=650.$ The posterior beta distribution has mean about 0.59 and puts about 95% of its probability in the interval $(0.57, 0.62),$ which we take as our posterior probability interval for $\psi$, a somewhat more optimistic outlook for the candidate than given by the prior.

Here again, the information in the prior distribution and the data (as reflected in the likelihood function) have been combined to give a posterior distribution. Very roughly speaking, it is as if the consultant's prior distribution contributed information equivalent to that in a poll of 600 prospective voters of whom 330 favored the candidate.

Note: I have chosen these two examples, so that the math (if you care to carry it through) is quite simple. In some cases, much more computational effort is required to find and use the posterior distribution. But the computation needs to be viewed as a means to an end: to combine the information in the prior with the information in the data in order to make inferences based on both.

Acknowledgment: Numbers and distributions in these examples are the same as for ones in Ch 5 & 8, respectively, of Suess and Trumbo (2010), Springer.

Start with an example. Say you have a test that discovers some disease. Even if a person tests positive for the disease, all is not lost, since the test may not be accurate.

What are the person's chances of actually having the disease? There are three factors involved:

1. What are the overall chances of contracting the disease? It would be very unlikely for townsfolk of some small town in Europe who never left it to contract Ebola for instance, irrespective of the test results.
2. What is the test accuracy: Given that a person has the disease, what are the chances that you test positive?
3. What is the test accuracy II: How often does the test give positive results? If it almost always gives a negative result, but in your patient's case it gave a positive result, maybe you should be worried.

Combining these factors gives us Bayes's theorem - Factors 1 and 2 increase the probability and should therefore be multiplied, while Factor 3 decreases the probability, and should be divided:

$$P(\text{Is Sick} |\text{Pos. Test}) =\frac{P(\text{Pos. Test} | \text{Is Sick})\cdot P(\text{Is Sick})}{P(\text{Pos. Test})}$$

• nice explanation, thanks a lot :) – NLK511 Mar 28 '15 at 15:31

The answers here are good, but if you're like me, you learn better with visual aids and actual numbers. I have one for Bayes's Theorem using the same testing idea. Here's the setup:

• You have a population of 100,000 people
• 0.5% of this population use a drug
• We have a test that tells us with 99% accuracy if a person is or is not a drug user
• You tested positive for this drug. What are the odds that you're a user given this information?

Let's set this up in a simple grid:

In the columns, I show counts of users and non-users. We are told that 0.5% of the population of 100,000 use this drug, so there are 500 users and 99,500 non-users.

In the rows, I show counts of test results. I will fill in these values in a moment.

I've highlighted cells inside the table in a certain way:

• Green cells are accurate test results. If a person is a user and the test result is positive, the test is accurate and the cell is green. Likewise, if a person is not a user and the test result is negative, the test is accurate and the cell is green.
• Red cells are inaccurate test results. If a person is a user but the test is negative, this is a false-negative and the cell is red. Likewise, if a person is not a user but the test is positive, this is a false-positive and the cell is red.

Let's start filling in the table. Given that we have 500 users, how many positive and negative test results will we have among this subset of the population? We're told the test is 99% accurate, so that's 495 positive results and 5 false-negatives:

Following the same process for the non-users: given that we have 99,500 non-users, how many positive and negative results will we have among this subset of the population? Again, the test is 99% accurate, so that's 98,505 negative results and 995 false-positives.

Time to analyze the results. We're told that you tested positive for the drug, so let's throw out the results pertaining to negative test results and look at only the "Positive" row in the grid. 1,490 people tested positive for the drug, of which 495 actually are users, and 995 are false-positives. From here it's easy, the probability that you're actually a user given that you're in the "Positive" row is $\frac{495}{1490} = 33.2\%$

Turning this into a formula:

$P(Drug User Given Positive Result) = \frac{P(DrugUser)*N*TestAccuracy}{P(DrugUser)*N*TestAccuracy + P(NotDrugUser)*N*(1 - TestAccuracy)}$

This is exactly the formula I calculated to get 33.2%. Here, N is the 100,000 number that I made up. N cancels out and thus my 100,000 is extraneous information, but I thought it'd be helpful to include it.

With this grid I hope it's clear why, even with a test accuracy so high, your chances of actually being a drug user given a positive test result are so much lower. The number of false-positives is large enough to skew the naive intuition of the result.

• @Canada-Area51Proposal, Thanks. Why did you make a recent edit to my answer? Your edits don't add or elucidate any parts of it. After your edits, we're left with some awkward sentences like In the rows, I show counts of test results that I will fill, after I explain how I highlighted the cells in a certain way There are inconsistencies with how you refer to the four setup steps. And what's the point of changing Turning this into a formula to This can be turned into a formula? All that does is make the sentence passive. A highschool English teacher would scold you for that. – user2023861 Dec 27 '17 at 15:29
• Sorry for any mistake. I corrected the 2 sentences in the blockquotes. I edited because some sentences appear redundant and repeated? – Greek - Area 51 Proposal Dec 28 '17 at 1:08
• @Canada Ultimately it's between you and the answerer, but I think you're well overstepping the norm of reasonable editing of someone else's answer. – Rahul Dec 28 '17 at 1:12

It helps to disambiguate the meaning of "accuracy" more precisely like this Reddit comment, in which there's a typo: "1485" (in "out of 1485 people who test positive") ought be "1495". I rewrote it with whole numbers (rather than 0.5% as the disease rate).

To understand the theorem, you need to understand the vocabulary. "99% accurate" doesn't really give us information about the disease. We ought use the following terms:

Sensitivity - the odds that the test will be positive if you have the disease.

Specificity - the odds that the test will be negative if you lack the disease.

Positive predictive value - the odds that the test will correctly predict you have the disease, if you test positive.

Negative predictive value - the odds that the test will correctly predict you lack the disease, if you test negative.

Our population of 10,000 people has a 1% disease rate. So $$\color{springgreen}{1000}$$ people have the disease, and $$\color{forestgreen}{99,000}$$ don't.

We introduce a test that is 98% sensitive and 99% specific. It will correctly identify $$\color{deepskyblue}{980}$$ of 1000 people with the disease and $$\color{red}{98,010}$$ of 99,000 without the disease. It will incorrectly claim $$\color{red}{20}$$ people ($$= 1000 - 980$$) with the disease don't have it, and $$\color{deepskyblue}{990}$$ people ($$= 99,000 - 98,010$$) without the disease have it.

So out of $$\color{deepskyblue}{1970 \; (= 980 + 990)}$$ people who test positive, 980 have the disease. Thus, our positive predictive value is $$\dfrac{\color{springgreen}{1000}}{\color{deepskyblue}{1970}} = 50.76\%$$.

Out of $$\color{red}{98,030 \; (= 98,010 + 20)}$$ who test negative, $$\color{forestgreen}{99,000}$$ do not have the disease. Thus, our negative predictive value is $$\dfrac{\color{red}{98,030}}{\color{forestgreen}{99,000}} = 99.02\%$$.

In this case, this test is first-rate for determining who lacks the disease. The $$\color{deepskyblue}{1970}$$ who test positive can be tested to confirm they do have the disease, whereas those who tested negative need no further tests.

There are many professions where useful probability estimates are made without explicitly employing Bayes's formula, although the 'calculations' are still all about changing estimates given new information. If you believe in this, then you can accept that a math model can also accurately be employed in some situations.

Example 1: A specialist has been working for almost a year with a patient that has a serious form of cancer. The doctor now estimates that the patient has only a 10% chance of living more than a month. He shares the diagnosis and advises the patient that he can't recommend any further aggressive treatments. So, throughout his interactions with his patient, the specialist observed how events unfolded and was able to make 'soft assessment/estimates' while trying several treatments.

Example 2: A detective is meeting with a murder suspect. From his experience, he asks the suspect an indirect question that could shed light on the investigation. He knows that certain 'weird' responses will bring things into sharper focus. The suspect comes up with an offbeat response that the detective figures an innocent person couldn't even come up with. Yes, just circumstantial evidence, but the detective now has a prime suspect.


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# Isomorphisms preserve the property of being a principal ideal ring

Let $f:R\rightarrow S$ be an isomorphism. Prove that if we let $R$ be a principal ideal ring it follows that $S$ is a principal ideal ring too.

How should I begin the proof?

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What is isomorphic? – Phira May 11 '11 at 21:35
If the map is an isomorphism, then everything is obvious (the ideals are the same, the generators are the same, everything you want is preserved by isomorphisms). What you probably want is for the map to be surjective. – Aaron May 11 '11 at 21:41
@Zev Chonoles: I do not think that it is certain that the OP wanted to talk about isomorphisms. – Phira May 11 '11 at 21:41
@user9325: The original text was "Let R --> S be an isomorphic". It would make sense to have "Let $R$ and $S$ be isomorphic", but the existence of both the word "an" and the arrow notation, indicates to me that the isomorphism was intended to be an explicit part of this question. – Zev Chonoles May 11 '11 at 21:49
Also I have no compunctions about changing the title to something that is descriptive of what the question is about. – Zev Chonoles May 11 '11 at 21:50

$S$ is a principal ideal ring, by definition, when every ideal $I\subseteq S$ is principal, that is, $$I=(s)=\{sx\mid x\in S\}$$ for some $s\in S$.

You should know from class (or, if not, prove on your own) that, for any homomorphism $g:A\rightarrow B$ where $A$ and $B$ are rings, then if $J\subseteq B$ is an ideal of $B$, then $g^{-1}(J)$ is an ideal of A.

Let $I$ be any ideal of $S$. We want to show $I=(s)$ for some $s\in S$. Using the above fact, we know that $f^{-1}(I)$ is an ideal of $R$. Do you see how to use what we've assumed to be true about $R$ (that $R$ is a principal ideal ring), combined with the ability to send elements of $R$ to elements of $S$ by $r\mapsto f(r)$ and vice versa by $s\mapsto f^{-1}(s)$ (because $f$ is an isomorphism), to prove what we want?

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Isomorphism is a very strong condition. A ring $R$ has every property an isomorphic ring has as a ring, such as being principal ideal domain or being a field.

I think the condition of this problem can be weakened a little bit.

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## Question sentences involving negation

9

4

(1) Does he not know?

(2) Doesn't he know?

I don't usually see and hear questions formed in the first style. I was even surprised to know that it is grammatically correct. And actually it is the most "proper" way to form an interrogative sentence. Ngram shows that the first style was much more popular before than now.

What are your suggestions on choosing between these styles? And I will probably add the third one.

(3) He does not know?

EDIT:

The answers have explained how to form a question, but have not actually provided an account on the usage of the first two forms. Especially I'm interested in the first form: does it sound formal, is it used in speech or only in writing, et cetera.

1Ngrams expands "Doesn't he know?" to "Does not he know?", despite the fact that the second form is not used much and is generally considered ungrammatical. You can't trust Ngrams here. "Does not he know?" actually appears relatively commonly in 19th century literature, but I suspect that was because contractions were avoided then in written language, and that it usually represents a spoken "Doesn't he know?". – Peter Shor – 2014-03-21T21:03:19.643

1So you are really asking about when to use contracted negatives? I'd say you simply use "doesn't he know" when you would use "he doesn't know", and you use "does he not know" when you would write "he does not know". But that has little to do with question sentences. – oerkelens – 2014-03-23T12:54:21.843

"If you tickle us, do we not laugh?", in a well known play by Shakespeare. – QuentinUK – 2014-03-26T18:16:33.903

Well, the question was quite different when I answered it. It asked about "Does he not know?" versus "Does not he know?", and the main difference between these is that the latter is ungrammatical. – snailplane – 2014-03-27T05:04:17.970

## Answers

13

The basic (declarative) sentence is:

He does not know.

You can turn this into an interrogative clause with subject-auxiliary inversion. Just switch he and does:

Does he not know?

You can optionally replace does not with doesn't in the original sentence, using the suffix -n't rather than the word not:

He doesn't know.

Now the auxiliary is the single word doesn't, so if we apply subject-auxiliary inversion, we get:

Doesn't he know?

Lastly, you can turn the original sentence into a question by using rising intonation at the end, which you indicate in writing by replacing the period . with a question mark ?:

He does not know?

However, this is only appropriate in certain situations, for example as an echo question—repeating what you heard as a question to express incredulity or to confirm that you heard the speaker correctly. Most (but not all) situations where this would be appropriate are informal, and as a result "He does not know?" sounds somewhat unusual. Instead, you can use the contracted form:

He doesn't know?

But most of the time the interrogative form is more appropriate.

3

Actually, does not he know sounds strange, but when you use contraction, doesn't he know sounds perfectly fine.

It seems that when forming a question from a negative sentence, we do move the verb, but normally we do not move the not of the negation:

He does not know > Does he not know?
She does not play > Does she not play?
I do not read a lot > Do I not read a lot?

However, when we use a contraction to form the negative, the not gets "reunited" with the verb that it negates:

He doesn't know > Doesn't he know?
She doesn't play > Doesn't she play?
I don't read a lot > Don't I read a lot?

2

When you get confused, a good clue will be how a question tag is formed.

He does know, doesn't he? OR He knows, doesn't he?

Also, the contraction is doesn't. So, my vote goes to...

Doesn't he know?

Your third suggestion is also correct but then it's called as a declarative question where you are somewhat sure that he knows! I read it in Swan's Book, Practical English Usage.

Example -

She's your girlfriend?

2

"Does he not know?" is current usage. This is a bit counterintuitive, given that "Doesn't he know?" deconstructs to "Does not he know?" which of course isn't "correct". But that's English. :)

As Maulik says: "She doesn't know?" is a declarative question. It is often used to convey a strong expression of surprise. For example, suppose you had heard that your best friend had gotten a new girlfriend. You were getting together with them to meet her for the first time. It turns out that the girlfriend is your cousin's ex-wife; the divorce went through just a week ago. The conversation might go something like this:

You: She's your girlfriend?
Best Friend: Yep.
You: Seems a little early for you to be dating my cousin's ex-wife, since she only got divorced a week ago. How long have you been dating?
BF: She's your cousin's ex-wife?
You: Yes. You didn't know?
BF: No, I didn't know. (to girlfriend) And you only got divorced last week? Why didn't you tell me?
You (interrupting, to GF): You didn't tell him?
GF: It's complicated...

They are expressing surprise to the point of incredulity about the situation.

1

Entirely a point of view, but I've noticed the choice of one of those statements can depend on the point of the sentence.

If there is some specific information at issue eg, some piece of news or gossip, then the first example will generally be used. "I'm telling Steve that Michelle has been seeing Peter." "OMG, does he not know?"

On the other hand, if the lack of intelligence is the point of the sentence in some form, then the second seems more common. "Michelle needed to use her calculator to find three multiplied by eight." "OMG doesn't she know her times tables?"

This aligns somewhat with the negation in the first being applied to 'knowing' and in the second to 'doing'. It's also not a hard and fast rule, just how I've seen it used.

PS: It's rather archaic, but you could throw a fourth form in there: Does he know not?

And you can add the Yodaic form: "Know not, does he?" It's very important for ESL learners to know about this form which is never taught in ESL classes for some reason. – CoolHandLouis – 2014-03-27T21:49:56.593

Is not the form Yodaic: 'Know he, does not'? – mcalex – 2014-03-31T08:00:58.417


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## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 11 - Counting Methods and Probability Theory - 11.3 Combinations - Exercise Set 11.3 - Page 707: 34

#### Answer

3.066 x $10^{19}$

#### Work Step by Step

The number of possible combinations if r items are taken from n items is nCr = $\frac{n!}{r!(n-r)!}$ The number of ways to select Senate members for a Foreign Relations committee is 100C18= $\frac{100!}{18!(100-18)!}$ = $\frac{100!}{18!82!}$ = 3.066 x $10^{19}$ ways

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# “Peakedness” of a skewed probability density function

I would like to describe the "peakedness" and tail "heaviness" of several skewed probability density functions.

The features I want to describe, would they be called "kurtosis"? I've only seen the word "kurtosis" used for symmetric distributions?

• Indeed, the measures of kurtosis are typically applied to symmetric distributions. You can calculate it for skewed ones as well but the interpretation changes since this value varies when the asymmetry is introduced. In fact, these two concepts are difficult to separate. Recently, a skewness-invariant measure of kurtosis was proposed in this paper. – user10525 Jan 3 '13 at 16:22
• High kurtosis is associated with peakedness and with heavy tailedness (it's also characterized as 'lack of shoulders'). One of the volumes of Kendall and Stuart discuss these issues at some length. But such interpretations, are, as you note, generally given in the situation of near-symmetry. In nonsymmetric cases, the standardized 4th moment is usually highly correlated with the square of the standardized third moment, so they're mostly measuring much the same kind of thing. – Glen_b -Reinstate Monica Jan 3 '13 at 23:44
• Indeed, given the particular way I phrased it in my earlier comment, it's true even of symmetric distributions - the square of the sample standardized third moment (squared moment skewness) is highly correlated with the sample standardized fourth moment ('kurtosis'), even at say the normal. – Glen_b -Reinstate Monica Jun 19 '13 at 0:27

With variance being defined as the second moment $\mu_{2}$, skewness being defined as the third moment $\mu_{3}$ and the kurtosis being defined as the fourth moment $\mu_{4}$, it is possible to describe the properties of a wide range of symmetric and non-symmetric distributions from the data.

This technique was originally described by Karl Pearson in 1895 for the so-called Pearson Distributions I to VII. This has been extended by Egon S Pearson (date uncertain) as published in Hahn and Shapiro in 1966 to a wide range of symmetric, asymmetric and heavy tailed distributions that include Uniform, Normal, Students-t, Lognormal, Exponential, Gamma, Beta, Beta J and Beta U. From the chart of p. 197 of Hahn and Shapiro, $B_{1}$ and $B_{2}$ can be used to establish descriptors for skewness and kurtosis as:

$\mu_{3} = \sqrt {B_{1}\ \mu_{2}^{3}}$
$\mu_{4} = B_{2}\ \mu_{2}^{2}$

If you just wanted simple relative descriptors then by applying a constant $\mu_{2} = 1$ the skewness is $\sqrt {B_{1}}$ and the kurtosis is $B_{2}$.

We have attempted to summarize this chart here so that it could be programmed, but it is better to review it in Hahn and Shapiro (pp 42-49,122-132,197). In a sense we are suggesting a little bit of reverse engineering of the Pearson chart, but this could be a way to quantify what you are seeking.

The main issue here is, what is "peakedness"? Is it curvature at the peak (2nd derivative?) Does it require standardization first? (You would think so, but there is a stream of literature starting with Proschan, Ann. Math. Statist. Volume 36, Number 6 (1965), 1703-1706, that defines peakedness in a way that normal with smaller variance are more "peaked"). Or is it probability concentration within a standard deviation of the mean, as implicit in Balanda and Macgillivray (The American Statistician, 1988, Vol 42, 111-119)? Once you settle on a definition, then it should be trivial to apply it. But I would ask, "why do you care?" Of what relevance is "peakedness", however defined?

BTW, Pearson's kurtosis measures tails only, and does not measure any of the above mentioned "peakedness" definitions. You can change the data or distribution within a standard deviation of mean as much as you want (keeping the mean=0 and variance=1 constraint), but the kurtosis can only change within a maximum range of 0.25 (usually much less). So you can rule out using kurtosis to measure peakedness for any distribution, even though kurtosis is indeed a measure of tails for any distribution, no matter whether the distribution is symmetric, asymmetric, discrete, continuous, discrete/continuous mixture, or empirical. Kurtosis measures tails for all distributions, and virtually nothing about peak (however defined).

A possible very practical approach could be calculate the ratio of the survival function of the distribution $\Pr\left(\tilde X \gt 1- \alpha \right)$ against the normal one, showing it is quite far greater. Another approach can be calculating the ratios of percentiles $w_1=\frac{\tilde{x_{99}}-\tilde{x_{50}}}{\tilde{x_{75}}-\tilde{x_{50}}}$ of the distribution $\tilde x$ under interest and dividing it against the normal one quantile values, $w_2=\frac{\tilde{\Phi_{99}}-\tilde{\Phi_{50}}}{\tilde{\Phi_{75}}-\tilde{\Phi_{50}}}$, $\tau=\frac{w_1}{w_2}$.

I am not sure I get your understanding of peakedness and heaviness. Kurtosis means "Excess" in German, so it describes the "head" or "peak" of a distribution, describing whether it is very wide or very narrow. Wikipedia states that the "peakedness" is actually described by the "kurtosis", whereas peakedness does not to appear to be a real word and you should use the term "Kurtosis".

So I think you might have gotten everything right, the head is the Kurtosis, The "heaviness" of the tail might be the Skewness":

Here is how you find it:

$$a_3 = \frac{\Sigma^{N}_{i=1}(x_i - \overline x)^3}{N * s^3_x}$$

with s as the standard deviation for x.

The values indicate:

Negative Skew: $$a_3 < 0$$

Positive Skew: $$a_3 > 0$$

No Skew $$a_3 = 0$$

You can get a value for the kurtosis with: $$a_4 = \frac{\Sigma^{N}_{i=1}(x_i - \overline x)^4}{N * s^4_x}$$

The values indicate:

Platycurtic: $$a_4 < 3$$

Leptocurtic: $$a_4 > 3$$

Normal: $$a_4 = 3.0$$

Did that help?

• I'm afraid this answer in its current form may be less than helpful due to errors in it. Skewness is a standard measure of asymmetry. It is not closely related to heaviness of tails: it is possible for the tails to be extremely heavy and the skewness to be zero (which is the case for any symmetric distribution, for instance). Please note, too, that it is impossible for $a_4$ to be negative, so the second half of this answer makes little sense. (Perhaps you confused kurtosis with excess kurtosis?) – whuber Jan 3 '13 at 17:53
• Thank you for clarifying. There might indeed be some errors in the formulas, I just copied them from the scripts they provide at uni. I oversaw the fact that a4 can't be negative. – Johannes Hofmeister Jan 3 '13 at 22:46
• I looked up why my answer is wrong - it is a translational error, I apologize for that. My slides are all in German, mixing Kurtosis and Excess. – Johannes Hofmeister Jan 3 '13 at 22:52
• @Peter As Peter Westfall keeps pointing out, your comment is incorrect: "peakedness" (of any mode), thought of vaguely as pointiness or height, has absolutely nothing to do with the tails of any distribution, nor is it measured by any finite combination of moments (such as the kurtosis). It may happen to be connected to heaviness of tails for a family of distributions, but that's a completely different matter. – whuber Jan 3 '18 at 13:44

Kurtosis is definitely associated with the peakedness of the curve. I henceforth believe that you are really looking for kurtosis which does exist whether the distribution is symmetric or not. (user10525) has definitely said it right ! I hope your problem is resolved by now. Do share its outcome, all opinions are welcome.

• I'm not sure how this constitutes a helpful answer beyond what was already written here. How about you expand more on kurtosis and peakedness of the curve? – Momo Oct 18 '13 at 18:58
• Wanted to give clear cut clarification to the query. The discussion seemed to be confusing @Momo – Vani Oct 18 '13 at 19:00


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# If $PAP^{-1} = B$, does there exist $Q$ with positive determinant such that $QAQ^{-1} = B$?

Suppose $$A,B$$ are conjugate real $$n \times n$$-matrices with $$B = PAP^{-1}$$ for some real matrix $$P$$. Does there exist a real matrix $$Q$$ with positive determinant such that $$B = QAQ^{-1}$$? The answer to the question is obviously yes if $$n$$ is odd since we can just take $$Q = -P$$. What about even $$n$$?

I asked this question because I wanted to show that the conjugacy classes of real matrices are path-connected (which is not true, as it turns out) and I had reduced the problem down to the above question.

A counterexample for $$n = 2$$ is $$A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$ and $$B = -A$$. Then, the only $$2\times 2$$-matrices $$P$$ that satisfy $$PA = BP$$ are those of the form $$\begin{pmatrix} a & b \\ b & -a \end{pmatrix}$$. Obviously, if such a matrix $$P$$ has real entries, then its determinant is $$\leq 0$$.


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# Density Operators, Trace and Partial Trace

#### skynelson

I have some math questions about quantum theory that have been bugging me for a while, and I haven't found a suitable answer in my own resources. I'll start with the Trace operation.

Question A) My understanding is that if we take system A and perform the partial trace over system B, we essentially remove any dependence or correlation of the system A on system B. For example, I have heard reference in the literature to 'performing a partial trace over the environment' in order to essentially remove the influence of the environment on the system being analyzed.

Is this intuitive understanding correct?

Can anyone provide a more concrete understanding of why the trace is interesting? Since it is simply the sum of the diagonal elements, can you explain why it is interesting/useful in the following two cases, and what it represents?
1) A diagonalized matrix (the diagonals are the eigenvalues of the system)
2) A Hermitian but non-diagonalized matrix

Question B)
Furthermore, let's say we have a density operator, which may have non-diagonal elements, but has trace equal to 1 (because it is normalized). (Am I correct so far?)

1) What is the meaning of taking the trace, and disregarding the off-diagonal elements? Why are we allowed to disregard the rest of the matrix?
2) Will the trace change if we change the basis of the density operator?

You may find that my questions belie misunderstandings in the nature of some of these concepts, which is why I appreciate any help that can be provided!

Related Quantum Physics News on Phys.org

#### DrClaude

Mentor
Question A) My understanding is that if we take system A and perform the partial trace over system B, we essentially remove any dependence or correlation of the system A on system B. For example, I have heard reference in the literature to 'performing a partial trace over the environment' in order to essentially remove the influence of the environment on the system being analyzed.

Is this intuitive understanding correct?
Yes.

Can anyone provide a more concrete understanding of why the trace is interesting? Since it is simply the sum of the diagonal elements, can you explain why it is interesting/useful in the following two cases, and what it represents?
1) A diagonalized matrix (the diagonals are the eigenvalues of the system)
2) A Hermitian but non-diagonalized matrix
The trace is interesting because it is the way to calculate useful results. For instance, the expectation value of operator $A$ over the density operator $\rho$ is $Tr(A \rho)$. It is also used to calculate the norm of a density operator, $Tr(\rho)$, and to check if it represents a pure state, $Tr(\rho^2) = 1$, or a mixed state, $Tr(\rho^2) < 1$.

Question B)
Furthermore, let's say we have a density operator, which may have non-diagonal elements, but has trace equal to 1 (because it is normalized). (Am I correct so far?)
Correct.

1) What is the meaning of taking the trace, and disregarding the off-diagonal elements? Why are we allowed to disregard the rest of the matrix?
It gives the norm. It is not that the off-diagonal elements are disregarded, it is that they are not relevant to the norm.

2) Will the trace change if we change the basis of the density operator?
No. Note that the above examples I gave are in terms of the density operator, without reference to a particular representation (which would give a density matrix).

"Density Operators, Trace and Partial Trace"

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text:
#### The Lazy Online Subgradient Algorithm is Universal on Strongly Convex Domains

We study Online Lazy Gradient Descent for optimisation on a strongly convex domain. The algorithm is known to achieve $O(\sqrt N)$ regret against adversarial opponents; here we show it is universal in the sense that it also achieves $O(\log N)$ expected regret against i.i.d opponents. This improves upon the more complex meta-algorithm of Huang et al \cite{FTLBall} that only gets $O(\sqrt {N \log N})$ and $O(\log N)$ bounds. In addition we show that, unlike for the simplex, order bounds for pseudo-regret and expected regret are equivalent for strongly convex domains.


text:
# Facebook Birthday Problem: Part 5

Recently, I devised the following problem:

Suppose that you have n friends, and you always say “Happy Birthday” to each friend on his/her birthday. On how many days of the year will you not say “Happy Birthday” to one of your friends?

Until somebody tells me otherwise, I’m calling this the Facebook birthday problem in honor of Facebook’s daily alerts to say “Happy Birthday” to friends.

Here’s how I solved this problem. Let $I_k$ be an indicator random variable for “no friend has a birthday on day $k$, where $k = 366$ stands for February 29 and $k = 1, \dots, 365$ stand for the “usual” 365 days of the year. Therefore, the quantity $N$, representing the number of days of the year on which no friend has a birthday, can be written as

$N = I_1 + \dots + I_{365} + I_{366}$

In yesterday’s post, I began the calculation of the standard deviation of $N$ by first computing its variance. This calculation is complicated by the fact that $I_1, \dots, I_{366}$ are dependent. Yesterday, I showed that

$\hbox{Var}(N) = 365\displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right] + \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]$

$+ \displaystyle (365)(364) \left[ \displaystyle \left( \frac{1453}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^{2n} \right] + 2 \sum_{k=1}^{365}\hbox{Cov}(I_k,I_{366})$.

To complete this calculation, I’ll now find $\hbox{Cov}(I_k,I_{366})$, where $1 \le k \le 365$. I’ll use the usual computation formula for a covariance,

$\hbox{Cov}(I_k,I_{366}) = E(I_k I_{366}) - E(I_k) E(I_{366})$.

We have calculated $E(I_k)$ earlier in this series. In any four-year span, there are $4 \times 365 + 1 = 1461$ days, of which only one is February 29. Assuming the birthday’s are evenly distributed (which actually doesn’t happen in real life), the chance that someone’s birthday is not on day $k$ is

$\displaystyle 1 - \frac{4}{1461} = \displaystyle \frac{1457}{1461}$,

so that the probability that no friend has a birthday on day $k$ is

$\displaystyle \left( \frac{1457}{1461} \right)^n$.

Therefore, since the expected value of an indicator random variable is the probability that the event happens, we have

$E(I_k) = \displaystyle \left( \frac{1457}{1461} \right)^n$

for $k = 1, \dots, 365$. Similarly,

$E(I_{366}) = \displaystyle \left( \frac{1460}{1461} \right)^n$,

so that

$\hbox{Cov}(I_k,I_{366}) = E(I_k I_{366}) - \displaystyle \left( \frac{1457}{1461} \right)^n \left( \frac{1460}{1461} \right)^n$.

To find $E(I_k I_{366})$, we note that since $I_k$ is equal to either 0 or 1 and $I_{366}$ is equal to either 0 or 1, the product $I_k I_{366}$ can only equal 0 and 1 as well. Therefore, $I_k I_{366}$ is itself an indicator random variable. Furthermore, $I_k I_{366} = 1$ if and only if $I_k = 1$ and $I_{366} = 1$, which means that no friends has a birthday on either day $k$ or day $366$ (that is, February 29). The chance that someone doesn’t have a birthday on day $k$ or February 29 is

$\displaystyle 1 - \frac{4}{1461} - \frac{1}{1461} = \displaystyle \frac{1456}{1461}$,

so that the probability that no friend has a birthday on day $k$ or February 29 is

$\displaystyle \left( \frac{1456}{1461} \right)^n$.

Therefore, as before,

$E(I_k I_{366}) = \displaystyle \left( \frac{1456}{1461} \right)^n$,

so that

$\hbox{Cov}(I_k,I_{366}) = \displaystyle \left( \frac{1456}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^n \left( \frac{1460}{1461} \right)^n$.

Therefore,

$\hbox{Var}(N) = 365\displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right] + \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]$

$+ \displaystyle (365)(364) \left[ \displaystyle \left( \frac{1453}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^{2n} \right] + 2(365) \left[ \left( \frac{1456}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^n \left( \frac{1460}{1461} \right)^n \right]$,

and we find the standard deviation of $N$ using

$\hbox{SD}(N) = \sqrt{\hbox{Var}(N)}$.

The graph below shows the expected value of $N$, which was shown earlier to be

$E(N) = 365 \displaystyle \left( \frac{1457}{1461} \right)^n + \left( \frac{1460}{1461} \right)^n$,

along with error bars representing two standard deviations.

Interestingly, the standard deviation of $N$ changes for different values of $n$; a direct calculation shows that the $\hbox{SD}(N)$ is maximized at $n = 459$ with maximum value of approximately $6.1$. Accordingly, for $n = 450$ and $n = 500$, the error bars in the above figure have a total width of approximately 24 days (two standard deviations both above and below the expected value).

# Facebook Birthday Problem: Part 4

Recently, I devised the following problem:

Suppose that you have n friends, and you always say “Happy Birthday” to each friend on his/her birthday. On how many days of the year will you not say “Happy Birthday” to one of your friends?

Until somebody tells me otherwise, I’m calling this the Facebook birthday problem in honor of Facebook’s daily alerts to say “Happy Birthday” to friends.

Here’s how I solved this problem. Let $I_k$ be an indicator random variable for “no friend has a birthday on day $k$, where $k = 366$ stands for February 29 and $k = 1, \dots, 365$ stand for the “usual” 365 days of the year. Therefore, the quantity $N$, representing the number of days of the year on which no friend has a birthday, can be written as

$N = I_1 + \dots + I_{365} + I_{366}$

In yesterday’s post, I began the calculation of the standard deviation of $N$ by first computing its variance. This calculation is complicated by the fact that $I_1, \dots, I_{366}$ are dependent. Yesterday, I showed that

$\hbox{Var}(N) = 365\displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right] + \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]$

$+ \displaystyle 2 \!\!\!\!\! \sum_{1 \le j < k \le 365} \!\!\!\!\! \hbox{Cov}(I_j,I_k) + 2 \sum_{k=1}^{365} \hbox{Cov}(I_k,I_{366})$

To complete this calculation, I’ll now find the covariances. I’ll begin with $\hbox{Cov}(I_j,I_k)$ if $1 \le j < k \le 365$; that is, if $j$ and $k$ are days other than February 29. I’ll use the usual computation formula for a covariance,

$\hbox{Cov}(I_j,I_k) = E(I_j I_k) - E(I_j) E(I_k)$.

We have calculated $E(I_k)$ earlier in this series. In any four-year span, there are $4 \times 365 + 1 = 1461$ days, of which only one is February 29. Assuming the birthday’s are evenly distributed (which actually doesn’t happen in real life), the chance that someone’s birthday is not on day $k$ is

$\displaystyle 1 - \frac{4}{1461} = \displaystyle \frac{1457}{1461}$,

so that the probability that no friend has a birthday on day $k$ is

$\displaystyle \left( \frac{1457}{1461} \right)^n$.

Therefore, since the expected value of an indicator random variable is the probability that the event happens, we have

$E(I_k) = \displaystyle \left( \frac{1457}{1461} \right)^n$

for $k = 1, \dots, 365$. Therefore,

$\hbox{Cov}(I_j,I_k) = E(I_j I_k) - \displaystyle \left( \frac{1457}{1461} \right)^n \left( \frac{1457}{1461} \right)^n = E(I_j I_k) - \displaystyle \left( \frac{1457}{1461} \right)^{2n}$.

To find $E(I_j I_k)$, we note that since $I_j$ is equal to either 0 or 1 and $I_k$ is equal to either 0 or 1, the product $I_j I_k$ can only equal 0 and 1 as well. Therefore, $I_j I_k$ is itself an indicator random variable, which I’ll call $I_{jk}$. Furthermore, $I_{jk}$ if and only if $I_j = 1$ and $I_k = 1$, which means that no friends has a birthday on either day $j$ or day $k$. The chance that someone doesn’t have a birthday on day $j$ or day $k$ is

$\displaystyle 1 - \frac{4}{1461} - \frac{4}{1461} = \displaystyle \frac{1453}{1461}$,

so that the probability that no friend has a birthday on day $j$ or $k$ is

$\displaystyle \left( \frac{1453}{1461} \right)^n$.

Therefore, as before,

$E(I_j I_k) = \displaystyle \left( \frac{1453}{1461} \right)^n$,

so that

$\hbox{Cov}(I_j,I_k) = \displaystyle \left( \frac{1453}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^{2n}$.

Since there are $\displaystyle {365 \choose 2} = \displaystyle \frac{365\times 364}{2}$ pairs $(j,k)$ so that $1 \le j < k \le 365$, we have

$\hbox{Var}(N) = 365\displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right] + \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]$

$+ \displaystyle 2 \times \displaystyle \frac{365\times 364}{2} \left[ \displaystyle \left( \frac{1453}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^{2n} \right] + 2 \sum_{k=1}^{365}\hbox{Cov}(I_k,I_{366})$,

or

$\hbox{Var}(N) = 365\displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right] + \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]$

$+ \displaystyle (365)(364) \left[ \displaystyle \left( \frac{1453}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^{2n} \right] + 2 \sum_{k=1}^{365}\hbox{Cov}(I_k,I_{366})$.

The calculation of $\hbox{Cov}(I_k,I_{366})$ is similar to the above calculation; I’ll write this up in tomorrow’s post.

# Facebook Birthday Problem: Part 3

Recently, I devised the following problem:

Suppose that you have n friends, and you always say “Happy Birthday” to each friend on his/her birthday. On how many days of the year will you not say “Happy Birthday” to one of your friends?

Until somebody tells me otherwise, I’m calling this the Facebook birthday problem in honor of Facebook’s daily alerts to say “Happy Birthday” to friends.

Here’s how I solved this problem. Let $I_k$ be an indicator random variable for “no friend has a birthday on day $k$, where $k = 366$ stands for February 29 and $k = 1, \dots, 365$ stand for the “usual” 365 days of the year. Therefore, the quantity $N$, representing the number of days of the year on which no friend has a birthday, can be written as

$N = I_1 + \dots + I_{365} + I_{366}$

In yesterday’s post, I showed that

$E(N) = E(I_1) + \dots + E(I_{365}) + E(I_{366}) = 365 \displaystyle \left( \frac{1457}{1461} \right)^n + \left( \frac{1460}{1461} \right)^n$.

The calculation of the standard deviation of $N$ is considerably more complicated, however, since the $I_1, \dots, I_{366}$ are dependent. So we will begin by computing the variance of $N$:

$\hbox{Var}(N) = \displaystyle \sum_{k=1}^{366} \hbox{Var}(I_k) + 2 \!\!\!\!\! \sum_{1 \le j < k \le 366} \!\!\!\!\! \hbox{Cov}(I_j,I_k)$,

or

$\hbox{Var}(N) = \displaystyle \sum_{k=1}^{365} \hbox{Var}(I_k) + \hbox{Var}(I_{366}) + 2 \!\!\!\!\! \sum_{1 \le j < k \le 365} \!\!\!\!\! \hbox{Cov}(I_j,I_k) + 2 \sum_{k=1}^{365} \hbox{Cov}(I_k,I_{366})$

For the first term, we recognize that, in any four-year span, there are $4 \times 365 + 1 = 1461$ days, of which only one is February 29. Assuming the birthday’s are evenly distributed (which actually doesn’t happen in real life), the chance that someone’s birthday is not on day $k$ is

$\displaystyle 1 - \frac{4}{1461} = \displaystyle \frac{1457}{1461}$.

Therefore, the chance that all $n$ friends don’t have a birthday on day $k$ is

$\displaystyle \left( \frac{1457}{1461} \right)^n$.

Using the formula $\hbox{Var}(I) = p(1-p)$ for the variance of an indicator random variable, we see that

$\hbox{Var}(I_k) = \displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right]$

for $k = 1, \dots, 365$. Similarly, for the second term,

$\hbox{Var}(I_{366}) = \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]$

Therefore, so far we have shown that

$\hbox{Var}(N) = 365\displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right] + \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]$

$+ \displaystyle 2 \!\!\!\!\! \sum_{1 \le j < k \le 365} \!\!\!\!\! \hbox{Cov}(I_j,I_k) + 2 \sum_{k=1}^{365} \hbox{Cov}(I_k,I_{366})$

In tomorrow’s post, I’ll complete this calculation by finding the covariances.

# Facebook Birthday Problem: Part 2

Recently, I devised the following problem:

Suppose that you have n friends, and you always say “Happy Birthday” to each friend on his/her birthday. On how many days of the year will you not say “Happy Birthday” to one of your friends?

Until somebody tells me otherwise, I’m calling this the Facebook birthday problem in honor of Facebook’s daily alerts to say “Happy Birthday” to friends.

Here’s how I solved this problem. Let $I_k$ be an indicator random variable for “no friend has a birthday on day $k$, where $k = 366$ stands for February 29 and $k = 1, \dots, 365$ stand for the “usual” 365 days of the year. Therefore, the quantity $N$, representing the number of days of the year on which no friend has a birthday, can be written as

$N = I_1 + \dots + I_{365} + I_{366}$

Let’s start with any of the “usual” days. In any four-year span, there are $4 \times 365 + 1 = 1461$ days, of which only one is February 29. Assuming the birthday’s are evenly distributed (which actually doesn’t happen in real life), the chance that someone’s birthday is not on day $k$ is

$\displaystyle 1 - \frac{4}{1461} = \displaystyle \frac{1457}{1461}$.

Therefore, the chance that all $n$ friends don’t have a birthday on day $k$ is

$\displaystyle \left( \frac{1457}{1461} \right)^n$.

Since the expected value of an indicator random variable is the probability of the event, we see that

$E(I_k) = \displaystyle \left( \frac{1457}{1461} \right)^n$

for $k = 1, \dots, 365$. Similarly, the expected value for the indicator for February 29 is

$E(I_{366}) = \displaystyle \left( \frac{1460}{1461} \right)^n$.

Since $E(X+Y) = E(X) + E(Y)$ even if $X$ and $Y$ are dependent, we therefore conclude that

$E(N) = E(I_1) + \dots + E(I_{365}) + E(I_{366}) = 365 \displaystyle \left( \frac{1457}{1461} \right)^n + \left( \frac{1460}{1461} \right)^n$.

This function is represented by the red dots on the graph below.

In tomorrow’s post, I’ll calculate of the standard deviation of $N$.

# Facebook Birthday Problem: Part 1

The “birthday problem” is one of the classic problems in elementary probability because of its counter-intuitive solution. From Wikipedia:

In probability theory, the birthday problem or birthday paradox concerns the probability that, in a set of n randomly chosen people, some pair of them will have the same birthday. By the pigeonhole principle, the probability reaches 100% when the number of people reaches 367 (since there are only 366 possible birthdays, including February 29). However, 99.9% probability is reached with just 70 people, and 50% probability with 23 people. These conclusions are based on the assumption that each day of the year (except February 29) is equally probable for a birthday.

Recently, I devised the following different birthday problem:

Suppose that you have n friends, and you always say “Happy Birthday” to each friend on his/her birthday. On how many days of the year will you not say “Happy Birthday” to one of your friends?

Until somebody tells me otherwise, I’m calling this the Facebook birthday problem in honor of Facebook’s daily alerts to say “Happy Birthday” to friends.

In this series, I will solve this problem. While this may ruin the suspense, here’s a graph of the solution for $100 \le n \le 1000$ along with error bars indicating two standard deviations.

Before deriving this solution, I’ll start with a thought bubble if you’d like to take some time to think about how to do this.

# What is a Random Walk?

From PBS Infinite Series:

# A Long-Sought Proof, Found and Almost Lost

I enjoyed this article from Quanta Magazine, both for its mathematical content as well as the human interest story.

# A Long-Sought Proof, Found and Almost Lost

From the opening paragraphs:

Known as the Gaussian correlation inequality (GCI), the conjecture originated in the 1950s, was posed in its most elegant form in 1972 and has held mathematicians in its thrall ever since. “I know of people who worked on it for 40 years,” said Donald Richards, a statistician at Pennsylvania State University. “I myself worked on it for 30 years.”

[Thomas] Royen hadn’t given the Gaussian correlation inequality much thought before the “raw idea” for how to prove it came to him over the bathroom sink… In July 2014, still at work on his formulas as a 67-year-old retiree, Royen found that the GCI could be extended into a statement about statistical distributions he had long specialized in. On the morning of the 17th, he saw how to calculate a key derivative for this extended GCI that unlocked the proof. “The evening of this day, my first draft of the proof was written,” he said.

Not knowing LaTeX, the word processer of choice in mathematics, he typed up his calculations in Microsoft Word, and the following month he posted his paper to the academic preprint site arxiv.org. He also sent it to Richards, who had briefly circulated his own failed attempt at a proof of the GCI a year and a half earlier. “I got this article by email from him,” Richards said. “And when I looked at it I knew instantly that it was solved” …

Proofs of obscure provenance are sometimes overlooked at first, but usually not for long: A major paper like Royen’s would normally get submitted and published somewhere like the Annals of Statistics, experts said, and then everybody would hear about it. But Royen, not having a career to advance, chose to skip the slow and often demanding peer-review process typical of top journals. He opted instead for quick publication in the Far East Journal of Theoretical Statistics, a periodical based in Allahabad, India, that was largely unknown to experts and which, on its website, rather suspiciously listed Royen as an editor. (He had agreed to join the editorial board the year before.)

With this red flag emblazoned on it, the proof continued to be ignored… No one is quite sure how, in the 21st century, news of Royen’s proof managed to travel so slowly. “It was clearly a lack of communication in an age where it’s very easy to communicate,” Klartag said.

# Happy Independence Day

From last year’s Math With Bad Drawings (I recommend reading the whole article):


text:
# Secret of the universe

There once was a mythical stone tablet on which was said to be written the secret of the universe.

If the secret was to fall within the wrong hands, it was said that the consequences would be catastrophic.

This is why the message on the tablet had been encrypted since its creation.

A MAGE in charge of the task of protecting the tablet was one day faced by a powerful demon who claimed he could solve the encryption and so wanted to steal the tablet.

The two fought for a long time but when the MAGE was about to lose, he decided to sacrifice himself with a powerful spell that merged the MAGE's entire being within the tablet.

The content was therefore changed and prevented the demon from understanding its content.

Many years later, a famous leader found the tablet.

Curious about its content, he tried to solve its encryption.

He then tried to decipher the message on a piece of paper by using a cipher that he created himself.

Frustrated that it did not work, he smashed the tablet into pieces.

If only he knew that his cipher would have worked if the MAGE had not intervened a long time ago.

Many more years later, you found the piece of paper on which was written his attempt at deciphering the message.

The message on the paper is

dbjxbdhbexjdlxldjdboxxbd jxdkbidtxjtjbxjyd

Will you be able to decipher the secret of the universe?

• Love the story. +1 for it. So the message that we currently have is the leader's attempt which failed but would have solved it in case the MAGE hadn't intervened? What about the actual encrypted text on the tablet? – Abhirath Mahipal Oct 28 '16 at 8:19
• @AbhirathMahipal It's hard to not give too much away by answering. Let's just say that all that is left of the original is the failed attempt you hold in your hands now. – stack reader Oct 28 '16 at 8:24
• @stackreader, no problem hehe neither the 48 people that viewed xD – lois6b Oct 28 '16 at 9:44
• He then tried to decipher the message on a piece of paper by using a cipher that he created himself. -- Cryptoanalyst of the Century :))))))) – mr23ceec Oct 28 '16 at 10:44

The secret of the universe is

Hello Kitty

Because the famous leader who found the tabelet is

Caesar.

He invented a code named

Caesar code, which is shifting all the letters by 3 places in alphabet.

The message on the piece of paper has been decoded using

Caesar cipher

so if we use the cipher above to encode it back we get

gemaegkehamgoaogmgeraaeg magnelgwamwmeambg

When we undo what the MAGE has done to it

the MAGE merged all of himself to the message - that is the letters 'm', 'a', 'g' and 'e'. As we take out all the letters 'm', 'a', 'g' and 'e'

we get

khoor nlwwb

which again decoded by the cipher of the great leader gives us

hello kitty

• Everytime I skip solving cipher's, the answers are very easy :D +1 – Techidiot Oct 28 '16 at 11:17
• Too easy was it? Maybe I should add 2-3 more layers of encryption or give less hints XD. Kinda frustrating when the puzzle get solved faster than the time it took to make it ;P. I hope it was fun though. That is the only reason I make those after all. – stack reader Oct 28 '16 at 11:25
• The (no-so-high) difficulty level was adjusted perfectly to my deciphering skills :P I had fun solving this so thanks for the puzzle. – oleslaw Oct 28 '16 at 12:13
• @stackreader Don't get discouraged. Many of my puzzles on this site have been solved very quickly. It's not the sign of an inferior puzzle; just a sign that it's being seen by a lot of very clever people. – GentlePurpleRain Oct 28 '16 at 14:26
• @GentlePurpleRain Thanks for the link, was an interesting read. – stack reader Oct 28 '16 at 14:59

Hello Kitty

My solution probably isn't going to earn any points, because I solved it within a couple of minutes by

grouping identical letters while reading them: dddddddd bbbbbbb jjjjjjj xxxxxxxx h e ll o k i tt y

without even thinking of

Ceasar's or any other cipher,

and the Mage was nowhere to be seen either. At first I immediately noticed that

the letters are close together on the keyboard, like what you get when quickly typing random nonsense,

but that was a false clue.

Seems like I did stumble on a backdoor to the universe, though...

• This is a good answer because it shows practical cryptanalysis. – A. I. Breveleri Oct 29 '16 at 13:08
• To improve this answer, notice that "dddddddd bbbbbbb jjjjjjj xxxxxxxx" is Caesarean rotation -3 of "gggggggg eeeeeee mmmmmmm aaaaaaaa", which is enough to spell "mage mage mage mage mage mage mage ag". I assume the mage's name was "Ag". – A. I. Breveleri Oct 29 '16 at 13:14
• @A.I.Breveleri Haha! In fact I didn't notice, so it would by cheating to add it to my answer. I'm happy as it is. Thanks for the upvotes everyone, but oleslaw deserves full credit for explaining the whole story. – Zaaikort Oct 29 '16 at 13:43
• @A.I.Breveleri Maybe it was a silver mage. – jpmc26 Oct 29 '16 at 21:46

The true solution to the cipher is:

Goodbye Dog

Because

The demon could understand falsehood and lie. Therefore, it could break the encryption.

Also,

"Hello Kitty" is garnered by using a caeser cipher, removing all the letters in "mage" and then undoing the caesar cipher.

This holds true with what I said, because:

The caesar cipher was only used in obscuring the mage. Therefore, the tablet must have originally said "Hello Kitty". Hence, the demon could solve the encryption by simply assuming the tablet to be lying.

Also:

I'm so sorry. I couldn't resist.

• So enjoyed to see someone trolling puzzling for questions that allow lots of potential answers. – polkovnikov.ph Oct 30 '16 at 14:29


text:
i need it step by step thanks!

Image text transcribed for accessibility: Consider a stream of binary data consisting of a long sequence of Is, followed by a zero, followed by a long sequence of Is. Preceding bit and level is indicated within parentheses. Draw the waveforms for NRZI (high), AMI (1 as negative voltage), and pseudo-ternary (0 as negative voltage). (6 points) The analog waveform shown in the following figure is to be delta modulated. The sampling period and the step size are indicated by the grid. The first DM output is also shown. Give the DM output for the complete signal. (4 points


text:
# On Uniformization of Complex Manifolds: The Role of Connections. (MN-22)

ROBERT C. GUNNING
Pages: 150
https://www.jstor.org/stable/j.ctt130hk2c

1. Front Matter
(pp. None)
2. PREFACE
(pp. i-i)
R. C. Gunning
(pp. ii-ii)
4. §1. Introduction
(pp. 1-5)

The general uniformization theorem for Riemann surfaces is one of the most remarkable results in complex analysis, and is at the center of a circle of problems which are still very actively being investigated. An interest in extending this theorem to complex manifolds of higher dimensions has long been manifest, and indeed there have been several extensions of one or another aspect of the general uniformization theorem. As has been observed in other cases, some theorems in classical complex analysis appear as the accidental concurrence in the one-dimensional special case of rather separate phenomena in the general case; so a...

5. Part I: Description of the pseudogroups
• §2. The group of k-jets and its Lie algebra.
(pp. 6-15)

Consider the set of all germs of complex analytic mappings from the origin to the origin in the space ℂnof n complex variables. The k-jet of such a germ f, denoted by jkf, is defined to consist of the terms of order ≤ k in the Taylor expansion of the germ f ; but since all these germs are assumed to take the origin to the origin the conventional usage will be slightly modified in that the constant terms in the Taylor expansion, the terms of order = 0, will not be considered as part of the k-jet. Upon...

• §3. The pseudogroups defined by partial differential equations.
(pp. 16-20)

The definition and classification of the pseudogroups defined by families of partial differential equations are rather straightforward matters once the preceding general machinery has been developed. An analytic family of partial differential equations of order k in the analytic local homeomorphisms from ℂnto ℂncan be thought of merely as being an analytic subvariety A ⊆ Gk(n,ℂ). Of course this is a somewhat restrictive definition, since such families of partial differential equations do not involve the actual values of the mappings but only the derivatives of orders 1 through k of the component functions of the mappings, and the...

• §4. The classification of tangentially transitive pseudogroups: algebraic aspects
(pp. 21-38)

The detailed classification of pseudogroups will only be attempted here for the special case of the tangentially transitive Lie pseudogroups, those for which all the defining groups A ⊆ Gk(n,ℂ) have the property that j1A = G1(n,ℂ); these are the pseudogroups for which there are no restrictions imposed on the values of the Jacobian matrices of the mappings. The classification apparently involves determining all the integrable subgroups A ⊆ Gk(n,ℂ) with j1A = G1(n,ℂ), for all k ≥ 1, and then determining which of these subgroups describe the same pseudogroups; but it is actually a considerably simpler matter than might...

• §5. The classification of tangentially transitive pseudogroups: analytic aspects
(pp. 39-52)

There are the four general classes of possible defining groups for tangentially transitive Lie pseudogroups, described by the four classes of equations (51), (52), (53), and (54) respectively; and within each class the defining groups are parametrized by a linear space of tensors A. It is useful to introduce a notion of equivalence among the defining groups in each class separately; but the definition and elementary properties of this relation are formally almost the same in the different cases, so for convenience will only be discussed in detail for the class given by equation (51). In that case the defining...

6. Part II: Description of the connections
• §6. Pseudogroup structures and their associated connections
(pp. 53-67)

As is of course well known, an m-dimensional topological manifold M is a Hausdorff space each point of which has an open neighborhood homeomorphic to an open subset of ℝm. A coordinate covering [symbol] = {U2} of such a manifold is a covering of M by open subsets Uα⊆ M, for each of which there is a homeomorphism zα: Uα→ Vαbetween Uαand an open subset Vα⊆ ℝm; the sets Uαare called coordinate neighborhoods, and the homeomorphisms zαare called local coordinates. The compositions${{\text{f}}_{\alpha\beta}}={{\text{z}}_{\alpha}}\circ \text{z}_{\beta}^{-1}:{{\text{z}}_{\beta}}({{\text{U}}_{\alpha}}\cap {{\text{U}}_{\beta}})\to {{\text{z}}_{\alpha}}({{\text{U}}_{\alpha}}\cap {{\text{U}}_{\beta}})$are then homeomorphisms between subsets of Vβand...

• §7. Complex analytic affine connections
(pp. 68-78)

To begin the more detailed discussion of some properties and applications of the complex analytic connections associated with the various pseudogroup structures, consider the complex analytic affine connections. As might be expected from the terminology, these are essentially just the complex analytic analogues of the classical affine connections in differential geometry; but there is one point of difference which must be kept in mind. If$\{{{\text{s}}_{\alpha}}\}=\{\text{s}_{\alpha{{\text{j}}_{1}}{{\text{j}}_{2}}}^{\text{i}}\}$is a complex analytic affine connection then recalling (24) and (44) the defining equation (77) can be written out explicitly in the form$\caption {(89)} {{\Sigma }_{\text{k}}}\frac{\partial {{\text{z}}_{\beta\text{i}}}}{\partial {{\text{z}}_{\alpha\text{k}}}}\frac{{{\partial }^{2}}{{\text{z}}_{\alpha\text{k}}}}{\partial {{\text{z}}_{\beta{{\text{j}}_{\text{1}}}}}\partial {{\text{z}}_{\beta{{\text{j}}_{\text{2}}}}}}=\text{s}_{\beta{{\text{j}}_{\text{1}}}{{\text{j}}_{2}}}^{\text{i}}-{{\Sigma }_{\text{k}}}\frac{\partial {{\text{z}}_{\beta\text{i}}}}{\partial {{\text{z}}_{\alpha\text{k}}}}\text{s}_{\alpha{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{2}}}}^{\text{k}}\frac{\partial {{\text{z}}_{\alpha{{\text{k}}_{1}}}}}{\partial {{\text{z}}_{\beta{{\text{j}}_{1}}}}}\frac{\partial {{\text{z}}_{\alpha{{\text{k}}_{2}}}}}{\partial {{\text{z}}_{\beta{{\text{j}}_{2}}}}};$and that is the complex analytic analogue of the familiar...

• §8. Complex analytic projective connections.
(pp. 79-94)

Turning next to the complex analytic projective connections, there is a well developed but perhaps not so well known classical theory of projective connections [11]; a particularly readable recent survey of that theory can be found in [28]. Here too the complex analytic projective connections are just the complex analytic analogues of the classical projective connections, indeed more so than in the case of the complex analytic affine connections since for the projective connections symmetry is normally presumed. For the case n = 1 complex analytic projective connections were discussed in [20], and it was shown there that all such...

• §9. Complex analytic canonical connections
(pp. 95-100)

The complex analytic canonical connections are in many ways the simplest of the three classes of connections considered here. As was already essentially noted, a complex analytic canonical connection$\{\text{s}_{\alpha}^{'}\}=\{\text{s}_{\alpha{{\text{j}}_{1}}{{\text{j}}_{2}}}^{{{'}^{\text{i}}}}\}$can be written in the form (116) in each coordinate neighborhood Uαwhere the functions${{\text{x}}_{\alpha\text{j}}}={{\Sigma }_{\text{k}}}\text{s}_{\alpha\text{kj}}^{{{'}^{\text{k}}}}$are complex analytic in Uα; and the condition that$\text{s}_{\alpha}^{'}$is a canonical connection is equivalent to (117). Introducing the complex analytic 1-form φαin the coordinate neighborhood Uαdefined by$\caption {(132)} {{\varphi}_{\alpha}}({{\text{z}}_{\alpha}})={{\Sigma }_{\text{k}}}{{\text{ x}}_{\alpha\text{k}}}({{\text{z}}_{\alpha}})\text{d}{{\text{z}}_{\alpha}},$the condition (117) can be rewritten$\caption {(138)} \text{d}\ \log \,{{\vartriangle }_{\alpha\beta}}={{\varphi}_{\beta}}-{{\varphi}_{\alpha}}\quad \text{in}\ {{\text{U}}_{\alpha}}\cap {{\text{U}}_{\beta}}$where Δαβ= det{∂zαi/∂zβj} ; thus a complex analytic canonical connection can be described equivalently...

7. Part III: Complex analytic surfaces
• §10. Complex flat canonical structures on surfaces
(pp. 101-108)

A discussion of complex analytic pseudogroup structures on one-dimensional compact complex manifolds can be found in [20] ; although there remain several open problems in that case, they mostly have to do with more detailed properties of complex projective structures, and will not be considered here. Turning next to two-dimensional compact complex manifolds, it is convenient to change from the order in which the various pseudogroups have been discussed and to begin with the simplest case, that of the complex flat canonical structures.

If the two-dimensional compact complex manifold M admits a complex flat canonical structure then as already noted...

• §11. Complex affine structures on surfaces.
(pp. 109-122)

Although complex affine structures are more complicated than flat canonical structures, in part because of the nonlinearity of the defining partial differential equations, there is nonetheless a great deal known about such structures. Complex affine structures are of course subordinate to complex flat canonical structures, so that to determine which complex manifolds admit complex flat canonical structures it is only necessary to run through the list of complex manifolds with complex flat canonical structures and see which admit this finer structure. Thus the only compact complex surfaces that can possibly admit complex affine structures are those with Chern classes c1...

• §12. Complex projective structures on surfaces
(pp. 123-136)

Complex projective structures, the general structures in the case of one-dimensional manifolds, have been the least investigated of the structures considered here in the two-dimensional case. Turning now to the discussion of these structures, recall from Theorem 5 that on a two-dimensional manifold which admits a complex analytic projective connection the Chern classes must satisfy$\caption {(152)} {{\text{c}}_{2}}=\frac{1}{3}{{\text{c}}_{1}}^{2}.$

The first stage of the discussion is to determine in general terms which two-dimensional compact complex manifolds M satisfy this topological restriction (152). Unlike the conditions considered previously this topological restriction does not preclude the possibility that the surface M contains exceptional curves of...

8. Bibliography
(pp. 137-142)
9. Back Matter
(pp. 143-143)


text:
# mathr

## Plastic rectangles

I was intrigued by something I spotted on Wikipedia about the plastic number:

There are two ways of partitioning a square into three similar rectangles: the trivial solution given by three equal rectangles with aspect ratio 1:3, and another solution in which the three rectangles all have different sizes, but the same shape, with the square of the plastic number as their aspect ratio.

Wikipedia lacked a diagram so I made one (above), and here follows a proof that the aspect ratio is as claimed. The plastic number $$p$$ is the unique real root of $$x^3=x+1$$. The sides labeled in the diagram satisfy:

\begin{aligned} b &= 1 - a \\ c &= a (1 - a) \\ d &= 1 - a (1 - a) \end{aligned}

The rectangles are all similar, meaning they have the same aspect ratio, so:

$\frac{a}{1} = \frac{b}{d} = \frac{1 - a}{1 - a (1 - a)}$

Multiplying out the equation gives:

$a^3 - a^2 + 2a - 1 = 0$

The claim is that $$a = \frac{1}{p^2}$$ because $$p > 1$$ and clearly $$a < 1$$, so the aspect ratio convention was switched in my diagram. Substituting this for $$a$$ gives:

$p^{-6} - p^{-4} + 2p^{-2} - 1 = 0$

which multiplies out to:

$1 - p^2 + 2p^4 - p^6 = 0$

Now, substituting $$p^3 = p + 1$$ gives:

$1 - p^2 + 2p(p+1) - (p+1)^2 = 0$

and multiplying this out gives $$0 = 0$$ as all the terms cancel.