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# zbMATH — the first resource for mathematics

On the oscillation and periodic character of a third order rational difference equation. (English) Zbl 1014.39010
The authors prove that every positive solution of the following difference equation $x_n=1+{x_{n-2}\over x_{n-3}},\;n=0,1,\dots,$ converges to a periodic solution. As to some related results, see El-Metwally, E. A. Grove, G. Ladas and H. D. Voulov [J. Difference Equ. Appl. 7, No. 6, 837-850 (2001; Zbl 0993.39008)].

##### MSC:
 39A11 Stability of difference equations (MSC2000) 39B05 General theory of functional equations and inequalities
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##### References:
 [1] R. DeVault, G. Ladas, and S. W. Schultz, On the recursive sequence \?_\?+1=\?/\?_\?+1/\?_\?-2, Proc. Amer. Math. Soc. 126 (1998), no. 11, 3257 – 3261. · Zbl 0904.39012 [2] H. El-Metwally, E. A. Grove, and G. Ladas, A global convergence result with applications to periodic solutions, J. Math. Anal. Appl. 245 (2000), no. 1, 161 – 170. · Zbl 0971.39004 · doi:10.1006/jmaa.2000.6747 · doi.org [3] H. El-Metwally, E.A. Grove, G. Ladas, and H.D. Voulov, On the global attractivity and the periodic character of some difference equations, J. Diff. Eqn. Appl. 7 (2001), 837-850. · Zbl 0993.39008 [4] George Karakostas, Asymptotic 2-periodic difference equations with diagonally self-invertible responses, J. Differ. Equations Appl. 6 (2000), no. 3, 329 – 335. · Zbl 0963.39020 · doi:10.1080/10236190008808232 · doi.org [5] George Karakostas, Convergence of a difference equation via the full limiting sequences method, Differential Equations Dynam. Systems 1 (1993), no. 4, 289 – 294. · Zbl 0868.39002 [6] M.R.S. Kulenovic and G. Ladas, Dynamics of Second-Order Rational Difference Equations with Open Problems and Conjectures, Chapman and Hall/CRC, 2002. [7] G. Ladas, Open problems and conjectures, AMS Joint Math. Meetings, January 2001 (New Orleans), Program #364. · Zbl 1057.39505
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# zbMATH — the first resource for mathematics

Towards mirror symmetry as duality for two-dimensional abelian gauge theories. (English) Zbl 0957.81656
Summary: Superconformal sigma models with Calabi-Yau target spaces described as complete intersection subvarieties in toric varieties can be obtained as the low-energy limit of certain abelian gauge theories in two dimensions. We formulate mirror symmetry for this class of Calabi-Yau spaces as a duality in the abelian gauge theory, giving the explicit mapping relating the two Lagrangians. The duality relates inequivalent theories which lead to isomorphic theories in the low-energy limit. This formulation suggests that mirror symmetry could be derived using abelian duality. The application of duality in this context is complicated by the presence of nontrivial dynamics and the absence of a global symmetry. We propose a way to overcome these obstacles, leading to a more symmetric Lagrangian. The argument, however, fails to produce a derivation of the conjecture.

##### MSC:
 81T40 Two-dimensional field theories, conformal field theories, etc. in quantum mechanics 14J32 Calabi-Yau manifolds (algebro-geometric aspects) 32G81 Applications of deformations of analytic structures to the sciences 32J81 Applications of compact analytic spaces to the sciences
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##### References:
 [1] Dixon, L.J., Some world-sheet properties of superstring compactifications, on orbifolds and otherwise, (), 67-126, Singapore, New Jersey, Hong Kong [2] Lerche, W.; Vafa, C.; Warner, N.P., Chiral rings in N=2 superconformal theories, Nucl. phys., B324, 427-474, (1989) [3] Candelas, P.; Lynker, M.; Schimmrigk, R., Calabi-Yau manifolds in weighted ℙ_{4}, Nucl. phys., B341, 383-402, (1990) · Zbl 0962.14029 [4] Greene, B.R.; Plesser, M.R., Duality in Calabi-Yau moduli space, Nucl. phys., B338, 15-37, (1990) [5] Batyrev, V.V., Dual polyhedra and mirror symmetry for Calabi-Yau hypersurfaces in toric varieties, J. algebraic geom., 3, 493-535, (1994) · Zbl 0829.14023 [6] Borisov, L., Towards the mirror symmetry for Calabi-Yau complete intersections in toric Fano varieties, (1993), University of Michigan, Preprint [7] Batyrev, V.V.; Borisov, L.A., Dual cones and mirror symmetry for generalized Calabi-Yau manifolds, To appear in “Essays on Mirror Manifolds II” · Zbl 0927.14019 [8] Witten, E., Phases of N=2 theories in two dimensions, Nucl. phys., B403, 159-222, (1993) · Zbl 0910.14020 [9] Silverstein, E.; Witten, E., Global U(1) R-symmetry and conformal invariance of (0,2) models, Phys. lett., 328B, 307-311, (1994) [10] Morrison, D.R.; Plesser, M.R., Summing the instantons: quantum cohomology and mirror symmetry in toric varieties, Nucl. phys., B440, 279-354, (1995) · Zbl 0908.14014 [11] Candelas, P.; de la Ossa, X.; Katz, S., Mirror symmetry for Calabi-Yau hypersurfaces in weighted ℙ_{4} and extensions of Landau-Ginzburg theory, Nucl. phys., B450, 267-356, (1995) · Zbl 0896.14023 [12] Berglund, P.; Hübsch, T., A generalized construction of mirror manifolds, Nucl. phys., B393, 377-391, (1993) · Zbl 1245.14039 [13] Aspinwall, P.S.; Greene, B.R., On the geometric interpretation of N=2 superconformal theories, Nucl. phys., B437, 205-230, (1995) · Zbl 1052.32502 [14] Aspinwall, P.S.; Greene, B.R.; Morrison, D.R., Calabi-Yau moduli space, mirror manifolds and spacetime topology change in string theory, Nucl. phys., B416, 414-480, (1994) · Zbl 0899.32006 [15] Aspinwall, P.S.; Greene, B.R.; Morrison, D.R., The monomial-divisor mirror map, Internat. math. res. notices, 319-337, (1993) · Zbl 0798.14030 [16] Batyrev, V.V., Quantum cohomology rings of toric manifolds, (), 9-34, of Astérisque · Zbl 0806.14041 [17] S. Katz and D. R. Morrison, The Multinomial-Divisor Mirror Map, In preparation. [18] Seiberg, N.; Witten, E.; Seiberg, N.; Witten, E., Electric-magnetic duality, monopole condensation, and confinement in N=2 supersymmetric Yang-Mills theory, Nucl. phys., Nucl. phys., B430, 485-486, (1994), Erratum · Zbl 0996.81511 [19] Seiberg, N.; Witten, E., Monopoles, duality and chiral symmetry breaking in N=2 supersymmetric QCD, Nucl. phys., B431, 484-550, (1994) · Zbl 1020.81911 [20] Seiberg, N., Electric-magnetic duality in supersymmetric nonabelian gauge theories, Nucl. phys., B435, 129-146, (1995) · Zbl 1020.81912 [21] Gates, S.J.; Hull, C.M.; Roček, M., Twisted multiplets and new supersymmetric nonlinear sigma models, Nucl. phys., B248, 157-186, (1984) [22] Giveon, A., Target space duality and stringy black holes, Modern phys. lett., A6, 2843-2854, (1991) · Zbl 1020.81842 [23] Roček, M.; Verlinde, E., Duality, quotients, and currents, Nucl. phys., B373, 630-646, (1992) [24] Giveon, A.; Witten, E., Mirror symmetry as a gauge symmetry, Phys. lett., 332B, 44-50, (1994) [25] Berglund, P.; Henningson, M., Landauginzhurg orbifolds, mirror symmetry and the elliptic genus, Nucl. phys., B433, 311-332, (1995) · Zbl 0899.58068 [26] Giveon, A.; Roček, M., Introduction to duality, To appear in “Essays on Mirror Manifolds II” · Zbl 1129.81341
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.

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How common are probability distributions with a finite variance?

It's always very surprising to learn that some of the entities one has been assiduously studying actually represent negligibly tiny minorities (e.g. continuous functions vis-à-vis all functions)...

Now, the Central Limit Theorem, for one, holds only for probability distributions with a finite variance.

How common are such distributions in the space of all probability distributions?

More formally, let $U$ be the set of all probability distributions on $\mathbb{R}$ (say), and $F \subset U$ be the subset of those probability distributions that have a finite variance. I expect that, of the cardinalities $|F|$ and $|U\setminus F|$, one will be a strictly larger infinity than the other, but I have no intuition as to which.

(I guess this is a "meta-measure theory" question.)

• What's stopping you there -- why not asking about those with an expectation? – Clement C. Aug 22 '17 at 11:58
• @ClementC: the motivation for my question was wanting to gauge the CLT's generality, but you bring up an excellent point. – kjo Aug 22 '17 at 12:01
• A small note, but you'll want to consider probability distributions up to a.e. equality at most, because there are $\mathfrak c$-sized subsets of measure zero on which you can choose any function value you want, so that any probability distribution already has $\mathfrak{c}^\mathfrak{c}$-many probability distributions a.e. equal to it. – Mees de Vries Aug 22 '17 at 12:08
• @kjo a different, but related point: $L_2$ is dense in $L_1$. – Clement C. Aug 22 '17 at 12:10

I presume that you speak about distributions on the Borel $\sigma$-algebra $\mathcal B(\mathbb R)$. In terms of cardinality, there are quite few of them, namely, $\mathfrak{c} = |\mathbb{R}|$. Indeed, each probability distribution $\mathcal B(\mathbb R)$ is uniquely determined by its values on the intervals $(-\infty,q)$ with $q\in\mathbb{Q}$. Therefore, the number of probability measures does not exceed the number of sequences of real numbers, which is $\mathfrak{c}$. Since, obviously, there are both $\mathfrak{c}$ distributions with finite and infinite variance, so your guess is wrong.
• @MeesdeVries, yes, there are plenty. Take $\arctan$ of any infinite variance variable to get a finite variance one. But this is a good point. – zhoraster Aug 22 '17 at 12:30
• @zhoraster It's early here, so I may very well spout nonsense. But in your latest comment: if $X$ has an infinite variance due to non-integrability at $0$, wouldn't $\arctan X$ have exactly the same problem? – Clement C. Aug 22 '17 at 12:49


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# Triangle inequality Clebsch-Gordan coeffcients

The Clebsch-Gordan coefficients can only be non-zero if the triangle inequality holds: $$\vert j_1-j_2 \vert \le j \le j_1+j_2$$ In my syllabus they give the following proof: $$-j \le m \le j$$ $$-j_1 \le m_1 \le j_1$$ and $$-j_2 \le m_2 \le j_2$$

When $m$ takes its maximal value, $m = j$, $m_1 = j_1$ and $m_2 = j_2$, and we get:

1) $-j_1 \le j-j_2 \le j_1$ which implies $j_2-j_1 \le j \le j_1+j_2$

2) $-j_2 \le j-j_1 \le j_2$ which implies $j_1-j_2 \le j \le j_1+j_2$

which should prove the triangle inequality.

This proof looks really simple, but I don't completely understand it though. It seems that I'm missing some essential reasoning, and I can't find where. Why for instance do they take for $m_1$, $m_2$ and $m$ all maximal values? Can't I also take $m$ maximal and $m_1$ minimal? This would give bad results though. So I really don't understand it, and I hope that someone can clarify it.

First of all, if $m$ is maximal, $m_1$ cannot be minimal because $m=m_1+m_2$ so $m_{max}=m_{1, max}+m_{2, max}$ by definition.

The reason they need to use maximal values for $m_i$ is because they need to find a relationship between only the $j$-values, but they only have a relationships between $m$-values and $j$-values (namely, $-j_i \le m_i \le j_i$).

When $m_i$ is maximal, $m_i=j_i$ (by virtue of the inequality relationship above). Therefore, $j=m_{max}=m_{1, max}+m_{2, max}=j_1+j_2$ or, in other words, $j_2=m_2=j-j_1$ (and vice versa for $m_1$).

By substituting into the inequality $-j_2\le m_2\le j_2$, we get $-j_2\le j-j_1\le j_2$. The rest of the proof is a trivial extension of what is done here. The arguments are exactly the same.

This is only to add on the other answer:

Note that we have $m = m_1 + m_2$ and $-j_1 \le m_1 \le j_1$, so $$-j_1 \le m - m_2 \le j_1$$ In particular $m$ can take its maximum value $j$ and $m_2$ can take its maximum value $j_2$, which gives $$-j_1 \le j - j_2 \le j_1$$ In fact one can consider the situation where $m$ takes its minimum value $-j$ and $m_2$ takes its minimum value $-j_2$, which gives $$-j_1 \le -j + j_2 \le j_1$$ But this is in fact the same inequality after rearrangement.

The other two situations when one takes its minimum and the other takes its maximum are forbidden.


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# Chaitanya's Random Pages

## December 5, 2011

### The MMSE estimator is orthogonal projection

Filed under: mathematics — ckrao @ 9:36 pm

Suppose we are given the value of the random vector $Y$ from which we wish to estimate the value of another random vector $X$. In other words, given $Y$ we want to find a function $f$ such that $f(Y)$ is a good approximation of $X$. How might we go about this? If our definition of “good approximation” is the MMSE (minimum mean squared error) $E \|f(Y)-X\|^2$ and the random vectors $X$ and $Y$ have finite second order moments (i.e. $E \|X\|^2 = E \sum_i |X_i|^2$ is finite), then the language of Hilbert spaces helps us greatly.

Let the dimensions of $X$ and $Y$ be $m$ and $n$ respectively (they need not be equal). The convenience of the metric $\|f(Y) - X\|^2$ is that we may minimise it by finding the component functions  $f_1(Y), \ldots, f_m(Y)$ that minimise each term $E|f_i(Y) - X_i|^2$ of the sum separately, and then our required $f$ is $f(Y) = [f_1(Y), \ldots, f_m(Y)]^T$.

Imagine a component $X_i$ (now a random variable as opposed to a random vector) living in the space of random variables with finite second order moment (this space is denoted by $L^2(\cal{F})$ where $\cal{F}$ is a collection of events whose probability can be measured). The space of measurable functions of $Y$ from $\mathbb{C}^n$ to $\mathbb{C}$ forms a subspace of $L^2(\cal{F})$ which we denote by $L^2(\sigma(Y))$. (By measurable we mean that we can still compute the probability of events based on values of the function.) If $X$ is in this subspace, it means that $X_i$ can be entirely determined from knowledge of $Y$, and an error of zero is possible. If $X$ is not in the subspace, we imagine $X_i$ as an arrow from $0$ sticking out from the subspace.

With inner product on $L^2(\cal{F})$ given by $\langle U,V \rangle = E UV^*$ (noting that $\langle U,U \rangle = 0$ if and only if $U = 0$ with probability one), we see that $E|f_i(Y) - X_i|^2$ is minimised when the error $e_i = f_i(Y) - X_i$ is orthogonal to any measurable function of $Y$. In our Hilbert space we think of $f_i(Y)$ being the projection of $X$ onto the subspace $L^2(\sigma(Y))$:

$\displaystyle \langle g(Y), f_i(Y) - X_i \rangle = 0$ for any random variable $g(Y) \quad \quad (*)$

Note that if this were not true, but instead there existed $g(Y) \in L^2(\sigma(Y))$ with $\langle g(Y),g(Y) \rangle = E|g(Y)|^2 = 1$ (i.e. normalised) and $\langle f_i(Y) - X_i, g(Y) \rangle = \alpha \neq 0$, we may write $h(Y) = f_i(Y) - \alpha g(Y) \in L^2(\sigma(Y))$ and so

$\begin{array}{lcl} E|h(Y) - X_i|^2 &=& E|f_i(Y) - \alpha g(Y) - X_i|^2 \\ &=& E| f_i(Y) - X_i|^2 + |\alpha|^2 E |g(Y)|^2 - \alpha^* \langle f_i(Y) - X_i , g(Y) \rangle - \alpha \langle g(Y), f_i(Y) - X_i \rangle \\&=& E|f_i(Y) - X_i|^2 + |\alpha|^2 - |\alpha|^2 - |\alpha|^2 \\ &<& E| f_i(Y) - X_i|^2,\end{array}$

contradicting the minimality of $E|f_i(Y) - X_i|^2$. Hence the orthogonality condition $(*)$ must hold. Such a projection is unique in the almost-sure sense (i.e. any other random variable is equal to $f_i(Y)$ with probability one).

By $(*)$, $f_i(Y) - X_i$ is also orthogonal to any constant random variable, or in other words, $Ec(f_i(Y) - X_i) = 0$ for any constant $c$ showing that the error $f_i(Y) - X_i$ has zero mean (worth remembering: zero-mean random variables are orthogonal to constants!). In vector notation, with $e := f(Y) - X$,

$Ee = 0$  or $Ef(Y) = EX \quad \quad (1)$

For each $j$, $f_j(Y)$ is in $L^2(\sigma(Y))$ so by $(*)$, $\langle f_j(Y), f_i(Y) - X_i \rangle = 0$. Switching to vector notation, this gives us

$\displaystyle {\rm Cov}(f(Y), f(Y)-X) = Ef(Y)(f(Y) - X)^* = 0 \quad \quad (2)$,

where ${\rm Cov}(U,V) :=: \Sigma_{UV} := E(U-E(U))(V-E(V))^*$ (we shall use the notation ${\rm Cov}(U) = \Sigma_U$ to mean ${\rm Cov}(U,U)$).

This is the orthogonality condition in our vector case. By Pythagoras’s theorem, the minimum error is

$\sum_i E|f_i(Y) - X_i|^2 = \sum_i \left(E|X_i|^2 - E|f_i(Y)|^2\right) = {\rm tr}({\rm Cov}(X) - {\rm Cov}(f(Y))),$

where ${\rm tr}$ represents the sum of the diagonal entries of a square matrix (the trace).

#### Linear Case

If we take the example of a linear estimator of the form $f(Y) = AY + b$ ($A$ a matrix, $b$ a vector), we may use (1) and (2) to identify $A$ and $b$ that minimises $E\|f(Y) - X\|^2$. By (1), $AEY + Eb = EX$, from which $b = Eb = EX - AEY$ and $f(Y) = A(Y-EY) + EX$.

Note that any component of $Y$ is in itself a linear transform of $Y$. By (*), it must be orthogonal to any component of the error vector $f(Y) - X$, giving us

$\displaystyle 0 = E(f(Y)-X)Y^* = E[(A(Y-EY) + EX - X)Y^*] = A\Sigma_Y - \Sigma_{XY}.$

If ${\rm Cov}(Y)$ is invertible, this gives $A = \Sigma_{XY}\Sigma_Y^{-1}$, leading to the following expressions:

$\displaystyle f(Y) = EX + \Sigma_{XY}\Sigma_Y^{-1}(Y-EY) \quad \quad (3)$

$\begin{array} {lcl} {\rm Cov}(f(Y) - X) &=& {\rm Cov}(X) - {\rm Cov}(EX + {\rm Cov}(X,Y){\rm Cov}(Y)^{-1}(Y-EY))\\ &=& {\rm Cov}(X) - {\rm Cov}(X,Y){\rm Cov}(Y)^{-1}{\rm Cov}(Y-EY) {\rm Cov}(Y)^{-1}{\rm Cov}(Y,X)\\ &=& \Sigma_X - \Sigma_{XY}\Sigma_Y^{-1} \Sigma_{YX} \quad \quad (4) \end{array}$

(here using the fact that ${\rm Cov}(AX) = A{\rm Cov}(X)A^*$),

$E\|f(Y) - X\|^2 = {\rm tr}( {\rm Cov}(f(Y)) - X) = {\rm tr} (\Sigma_X - \Sigma_{XY}\Sigma_Y^{-1} \Sigma_{YX}) \quad \quad (5)$.

Note that this precise argument is made in least squares problems: if we wish to find a matrix $W$ so that $\|x - Wy\|^2$ is minimised and $Wy$ is only permitted in a space spanned by the columns of some matrix $C$, we find the projection of $x$ onto the subspace whose columns are formed by $C$. This gives the orthogonality condition $0 = C^* (x - Wy) = C^* (x - Cz)$ (for some $z$, since $Wy$ is in the column space of $C$), from which

$\displaystyle Wy = Cz = C(C^*C)^{-1}C^*x.$

In the special case when $C =c$ is a single column vector, the projection operator has the attractive form $(cc^*)/(c^*c)$ which is simply the outer product $cc^*$ if $c$ has unit length.

#### Conditional Expectation

The post so far has made no mention of the conditional expectation $E[X|Y]$, which is what in fact the minimum mean squared estimator turns out to be. In brief $E[X|Y]$ is defined as a measurable function of $Y$ satisfying

$\displaystyle E[g(Y)E[X|Y]] = E[g(Y)X] \quad \quad (+)$

for any measurable function $g(Y)$. This is a measure-theoretic definition that has the advantage of unifying the discrete and continuous cases, and avoids division-by-zero possibilities that can arise when dealing with probability densities. (Actually conditional expectation is first defined with respect to a set of events called a sigma algebra which may be considered as an information source for that random variable. The larger the set, the more information we have about that random variable.)

If expectation is thought of as an averaging process, then conditional expectation is an average with respect to information or uncertainty. Formula (+) says that this average should be equal to the unconditioned average on any measurable set (to see this take $g(Y)$ to be $1$ on that set and $0$ otherwise). There are details missing here, but the interested reader is encouraged to see the references for more.

In the Hilbert space of random variables with finite second moment, this condition is equivalent to $(X-E[X|Y])$ being orthogonal to $g(Y)$, so $E[X|Y]$ is our best estimate $f(Y)$ found earlier. Hence the conditional expectation $E[X_i|Y]$ can be viewed as the vector projection of $X_i$ onto $L^2(\sigma(Y))$, i.e. the best estimate of $X_i$ in the expected least squares sense. Conditional probability may be defined in terms of conditional expectation via $P(A|B) = E[1_A|B]$, where $1_A$ is the random variable equal to $1$ if our event $A$ occurs, and $0$ otherwise.

#### Gaussian Case

In the particular case of $X$ and $Y$ being Gaussian vectors, it turns out that the best linear estimate is also the best overall estimate in the least squares sense. To see this, let $f(Y)$ be the best linear estimate of $X$ given $Y$. Since uncorrelated and independent are equivalent notions in the Gaussian world, it follows from the independence of $(X-f(Y))$ and $f(Y)$ that given $Y=y$ the conditional distribution of $(X-f(Y))$ does not depend on $y$. It is in fact zero-mean Gaussian with variance given in (4)). Hence the distribution of $X$ $(= (X-f(Y)) + f(Y))$ given $Y=y$ is Gaussian with mean $f(Y)$ (the linear estimate of $X$) and the same covariance matrix. As a result the conditional expectation $E[X|Y=y]$ is equal to the mean of that Gaussian distribution, which is $f(y)$, and as this is true for all $y$, $E[X|Y] = f(Y)$.

For example, if $Y = HX + V$, where $X$ is zero-mean Gaussian, $H$ is a deterministic $n$ by $m$ matrix and $V$ is a zero-mean $n$-dimensional Gaussian noise vector uncorrelated with $X$, we have

$\displaystyle \Sigma_{XY} ={\rm Cov}(X,HX +V) = {\rm Cov}(X)H^* + {\rm Cov}(X,V) = \Sigma_X H^*$

$\displaystyle \Sigma_Y = {\rm Cov}(HX+V) = H\Sigma_X H^*+ \Sigma_V,$

and as $X$ and $Y$ are jointly Gaussian, our MMSE estimator is also the best linear estimator:

$\displaystyle E[X|Y] = f(Y) =\Sigma_{XY}\Sigma_Y^{-1}Y = \Sigma_X H^*(H\Sigma_X H^* + \Sigma_V)^{-1} Y.$

This solution is used in a wide variety of linear estimation applications, ranging from regression analysis in statistics to communication theory (estimating signals passing through a channel $H$ and corrupted by noise $V$).

#### References

[1] Williams, Probability with Martingales, Cambridge University Press, 2001.

[2] Hajek, Notes for ECE 534: An Exploration of Random Processes for Engineers, July 2011, available here.


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Timezone: »

Spotlight
Knowledge Base Question Answering by Case-based Reasoning over Subgraphs
Rajarshi Das · Ameya Godbole · Ankita Rajaram Naik · Elliot Tower · Manzil Zaheer · Hannaneh Hajishirzi · Robin Jia · Andrew McCallum

Tue Jul 19 11:45 AM -- 11:50 AM (PDT) @ Ballroom 1 & 2
Question answering (QA) over knowledge bases (KBs) is challenging because of the diverse, essentially unbounded, types of reasoning patterns needed.However, we hypothesize in a large KB, reasoning patterns required to answer a query type reoccur for various entities in their respective subgraph neighborhoods.Leveraging this structural similarity between local neighborhoods of different subgraphs, we introduce a semiparametric model (CBR-SUBG) with (i) a nonparametric component that for each query, dynamically retrieves other similar $k$-nearest neighbor (KNN) training queries along with query-specific subgraphs and (ii) a parametric component that is trained to identify the (latent) reasoning patterns from the subgraphs of KNN queries and then apply them to the subgraph of the target query. We also propose an adaptive subgraph collection strategy to select a query-specific compact subgraph, allowing us to scale to full Freebase KB containing billions of facts. We show that CBR-SUBG can answer queries requiring subgraph reasoning patterns and performs competitively with the best models on several KBQA benchmarks. Our subgraph collection strategy also produces more compact subgraphs (e.g. 55\% reduction in size for WebQSP while increasing answer recall by 4.85\%)\footnote{Code, model, and subgraphs are available at \url{https://github.com/rajarshd/CBR-SUBG}}.

#### Author Information

##### Ankita Rajaram Naik (University of Massachusetts Amherst)

I'm a final year master student at the University of Massachusetts Amherst with 4+ years of experience working in Natural Language Processing. I am actively looking for full-time positions in Natural Language Processing. My most recent projects are focused on Knowledge Graph Reasoning for Question Answering and Graph Completion; Retrieval models for knowledge bases and tables; and Biomedical Applications of NLP with research publications in leading conference such as AAAI, NAACL and ICML. At UMass Amherst, I primarily worked with Prof. Andrew Mccallum and Rajarshi Das at Information Extraction and Synthesis Lab (IESL). I also interned with the Knowledge Induction Group at IBM Research AI under the mentorship of Michael Glass and Alfio Gloizzo. Moreover, I have also worked with researchers at GE Healthcare and Walmart while receiving my MS in Computer Science.


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Finance - Maple Programming Help

Home : Support : Online Help : Mathematics : Finance : Financial Instruments : Finance/ImpliedVolatilitySurface

Finance

 ImpliedVolatilitySurface
 create new implied volatility term structure

 Calling Sequence ImpliedVolatilitySurface(volatility, T, S, opts) ImpliedVolatilitySurface(riskfreerate, dividendyield, spot, times, strikes, prices, opts) ImpliedVolatilitySurface(riskfreerate, dividendyield, spot, dates, strikes, prices, opts) ImpliedVolatilitySurface(riskfreerate, dividendyield, price, T, K, opts) ImpliedVolatilitySurface(times, volatilities, opts) ImpliedVolatilitySurface(dates, volatilities, opts) ImpliedVolatilitySurface(times, values, volatilities, opts) ImpliedVolatilitySurface(dates, values, volatilities, opts)

Parameters

 volatility - algebraic expression, operator or procedure; implied volatility T - name; variable representing time to maturity S - name; variable representing the value of the underlying riskfreerate - non-negative constant, algebraic expression or a yield term structure; risk-free rate dividendyield - non-negative constant, algebraic expression or a yield term structure; dividend yield spot - algebraic expression; initial value of the underlying asset times - list or Vector; times (in years) strikes - list, Vector, listlist or Matrix; strike prices of European call options prices - list, Vector, listlist or Matrix; prices of European call options dates - list; dates price - algebraic expression or procedure; price of a European call option K - name; variable representing the strike price volatilities - list, Vector, listlist or Matrix; volatilities values - list or Vector; values of the underlying opts - equations of the form option = value where option is one of daycounter, interpolation, or referencedate; specify options for the ImpliedVolatilitySurface command

Options

 • daycounter = Actual360, Actual365Fixed, AFB, Bond, Euro, Historical, ISDA, ISMA, OneDay, Simple, Thirty360BondBasis, Thirty360EuroBondBasis, Thirty360European, Thirty360Italian, Thirty360USA, or a day counter data structure -- This option specifies the convention used to convert the amount of time between two dates to year fractions.
 • interpolation = Bilinear or Bicubic -- This option specifies the type of interpolation used to build a volatility surface from a discrete set of rates. The Bilinear interpolation is used by default.
 • referencedate = date in any of the formats recognized by the Finance[ParseDate] command -- This option specifies the reference date (date when the discount factor is equal to 1).

Description

 • The ImpliedVolatilitySurface command creates new implied volatility term structure with the given parameters. This term structure is represented as a module. This module can be passed to other commands of the Finance package that expect an implied volatility term structure as one of the parameters; it can also be used as if it were a procedure. Assume, for example, that the module returned by ImpliedVolatilitySurface was assigned to the name $V$. Then for any positive constants $T$ and $S$, $V\left(T,S\right)$ will return the corresponding value of the implied volatility. If one of the parameters $T$ or $S$ is non-numeric, $V\left(T,S\right)$ will return unevaluated.
 • The ImpliedVolatilitySurface(times, volatilities, opts) and  ImpliedVolatilitySurface(dates, volatilities, opts) calling sequences create an implied-volatility term structure based on a piecewise interpolation of the specified implied volatilities at the given times or dates. The implied volatilities are assumed to be strike-independent. The times parameter is a list or Vector of times for which implied volatilities are available. In this case the value of the implied volatility at any time is assumed to be independent of the value of the underlying. The dates parameter must be a list of dates given in any of the formats recognized by the Finance[ParseDate] command. The volatilities parameter is a list or Vector of known volatilities.
 • The ImpliedVolatilitySurface(times, values, volatilities, opts) and ImpliedVolatilitySurface(dates, values, volatilities, opts) calling sequences create an implied-volatility term structure based on a piecewise interpolation of the specified implied volatilities at the given times or dates. The values parameter is a list or a Vector of values of the underlying process for which implied volatilities are available. In this case the volatilities parameter must be a Matrix. The number of rows of the volatilities matrix must be equal to the number of elements in the values vector. Columns of the volatilities matrix correspond to elements of times or dates. So, ${\mathrm{volatility}}_{i,j}$ is the implied volatility at ${\mathrm{times}}_{j}$ (or ${\mathrm{dates}}_{j}$) if the value of the underlying is ${\mathrm{values}}_{i}$.
 • The ImpliedVolatilitySurface(volatility, T, S, opts) calling sequence constructs an implied-volatility term structure based on the given volatility function. The parameter volatility is the implied volatility given as a function of time from reference and value of the underlying. It can be given either as a Maple procedure or as an algebraic expression. If volatility is given as a procedure it must accept two parameters (time and value of the underlying) and return the corresponding implied volatility as a floating-point number. Finally, if volatility is an algebraic expression, it must depend on two variables. In this case the parameters T and S will be used to determine which variable represents time and which variable represents the value of the underlying.
 • The ImpliedVolatilitySurface(riskfreerate, dividendyield, callprice, T, K, opts) calling sequence constructs an implied-volatility term structure implied by the available prices of European call options. The parameter riskfreerate is the instantaneous interest rate, which may be time-dependent. The parameter spot is the initial value of the underlying asset. The parameter callprice is the price of the European call option given as a function of time to maturity and strike price. This parameter can be given either as a procedure or as an algebraic expression. In the first case, callprice must accept two parameters: time to maturity and strike price and return the corresponding option price. In the second case, callprice must be an algebraic expression depending on at most two variables. The parameters T and K will specify which variable represents time to maturity and which variable represents the strike price.
 • The ImpliedVolatilitySurface(riskfreerate, dividendyield, spot, times, strikes, prices, opts) and ImpliedVolatilitySurface(riskfreerate, dividendyield, dates, strikes, prices, opts) calling sequences are similar to the above except that the price function will be determined based on an interpolation of known prices.

Examples

 > $\mathrm{with}\left(\mathrm{Finance}\right):$
 > $r≔0.05$
 ${r}{:=}{0.05}$ (1)
 > $d≔0.01$
 ${d}{:=}{0.01}$ (2)
 > $\mathrm{σ}≔0.3$
 ${\mathrm{σ}}{:=}{0.3}$ (3)

First construct an implied volatility term structure based on:

 > $C≔\mathrm{BlackScholesPrice}\left(100,K,T,r,\mathrm{σ},d,\mathrm{call}\right)$
 ${C}{:=}{100.}{}{{ⅇ}}^{{-}{0.01000000000}{}{T}}{}\left({0.5000000000}{+}{0.5000000000}{}{\mathrm{erf}}{}\left(\frac{{14.14213562}{}\left({\mathrm{ln}}{}\left(\frac{{100.}}{{K}}\right){+}{0.2912500000}{}{T}\right)}{\sqrt{{T}}}\right)\right){-}{1.}{}{{ⅇ}}^{{-}{0.3000000000}{}{T}}{}{K}{}\left({0.5000000000}{+}{0.5000000000}{}{\mathrm{erf}}{}\left(\frac{{14.14213562}{}\left({\mathrm{ln}}{}\left(\frac{{100.}}{{K}}\right){+}{0.2912500000}{}{T}\right)}{\sqrt{{T}}}{-}{0.03535533905}{}\sqrt{{T}}\right)\right)$ (4)
 > $\mathrm{LV}≔\mathrm{ImpliedVolatilitySurface}\left(r,d,C,T,K\right)$
 ${\mathrm{LV}}{:=}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}$ (5)
 > $\mathrm{LV}\left(1.0,50\right)$
 ${1.861242223}{}{{10}}^{{41}}$ (6)
 > $\mathrm{LV}\left(1.0,150\right)$
 ${0.1088801580}$ (7)

The following example is based on a piecewise approximation of known implied volatilities.

 > $V≔⟨⟨0.30,0.29,0.3⟩|⟨0.31,0.30,0.31⟩|⟨0.32,0.31,0.33⟩|⟨0.33,0.32,0.35⟩⟩$
 ${V}{:=}\left[\begin{array}{cccc}{0.30}& {0.31}& {0.32}& {0.33}\\ {0.29}& {0.30}& {0.31}& {0.32}\\ {0.3}& {0.31}& {0.33}& {0.35}\end{array}\right]$ (8)
 > $T≔⟨0.5,1.0,1.5,2.0⟩$
 ${T}{:=}\left[\begin{array}{c}{0.5}\\ {1.0}\\ {1.5}\\ {2.0}\end{array}\right]$ (9)
 > $S≔⟨95,100,105⟩$
 ${S}{:=}\left[\begin{array}{r}{95}\\ {100}\\ {105}\end{array}\right]$ (10)
 > $\mathrm{LV2}≔\mathrm{ImpliedVolatilitySurface}\left(T,S,V,\mathrm{Bicubic},\mathrm{referencedate}="Jun-7-2006"\right)$
 ${\mathrm{LV2}}{:=}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}$ (11)
 > $\mathrm{LV2}\left(0.5,105\right)$
 ${0.3000000000}$ (12)
 > $\mathrm{LV2}\left(0.5,100\right)$
 ${0.2900000000}$ (13)
 > $\mathrm{LV2}\left("Jun-7-2007",105\right)$
 ${0.3100000000}$ (14)
 > $\mathrm{LV2}\left("Jun-7-2007",110\right)$
 ${0.3196873473}$ (15)
 > $t≔\mathrm{YearFraction}\left("Jun-7-2006","Jun-7-2017"\right)$
 ${t}{:=}{11.}$ (16)
 > $\mathrm{LV2}\left(t,110\right)$
 ${0.3776241518}$ (17)

In this example you use bi-linear and bi-cubic interpolation of known prices.

 > $T≔⟨\mathrm{seq}\left(0.1i,i=1..20\right)⟩$
 ${T}{:=}\left[\begin{array}{c}{\mathrm{1 .. 20}}{{\mathrm{Vector}}}_{{\mathrm{column}}}\\ {\mathrm{Data Type:}}{\mathrm{anything}}\\ {\mathrm{Storage:}}{\mathrm{rectangular}}\\ {\mathrm{Order:}}{\mathrm{Fortran_order}}\end{array}\right]$ (18)
 > $K≔⟨\mathrm{seq}\left(90+i,i=1..20\right)⟩$
 ${K}{:=}\left[\begin{array}{c}{\mathrm{1 .. 20}}{{\mathrm{Vector}}}_{{\mathrm{column}}}\\ {\mathrm{Data Type:}}{\mathrm{anything}}\\ {\mathrm{Storage:}}{\mathrm{rectangular}}\\ {\mathrm{Order:}}{\mathrm{Fortran_order}}\end{array}\right]$ (19)
 > $A≔\mathrm{Matrix}\left(\left[\mathrm{seq}\left(\left[\mathrm{seq}\left(\mathrm{BlackScholesPrice}\left(100,{K}_{i},{T}_{j},r,\mathrm{σ},d\right),j=1..20\right)\right],i=1..20\right)\right],\mathrm{datatype}=\mathrm{float}[8]\right)$
 ${A}{:=}\left[\begin{array}{c}{\mathrm{20 x 20}}{\mathrm{Matrix}}\\ {\mathrm{Data Type:}}{{\mathrm{float}}}_{{8}}\\ {\mathrm{Storage:}}{\mathrm{rectangular}}\\ {\mathrm{Order:}}{\mathrm{Fortran_order}}\end{array}\right]$ (20)
 > $\mathrm{LV2}≔\mathrm{ImpliedVolatilitySurface}\left(r,d,100,T,K,A,\mathrm{Bilinear}\right)$
 ${\mathrm{LV2}}{:=}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}$ (21)
 > $\mathrm{LV3}≔\mathrm{ImpliedVolatilitySurface}\left(r,d,100,T,K,A,\mathrm{Bicubic}\right)$
 ${\mathrm{LV3}}{:=}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}$ (22)
 > $\mathrm{LV2}\left(1,100\right)$
 ${0.6011620097}$ (23)
 > $\mathrm{LV2}\left(1,105\right)$
 ${0.5527749264}$ (24)
 > $\mathrm{LV2}\left(1.5,100\right)$
 ${0.7016426945}$ (25)
 > $\mathrm{LV2}\left(1.5,105\right)$
 ${0.6658347722}$ (26)
 > $\mathrm{LV3}\left(1,100\right)$
 ${0.6011620097}$ (27)
 > $\mathrm{LV3}\left(1,105\right)$
 ${0.5527749264}$ (28)
 > $\mathrm{LV3}\left(1.5,100\right)$
 ${0.7016426945}$ (29)
 > $\mathrm{LV3}\left(1.5,105\right)$
 ${0.6658347722}$ (30)

In this example you plot implied volatility surface obtained using a piecewise interpolation of known prices.

 > $P≔\mathrm{Matrix}\left(\left[\left[7.20,7.60,8.60,10.0\right],\left[2.85,3.52,5.00,6.30\right],\left[0.50,1.10,2.25,3.50\right],\left[0.05,0.25,0.85,1.85\right],\left[0.05,0.10,0.35,0.85\right]\right]\right)$
 ${P}{:=}\left[\begin{array}{cccc}{7.20}& {7.60}& {8.60}& {10.0}\\ {2.85}& {3.52}& {5.00}& {6.30}\\ {0.50}& {1.10}& {2.25}& {3.50}\\ {0.05}& {0.25}& {0.85}& {1.85}\\ {0.05}& {0.10}& {0.35}& {0.85}\end{array}\right]$ (31)
 > $T≔⟨0.05,0.10,0.35,0.65⟩$
 ${T}{:=}\left[\begin{array}{c}{0.05}\\ {0.10}\\ {0.35}\\ {0.65}\end{array}\right]$ (32)
 > $K≔⟨80.0,85.0,90.0,95.0,100.0⟩$
 ${K}{:=}\left[\begin{array}{c}{80.0}\\ {85.0}\\ {90.0}\\ {95.0}\\ {100.0}\end{array}\right]$ (33)
 > $V≔\mathrm{Finance}:-\mathrm{ImpliedVolatilitySurface}\left(0.05,0.00,87,T,K,P,\mathrm{Bicubic}\right)$
 ${V}{:=}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}$ (34)
 > $\mathrm{plot3d}\left(V,0.05..0.65,80..100,\mathrm{orientation}=\left[-145,75\right],\mathrm{axes}=\mathrm{BOXED}\right)$

References

 Brigo, D., Mercurio, F., Interest Rate Models: Theory and Practice, Springer Verlag, 2001.
 Gatheral, J., The Volatility Surface: A Practitioner's Guide, (with foreword by Nassim Taleb), Wiley, 2006.
 Hull, J., Options, Futures, and Other Derivatives, 5th. edition, Prentice Hall, 2003.

Compatibility

 • The Finance[ImpliedVolatilitySurface] command was introduced in Maple 15.


text:
# Check my solution to $x^2 + x + 1 > 0$

I spent an hour or so yesterday trying to solve the inequality $x^2 + x + 1 > 0$. Since I'd spent so long on a problem didn't seem like it should be that difficult, I decided I'd call it a day and try it again later.

I just had another look at it and this solution became immediately obvious:

$$x^2 + x + 1 > 0 \ \ \forall \ \ x \in \mathbb{R}$$

I'd justify this by stating that $x^2 > x \ \ \forall \ \ x \in \mathbb{R}$. Because of this, even if $x < 0$, the right hand side of the inequality will always be positive. Am I correct?

-
actually $x^2 < x$ for $0 <x <1$. –  Hardy Mar 9 '12 at 23:15
Your claim is false $(1/2)^2< 1/2$. –  azarel Mar 9 '12 at 23:16
Didn't think about that .. –  jamesbrewr Mar 9 '12 at 23:16
It is true that $x^2 + x + 1 > 0$ for all $x \in \mathbb{R},$ but not for the resons you give. If $0 < x < 1,$ we have $x^2 < x.$ –  Geoff Robinson Mar 9 '12 at 23:16

Hint $\rm\ \ 4\:f(x)\: =\: 4\:(x^2+x+1)\ =\ (2x+1)^2 + 3\: \ge\: 3\$ thus $\rm\:f(x)\ge 3/4$

More generally one can decide polynomial (in)equations by partitioning the real line into intervals based on the finite number of roots, and testing on each interval. Here there are no real roots, so it has constant sign, so has the sign of $\rm\:f(0) = 1.\:$

This is a special case of Collin's cylindrical algebraic decomposition algorithm, an effective form of Tarski-Seidenberg quantifier elimination for the reals. For deep generalizations, search on "semialgebraic geometry" and "o-minimal".

-
Is there any special significance in choosing to multiply by 4 and add 3, or is this just a case that works out? –  jamesbrewr Mar 9 '12 at 23:25
@James It's a convenient way to complete the square, just as in the classical quadratic formula. –  Bill Dubuque Mar 9 '12 at 23:32

$$x^2 + x + 1 > 0 \Leftrightarrow x^2 + x + \frac{1}{4} > -1 + \frac{1}{4} \Leftrightarrow (x+\frac{1}{2})^2 > -\frac{3}{4}$$ And we have the desired result by the positivity of the square. (i.e. $\forall x \in \mathbb{R}: x^2 \geq 0$)

-

First I will show that with small modifications, your approach can be made to work. (It is, however, simpler to use one of the methods in other answers.) At the end, I suggest an additional simple approach.

It is clear that $x^2+x+1 \ge 1$ if $x \ge 0$. There is potential trouble only when $x<$, so suppose that $x<0$. Because it is all too easy to make errors when handling negative numbers, let $w=-x$. Then $x^2+x+1=w^2-w+1$, and $w$ is positive.

If $w \ge 1$ (that is, if $x\le -1$), then $w^2 \ge w$, so $w^2 -w+1\ge 0$. (Here we used the method that you proposed.)

That method doesn't work when $0<w<1$, for then $w^2<w$. But $w^2>0$ and $w<1$, so $w^2-w>-1$, and therefore $w^2-w+1>0$.

Another way: Note that $(x-1)(x^2+x+1)=x^3-1$, and that $x-1$ and $x^3-1$ are positive if and only if $x>1$, and are negative if and only if $x<1$.

If $x>1$, then $x-1$ and $x^3-1$ are both positive, so the ratio $\frac{x^3-1}{x-1}$, that is, $x^2+x+1$, is positive.

If $x-1$ and $x^3-1$ are both negative, then again their ratio $x^2+x+1$ is positive.

And finally if $x=1$, then $x^2+x+1$ is positive.

Note that in exactly the same way, we can show that $x^n+x^{n-1}+\cdots +x+1$ is always positive if $n$ is even.

-

I bring x and 1 to the other side to change the problem to $x^2 > -1 - x$.

Note that $x^2$ is always positive for real $x$, so the equality can only be false when $x < -1$.

But, if $x < -1$, $x^2$ is still positive, and so is $- x$. So we need to prove that when $x > 1$, then $x^2 > x - 1$ holds. But since $x^2 > x$ when $x > 1$ this holds.

-

There's also this answer (but I'm not sure if it qualifies as a proof):

This is a plot of $y=x^2+x+1$, notice it never touches the $x$-axis.

-
It may be good to add some words to this chart to show how it solves the problem. –  Emmad Kareem Mar 10 '12 at 11:37
Graphing your data is a good way to get started to solve a problem. –  Frank Meulenaar Mar 15 '12 at 18:30

One more way to solve this (actually a different approach):

Since $1^2+1+1=3>0$ any y for which $y^2+y+1 \leq 0$ holds would yield a root of the polynomial $p(x)=x^2+x+1$ by the intermediate value theorem. Therefore you get a second root (besides $1$) of the polynomial $q(x)=p(x)(x-1)=x^3-1$. But this is a monotonically increasing function and hence has at most one zero.

-


text:
# Solving ODEs vs VoronoiMesh

I'm using VoronoiMesh to model cells in an epithelium. The concentration of certain proteins within the cells are determined by solving a system of ODEs per cell. My goal is to model emerging patterns based on the concentrations of such proteins in each cell, where the colour of each cell is determined by the concentration of a specific protein.

I have two questions regarding the implementation of this simulation:

1. Is it possible to change the colour of each cell in a Voronoi mesh, given, for example, an array of the solutions of such concentrations? I'm aware of the MeshPrimitives command, but I'm not sure how to efficiently use it.

2. Also, these concentrations depend on the protein concentrations of neighbouring cells (these are cell-cell signalling dynamics). Is it possible to detect neighbouring cells so that I can incorporate their protein concentrations in the ODE system for a specific cell? Say, something like NeighbourQ? Maybe changing the whole thing into a graph could help, but I'd appreciate any suggestions.

Any ideas? Thank you.

The following example makes use of my package, IGraph/M.

Needs["IGraphM"]

First we generate some random points to use as Voronoi centres. We apply a few steps of Lloyd relaxation to make them more equi-distant.

pts = RandomReal[{-1, 1}, {100, 2}];
Do[
pts = PropertyValue[{VoronoiMesh[pts, {{-1, 1}, {-1, 1}}], {2, All}}, MeshCellCentroid],
{2}
]

This is our mesh, representing the cells:

mesh = VoronoiMesh[pts, {{-1, 1}, {-1, 1}},
MeshCellStyle -> {1 -> White}]

We are going to simulate diffusion between the cells. For this, we need the Laplacian matrix (also called Kirchhoff matrix) of their neighbourhood graph.

ag = IGMeshCellAdjacencyGraph[mesh, 2];
km = IGKirchhoffMatrix[ag]

IGraph/M provides IGMeshCellAdjacencyGraph to build the neighbourhood graph of mesh cells of various dimensions, IGMeshCellAdjacencyMatrix to build the corresponding adjacency matrix (faster), and IGKirchhoffMatrix to get the graph Laplacian. There is more information about these functions in the package's documentation (IGDocumentation[]).

Let us create random initial concentrations:

initC = RandomReal[{0, 1}, MeshCellCount[mesh, 2]];

Then solve the diffusion equation using NDSolve:

diffConst = 1.0;
solfun = NDSolveValue[{c'[t] == - diffConst * km.c[t], c[0] == initC}, c, {t, 0, 1}]

We can animate the solution like this:

Animate[
SetProperty[{mesh, {2, All}}, MeshCellStyle -> ColorData["Rainbow"] /@ solfun[t]],{t, 0, 1}
]

• Thank you. This looks promising, I will take a deeper look. Just one quick question: my ultimate goal is to deploy an interactive interface (using Manipulate) to the Cloud and be able to share it with other researchers, most of which will only install the WolframPlayer to use it. How compatible is this package with this player? Do people need to install it manually on Mathematica (this is something I'd like to avoid)? I wonder if commands like Get or Need work inside a Manipulate. Dec 28 '19 at 15:45
• @samwolfe It does not work with the free version of the Wolfram Player. It requires a full version of Mathematica. I have no experience with the paid version of the Wolfram Player. It does not work in the cloud because they do not allow arbitrary binaries there. Installing it into a full version of Mathematica is quite trivial though (see the instructions). Dec 28 '19 at 16:05
• I will let someone else post an answer that does not depend on my package. Dec 28 '19 at 16:07
• Thanks! One more thing: instead of MeshCellStyle -> ColorData["Rainbow"]/@ solfun[t], I wish the solution determines the opacity of a certain RGB colour, that is, something like RGBColor[0.5, 0.34, 0.5,solfun[t]]`, which doesn't work. How can I do that? Dec 28 '19 at 19:22
• Also, how do you guarantee that the right cell is coloured? It seems I'm getting the colouring in the wrong cells. Dec 28 '19 at 20:50


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# Consider a box of mass 10 kg being lifted upward by a person standing on the surface of the earth. At a given instant the box is being accelerated upwards with an acceleration of size 5 m/s$^2$. Find the force that the person exerts on the box at this instant.

Can you answer this question?

## 1 Answer

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Answer: 148 N
Given, the mass of the object is m = 1 0 kg and that the object is being accelerated upward. We need to find the Force exerted.
Let $\vec{F_P}$ be the force of the person lifting the box, and $\vec{F_g}$ be the gravitational pull of the earth.
The forces are in the opposite direction along the y-axis:
$\Rightarrow \vec{F_P} = F_{PY} \hat{y} = -mg \hat{y}$ (negative sign as the earth's gravitational pull is downwards).
$\Rightarrow \vec{F} = \vec{F_P} + \vec{F_g} = ma_y$
$\Rightarrow F_{PY} = ma_y+mg = m (a_y+g) = 10 Kg (5 + 9.8) m/s^2 =148\;N$
answered Aug 19, 2014

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text:
The test result for maths are normally destributed . The mean is 46 % and the standerd deviation is 4. What percentage of student obtains a mark :

a) less than 49
B) within 1.8 standard deviations from the mean

2. You are being asked to computer z-scores and interpret the resultant area associated with that z-score as a probability. Do you know how to do this?

3. Originally Posted by ANDS!
You are being asked to computer z-scores and interpret the resultant area associated with that z-score as a probability. Do you know how to do this?

I know the first bit a) Mean 46 . standard deviation 4

So Z= 49-46 / 4 = 0.75 and from the stats table will give me Z= 0.2734 = 27.34 %

But the other bit i have no clue ...

4. A. The math is right, interpretation is not. If your mean is 46, and you are being asked for scores LESS THAN a statistic that is higher than the mean (49), then your answer should be at least 50%. Do you see why? If the test scores are normal (and they are), then 50% of the scores will be less than the mean, and 50% will be higher than the mean. You are being asked for scores less than a score higher than the mean, which means you are being asked for the area under the normal curve to the left of your score (the 49). What kind of z-table are you using. If you are using one that just has one half of the table (meaning it goes from 0 to 3.5), then you have to be careful on how you use that, as it is not a CUMMULATIVE z-score table. In your case, you need to find 0.75 and the percentage associated with it (which is indeed 0.2734) and then ADD .5 to it.

B. For part B we are being asked to find the percentage of scores that are within 1.8 standard deviations from the mean. Note that they did not tell us above or below, but simply within. That means we have to find two statistics. Remember what a z-score is; it is the number of standard deviations a statistic is away from the mean, a mean that is part of a normal population. So if we are being asked for test statistics 1.8 within the mean, that means we need to find the statistics that will give us a z-score of plus or minus 1.8:

$-1.8=\frac{\overline{X}-46}{4}$;

$1.8=\frac{\overline{X}-46}{4}$

You can take it from here.

5. For the question a i understood why add the .5 because of the 50 50 around the mean

for the question b i need the stats X so rearaging the furmula around the x the question gives the z score of 1,8

So for the Z= 1.8 the Stats will be 53.2

and the Z-1.8 the stats will be 38.8

thanks for the help .. very much appreciated ..

6. For problem b, you are already given the z score! Some normal distribution tables just give the probability from 0 to z. With one of those, look up the probability for z between 0 and 1.8 and double it. That uses the fact that the normal curve is symmetric about z= 0.

Others give the probability from negative infinity to z, but only for z positive. With one of those, look up the probability for z= 1.8, double it and subtract 1 from that. That's because, if the probability from negative infinity to z is P, then the probability from z to positive infinity is 1- P. Because the normall distribution is symmetric, that is also the probability from negative infinity to -z and that you want to subtract off: P-(1-P)= 2P- 1.


text:
# Nobel Prize 2018 Announcements

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Gold Member
That is the question that, I believe, only the Nobel organization can address. Without in-depth investigation, all we can do is speculate, and it requires no knowledge or skill in anything to do that.

Maybe we will see some female prize winner this year?
In 2017, the Noble Prize committees have made efforts to obtain nominations from more women by inviting more female researchers to nominate for the Nobel Prize 2018. According to the permanent secretary of the Royal Academy of Sciences.

Mentor
Not yet, but it’s discussed at least in the Nobel Committee for Physics.
One reason for the delay is that the Nobel Foundation has been traditionally close to Alfred Nobel's will. But it is now 100 years later and the scientific research has shifted to large research groups often in global collaboration with other groups. The Nobel Foundation is very aware of the situation.
2013 (Higgs: ATLAS and CMS) and 2017 (gravitational waves: LIGO) would have been two great opportunities for that. Big discoveries made by big collaborations.

Gold Member
I would be embarrassed if I was a woman and won the award because some committee determined that they needed to give the award to more women.

dyn
Staff Emeritus
Homework Helper
Gold Member
2021 Award
I would be embarrassed if I was a woman and won the award because some committee determined that they needed to give the award to more women.
This is not going to happen. The prize would never be given to someone who did not deserve it. The question discussed is to remove a negative gender bias, not to introduce a positive gender bias.

Gold Member
2013 (Higgs: ATLAS and CMS) and 2017 (gravitational waves: LIGO) would have been two great opportunities for that. Big discoveries made by big collaborations.
It would have been nice and a significant milestone. It will be interesting to see how the Nobel Foundation solves this group problem in the coming years.

If the foundation applied the Nobel testament exactly, we would see one (1) prize winner in each group this year; physics, chemistry and physiology or medicine for discoveries made in 2017. When did it happen last?
Obviously there is already some flexibility in the application. Nobel Prize 2.0.

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It would have been nice and a significant milestone. It will be interesting to see how the Nobel Foundation solves this group problem in the coming years.
Are you suggesting that the foundation issues new directives to the Royal Academy of Sciences and the other bodies deciding the laureates? Do you think their freedom in actually awarding the prize to a collaboration is restricted at the moment? After all, as a European citizen, I am a part of the EU and therefore essentially a Nobel Peace prize laureate ...

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Yakir Aharonov and Michael Berry.

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Are you suggesting that the foundation issues new directives to the Royal Academy of Sciences and the other bodies deciding the laureates?

I'm not competent to have a good opinion. The primary role of the Nobel Foundation is to manage capital and maintain the spirit of Nobel's will. Directives? The Nobel Foundation does not care how the institutions choose their prize winners. They have different focuses. There is of course a close cooperation. The Managing Directors of the Institutions appoint the majority of the members of the Nobel Foundation Board.

Some legal aspects. Nobel's will can never be changed, only "interpreted" to adapt to a changed reality.
The foundation can not change its own statutes, but apply for so-called "permutation" at the government or Authority.

Do you think their freedom in actually awarding the prize to a collaboration is restricted at the moment?

Restricted to 1-3 individual laurates. A prize to a research group or collaboration between groups is currently more complicated.

After all, as a European citizen, I am a part of the EU and therefore essentially a Nobel Peace prize laureate ...

Since EU won the Nobel Peace Prize 2012... of course I think you are a kind of Nobel laureate. I hope you received your share of the prize money - about 2 cents.

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Restricted to 1-3 individual laurates. A prize to a research group or collaboration between groups is currently more complicated.
This misses the point of my question and humorously intended interjection about the EU. The will is worded the same way for the physics prize as for the peace prize, yet clearly the Norwegian Nobel Committee has had no problems in awarding organizations so my question is how the physics prize differs in this regard.

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The will is worded the same way for the physics prize as for the peace prize...
Common to the physics and peace prize is the formulation: "one part to the person".
Then the will is worded a little different:

Physics: "the most important discovery or invention.”
Peace: “the most or best to advance fellowship among nations,
the abolition or reduction of standing armies, and
the establishment and promotion of peace congresses.”
... yet clearly the Norwegian Nobel Committee has had no problems in awarding organizations so my question is how the physics prize differs in this regard.
The reason is that it is already established by the Nobel Foundation for the Peace Prize.

"The candidates eligible for the Nobel Peace Prize are those
persons or organizations nominated by qualified individuals"

Scroll down to Candidacy Criteria:
https://www.nobelprize.org/nomination/peace/

This does not answer your question why the physics prize can’t be given to an organization.
I have no concrete answer, but have done some research and found something
that surprised me:

“Each prize-awarding body shall be competent to decide whether the prize
it is entitled to award may be conferred upon an institution or association.”

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The Nobel Prize in Physics 2018 has been awarded to
Arthur Ashkin (1/2) "for the optical tweezers and their application to biological systems"
Gerard Mourou (1/4) and Donna Strickland (1/4) "for their method of generating high-intensity, ultra-short optical pulses"

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dlgoff, berkeman, StatGuy2000 and 3 others
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Dr. Arthur Ashkin - Harvey Prize Recipient 2004

FiO+LS 2018 - Gerard Mourou Plenary Session

Donna Strickland - Multi-frequency Raman Generation for Intense Ultrashort Pulses

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Astronuc, Choppy, StatGuy2000 and 4 others
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Something is going on. The dawn of a Nobel makeover?

"For groundbreaking inventions in the field of laser physics"
1. Inventions, not discoveries. Old Alfred would have been happy, as an inventor and entrepreneur himself.
2. Donna Strickland - A Female Physics Prize winner. The first since 1963; 55 long years.
3. Arthur Ashkin 96 - The oldest Nobel Prize winner ever, not just within physics. New hope for older retired physicists.

It's never too late.

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phoenix95 and bob012345
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3. Arthur Ashkin 92 - The oldest Nobel Prize winner ever, not just within physics. New hope for older retired physicists.
It's never too late.
Ashkin it actually 96.

Something is going on. The dawn of a Nobel makeover?

"For groundbreaking inventions in the field of laser physics"
1. Inventions, not discoveries. Old Alfred would have been happy, as an inventor and entrepreneur himself.
2. Donna Strickland - A Female Physics Prize winner. The first since 1963; 55 long years.
3. Arthur Ashkin 92 - The oldest Nobel Prize winner ever, not just within physics. New hope for older retired physicists.

It's never too late.

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Ashkin it actually 96.
Corrected. Thanks!
(To quote myself: "It's never too late.")

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The Nobel Prize in Physics 2018 has been awarded to
Arthur Ashkin (1/2) "for the optical tweezers and their application to biological systems"
Gerard Mourou (1/4) and Donna Strickland (1/4) "for their method of generating high-intensity, ultra-short optical pulses"

Today's Physics prize, especially the prize to Ashkin, highlight inventions that have been important field of biophysics. In graduate school, I worked in the field of single molecule biophysics, which aims to watch the action of single enzymes in order to gain greater knowledge of their mechanisms of action.

This field was primarily built on two different technologies. One (which I used) was the detection of fluorescence from single fluorescent molecules (pioneered, among others, by W.E. Moerner, a co-recipient of the 2014 Nobel Prize in Chemistry), and the other was optical tweezer technology. Optical tweezers give biologists not only the ability to manipulate biological molecules (e.g. to study the mechanical properties of DNA), but also to very sensitively monitor the motion of molecular motors.

In now classic studies in biophysics, researchers were able to watch DNA polymerase synthesize DNA, RNA polymerase make RNA from DNA with base pair resolution, and watch the ribosome move codon-by-codon on an mRNA as it synthesizes protein (the three key steps in the "central dogma" of biology). While Askin was key to developing optical tweezer technology, many biophysicsts (principally, Steven Block at Stanford and Carlos Bustamante at Berkeley) have been at the forefront of applying optical tweezer technology to the study of biological systems (indeed, their work is cited heavily in the Nobel prize material describing the applications of optical tweezers).

In fact, one could make a strong argument that a prize for optical tweezers and their application to biological systems probably should have been its own prize to Ashkin, Block and Bustamante. While there are still many questions left in the field, these optical tweezer studies have really helped us learn how molecular motor proteins can couple a chemical reaction (such as the hydrolysis of ATP) to the directed motion of protein molecules on their substrates.

Pulsed lasers have also had numerous applications in biophysics, including multiphoton microscopy (most famously, two-photon microscopy). Two photon microscopy is particularly useful for imaging deep into tissue (use of two photon excitation allows with IR lasers that are absorbed and scattered less than visible light by tissue) and has found important applications in neurobiology. Other multiphoton techniques (such as those based on stimulated raman scattering) offer the promise of being able to image specific molecules in living samples without having to label them with an external fluorophore.

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pinball1970, DrClaude, robphy and 4 others
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While Askin was key to developing optical tweezer technology, many biophysicsts (principally, Steven Block at Stanford and Carlos Bustamante at Berkeley) have been at the forefront of applying optical tweezer technology to the study of biological systems (indeed, their work is cited heavily in the Nobel prize material describing the applications of optical tweezers).

In fact, one could make a strong argument that a prize for optical tweezers and their application to biological systems probably should have been its own prize to Ashkin, Block and Bustamante. .

https://www.ibiology.org/talks/optical-traps/

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I'm glad this prize was about technology and usefulness and not about something so esoteric even most educated people couldn't begin to comprehend it.

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I'm glad this prize was about technology and usefulness and not about something so esoteric even most educated people couldn't begin to comprehend it.
I would've been fine with either.

phoenix95 and Ygggdrasil
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I would've been fine with either.
I agree, I think both types of prizes are ultimately very important and in the spirit of Alfred Nobel's will. However, it has been some time since there was last a physics prize to an invention of this sort as the last couple of years have been full of very important discoveries in fundamental physics.

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The Nobel awards one million dollars. The Wolf prize (my favorite since the term Wolf Laureate just sounds neat) awards one tenth that amount. There is an opportunity to create a prize either greater than a Nobel or in between. I wonder why the U.S. has never created an American Prize even greater than the Nobel?

.

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The Nobel awards one million dollars. The Wolf prize (my favorite since the term Wolf Laureate just sounds neat) awards one tenth that amount. There is an opportunity to create a prize either greater than a Nobel or in between. I wonder why the U.S. has never created an American Prize even greater than the Nobel?
I do not think you can just create a prize that awards more money and call it "greater". The prize money is of course a nice touch, but if you ask the laureates what meant the most to them, I am pretty sure that the recognition and prestige would come first. Sure, you can throw in the money and make a huge prize, but the Nobel Prize would still have a head start of over 100 years of history and tradition.

Greg Bernhardt
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I do not think you can just create a prize that awards more money and call it "greater". The prize money is of course a nice touch, but if you ask the laureates what meant the most to them, I am pretty sure that the recognition and prestige would come first. Sure, you can throw in the money and make a huge prize, but the Nobel Prize would still have a head start of over 100 years of history and tradition.
Yes, that's true but if the new prize was known to be administered by the best minds in American science and the prize large, the prestige would grow. The Wolf prize has only been around since the late 70's. There could be an added incentive such as any American Prize winner could just walk into the best American institutions and be given whatever they want.

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The Nobel awards one million dollars. The Wolf prize (my favorite since the term Wolf Laureate just sounds neat) awards one tenth that amount. There is an opportunity to create a prize either greater than a Nobel or in between. I wonder why the U.S. has never created an American Prize even greater than the Nobel?

The Breakthrough Prize gives a 3 million dollar award to its laureates. Some of the awards seem to be going to scientists initially passed over for Nobels (e.g. this year's award to Jocelyn Bell Burnell)

bob012345
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Yes, that's true but if the new prize was known to be administered by the best minds in American science and the prize large, the prestige would grow.
You seem to assume that there would be some inherent additional prestige in the "best minds in American science". I do not think that this is true.

There could be an added incentive such as any American Prize winner could just walk into the best American institutions and be given whatever they want.
Institutions would never agree to that.

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The Fields medal is - together with the Abel prize - the most prestigious award in mathematics, with a prize money of just ~$10,000. Yes, that's true but if the new prize was known to be administered by the best minds in American science and the prize large, the prestige would grow. The Wolf prize has only been around since the late 70's. There could be an added incentive such as any American Prize winner could just walk into the best American institutions and be given whatever they want. A national prize will never get the reputation of the big international prizes. And that reward is unrealistic as well. Science Advisor Gold Member Worth noting that Donna Strickland is Canadian too! And her BS was in Engineering Physics. Oh yea. bob012345, StatGuy2000 and atyy Science Advisor Gold Member 2021 Award The Royal Swedish Academy of Sciences has decided to award the Nobel Prize in Chemistry 2018 with one half to Frances H. Arnold ”for the directed evolution of enzymes” and the other half jointly to George P. Smith and Sir Gregory P. Winter ”for the phage display of peptides and antibodies”. https://www.nobelprize.org/prizes/chemistry/2018/press-release/ Definitely a well deserved prize. Both techniques are very useful and are widely used in a number of applications. Astronuc, BillTre and Greg Bernhardt diogenesNY Some fun with Nobel Numbers: The cash prize that goes along with the Nobel is 9,000,000 Swedish Krona/Kronor (SEK) this is just shy of$1 million USD

Between 1901 and 2017, the Nobel Prizes and the Prize in Economic Sciences were awarded 585 times

• Awards: 585
• Laureates: 923
• Prize categories: 6
• Awarded women: 48
• Awarded organisations: 48
• Average age: 60
• Youngest laureate: 17
• Oldest laureate: 90

Nobel information from https://www.nobelprize.org/
currency rates from https://finance.yahoo.com/currency-converter?.tsrc=fin-srch

diogenesNY

Mentor
Oldest laureate: 90
Arthur Ashkin is 96 and will get the prize in December.

diogenesNY


text:
# Welcome

This is the website for Data Science for Economists and Other Animals. The book is very much in the early development stages, but draws from lecture material that we have been refining over the last several years. We’re not actively looking for feedback yet, but hope to once we’ve managed to build out enough of the chapters and basic book structure.

If you’re interested in the motivation for this book, here is some placeholder text from Grant’s syllabus:

This seminar (ED: book) is targeted at economics PhD students (ED: and other animals) and will introduce you to the modern data science toolkit. While some material will likely overlap with your other quantitative and empirical methods courses, this is not just another econometrics course. Rather, my goal is bring you up to speed on the practical tools and techniques that I feel will most benefit your dissertation work and future research career. This includes many of the seemingly forgotten skills — like where to find interesting data sets in the “wild” and how to actually clean them — that are crucial to any successful scientific project, but are typically excluded from core econometrics and statistics classes. We will cover topics like version control and effective project management; programming; data acquisition (e.g. web-scraping), cleaning and visualization; GIS and remote sensing products; and tools for big data analysis (e.g. relational databases, cloud computation and machine learning). In short, we will cover things that I wish someone had taught me when I was starting out in graduate school.

## 0.1 Install necessary software packages

The primary programming language used throughout this book is R (https://www.r-project.org/). The book was built against R version 4.1.0.

Each chapter of the book begins with a description of the software — e.g. R packages and external libraries — needed to complete that section. However, users may find it convenient to install all of the necessary packages (and versions thereof) in a single go. To that end, we have used renv (link) to snapshot the project environment of the book. This means that you can install all of the R packages needed to run the code in this book, simply by cloning the companion GitHub repo (https://github.com/grantmcdermott/ds4e) and running:

# renv::init() ## Only run this line if the next line returns an error
renv::restore(prompt = FALSE)

### 0.1.1 Special requirements for Linux users

Linux users will likely need to install a few system requirements before running the renv::restore() command above, mostly related to spatial libraries. (More information about individual distro requirements here.) For Ubuntu users on 20.04 and above, the following commands should suffice:

sudo apt update && sudo apt upgrade -y
sudo apt install -y libudunits2-dev libgdal-dev gdal-bin \
libgeos-dev libproj-dev libcairo2-dev

As an aside, our renv configuration is set up so that it will install pre-compiled R package binaries across operating systems, including most Linux distros. This should make package installation much faster and less error prone. If you are on a Linux system, but don’t plan to use our recommended renv approach for simultaneously installing all of the needed packages, we would still recommend configuring your system to install R package binaries from RStudio Package Manager. (RSPM). Again, assuming a Linux user on Ubuntu 20.04, this would require adding the following line to your ~/.Rprofile dotfile:

options(repos = c(RSPM = "https://packagemanager.rstudio.com/all/__linux__/focal/latest"))

An alternative approach for installing pre-compiled R package binaries on Linux is available in the form of the bspm package (link). The benefit of this approach (compared to our recommended renv/RSPM approach) is that it will automatically handle system dependencies too. The downside is that there is no way to snapshot packages and it will simply (and automatically) pull in the most recently available package versions. This may inadvertently introduce breaking changes with respect to the code presented here.

## 0.2 Building the book locally

Finally, note that you might need a few extra packages if you want to build the book locally (unlikely):

library(bookdown)
library(bslib)
library(downlit)
library(usethis)