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# 1.4: General Aspects of Probability Distributions

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## Discrete and Continuous Distributions

Consider a system whose possible configurations $$\sket{n}$$ can be labeled by a discrete variable $$n\in \CC$$, where $$\CC$$ is the set of possible configurations. The total number of possible configurations, which is to say the order of the set $$\CC$$, may be finite or infinite. Next, consider an ensemble of such systems, and let $$P\ns_n$$ denote the probability that a given random element from that ensemble is in the state (configuration) $$\sket{n}$$. The collection $$\{P\ns_n\}$$ forms a discrete probability distribution. We assume that the distribution is normalized, meaning $\sum_{n\in\CC} P\ns_n=1\ .$

Now let $$A\ns_n$$ be a quantity which takes values depending on $$n$$. The average of $$A$$ is given by $\langle A\rangle =\sum_{n\in\CC} P\ns_n\,A\ns_n\ .$ Typically, $$\CC$$ is the set of integers ($$\MZ$$) or some subset thereof, but it could be any countable set. As an example, consider the throw of a single six-sided die. Then $$P\ns_n=\frac{1}{6}$$ for each $$n\in\{1,\ldots,6\}$$. Let $$A\ns_n=0$$ if $$n$$ is even and $$1$$ if $$n$$ is odd. Then find $$\langle A\rangle=\half$$, on average half the throws of the die will result in an even number.

It may be that the system’s configurations are described by several discrete variables $$\{n\ns_1,n\ns_2,n\ns_3,\ldots\}$$. We can combine these into a vector $$\Bn$$ and then we write $$P\ns_\Bn$$ for the discrete distribution, with $$\sum_\Bn P\ns_\Bn=1$$.

Another possibility is that the system’s configurations are parameterized by a collection of continuous variables, $$\Bvphi=\{\varphi\ns_1,\ldots,\varphi\ns_n\}$$. We write $$\Bvphi\in\ROmega$$, where $$\ROmega$$ is the phase space (or configuration space) of the system. Let $$d\mu$$ be a measure on this space. In general, we can write $d\mu=W(\varphi\ns_1,\ldots,\varphi\ns_n)\, d\varphi\ns_1\,d\varphi\ns_2\cdots d\varphi\ns_n\ .$ The phase space measure used in classical statistical mechanics gives equal weight $$W$$ to equal phase space volumes: $d\mu=\CC\prod_{\sigma=1}^r dq\ns_\sigma\,dp\ns_\sigma\ ,$ where $$\CC$$ is a constant we shall discuss later on below8.

Any continuous probability distribution $$P(\Bvphi)$$ is normalized according to $\int\limits_\ROmega\!\!d\mu\,P(\Bvphi)=1\ .$ The average of a function $$A(\Bvphi)$$ on configuration space is then $\langle A\rangle =\int\limits_\ROmega\!\!d\mu\, P(\Bvphi)\,A(\Bvphi)\ .$ For example, consider the Gaussian distribution $P(x)={1\over\sqrt{2\pi\sigma^2}}\,e^{-(x-\mu)^2/2\sigma^2}\ . \label{pgauss}$ From the result9 $\impi dx\>e^{-\alpha x^2}\,e^{-\beta x}=\sqrt{\pi\over\alpha} \ e^{\beta^2/4\alpha}\ ,$ we see that $$P(x)$$ is normalized. One can then compute $\begin{split} \langle x\rangle&=\mu\\ \langle x^2\rangle-\langle x\rangle^2 &=\sigma^2\ . \end{split}$ We call $$\mu$$ the mean and $$\sigma$$ the standard deviation of the distribution, Equation [pgauss].

The quantity $$P(\Bvphi)$$ is called the distribution or probability density. One has $P(\Bvphi)\,d\mu = \hbox{probability that configuration lies within volume d\mu centered at \Bvphi}$ For example, consider the probability density $$P=1$$ normalized on the interval $$x\in\big[0,1\big]$$. The probability that some $$x$$ chosen at random will be exactly $$\half$$, say, is infinitesimal – one would have to specify each of the infinitely many digits of $$x$$. However, we can say that $$x\in\big[0.45\,,\,0.55\big]$$ with probability $$\frac{1}{10}$$.

If $$x$$ is distributed according to $$P\ns_1(x)$$, then the probability distribution on the product space $$(x\ns_1\,,\,x\ns_2)$$ is simply the product of the distributions: $$P\ns_2(x\ns_1,x\ns_2)=P\ns_1(x\ns_1)\,P\ns_1(x\ns_2)$$. Suppose we have a function $$\phi(x\ns_1,\ldots,x\ns_N)$$. How is it distributed? Let $$P(\phi)$$ be the distribution for $$\phi$$. We then have $\begin{split} P(\phi)&=\impi dx\ns_1\cdots\impi dx\ns_N\,P\ns_N(x\ns_1,\ldots,x\ns_N)\> \delta\Big(\phi(x\ns_1,\ldots,x\ns_N)-\phi\Big)\\ &=\impi dx\ns_1\cdots\!\impi dx\ns_N\,P\ns_1(x\ns_1)\cdots P\ns_1(x\ns_N)\> \delta\Big(\phi(x\ns_1,\ldots,x\ns_N)-\phi\Big)\ , \end{split}$ where the second line is appropriate if the $$\{x\ns_j\}$$ are themselves distributed independently. Note that $\impi d\phi\>P(\phi) = 1\ ,$ so $$P(\phi)$$ is itself normalized.

## Central limit theorem

In particular, consider the distribution function of the sum $$X=\sum_{i=1}^N x\ns_i$$. We will be particularly interested in the case where $$N$$ is large. For general $$N$$, though, we have $P\ns_N(X)=\impi dx\ns_1\cdots\!\impi dx\ns_N\,P\ns_1(x\ns_1)\cdots P\ns_1(x\ns_N)\> \delta\big(x\ns_1+x\ns_2+\ldots+x\ns_N-X\big)\ .$ It is convenient to compute the Fourier transform10 of $$P(X)$$: $\begin{split} {\hat P}\ns_N(k)&=\impi dX\,P\ns_N(X)\,e^{-ikX}\\ &=\impi dX\!\!\impi dx\ns_1\cdots\!\impi dx\ns_N\,P\ns_1(x\ns_1)\cdots P\ns_1(x\ns_N) \>\delta\big(x\ns_1+\ldots+x\ns_N-X)\,e^{-ikX}=\big[{\hat P\ns_1}(k)\big]^N\ , \end{split}$ where ${\hat P}\ns_1(k)=\impi dx\,P\ns_1(x)\,e^{-ikx}$ is the Fourier transform of the single variable distribution $$P\ns_1(x)$$. The distribution $$P\ns_N(X)$$ is a convolution of the individual $$P\ns_1(x\ns_i)$$ distributions. We have therefore proven that the Fourier transform of a convolution is the product of the Fourier transforms.

OK, now we can write for $${\hat P}\ns_1(k)$$ $\begin{split} {\hat P}\ns_1(k)&=\impi dx\,P\ns_1(x)\,\big(1-ikx -\half \,k^2 x^2 +\frac{1}{6}\,i \,k^3\,x^3 + \ldots\big)\\ &=1-ik\langle x\rangle -\half\, k^2\langle x^2\rangle +\frac{1}{6} \, i\,k^3\langle x^3\rangle + \ldots \ . \end{split}$ Thus, $\ln{\hat P}\ns_1(k)=-i\mu k -\half\sigma^2 k^2 + \frac{1}{6}\,i\,\gamma^3\,k^3+\ldots\ ,$ where $\begin{split} \mu&=\langle x\rangle\\ \sigma^2&=\langle x^2\rangle - \langle x\rangle ^2\\ \gamma^3&=\langle x^3\rangle-3\,\langle x^2\rangle\,\langle x\rangle + 2\,\langle x\rangle^3 \end{split}$ We can now write $\big[{\hat P}\ns_1(k)\big]^N=e^{-iN\mu k}\, e^{-N \sigma^2 k^2/2}\, e^{iN\gamma^3 k^3/6}\cdots$ Now for the inverse transform. In computing $$P\ns_N(X)$$, we will expand the term $$e^{iN\gamma^3 k^3/6}$$ and all subsequent terms in the above product as a power series in $$k$$. We then have $\begin{split} P\ns_N(X)&=\impi{dk\over 2\pi}\ e^{ik(X-N\mu)}\,e^{-N\sigma^2 k^2/2}\>\Big\{1+\frac{1}{6}\, i\,N\gamma^3 k^3 + \ldots\Big\}\\ &=\bigg(1-{\gamma^3\over 6} N\, {\pz^3\over\pz X^3} + \ldots \bigg) {1\over\sqrt{2\pi N\sigma^2}}\, e^{-(X-N\mu)^2/2N\sigma^2}\bvph\\ &=\bigg(1-{\gamma^3\over 6}\,N^{-1/2}\, {\pz^3\over\pz \xi^3} + \ldots \bigg) {1\over\sqrt{2\pi N\sigma^2}}\, e^{-\xi^2/2\sigma^2}\ . \end{split}$ In going from the second line to the third, we have written $$X=N\mu+\sqrt{N}\,\xi$$, in which case $$\pz\ns_X=N^{-1/2}\,\pz\ns_\xi$$, and the non-Gaussian terms give a subleading contribution which vanishes in the $$N\to\infty$$ limit. We have just proven the central limit theorem: in the limit $$N\to\infty$$, the distribution of a sum of $$N$$ independent random variables $$x\ns_i$$ is a Gaussian with mean $$N\mu$$ and standard deviation $$\sqrt{N}\,\sigma$$. Our only assumptions are that the mean $$\mu$$ and standard deviation $$\sigma$$ exist for the distribution $$P\ns_1(x)$$. Note that $$P\ns_1(x)$$ itself need not be a Gaussian – it could be a very peculiar distribution indeed, but so long as its first and second moment exist, where the $$k^\ssr{th}$$ moment is simply $$\langle x^k\rangle$$, the distribution of the sum $$X=\sum_{i=1}^N x\ns_i$$ is a Gaussian.

## Moments and cumulants

Consider a general multivariate distribution $$P(x\ns_1,\ldots,x\ns_N)$$ and define the multivariate Fourier transform $\HP(k\ns_1,\ldots,k\ns_N)=\impi dx\ns_1\cdots\!\!\impi dx\ns_N\>P(x\ns_1,\ldots,x\ns_N)\,\exp\bigg(\!-i\sum_{j=1}^N k\ns_j x\ns_j\bigg)\ .$ The inverse relation is $P(x\ns_1,\ldots,x\ns_N)=\impi {dk\ns_1\over 2\pi}\cdots\!\!\impi {dk\ns_N\over 2\pi}\> \HP(k\ns_1,\ldots,k\ns_N)\,\exp\bigg(\!+i\sum_{j=1}^N k\ns_j x\ns_j\bigg)\ .$ Acting on $$\HP(\Bk)$$, the differential operator $$i\,{\pz\over\pz k\ns_i}$$ brings down from the exponential a factor of $$x\ns_i$$ inside the integral. Thus, $\Bigg[\bigg(\!i\,{\pz\over\pz k\ns_1}\bigg)^{\!m\ns_1}\!\!\cdots\bigg(\!i\,{\pz\over\pz k\ns_N}\bigg)^{\!m\ns_N}\,\HP(\Bk) \Bigg]\nd_{\Bk=0} \!\!\!\!=\blangle x_1^{m\ns_1}\cdots x_N^{m\ns_N}\brangle\ .$ Similarly, we can reconstruct the distribution from its moments, viz. $\HP(\Bk)=\sum_{m\ns_1=0}^\infty\cdots\sum_{m\ns_N=0}^\infty {(-i k\ns_1)^{m\ns_1}\over m\ns_1!}\cdots {(-i k\ns_N)^{m\ns_N}\over m\ns_N!}\, \blangle x_1^{m\ns_1}\cdots x_N^{m\ns_N}\brangle\ .$

The cumulants $$\langle\!\langle x_1^{m\ns_1}\cdots x_N^{m\ns_N} \rangle\!\rangle$$ are defined by the Taylor expansion of $$\ln \HP(\Bk)$$: $\ln\HP(\Bk)=\sum_{m\ns_1=0}^\infty\cdots\sum_{m\ns_N=0}^\infty {(-i k\ns_1)^{m\ns_1}\over m\ns_1!}\cdots {(-i k\ns_N)^{m\ns_N}\over m\ns_N!}\, \big\langle\!\big\langle x_1^{m\ns_1}\cdots x_N^{m\ns_N}\big\rangle\!\big\rangle\ .$ There is no general form for the cumulants. It is straightforward to derive the following low order results: $\begin{split} \langle\!\langle x\ns_i \rangle\!\rangle&= \langle x\ns_i \rangle \\ \langle\!\langle x\ns_i x\ns_j\rangle\!\rangle&= \langle x\ns_i x\ns_j \rangle - \langle x\ns_i\rangle \langle x\ns_j\rangle \\ \langle\!\langle x\ns_i x\ns_j x\ns_k \rangle\!\rangle&= \langle x\ns_i x\ns_j x\ns_k \rangle - \langle x\ns_i x\ns_j \rangle \langle x\ns_k \rangle - \langle x\ns_j x\ns_k \rangle \langle x\ns_i \rangle - \langle x\ns_k x\ns_i \rangle \langle x\ns_j \rangle + 2 \langle x\ns_i \rangle \langle x\ns_j \rangle \langle x\ns_k \rangle \ . \end{split}$

## Multidimensional Gaussian integral

Consider the multivariable Gaussian distribution, $P(\Bx)\equiv \bigg({\det\!A\over (2\pi)^n}\bigg)^{\!1/2}\exp\Big(-\half \, x\ns_i \, A\ns_{ij} \, x\ns_j\Big)\ ,$ where $$A$$ is a positive definite matrix of rank $$n$$. A mathematical result which is extremely important throughout physics is the following: $Z(\Bb)=\bigg({\det\!A\over (2\pi)^n}\bigg)^{\!1/2}\impi dx\ns_1\cdots\!\impi dx\ns_n \> \exp\Big(-\half \, x\ns_i \, A\ns_{ij} \, x\ns_j + b\ns_i\,x\ns_i\Big)=\exp\Big(\half \, b\ns_i\,A^{-1}_{ij}\,b\ns_j\Big)\ .$ Here, the vector $$\Bb=(b\ns_1\,,\,\ldots\,,\,b\ns_n)$$ is identified as a source. Since $$Z(0)=1$$, we have that the distribution $$P(\Bx)$$ is normalized. Now consider averages of the form $\begin{split} \langle \, x\ns_{j\ns_1}\!\!\cdots \,x\ns_{j\ns_{2k}} \, \rangle &= \int\!d^n\!x\>P(\Bx)\>x\ns_{j\ns_1}\!\!\cdots \,x\ns_{j\ns_{2k}} ={\pz^n\!Z(\Bb)\over\pz b\ns_{j\ns_1}\!\cdots\>\pz b\ns_{j\ns_{2k}}}\bigg|\nd_{\Bb=0} \\ &=\sum_{contractions} \!\!\! A^{-1}_{j\ns_{\sigma(1)} j\ns_{\sigma(2)}} \!\!\cdots A^{-1}_{j\ns_{\sigma(2k-1)} j\ns_{\sigma(2k)}}\ . \end{split}$ The sum in the last term is over all contractions of the indices $$\{j\ns_1\,,\,\ldots\,,\,j\ns_{2k}\}$$. A contraction is an arrangement of the $$2k$$ indices into $$k$$ pairs. There are $$C\ns_{2k}=(2k)!/2^k k!$$ possible such contractions. To obtain this result for $$C\ns_k$$, we start with the first index and then find a mate among the remaining $$2k-1$$ indices. Then we choose the next unpaired index and find a mate among the remaining $$2k-3$$ indices. Proceeding in this manner, we have $C\ns_{2k}=(2k-1)\cdot(2k-3)\cdots 3\cdot 1 = {(2k)!\over 2^k k!}\ .$ Equivalently, we can take all possible permutations of the $$2k$$ indices, and then divide by $$2^k k!$$ since permutation within a given pair results in the same contraction and permutation among the $$k$$ pairs results in the same contraction. For example, for $$k=2$$, we have $$C\ns_4=3$$, and $\langle \, x\ns_{j\ns_1} x\ns_{j\ns_2} x\ns_{j\ns_3} x\ns_{j\ns_4 \, }\rangle = A^{-1}_{j\ns_1 j\ns_2} A^{-1}_{j\ns_3 j\ns_4} + A^{-1}_{j\ns_1 j\ns_3} A^{-1}_{j\ns_2 j\ns_4} + A^{-1}_{j\ns_1 j\ns_4} A^{-1}_{j\ns_2 j\ns_3} \ .$ If we define $$b\ns_i=ik\ns_i$$, we have $\HP(\Bk)=\exp\Big(\!-\half\, k\ns_i \, A^{-1}_{ij}\,k\ns_j\Big)\ ,$ from which we read off the cumulants $$\langle\!\langle x\ns_i x\ns_j \rangle\!\rangle= A^{-1}_{ij}$$, with all higher order cumulants vanishing.