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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 |n can be labeled by a discrete variable nC, where C is the set of possible configurations. The total number of possible configurations, which is to say the order of the set C, may be finite or infinite. Next, consider an ensemble of such systems, and let Pn denote the probability that a given random element from that ensemble is in the state (configuration) |n. The collection {Pn} forms a discrete probability distribution. We assume that the distribution is normalized, meaning nCPn=1 .

Now let An be a quantity which takes values depending on n. The average of A is given by A=nCPnAn . Typically, C is the set of integers (Z) or some subset thereof, but it could be any countable set. As an example, consider the throw of a single six-sided die. Then Pn=16 for each n{1,,6}. Let An=0 if n is even and 1 if n is odd. Then find A=12, 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 {n1,n2,n3,}. We can combine these into a vector n and then we write Pn for the discrete distribution, with nPn=1.

Another possibility is that the system’s configurations are parameterized by a collection of continuous variables, φ={φ1,,φn}. We write φΩ, where Ω is the phase space (or configuration space) of the system. Let dμ be a measure on this space. In general, we can write dμ=W(φ1,,φn)dφ1dφ2dφn . The phase space measure used in classical statistical mechanics gives equal weight W to equal phase space volumes: dμ=Crσ=1dqσdpσ , where C is a constant we shall discuss later on below8.

Any continuous probability distribution P(φ) is normalized according to ΩdμP(φ)=1 . The average of a function A(φ) on configuration space is then A=ΩdμP(φ)A(φ) . For example, consider the Gaussian distribution P(x)=12πσ2e(xμ)2/2σ2 . From the result9 dxeαx2eβx=πα eβ2/4α , we see that P(x) is normalized. One can then compute x=μx2x2=σ2 . We call μ the mean and σ the standard deviation of the distribution, Equation [pgauss].

The quantity P(φ) is called the distribution or probability density. One has P(φ)dμ=probability that configuration lies within volume dμ centered at φ For example, consider the probability density P=1 normalized on the interval x[0,1]. The probability that some x chosen at random will be exactly 12, say, is infinitesimal – one would have to specify each of the infinitely many digits of x. However, we can say that x[0.45,0.55] with probability 110.

If x is distributed according to P1(x), then the probability distribution on the product space (x1,x2) is simply the product of the distributions: P2(x1,x2)=P1(x1)P1(x2). Suppose we have a function ϕ(x1,,xN). How is it distributed? Let P(ϕ) be the distribution for ϕ. We then have P(ϕ)=dx1dxNPN(x1,,xN)δ(ϕ(x1,,xN)ϕ)=dx1dxNP1(x1)P1(xN)δ(ϕ(x1,,xN)ϕ) , where the second line is appropriate if the {xj} are themselves distributed independently. Note that dϕP(ϕ)=1 , so P(ϕ) is itself normalized.

Central limit theorem

In particular, consider the distribution function of the sum X=Ni=1xi. We will be particularly interested in the case where N is large. For general N, though, we have PN(X)=dx1dxNP1(x1)P1(xN)δ(x1+x2++xNX) . It is convenient to compute the Fourier transform10 of P(X): ˆPN(k)=dXPN(X)eikX=dXdx1dxNP1(x1)P1(xN)δ(x1++xNX)eikX=[ˆP1(k)]N , where ˆP1(k)=dxP1(x)eikx is the Fourier transform of the single variable distribution P1(x). The distribution PN(X) is a convolution of the individual P1(xi) 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 ˆP1(k) ˆP1(k)=dxP1(x)(1ikx12k2x2+16ik3x3+)=1ikx12k2x2+16ik3x3+ . Thus, lnˆP1(k)=iμk12σ2k2+16iγ3k3+ , where μ=xσ2=x2x2γ3=x33x2x+2x3 We can now write [ˆP1(k)]N=eiNμkeNσ2k2/2eiNγ3k3/6 Now for the inverse transform. In computing PN(X), we will expand the term eiNγ3k3/6 and all subsequent terms in the above product as a power series in k. We then have PN(X)=dk2π eik(XNμ)eNσ2k2/2{1+16iNγ3k3+}=(1γ36N3X3+)12πNσ2e(XNμ)2/2Nσ2NN=(1γ36N1/23ξ3+)12πNσ2eξ2/2σ2 . In going from the second line to the third, we have written X=Nμ+Nξ, in which case X=N1/2ξ, and the non-Gaussian terms give a subleading contribution which vanishes in the N limit. We have just proven the central limit theorem: in the limit N, the distribution of a sum of N independent random variables xi is a Gaussian with mean Nμ and standard deviation Nσ. Our only assumptions are that the mean μ and standard deviation σ exist for the distribution P1(x). Note that P1(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 \boldsymbol{k^\ssr{th}} moment is simply xk, the distribution of the sum X=Ni=1xi is a Gaussian.

Moments and cumulants

Consider a general multivariate distribution P(x1,,xN) and define the multivariate Fourier transform ˆP(k1,,kN)=dx1dxNP(x1,,xN)exp(iNj=1kjxj) . The inverse relation is P(x1,,xN)=dk12πdkN2πˆP(k1,,kN)exp(+iNj=1kjxj) . Acting on ˆP(k), the differential operator iki brings down from the exponential a factor of xi inside the integral. Thus, [(ik1)m1(ikN)mNˆP(k)]k=0=xm11xmNN . Similarly, we can reconstruct the distribution from its moments, viz. ˆP(k)=m1=0mN=0(ik1)m1m1!(ikN)mNmN!xm11xmNN .

The cumulants xm11xmNN are defined by the Taylor expansion of lnˆP(k): lnˆP(k)=m1=0mN=0(ik1)m1m1!(ikN)mNmN!xm11xmNN . There is no general form for the cumulants. It is straightforward to derive the following low order results: xi=xixixj=xixjxixjxixjxk=xixjxkxixjxkxjxkxixkxixj+2xixjxk .

Multidimensional Gaussian integral

Consider the multivariable Gaussian distribution, P(x)(detA(2π)n)1/2exp(12xiAijxj) , where A is a positive definite matrix of rank n. A mathematical result which is extremely important throughout physics is the following: Z(b)=(detA(2π)n)1/2dx1dxnexp(12xiAijxj+bixi)=exp(12biA1ijbj) . Here, the vector b=(b1,,bn) is identified as a source. Since Z(0)=1, we have that the distribution P(x) is normalized. Now consider averages of the form xj1xj2k=dnxP(x)xj1xj2k=nZ(b)bj1bj2k|b=0=contractionsA1jσ(1)jσ(2)A1jσ(2k1)jσ(2k) . The sum in the last term is over all contractions of the indices {j1,,j2k}. A contraction is an arrangement of the 2k indices into k pairs. There are C2k=(2k)!/2kk! possible such contractions. To obtain this result for Ck, we start with the first index and then find a mate among the remaining 2k1 indices. Then we choose the next unpaired index and find a mate among the remaining 2k3 indices. Proceeding in this manner, we have C2k=(2k1)(2k3)31=(2k)!2kk! . Equivalently, we can take all possible permutations of the 2k indices, and then divide by 2kk! 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 C4=3, and xj1xj2xj3xj4=A1j1j2A1j3j4+A1j1j3A1j2j4+A1j1j4A1j2j3 . If we define bi=iki, we have ˆP(k)=exp(12kiA1ijkj) , from which we read off the cumulants xixj=A1ij, with all higher order cumulants vanishing.


This page titled 1.4: General Aspects of Probability Distributions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Daniel Arovas.

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