Blundell Thermal Physics · Problem 3.8

Probability

Physics — Statistical Mechanics · moment-generating-function

Problem

This question introduces a rather efficient method for calculating the mean and variance of probability distributions. We define the moment generating function M(t)M(t) for a random variable xx by M(t)=⟨etx⟩.(3.51)M(t) = \langle \mathrm{e}^{tx} \rangle. \tag{3.51} Show that this definition implies that ⟨xn⟩=M(n)(0),(3.52)\langle x^n \rangle = M^{(n)}(0), \tag{3.52} where M(n)(t)=dnM/dtnM^{(n)}(t) = \mathrm{d}^n M/\mathrm{d}t^n and further that the mean ⟨x⟩=M(1)(0)\langle x \rangle = M^{(1)}(0) and the variance σx=M(2)(0)−[M(1)(0)]2\sigma_x = M^{(2)}(0) - [M^{(1)}(0)]^2. Hence show that: (a) for a single Bernoulli trial, M(t)=pet+1−p;(3.53)M(t) = p\mathrm{e}^t + 1 - p; \tag{3.53} (b) for the binomial distribution, M(t)=(pet+1−p)n;(3.54)M(t) = (p\mathrm{e}^t + 1 - p)^n; \tag{3.54} (c) for the Poisson distribution, M(t)=em(et−1);(3.55)M(t) = \mathrm{e}^{m(\mathrm{e}^t - 1)}; \tag{3.55} (d) for the exponential distribution, M(t)=λλ−t.(3.56)M(t) = \frac{\lambda}{\lambda - t}. \tag{3.56} Hence derive the mean and variance in each case and show that they agree with the results derived earlier.

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