Moment Generating Functions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.436J/15.085J Lecture 13 Fall 2018 MOMENT GENERATING FUNCTIONS Contents 1. Moment generating functions 2. Sum of a random number of random variables 3. Transforms associated with joint distributions Moment generating functions, and their close relatives (probability generating functions and characteristic functions) provide an alternative way of representing a probability distribution by means of a certain function of a single variable. These functions turn out to be useful in many different ways: (a) They provide an easy way of calculating the moments of a distribution. (b) They provide some powerful tools for addressing certain counting and combinatorial problems. (c) They provide an easy way of characterizing the distribution of the sum of independent random variables. (d) They provide tools for dealing with the distribution of the sum of a random number of independent random variables. (e) They play a central role in the study of branching processes. (f) They play a key role in large deviations theory, that is, in studying the asymptotics of tail probabilities of the form P(X ≥ c), when c is a large number. (g) They provide a bridge between complex analysis and probability, so that complex analysis methods can be brought to bear on probability problems. (h) They provide powerful tools for proving limit theorems, such as laws of large numbers and the central limit theorem. 1 1 MOMENT GENERATING FUNCTIONS 1.1 Definition Definition 1. The moment generating function associated with a random variable X is a function MX : R → [0, ∞] defined by MX (s) = E[esX ]. The domain DX of MX is defined as the set DX = {s | MX (s) < ∞}. If X is a discrete random variable, with PMF pX , then X MX (s) = esx pX (x). x If X is a continuous random variable with PDF fX , then Z MX (s) = esx fX (x) dx. Note that this is essentially the same as the definition of the Laplace transform of a function fX , except that we are using s instead of −s in the exponent. 1.2 The domain of the moment generating function Note that 0 ∈ DX , because MX (0) = E[e0X ] = 1. For a discrete random variable that takes only a finite number of different values, we have DX = R. For example, if X takes the values 1, 2, and 3, with probabilities 1/2, 1/3, and 1/6, respectively, then MX (s) = 1 s 1 2s 1 3s e + e + e , 2 3 6 (1) which is finite for every s ∈ R. On the other hand, for the Cauchy distribution, fX (x) = 1/(ˇ(1 + x2 )), for all x, it is easily seen that MX (s) = ∞, for all s 6= 0. In general, DX is an interval (possibly infinite or semi-infinite) that contains zero. Exercise 1. Suppose that MX (s) < ∞ for some s > 0. Show that MX (t) < ∞ for all t ∈ [0, s]. Similarly, suppose that MX (s) < ∞ for some s < 0. Show that MX (t) < ∞ for all t ∈ [s, 0]. 2 Exercise 2. Suppose that lim sup x!1 log P(X > x) , − < 0. x Establish that MX (s) < ∞ for all s ∈ [0, ). 1.3 Inversion of transforms By inspection of the formula for MX (s) in Eq. (1), it is clear that the distribution of X is readily determined. The various powers esx indicate the possible values of the random variable X, and the associated coefficients provide the corresponding probabilities. At the other extreme, if we are told that MX (s) = ∞ for every s 6= 0, this is certainly not enough information to determine the distribution of X. On this subject, there is the following fundamental result. It is intimately related to the inversion properties of Laplace transforms. Its proof requires sophisticated analytical machinery and is omitted. Theorem 1. Inversion theoremSuppose that MX (s) is finite for all s in an interval of the form [−a, a], where a is a positive number. Then, MX determines uniquely the CDF of the random variable X . In particular, if MX (s) = MY (s) < ∞, for all s ∈ [−a, a], where a is a positive number, then the random variables X and Y have the same CDF. There are explicit formulas that allow us to recover the PMF or PDF of a random variable starting from the associated transform, but they are quite difficult to use (e.g., involving “contour integrals”). In practice, transforms are usually inverted by “pattern matching,” based on tables of known distribution-transform pairs. 1.4 Moment generating properties There is a reason why MX is called a moment generating function. Let us consider the derivatives of MX at zero. Assuming for a moment we can interchange the order of integration and differentiation, we obtain | | dMX (s) ds dm MX (s) dsm s=0 s=0 | | d E[esX ] = E[XesX ] = E[X], ds s=0 s=0 dm = m E[esX ] = E[X m esX ] = E[X m ] ds s=0 s=0 = | 3 | Thus, knowledge of the transform MX allows for an easy calculation of the moments of a random variable X. Justifying the interchange of the expectation and the differentiation does require some work. The steps are outlined in the following exercise. For simplicity, we restrict to the case of nonnegative random variables. Exercise 3. Suppose that X is a nonnegative random variable and that MX (s) < ∞ for all s ∈ (−∞, a], where a is a positive number. (a) Show that E[X k ] < ∞, for every k. (b) Show that E[X k esX ] < ∞, for every s < a. (c) Show that (ehX − 1)/h ≤ XehX . (d) Use the DCT to argue that h ehX − 1 i E[ehX ] − 1 E[X] = E lim = lim . h#0 h#0 h h 1.5 The probability generating function For discrete random variables, the following probability generating function is sometimes useful. It is defined by gX (s) = E[sX ], with s usually restricted to positive values. It is of course closely related to the moment generating function in that, for s > 0, we have gX (s) = MX (log s). One difference is that when X is a positive random variable, we can define gX (s), as well as its derivatives, for s = 0. So, suppose that X has a PMF pX (m), for m = 1, . . .. Then, 1 X gX (s) = sm pX (m), m=1 resulting in | dm gX (s) dsm s=0 = m! pX (m). (The interchange of the summation and the differentiation needs justification, but is indeed legitimate for small s.) Thus, we can use gX to easily recover the PMF pX , when X is a positive integer random variable. At the same time X d = mpX (m) = E[X]. gX (s) ds s=1 | m1 4 1.6 Examples d Example : X = Exp(). Then, ˆ Z 1 sx −x MX (s) = e e dx = 0  −s , ∞, s < ; otherwise. d Example : X = Ge(p) MX (s) = 1 X m=1 sm e m−1 p(1 − p) ( es p 1−(1−p)es , ∞, es < 1/(1 − p); otherwise. In this case, we also find gX (s) = ps/(1 − (1 − p)s), s < 1/(1 − p) and gX (s) = ∞, otherwise. d Example : X = N (0, 1). Then, Z 1 1 x2 MX (s) = √ exp(sx) exp(− )dx 2 2ˇ −1 Z 1 2 2 exp(s /2) x + 2sx − s2 √ = exp(− )dx 2 2ˇ −1 = exp(s2 /2). 1.7 Properties of moment generating functions We record some useful properties of moment generating functions. Theorem 2. (a) If Y = aX + b, then MY (s) = esb MX (as). (b) If X and Y are independent, then MX+Y (s) = MX (s)MY (s). (c) Let X and Y be independent random variables. Let Z be equal to X , with probability p, and equal to Y , with probability 1 − p. Then, MZ (s) = pMX (s) + (1 − p)MY (s). Proof: For part (a), we have MX (aX + b) = E[exp(saX + sb)] = exp(sb)E[exp(saX)] = exp(sb)MX (as). 5 For part (b), we have MX+Y (s) = E[exp(sX + sY )] = E[exp(sX)]E[exp(sY )] = MX (s)MY (s). For part (c), by conditioning on the random choice between X and Y , we have MZ (s) = E[esZ ] = pE[esX ] + (1 − p)E[esY ] = pMX (s) + (1 − p)MY (s). Example : (Normal random variables) (a) Let X be a standard normal random variable, and let Y = ˙X + µ, which we know to have a N (µ, ˙ 2 ) distribution. We then find that MY (s) = exp(sµ + 21 s2 ˙ 2 ). d (b) Let X = N (µ1 , ˙12 ) and Y = N (µ2 , ˙22 ). Then, n o 1 MX+Y (s) = exp( s(µ1 + µ2 ) + s2 (˙12 + ˙22 ) . 2 d Using the inversion property of transforms, we conclude that X + Y = N (µ1 + µ2 , ˙12 + ˙22 ), thus corroborating a result we first obtained using convolutions. 2 SUM OF A RANDOM NUMBER OF INDEPENDENT RANDOM VARIABLES Let X1 , X2 , . . . be a sequence of i.i.d. random variables, with mean µ and variance ˙ 2 . Let N be anotherPindependent random variable that takes nonnegative integer values. Let Y = N i=1 Xi . Let us derive the mean, variance, and moment generating function of Y . We have E[Y ] = E[E[Y | N ]] = E[N µ] = E[N ]E[X]. Furthermore, using the law of total variance,   �
var(Y ) = E var(Y | N ) + var E[Y | N ] = E[N ˙ 2 ] + var(N µ) = E[N ]˙ 2 + µ2 var(N ). Finally, note that n (s) = exp(n log MX (s)), E[exp(sY ) | N = n] = MX 6 implying that MY (s) = 1 X exp(n log MX (s))P(N = n) = MN (log MX (s)). n=1 The reader is encouraged to take the derivative of the above expression, and evaluate it at s = 0, to recover the formula E[Y ] = E[N ]E[X]. Example : Suppose that each Xi is exponentially distributed, with parameter , and that N is geometrically distributed, with parameter p ∈ (0, 1). We find that MY (s) = elog MX (s) p p/( − s) p = = log M (s) X 1 − (1 − p)/( − s) p − s (1 − p) 1−e which we recognize as a moment generating function of an exponential random variable with parameter p. Using the inversion theorem, we conclude that Y is exponentially distributed. In view of the fact that the sum of a fixed number of exponential random variables is far from exponential, this result is rather surprising. An intuitive explanation will be provided later in terms of the Poisson process. 3 TRANSFORMS ASSOCIATED WITH JOINT DISTRIBUTIONS If two random variables X and Y are described by some joint distribution (e.g., a joint PDF), then each one is associated with a transform MX (s) or MY (s). These are the transforms of the marginal distributions and do not convey information on the dependence between the two random variables. Such information is contained in a multivariate transform, which we now define. Consider n random variables X1 , . . . , Xn related to the same experiment. Let s1 , . . . , sn be real parameters. The associated multivariate transform is a function of these n parameters and is defined by   MX1 ,...,Xn (s1 , . . . , sn ) = E es1 X1 +···+sn Xn . The inversion property of transforms discussed earlier extends to the multivariate case. That is, if Y1 , . . . , Yn is another set of random variables and MX1 ,...,Xn (s1 , . . . , sn ), MY1 ,...,Yn (s1 , . . . , sn ) are the same functions of s1 , . . . , sn , in a neighborhood of the origin, then the joint distribution of X1 , . . . , Xn is the same as the joint distribution of Y1 , . . . , Yn . 7 Remarks: (a) Consider two random variables X and Y . Their joint transform is MX,Y (s, t) = E[esX etY ] = E[esX+tY ] = MZ (1), where Z = sX + tY . Thus, calculating a multivariate transform essentially amounts to calculating the univariate transform associated with a single random variable that is a linear combination of the original random variables. (b) If X and Y are independent, then MX,Y (s, t) = MX (s)MY (t). 8