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The Binomial Moment Generating Function MSc. Econ: MATHEMATICAL STATISTICS, 1996 The Moment Generating Function of the Binomial Distribution Consider the binomial function n! (1) b(x; n, p)= pxqnx with q =1p. x!(n x)! Then the moment generating function is given by Xn n! M (t)= ext pxqnx x x!(n x)! x=0 Xn (2) n! = (pet)x qnx x!(n x)! x=0 =(q+pet)n, where the nal equality is understood by recognising that it represents the expansion of binomial. If we dierentiate the moment generating function with respect to t using the function-of-a-function rule, then we get dM (t) x = n(q + pet)n1pet (3) dt = npet(q + pet)n1. Evaluating this at t = 0 gives (4) E(x)=np(q + p)n1 = np. Notice that this result is already familiar and that we have obtained it previ- ously by somewhat simpler means. To nd the second moment, we use the product rule duv dv du (5) = u + v dx dx dx to get d2M (t) x = npet (n 1)(q + pet)n2pet +(q+pet)n1 npet dt2 (6) = npet(q + pet)n2 (n 1)pet +(q+pet) = npet(q + pet)n2 q + npet . 1 MSc. Econ: MATHEMATICAL STATISTICS: BRIEF NOTES, 1996 Evaluating this at t = 0 gives E(x2)=np(q + p)n2(q + np) (7) = np(q + np). From this, we see that 2 V (x)=E(x2) E(x) (8) =np(q + np) n2p2 = npq. Theorems Concerning Moment Generating Functions In nding the variance of the binomial distribution, we have pursed a method which is more laborious than it need by. The following theorem shows how to generate the moments about an arbitrary datum which we may take to be the mean of the distribution. (9) The function which generates moments about the mean of a ran- dom variable is given by Mx(t) = exp{t}Mx(t) where Mx(t) is the function which generates moments about the origin. This result is understood by considering the following identity: t xt (10) Mx(t)=E exp{(x )t} = e E(e ) = exp{t}Mx(t). For an example, consider once more the binomial function. The moment generating function about the mean is then npt t n Mx(t)=e (q + pe ) pt t pt n (11) =(qe + pe e ) =(qept + peqt)n. Dierentiating this once gives dM (t) (12) x = n(qept + peqt)n1(pqept + qpeqt). dt At t = 0, this has the value of zero, as it should. Dierentiating a second time according to the product rule gives d2M (t) du (13) x = u(p2qept + q2peqt)+v , dt2 dt 2 MSc. Econ: MATHEMATICAL STATISTICS, 1996 where u(t)=n(qept + peqt) and (14) v(t)=(pqept + qpeqt). At t = 0 these become u(0) = n and v(0) = 0. It follows that (15) V (x)=n(p2q+q2p)=npq(p + q)=npq, as we know from a previous derivation. Another important theorem concerns the moment generating function of a sum of independent random variables: (16) If x f(x) and y f(y) be two independently distributed random variables with moment generating functions Mx(t) and My(t), then their sum z = x + y has the moment generating func- tion Mz(t)=Mx(t)My(t). This result is a consequence of the fact that the independence of x and y implies that their joint probability density function is the product of their individual marginal probability density functions: f(x, y)=f(x)f(y). From this, it follows that Z Z (x+y)t Mx+y(t)= e f(x, y)dydx Zx y Z (17) = extf(x)dx eytf(y)dy x y = Mx(t)My(t). 3.
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