STAT
A Probability Path.pdf

# Then and applying 95 with x replaced by tx we get n

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Then and applying (9.5) with x replaced by tX, we get n (it)k ( !tX!n+l 2!tX!n) l<t><t)- L -, E(Xk)! E '1\ -,- . k=O k. (n + 1). n. (9 . 6) Next, suppose all moments exist and that for all t E lR lim ltlnE(IXIn) =O . n-+oo n! (9.7) In (9.6), let n oo to get A sufficient condition for (9 .7) is (9.8) which holds if \ll(t) = Ee 1 x < oo, Yt E IR; that is, if the mgf exists on all of JR. (To check this last statement, note that if Ee 1 X < oo for all t, we have £elt11XI = E(eltiX l[X>OJ) + E(e-lt iX l[X<OJ) \ll(ltl) + \11(-ltl) < 00. This verifies the assertion.)

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300 9. Characteristic Functions and the Central Limit Theorem Example 9.3 .1 Let X be a random variable with N(O, 1) distribution. For any t E R, Ee'x = e'" --du i oo e-u 2 ;2 -00 ../2i = -- exp{--(u 2 - 2tu + t 2 )}du e' 1 2 i oo 1 1 2 -00 ../2i 2 (from completing the square) r2;2 e d i oo -!<u-t) 2 =e u -00 ../2i 2 2 = e' 1 2 n(t, 1, u)du = e' 1 2 . -00 Here, n(t, 1, u) represents the normal density with meant and variance 1 which integrates to 1. Thus we conclude that for all t E R Ee'x < oo . We may therefore expand the mgf as well as e' 2 1 2 to get that is, Equating coefficients yields that the E(X2n) e)n coefficient of t 2 n = = - 2 - (2n)! n! So we conclude that Thus, since
9.4 Moments and Derivatives 301 we get (9.9) This shows one way of computing the chf of the N(O, 1) distribution. Note that the chf of the normal density has the property that the chf and the density are the same, apart from multiplicative constants. This is a useful and unusual feature. 9.4 Moments and Derivatives When the kth absolute moment of a random variable exists, it can be computed by taking k-fold derivatives of the chf. Suppose X is a random variable with finite first absolute moment; that is, E(IXD < oo. Then .;._'f' ___ -'-'f'-'--- E(iXe 11 x) = E "'(t +h)_ "'(t) . (ei(t+h)X _ eitX _ ihX eitX) h h - E (it X {eihX - 1 - ihX}) - h 0 Apply (9.4) with n = 1 to get eihX 1 ihX leitXI I - h- I ::S 21XI E Lt . Since by (9.3) or (9.5) we have eihX -1- ihX h 2 X 2 X 2 I h I ::s 2h = hT--+ o as h .J.. 0, we get by dominated convergence that ( q,(t + q,(t) - E(iXeitX)) = E (lim eitX (ihX - 1 - ihX )) h =0.

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302 9. Characteristic Functions and the Central Limit Theorem Thus ¢'(t) = E(iXeitX). (9.10) In general, we have that if E(IXIk) < oo, ¢(k)(t) = E ((iXleitX), Vt E I. (9.11) and hence 9.5 Two Big Theorems: Uniqueness and Continuity We seek to prove the central limit theorem. Our program for doing this is to show that the chf of centered and scaled sums of independent random variables con- verges as the sample size increases to the chf of the N (0, 1) distribution which we know from Example 9.3.1 is exp{-t 2 /2}. To make this method work, we need to know that the chf uniquely determines the distribution and that when chf's of dis- tributions converge, their distributions converge weakly. The goal of this section is to prove these two facts. Theorem 9.5.1 (Uniqueness Theorem) The chf of a probability distribution uniquely determines the probability distribution. Proof. We use the fact that the chf and density of the normal distribution are the same apart from multiplicative constants.
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