Question

Problem . In class we saw that the even moments of the standard Gaussian Xx(0,1) are given by: EX2k = (2k – 1)!! = (2k – 1)(2k – 3)... 31 (2k)! 2kk! Meanwhile the odd moments vanish. The goal of this exercise is to prove the CLT by the method of moments. Suppose X1, X2, ..., Xn are independent identically distirbuted random variables with EX1 = 0 (mean zero), EX} = 1 (unit variance) and bounded higher moments EXK < M. Let's show that the moments E X1 + X2 + ... + Xn vn coincide with the corresponding k-th moment of the Gaussian. Under a technical assumption this is sufficient to prove the CLT. Parts (a), (b) and (c) here are just a warmup to see what's going on in the general case. Arguing part (d) alone will get full credit (a) Verify that the second moment is given by n n EX? +n-2EX;X; i<j What is the limiting value as n +00? i=1 (b) Verify that the third moment is given by n n 3/2 ŻEX +n+3/23EX{X; + 3EX;X; +n+3/2 6EX;X;Xk i=1 i<i<k i<j What is the limiting value as n +00? (c) Likewise, what is the expression for the forth moment of X1+X2+...Xn? What is its limiting value? Vn (d) Argue why in the general case the k-th moment vanishes if k is odd and equals (2m – 1)!! if k = 2m.

Answer 1

0 Let * be a dandom var jable having Normal distribution cofth mean it and variance o2, NC4,0). dhe density is f(x) = Van *(-)2 - Treko The centaal emoc theodem En the mathematical theory of probablety as expoessed as follovas "If Xi ilin be endependent sandom variable such that ECx) = pui and V(x) = 0;2, then under certain condi Hons/ the random vascablo Sn = x + x2+ + xn asymptotically normal weth mean and standard devcation wher H= 1=2 li and ir 1 The weak form of CLT is Suppose that xi's are independend, identically destoebuted pandong variables with zero mean and variance 2 Xi+X2+ +xn ~ NC00²). Here coo would like to poove the CLT method of the by moments. We compute the lim et of the momonts of Yo = X1 +82 t... txo we assume that is are Independent and edentically distributed roandom variables with zero mean ie, Elxi)=0 and and bounded unet Vaalance, vex:) - ECx+2) = hecgher moments Ecx, k) = M. It is easy to see that (xi's are ind] ECYn) = € x + x2- + Xp) - 3 ¿ECX) [since ECX) - ) O

The second moment of Yn is given by Econ) - El (x1+x2 + - +*n)? ] E + +(220x32) + 2 IECKIX)] -ECx2)+ - to Ecxix) E(X12)=1 ECKI) EC*) Dias Ecx) =o. 1 + no 1 As n as, the second moment tends to 1. we assume that ECx;3) SM then thEod moment is ECYn²= E (x + X2 + +Xn n 3/2 ******0)%) + [ 2013 +37**** **;*x] 2 ECx; 3) + 3 ova ECRİNj) the IACKI *3 *) ² E (x; 3) + 3 Ecki+xj) + € (x:x;) n2 IST 1 3/2 nh n2 izj +6 I ecrit; xk) nisjak Ś Mn M as no n 3½ Therd moment tends to zero as - OD

Fourth moment of Yn is given by El Yot) C ( ****2+ -+x0)* *] [22:+4,2*x +39 x1 *xj2 + 6 EX *** + I xixj *xxe] ijek itj #kte 1 tony terity next) + 4, 5, 6CK®x) + € *i**,25+6.5.2***) + Exi xj ***) na itj tkal 30 2. Ex:+) - 3 ECxi ²x; * )= E(x; ²) Ex 2 + 3 min-1 n? # 3n(n-1) + oc/n) as no ie 4th moment tends to 3 as 0 0 all terms of & in fact, what we were doing is classifying en several kends come | length 4. These ¿ ¿ . K u K com es in these varate زن Por og, the interesting toom 0 0 v All terms hich contain a singleton Ca vanable appearing only once) equal zero and can be dropped. of these be

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avaliable slot must be X2 Its counterpart nout been of the (m-3) available positions aod so on. Thus the number of possible formos (0-1) (-3)... ove of k is odd je. In general, the ith moment vaneshes And even moments are [lf to is even € (**) = (K-1) CK-3]... Let t=am, them ECX2M) = (2m-1) (2m-2)....