Question

Problem. In class we saw that the even moments of the standard Gaussian Xx(0,1) are given by: EX2 = (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, ..., X, are independent identically distirbuted random variables with EX; = 0 (mean zero), EX = 1 (unit variance) and bounded higher moments EX < M. Let's show that the moments X1 + X2 + ... + Xn E 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=1 What is the limiting value as no?

Answer 1

avariable slot must be *2. Its counterpart mout been one of the (n-3) available positions aod so on. Thus the number of possible teronos es (-1) (m-3). 1. ie, in generaal, the leth moment vaneshes of todd And even moments are [lf te is even E(X) = (K-1) CK-3]. Let k=am, them ECX2M) = (2m-1) (2m-2)...
Here we tries to prove CLT by the method of moments.