IND ENG 263A Lecture Notes - Lecture 2: Moment-Generating Function, Central Limit Theorem, Random Variable
14 Dec 2017
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Recall: cov(x, y) = e[(x e[x])(y e[y])] = e[xy] = e[x]e[y]. There are n men with n hats, and they randomly rearrange them. We want to show taht e[x] = 1 x 1, and. For n = 1, it is clearly a value of 1 and a variance of 0. For n = 2, either both men switch hats or both get their own. This gives a mean of 1 and a variance of 1. For n = 3, there are three possibilities. Let indicator variable ii be the indicator for the ith man matches. We notice that they are bernoulli random variables with. Var(xi) = var(p ii) +p cov(iiij) = n ( 1. For a large n, y = pn yi n. Suppose the yi"s are actually indicators, and are bernoulli(p). This is a better approximation if p is near 1] ) x bin(n, p) + nlarge (np, np(1 p)). n.
