I am wondering why Cov(X 1, X 2) in the following example is not zero? If Cov(X

Question

I am wondering why Cov(X1, X2) in the following example is not zero? If Cov(X1, X2 )=E(X1,*X2)-E(X1)E?X2?? Are E(X1)E?X2? and E(X1,*X2)= P2 since all the other terms in calculatingE(X1,*X2) are zero except the one when both X1 and X2 are one?

I really appreciate it if you can point it out where my reasoning goes wrong. Thanks!

Example 6.13 esn The results of the previous proposition allow for a straightforward derivation of the mean and variance of a hypergeometric rv, which were given without proof in Section 3.6. Recall that the distribution is defined in terms of a population with N items, of which M are successes and N - M are failures. A sample of size n is drawn of which X are successes. It is equivalent to view this as random arrangement of all N items, followed by selection of the first n·Let X, be 1 if the ith item is a success d arance d yergntn r/aed i wann and 0 if it is a failure, i-1.2,..., N. Thern According to the proposition, we can find the mean and variance of X if we can find the means, variances, and covariances of the terms in the sum.

Explanation / Answer

The give proof is correct, there is no corrections.

If the the two random variables are independent then cov(x1,x2)=0

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