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flyinghorse
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#1
Report Thread starter 14 years ago
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help

Hmm, probability just got difficult, so I thought I'd see if anyone has any ideas about this:

Let X_1, . . . ,X_N be independent identically distributed random variables, where N is a non-negative integer-valued random variable. Let Z = X_1 + . . . + X_N, (assuming that Z = 0 if N = 0).

a) Find E(Z)

b)Show that V(Z) = V(N)E(X_1)^2 + E(N)V(X_1)

???
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Jonny W
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Report 14 years ago
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E(Z)
= (sum over n) E(Z | N = n) P(N = n)
= (sum over n) n E(X_1) P(N = n)
= E(X_1) * (sum over n) n P(N = n)
= E(X_1) E(N)

For the next calculation we need to know that E(X_i X_j) = E(X_i)E(X_j) = E(X_1)^2 for all distinct i and j, which follows from the independence of X_i and X_j.

E(Z^2)
= (sum over n) E(Z^2 | N = n) P(N = n)
= (sum over n) E[(X_1 + ... + X_n)^2] P(N = n)
= (sum over n) [n E(X_1^2) + n(n - 1)E(X_1)^2] P(N = n)
= (sum over n) [n (E(X_1^2) - E(X_1)^2) + n^2 E(X_1)^2] P(N = n)
= (sum over n) [n V(X_1) + n^2 E(X_1)^2] P(N = n)
= V(X_1) * (sum over n) n P(N = n)
= + E(X_1)^2 * (sum over n) n^2 P(N = n)
= V(X_1) E(N) + E(X_1)^2 E(N^2)

V(Z)
= E(Z^2) - E(Z)^2
= V(X_1) E(N) + E(X_1)^2 E(N^2) - E(X_1)^2 E(N)^2
= V(X_1) E(N) + E(X_1)^2 (E(N^2) - E(N)^2)
= V(X_1) E(N) + E(X_1)^2 V(N)
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