Re: ?ans to 5.3.3
thanhnga la (tla@ucla.edu)
Mon, 06 Nov 95 18:19:24 -0800
This is to clarify Mike's question. The answers Var(W1) and Var(W2) in
the book are still right. Var(W1) is calculated as
Var(W1) = Var((1/n)*(Sum of Yi)) = (1/n^2) * Var(Sum of Yi)
= (1/n^2) * (Sum of (Var Yi)) because Y1,...,Yn are indpt.
= (1/n^2) * n * Var(Y1)
= (1/n) * Var(Y1)
and Var(Y1) = E(Y1^2) - (EY1)^2 = theta^2 if you carry out the whole
calculation
then Var(W1) = (1/n)*theta^2
What I meant was Var(W2) = Var (Y1) = theta^2, not Var(W2)=Var(W1),
because the exponential functions used to calculate the variances of W2,
and Y1 are similar. Therefore Var(W2) = theta^2.
I hope this would help to clarify Mike's confusion.
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