If X1, X2, Xn are iids, this means that:

Var(X1+X2+...+Xn) = Var(X1) + Var(X2) +...+Var(Xn), right??

but does : Var(X1^2 + X2^2 + ... + Xn^2) = Var(X1^2) + Var(X2^2)+...+Var(Xn^2)?

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- Aug 17th 2011, 12:42 AMfallingleaves75If X1, X2, ...., Xn are iid does that necessarily mean X1^2, X2^2, ..., Xn^2 are iids
If X1, X2, Xn are iids, this means that:

Var(X1+X2+...+Xn) = Var(X1) + Var(X2) +...+Var(Xn), right??

but does : Var(X1^2 + X2^2 + ... + Xn^2) = Var(X1^2) + Var(X2^2)+...+Var(Xn^2)? - Aug 17th 2011, 01:57 AMgirdavRe: If X1, X2, ...., Xn are iid does that necessarily mean X1^2, X2^2, ..., Xn^2 are
If $\displaystyle X_1,\ldots,X_n$ are independent then so are $\displaystyle X_1^2,\ldots,X_n^2$. To see that, take $\displaystyle a_1\leq b_1,\ldots,a_n\leq b_n$ and let $\displaystyle g:x\mapsto x^2$. We have $\displaystyle P\left(\bigcap_{j=1}^nX_j^2\in\left[a_j,b_j\right]\right) =P\left(\bigcap_{j=1}^ng(X_j)\in\left[a_j,b_j\right]\right) =P\left(\bigcap_{j=1}^nX_j\in g^{-1}(\left[a_j,b_j\right])\right),$

and since $\displaystyle X_1,\ldots,X_n$ are independent, we have $\displaystyle P\left(\bigcap_{j=1}^nX_j\in g^{-1}(\left[a_j,b_j\right])\right)=\prod_{j=1}^nP\left(X_j\in g^{-1}(\left[a_j,b_j\right])\right)$ and we are done.

We conclude that your formula is true if $\displaystyle Var(X_1^2)$ exists (but the problem doesn't come from the independence).