# Thread: If X1, X2, ...., Xn are iid does that necessarily mean X1^2, X2^2, ..., Xn^2 are iids

1. ## If 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)?

2. ## Re: If X1, X2, ...., Xn are iid does that necessarily mean X1^2, X2^2, ..., Xn^2 are

If $X_1,\ldots,X_n$ are independent then so are $X_1^2,\ldots,X_n^2$. To see that, take $a_1\leq b_1,\ldots,a_n\leq b_n$ and let $g:x\mapsto x^2$. We have $P\left(\bigcap_{j=1}^nX_j^2\in\left[a_j,b_j\right]\right) =P\left(\bigcap_{j=1}^ng(X_j)\in\left[a_j,b_j\right]\right) =P\left(\bigcap_{j=1}^nX_j\in g^{-1}(\left[a_j,b_j\right])\right),$
and since $X_1,\ldots,X_n$ are independent, we have $P\left(\bigcap_{j=1}^nX_j\in g^{-1}(\left[a_j,b_j\right])\right)=\prod_{j=1}^nP\left(X_j\in g^{-1}(\left[a_j,b_j\right])\right)$ and we are done.
We conclude that your formula is true if $Var(X_1^2)$ exists (but the problem doesn't come from the independence).