# independence of sums of normal random variables

Suppose $X_{i} \sim N(0,\sigma^{2})$, $i\leqslant 16$ are independent random variables. Let $Y=\sum\limits_{i=1}^{8} \frac{\sigma}{\sqrt{8}} X_{i}$ and $Z=\sum\limits_{i=9}^{16} \frac{X_{i}^{2}}{\sigma^{2}}$. Are $Y$ and $Z$ independent? Why?

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## 1 Answer

More generally, if $X_1,\ldots,X_{2k}$ are $2k$ independent random variables, then for all $f,g\colon \mathbb R^k\to\mathbb R$ measurable, $f(X_1,\ldots,X_k)$ and $g(X_{k+1},\ldots,X_{2k})$ are independent. Indeed, for $A,B\in\mathcal B(\mathbb R)$, putting $Y_1=(X_1,\ldots,X_k)$ and $Y_2=(X_{k+1},\ldots,X_{2k})$ \begin{align*} P((f(X_1,\ldots,X_k),g(X_{k+1},\ldots,X_{2k})\in A\times B)&=P(Y_1\in f^{-1}(A),Y_2\in g^{-1}(B))\\ &=P(Y_1\in f^{-1}(A))P(Y_2\in g^{-1}(B))\\ &=P((f(X_1,\ldots,X_k)\in A)P(g(X_{k+1},\ldots,X_{2k})\in B), \end{align*} since $f^{-1}(A)$ and $g^{-1}(B)$ are measurable. For the second $=$,take $B_1, B_2$ two elements of $\mathcal B(\mathbb R^n)$. The property $P(Y_1\in B_1,Y_2\in B_2)$ is easy to check if $B_1$ and $B_2$ are of the form $\prod_{j=1}^k[a_j,b_j]$. Since the collection of these sets generates $\mathcal B(\mathbb R^n)$ and is stable by finite intersection, we have the result.

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Thanks a lot :D –  beginner Jan 7 '12 at 14:28
You're welcome. –  Davide Giraudo Jan 7 '12 at 14:29
Excuse me, why are $Y_{1}$ and $Y_{2}$ independent? How can we get the second "="? –  beginner Jan 7 '12 at 14:34
In fact, this step should have been more detailed. I have added the details. –  Davide Giraudo Jan 7 '12 at 14:40
This result (which really concerns sigma-algebras) is routinely called théorème des coalitions in a language I will let you guess (but it seems the nickname is not used in other languages). –  Did Jan 7 '12 at 15:38
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