Stats/Probability Expectation:
The sum of $n$ independent squared standard normal random variables has a $χ^2$ distribution with $n$ degrees of freedom. Prove that if $X$ has a $χ^2$ distribution with $n$ degrees of freedom, then $E(X)=n$.

Question

Stats/Probability Expectation:
The sum of $n$ independent squared standard normal random variables has a $χ^2$ distribution with $n$ degrees of freedom. Prove that if $X$ has a $χ^2$ distribution with $n$ degrees of freedom, then $E(X)=n$.

kirbykirby · Answer

This is what I did:
Let $X_i$~$N(0,1),i=1,2,...,n$ and let $X=X_1^2+X_2^2+...+X_n^2$

So:
$E(X)=E(X_1^2+X_2^2+...+X_n^2)$

$E(X)=E(X_1^2)+E(X_2^2)+...+E(X_n^2)$

Since each $X_i^2$~$χ^2(1)$ <- 1 degree of freedom on chi-square, we know that $E(X_i^2)=1$ since it's the expectation of a Chi-Square on 1 df.

Hence, $E(X)=1+1+...+1$ (n times 1)
$E(X)=n$

Someone told me though this might not be great because you are using the fact that the expectation of a Chi-Square on 1 df is 1, but we're trying to prove the case for df=n (but I dunno I didn't realize the Chi-Square's mean was it's degrees of freedom, just that in another class we used a Chi-Square on 1 df a lot and was told its mean was 1 as well...)

Any other possible way to do it?