I wanted to know what the proof for the variance term in a central chi-squared distribution (degree n) is. I know that the answer is 2n, but I was wondering how to derive it.
Here's my attempt so far:
Let $X_2$ denote a variable governed by the central chi-squared distribution.
$Var(X_2) = E[(X_2)^2] - (E[X_2])^2$
$= E[(X_2)^2] - (n^2)$
I was able to prove that the mean of a central chi-squared distribution is it's degree (n), by using the formula:
$E[Za*Za] = Cov(Za, Za) + E[Za]^2$
$= sa*sa + 0$
$= sa^2 = 1^2 = 1.$
$E[X_2] = SUM(E[Za^2])$, a goes from 1 to n. (using linearity of expectation)
$= SUM(1)$, 1 to n
$X_2 = SUM(Za^2)$, a goes from 1 to n, and Za ~ N(0, 1) (this is the chi-squared definition)
sa = standard deviation, which in this case, is = 1.
So using this knowledge, the only term in the variance definition for the chi-squared distribution that I don't know, is $E[(X2)^2]$. However, I'm at a loss as to how to compute this. Any help here would be greatly appreciated.
N.B. This is my first post here, and I don't know a lot about mathematical typing. Pardon my poor notation and lack of detail in the question, if any. Would be happy to answer any counter-questions.