$\begingroup$Suppose $X_1, \cdots, X_n$ are i.d.d. samples from population $X \sim N(\mu,\sigma^2)$, and the sample variance is denoted by $T = \sum_{i = 1}^n \frac{(X_i - \overline{X})^2}{n}$.
I am curious about the expected-value of $T^2$, which is the square of $T$. Apparently the key problem is what the distribution of $T^2$ is ? According to my intuition, it may be some kind of F-distribution, but how to prove it ,especially to solve the cross term is the biggest problem that I have encountered.
1 Answer
$\begingroup$You might now this forumla: $$ \text{Var}[X] = E[X^2] - E[X]^2 $$ I.e. $$ E[X^2] = \text{Var}[X] + E[X]^2 $$ The variance is the expected value of the squared variable, but centered at its expected value.
In this case, the random variable is the sample distribution, which has a Chi-squared distribution – see the link in the comment.
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