The expected-value of the square of Sample Variance.

Question

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.

Answer 1

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.

Answer 2

Working from a random sample $\mathbf X=\{X_1,\dots,X_n\}$ with mutually independent $X_i\sim\mathcal N(\mu,\sigma^2)$, the distribution of the unbiased sample variance $$ S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar X)^2 $$ is distributed according to the gamma distribution (shape-rate parameterization) $$ S^2\sim\operatorname{Gamma}\left(\frac{n-1}{2},\frac{n-1}{2\sigma^2}\right). $$ Since we are using the rate parameterization of the gamma distribution and $T=\frac{n-1}{n}S^2$ is follows that $$ T\sim\operatorname{Gamma}\left(\underbrace{\frac{n-1}{2}}_{\alpha},\underbrace{\frac{n}{2\sigma^2}}_{\beta}\right). $$ To find $\mathsf E(T^2)$ we then write $$ \mathsf E(T^2)=\int_0^\infty t^2f_T(t)\,\mathrm dt=\frac{\beta^\alpha}{\Gamma(\alpha)}\int_0^\infty t^{\alpha+2-1}e^{-\beta t}\,\mathrm dt. $$ Substituting $u=\beta t$ then gives $$ \mathsf E(T^2)=\frac{\beta^\alpha}{\beta^{\alpha+2}\Gamma(\alpha)}\int_0^\infty u^{\alpha+2-1}e^{-u}\,\mathrm du=\frac{\Gamma(\alpha+2)}{\beta^2\Gamma(\alpha)}. $$ Using the property $x\Gamma(x)=\Gamma(x+1)$ this result simplfies to $$ \mathsf E(T^2)=\frac{(\alpha+1)\alpha}{\beta^2}=\frac{(n+1)(n-1)}{n^2}\sigma^4=\left(1-\frac{1}{n^2}\right)\sigma^4. $$ Note for large $n$ this becomes $$ \mathsf E(T^2)\sim\sigma^4+\mathcal O\left(\frac{1}{n^2}\right). $$