How to find values $h,k$ such that $T$ is unbiased for $\sigma^2?$

Question

Let $(x_1,...,x_n)$ be a realization of a random vector $(X_1,...,X_n)$ where $E(X_i)=\mu$ and $Var(X_i)=\sigma^2$ for all $i$ and $Cov(X_i,X_j)=\rho$ for all $i \neq j$. Further let $$T(x_1,...,x_n):=h\sum_{i=1}^nx_i^2 + k(\sum_{i=1}^n x_i)^2$$ be an estimator for $\sigma^2$. How can I find values of $h,k \in \mathbb R $ such that $T$ is unbiased?

Answer 1

For $\rho = 0$ you'll get the coefficients for the "regular" sample variance formula, that is $$ S^2 = \frac{1}{n-1}\sum (X_i - \bar{X}_n) $$
Just use the properties of expectation and the definition of variance, i.e., \begin{align} ET =& h E\sum_{i=1}^n X_i^2 + k E(n\bar{X}_n)^2 = hn(Var(X) + E^2(X)) + kn^2(Var(\bar{X}_n) + E^2(\bar{X}_n))\\ =& hn(\sigma^2 + \mu^2) +kn^2\frac{\sigma^2}{n} + kn^2\mu^2 = \sigma^2 (n k + hn) + n\mu^2 (h+kn) = \sigma^2. \end{align} So by equating coefficients we get $$ nk + hn=1, \quad h+kn=0 $$ hence, $h=\frac{1}{n-1}$ and $k=-\frac{1}{n^2 -n}$.

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For $\rho \neq 0$ note that $$ Var(\bar{X}_n) = \frac{1}{n^2}(n \sigma^2 + (n^2-n)\rho) = \frac{\sigma^2}{n} + \frac{n-1}{n}\rho, $$ thus \begin{align} ET =& h E\sum_{i=1}^n X_i^2 + k E(n\bar{X}_n)^2 = hn(Var(X) + E^2(X)) + kn^2(Var(\bar{X}_n) + E^2(\bar{X}_n))\\ =& hn(\sigma^2 + \mu^2) +kn^2(\frac{\sigma^2}{n} + \frac{n-1}{n} \rho - \mu^2)\\ =& \sigma^2(hn+kn) + \mu^2n(h-kn) + k(n^2-n)\rho \end{align} thus the two equations to b solved w.r.t $k$ and $h$ are $$ hn + kn =1, \quad \mu^2n(h-kn) + k(n^2-n)\rho = 0 $$ where $k=1-h$ and $h = 1 - \frac{\mu^2}{2\mu^2 - (n-1)\rho}$. Please check the calculations in the final raw. Cannot guarantee they have no typos.