I'm trying to do the following exercise
Let $(X_1,...,X_n)$ be a random sample from an $N(\mu, \sigma^2)$ distribution with $\mu$ and $\sigma$ unknown. Assuming $n>3$, find a minimum variance unbiased estimator for $\mu^2/\sigma^2$.
I think I need to find an unbiased estimator $g$ for $\mu^2/\sigma^2$ and then compute the conditional expectation $E(g| \sum x_i, \sum x_i^2)$. At the moment I'm not sure even how to write down the preliminary estimator.
By the LLN, the fraction $\theta^*=\frac{\left(\overline X\right)^2}{S^2}$ is a consistent estimate for $\theta=\frac{\mu^2}{\sigma^2}$. Here $$S^2=\frac{1}{n-1}\sum \left(X_i-\overline X\right)^2=\frac{n}{n-1}\left(\overline {X^2}-\left(\overline X\right)^2\right).$$
Let us find bias of this estimate: $\overline X$ and $S^2$ are independent, $$\overline X \sim N\left(\mu,\frac{\sigma^2}{n}\right) \text{ and } \frac{(n-1)S^2}{\sigma^2}\sim \chi^2_{n-1}. $$ Note also that $$\mathbb E\left[\frac{1}{\chi^2_{n-1}}\right]=\frac{1}{n-3}$$
Therefore $$ \mathbb E[\theta^*]=\mathbb E\left[\frac{\bigl(\overline X\bigr)^2}{S^2}\right] = \mathbb E\left[\bigl(\overline X\bigr)^2\right]\cdot \mathbb E\left[\frac{1}{\frac{(n-1)S^2}{\sigma^2}}\right]\cdot\frac{n-1}{\sigma^2} $$ $$ =\left(\mu^2+\frac{\sigma^2}{n}\right)\cdot \frac{1}{n-3}\cdot\frac{n-1}{\sigma^2} = \frac{n-1}{n-3}\left(\frac{\mu^2}{\sigma^2}+\frac1n\right). $$ Therefore $$\theta^{**}=\frac{n-3}{n-1}\theta^*-\frac1n$$ is the unbiased estimate for $\theta=\frac{\mu^2}{\sigma^2}$.