Suppose $X_1, \ldots , X_4$ are i.i.d $\mathsf N(\mu, \sigma^2)$ random variables. Give the UMVUE of $\frac{\mu^2}{\sigma}$ expressed in terms of $\bar{X}$, $S$, integers, and $\pi$.
Here is a relevant question.
I first note that if $X_1,\ldots,X_n$ are i.i.d $\mathsf N(\mu,\sigma^2)$ random variables having pdf
$$\begin{align*} f(x\mid\mu,\sigma^2) &=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\\\\ &=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{\mu^2}{2\sigma^2}}\exp\left(-\frac{1}{2\sigma^2}x^2+\frac{\mu}{\sigma^2}x\right) \end{align*}$$
where $\mu\in\mathbb{R}$ and $\sigma^2\gt0$, then
$$T(\vec{X})=\left(\sum_{i=1}^n X_i^2, \sum_{i=1}^n X_i\right)$$
are sufficient statistics and are also complete since $$\left\{\left(-\frac{1}{2\sigma^2},\frac{\mu}{\sigma^2}\right):\mu\in\mathbb{R}, \sigma^2\gt0 \right\}=(-\infty,0)\times(-\infty,\infty)$$
contains an open set in $\mathbb{R}^2$
I also note that the sample mean and sample variance are stochastically independent and so letting
$$\overline{X^2}=\frac{1}{n}\sum_{i=1}^n X_i^2$$
$$\overline{X}^2=\frac{1}{n}\sum_{i=1}^n X_i$$
we have
$$\mathsf E\left(\frac{\overline{X^2}}{S}\right)=\mathsf E\left(\overline{X^2}\right)\cdot\mathsf E\left(\frac{1}{S}\right)=\overline{X^2}\cdot\mathsf E\left(\frac{1}{S}\right)$$
It remains only to find $\mathsf E\left(\frac{1}{S}\right)$
We know that $$(n-1)\frac{S^2}{\sigma^2}\sim\chi_{n-1}^2$$
Hence
$$\begin{align*} \mathsf E\left(\frac{\sigma}{S\sqrt{3}}\right) &=\int_0^{\infty} \frac{1}{\sqrt{x}} \cdot\frac{1}{\Gamma(1.5)2^{1.5}}\cdot\sqrt{x}\cdot e^{-x/2}dx\\\\ &=\frac{4}{\sqrt{\pi}\cdot2^{1.5}} \end{align*}$$
So $$\mathsf E\left(\frac{1}{S}\right)=\frac{4\sqrt{3}}{\sqrt{\pi}\cdot 2^{1.5}\cdot \sigma}$$
But since $\mathsf E(S)\neq\sigma$ I don't think I can just plug in $S$ for $\sigma$ here.
I have that since $\mathsf E\left(\overline{X^2}\right)=\mathsf{Var}\left(\overline{X}\right)+\mathsf E\left(\bar{X}\right)^2=\frac{\sigma^2}{4}+\mu^2$
Hence
$$\sigma=\sqrt{4\left(E\left(\overline{X^2}\right)-E\left(\overline{X}\right)^2\right)}=\sqrt{4\left(\overline{X^2}-\overline{X}^2\right)}$$
Hence the UMVUE of $\frac{\mu^2}{\sigma}$ is
$$\frac{4\sqrt{3}\cdot\overline{X^2}}{\sqrt{\pi}\cdot 2^{1.5}\cdot \sqrt{4\left(\overline{X^2}-\overline{X}^2\right)}}=\frac{\sqrt{\frac{3}{2\pi}}\left(\frac{S^2}{4}+\bar{X}^2\right)}{\sqrt{\frac{S^2}{4}}}$$
Is this a valid solution?