Finding UMVUE from Lehmann-Sheffe Directly

Question

I am having some trouble with an example from the book I am following.

Let $X_1,X_2,...,X_n$ for $n>2$ be an iid set of $N(\mu,\sigma^2)$ random variables with, $\mu\in\mathbb{R}$ and $\sigma^2\in\mathbb{R}^+$, both unknown. Under this circumtance $T=(\overline{X}_n,S_n^2)=(Y,W)$, the bi-variate statistic containing the sample mean and sample variance, is sufficient and complete for $(\mu,\sigma^2)$, and $\overline{X}_n$ and $\frac{nS_n^2}{\sigma^2}$ are independent, $N(\mu,\sigma^2/n)$ and $\chi^2_{n-1}$ random variables, respectively.

Because of the Lehmann-Scheffe lemma, an unbiased function $h(Y,W)$ for $\mu$ is the almost surely unique UMVUE of $\mu$. Therefore, such a function $h$ must satisfy \begin{equation} \int_0^\infty f(w)\left[\int_{-\infty}^\infty h(y,w)f(y)dy\right]dw=\mu\nonumber \end{equation} where $f(y)$ and $f(w)$ are the pdf of $Y$ and $W$ respectively. I have tried several of the techniques like differentiating several times both sides by $\mu$, and using completeness in order to find the condition that leads to \begin{equation} h(Y,W)=Y\text{ } a.s. P_{\mu,\sigma^2}, \forall (\mu,\sigma^2)\in\mathbb{R}\times\mathbb{R}^+ \end{equation}

I can not seem to find the conditions required for the last equation to hold.

Any suggestions? is there any general rule that should be applied?

Best regards,

JMJulio

Answer 1

It was quite simple. Thanks to all who took the trouble to look at it.

Since the joint cdf of $Y$ and $Z$ is of bounded variation, the derivative wrt $\mu$ of the left hand side of \begin{equation} E_{\mu,\sigma^2}\left[h(Y,Z)\right]=\mu \end{equation} interchanges with both integrals, leading to \begin{equation} E_{\mu,\sigma^2}\left[h(Y,Z)\frac{n}{\sigma^2}(y-\mu)\right]=1 \end{equation}

Therefore, \begin{equation} E_{\mu,\sigma^2}\left[h(Y,Z)Y\right]=E_{\mu,\sigma^2}\left[Y^2\right] \end{equation} and applying completeness to \begin{equation} E_{\mu,\sigma^2}\left[h(Y,Z)Y-Y^2\right]=0 \end{equation} this leads to $h(Y,Z)Y=Y^2$ a.s. $P_{\mu,\sigma^2}$ $\forall\mu,\sigma^2\in\mathbb{R}\times\mathbb{R}^+$, from where \begin{equation} h(Y,Z)=Y \text{ a.s. } P_{\mu,\sigma^2} \forall\mu,\sigma^2\in\mathbb{R}\times\mathbb{R}^+ \end{equation}