Why is there no UMVUE for $\mu$ with two samples from $N(\mu, \sigma^2)$, $N(\mu, \tau^2)$? Question:


I have an attempted answer below; I don't think it's quite correct. I also have another sketch of a proof at the bottom.
My attempt:
(d) The family of independent samples $X_1, \dots, X_m$ from $N(\mu, \sigma^2)$ and $Y_1, \dots, Y_n$ from $N(\mu, \tau^2)$ have distributions having densities of the form
\begin{align*}
f_{X_1, \dots, X_m, Y_1, \dots, Y_n}(x_1, \dots, x_m, y_1, \dots, y_n; \mu, \sigma^2, \tau^2) =\\
\frac{1}{(\sqrt{2\pi})^m (\sqrt{2\pi})^n}\exp\Bigg[\frac{\mu}{\sigma^2}\sum_{i = 1}^m x_i - \frac{1}{2\sigma^2}\sum_{i = 1}^m x_i^2 - m\frac{\mu^2}{2\sigma^2} - m\log \sigma\\
+ \frac{\mu}{\tau^2}\sum_{i = 1}^n y_i - \frac{1}{2\tau^2}\sum_{i = 1}^n y_i^2 - n\frac{\mu^2}{2\tau^2} - n\log \tau\Bigg].
\end{align*}
According to the theory of exponential families, this is a four-dimensional exponential family. We have parameter space $\Theta = \mathbb R \times (0, \infty) \times (0, \infty)$. Let $\theta = (\mu, \sigma^2, \tau^2) \in \Theta$. Define the function
\begin{align*}
\eta'' \colon \Theta &\to \mathbb R^4 \text{ by}\\
\theta &\mapsto \Big(\frac{\mu}{\sigma^2}, -\frac{1}{2\sigma^2}, \frac{\mu}{\tau^2}, -\frac{1}{2\tau^2}\Big).
\end{align*}
We can write
\begin{align*}
\eta''(\theta) &=
\begin{pmatrix}
\frac{1}{\sigma^2} & 0 & 0\\ \\
0 & -\frac{1}{2\sigma^4} & 0\\ \\
0 & 0 &\frac{\mu}{\tau^4} \\ \\
0 & 0 & -\frac{1}{2\tau^4}
\end{pmatrix}
\begin{pmatrix}
\mu\\ \\
\sigma^2\\ \\
\tau^2
\end{pmatrix}\\ \\
&=
\begin{pmatrix}
\frac{\mu}{\sigma^2}\\ \\
-\frac{1}{2\sigma^2}\\ \\
\frac{\mu}{\tau^2}\\ \\
-\frac{1}{2\tau^2}
\end{pmatrix}.
\end{align*}
If I can argue that three vectors (the column vectors in the matrix) can't have an image in $\mathbb R^4$ containing an open ball, then I am done, right? This would prove that the conditions for Lehmann-Scheffé can't be satisified.
(e) The exponential family isn't full rank, so Lehmann-Scheffé doesn't apply.
Another outline of a proof:
Can anybody develop the argument below into a full proof? I don't get it. Why $\bar X - \bar Y$?

 A: Joint density of $\boldsymbol X=(X_i)_{1\le i\le m}$ and $\boldsymbol Y=(Y_j)_{1\le j\le n}$ is
$$f_{\theta}(\boldsymbol x,\boldsymbol y)\propto \frac1{\sigma^m \tau^n}\exp\left[-\frac1{2\sigma^2}\sum_i x_i^2+\frac{\mu}{\sigma^2}\sum_i x_i-\frac1{2\tau^2}\sum_j y_j^2+\frac{\mu}{\tau^2}\sum_j y_j\right]$$
So if $\theta=(\mu,\sigma^2,\tau^2)$ is the unknown parameter, a minimal sufficient statistic for $\theta$ is
$$T=\left(\sum_i X_i,\sum_j Y_j,\sum_i X_i^2,\sum_j Y_j^2\right)$$
Taking $g(T)=\overline X-\overline Y$, you have $E_{\theta}[g(T)]=0$ for every $\theta$ but $\overline X \ne \overline Y$ almost surely, which shows $T$ is not complete. In fact, there is no complete sufficient statistic in this probability model. This is because in the presence of a minimal sufficient statistic, if a complete sufficient statistic exists, then the complete sufficient statistic has to be minimal sufficient (details here). Hence Lehmann-Scheffé theorem is not applicable to find UMVUE (if it exists) of any function of $\theta$.
To prove that there is no UMVUE of $\mu$, use the hint given.
Suppose $\sigma=a\tau$ where $a(>0)$ is known. Now the parameter of interest can be taken to be $\theta^*=(\mu,\tau)$.
The joint pdf of $(\boldsymbol X,\boldsymbol Y)$ is therefore
\begin{align}
f_{\theta^*}(\boldsymbol x,\boldsymbol y)&\propto  \frac1{\tau^{m+n}}\exp\left[-\frac1{2a^2\tau^2}\sum_i x_i^2+\frac{\mu}{a^2\tau^2}\sum_i x_i-\frac1{2\tau^2}\sum_j y_j^2+\frac{\mu}{\tau^2}\sum_j y_j\right]
\\&=\frac1{\tau^{m+n}}\exp\left[-\frac1{2\tau^2}\left(\frac1{a^2}\sum_i x_i^2+\sum_j y_j^2\right)+\frac{\mu}{\tau^2}\left(\frac1{a^2}\sum_i x_i+\sum_j y_j\right)\right]
\end{align}
In this model, a (minimal) complete sufficient statistic for $\theta^*$ does exist and is given by
$$T_a=\left(\frac1{a^2}\sum_i X_i^2+\sum_j Y_j^2,\frac1{a^2}\sum_i X_i+\sum_j Y_j\right)$$
An unbiased estimator $\hat\mu_a$ (say) of $\mu$ based on $T_a$ will be the UMVUE of $\mu$ in this second model. Now if you assume that there exists a UMVUE of $\mu$ in the original model, you can relate it to $\hat\mu_a$ and that would give you a contradiction. A key point here is that UMVUE is unique whenever it exists.
A: Recall the definition of a complete statistic:
$T$ is a complete statistic for $\theta$ is $\mathbb{E}_{\theta}g(T)=0$ for all
$\theta$ implies $P_{\theta}\left(g(T)=0\right)=1$ for every measurable function $g$.
In your case, take $g(T)=\bar{X}-\bar{Y}$, then if $\mu=\nu=\theta$ then $\mathbb{E}_{\theta}g(T)=\mathbb{E}_{\theta}\bar{X}-\mathbb{E}_{\theta}\bar{Y}=0$ for all $\theta$ but $P\left(\bar{X}=\bar{Y}\right)\neq 1$. Since $g(T)$ is a measurable function then this means $T=\left(\bar{X},\bar{Y},\sigma^{2}_{X},\sigma^{2}_{Y}\right)$ is not a complete statistic for $\theta$. As a consequence, you cannot apply Lehmann-Scheffe.
By Lehmann-Scheffe I mean that if you have a complete and sufficient statistic $T$ and there exist a function $\varphi$ such that $\mathbb{E}_{\theta}\varphi(T)=\theta$ then $\varphi(T)$ is UMVUE for $\theta$.
As an intuition of what is going on, the fact that you don't know the relationship between the variances of $X$ and $Y$ difficults finding 'the best' estimator in terms of being UMVUE. If you knew them, then you could solve a minimization problem and find the optimal weighting of $\bar{X}$ and $\bar{Y}$, which would not depend on $\theta$.
