Find UMVUE of $\tau=(\lambda-\mu)e^{-(\lambda+\mu)}$ from samples $X_1, \dots, X_n\sim \rm{Pois}(\lambda)$ and $Y_1,\dots, Y_n\sim \rm{Pois}(\mu)$. Problem Statement
We have two independent Poisson samples $X_1, \ldots, X_n$ with means $\lambda$ and $Y_1, \ldots, Y_n$ with means $\mu$. We would like to estimate $\tau=(\lambda-\mu)e^{-(\lambda+\mu)}$.
(a) Find a function of $X_1$ and $Y_1$ that is an ubiased estimator of $\tau$.
(b) Find the UMVUE of $\tau$.
(c) Calculate the asymptotic variance of this estimator when $\lambda=\mu$.
Context
I am studying some old exams, and I came across this problem. I think I have the correct unbiased estimator, but my UMVUE is definitely wrong since it is a function of unknown parameters. I am not sure if I made a mistake with my complete sufficient statistic, or I did the conditional expectation wrong.
Attempted Solution
(a) Find some function $g(X_1,Y_1)$ such that $\mathbb{E}(g)=\tau$.
$$
\tau = \lambda e^{-(\lambda+\mu)}-\mu e^{-(\lambda+\mu)}=\mathbb{P}(X_1=1, Y_1=0)-\mathbb{P}(X_1=0, Y_1=1)=\mathbb{E}\big[I_{X_1}(1)I_{Y_1}(0)-I_{X_1}(0)I_{Y_1}(1)\big],
$$
therefore, $U=I_{X_1}(1)I_{Y_1}(0)-I_{X_1}(0)I_{Y_1}(1)$ is unbiased for $\tau$.
(b) The Lehman-Scheffe theorem tells us that, given a complete sufficient statistic $T$ and an unbiased estimate $U$, the UMVUE is $E(U|T)$. Let $S_j=\sum_{i=j}^nX_i+Y_i$. From previous results we know that $S_j\sim Pois((n+1-j)(\lambda+\mu))$, where Poisson is in the exponential family of distributions and $S_1$ is a complete sufficient statistic. Therefore,
\begin{equation*}
    \begin{split}
        E(U|S_1=t)=& P(X_1=1,Y_1=0|S_1=t)-P(X_1=0, Y_1=1|S_1=t)\\
        =& \frac{P(X_1=1,Y_1=0, S_2=t-1)-P(X_1=0,Y_1=1, S_2=t-1)}{P(S_1=t)}\\
        \overset{ind.}{=}& \frac{P(X_1=1)P(Y_1=0)P(S_2=t-1)-P(X_1=0)P(Y_1=1)P(S_2=t-1)}{P(S_1=t)}\\
        =&\bigg(\frac{n-1}{n}\bigg)^t\frac{t}{(n-1)(\lambda+\mu)}(\lambda-\mu),
    \end{split}
\end{equation*}
is the UMVUE for $\tau$.
 A: Joint pmf of the sample $(\boldsymbol X,\boldsymbol Y)$ for $x_i,y_i\in \{0,1,\ldots\}$ is
$$f(\boldsymbol x,\boldsymbol y\mid \lambda,\mu)=\frac{e^{-n(\lambda+\mu)}}{\prod_{i=1}^n x_i!y_i!}\cdot \lambda^{\sum\limits_{i=1}^n x_i}\mu^{\sum\limits_{i=1}^n y_i} \quad,\,\lambda,\mu>0$$
So a complete sufficient statistic is the pair $\left(\sum\limits_{i=1}^n X_i,\sum\limits_{i=1}^n Y_i\right)=(T_1,T_2)$, say.
We know that $T_1\sim \text{Poisson}(n\lambda)$, which is independent of $T_2\sim \text{Poisson}(n\mu)$.
So if $E\left[I(X_1=1,Y_1=0)\mid T_1=t_1,T_2=t_2\right]=g(t_1,t_2)$, then for $t_1,t_2\in \{0,1,2,\ldots\}$,
\begin{align}
g(t_1,t_2)&=\frac{P(X_1=1)P\left(\sum\limits_{i=2}^n X_i=t_1-1\right)P(Y_1=0)P\left(\sum\limits_{i=2}^n Y_i=t_2\right)}{P(T_1=t_1)P(T_2=t_2)}
\\&=\frac{(n-1)^{t_1-1}t_1!}{n^{t_1}(t_1-1)!}\cdot\left(\frac{n-1}{n}\right)^{t_2}
\\&=\frac{t_1}{n-1}\left(1-\frac1n\right)^{t_1+t_2}
\end{align}
Similarly, $$E\left[I(X_1=0,Y_1=1)\mid T_1=t_1,T_2=t_2\right]=g(t_2,t_1)$$
Hence the UMVUE of $\tau$ (for $n>1$) is $$g(T_1,T_2)-g(T_2,T_1)=\frac{T_1-T_2}{n-1}\left(1-\frac1n\right)^{T_1+T_2}$$
