Find UMVUE of $\theta^k= P(X_1=1,...,X_k=1)$, where $k = 1,...,n$, given a random sample from Bernoulli$(1,\theta)$

Question

I need to find the UMVUE of $\theta^k= P(X_1=1,...,X_k=1)$, where $k = 1,...,n$, given a random sample from $Bernoulli(1,\theta)$. I suppose that an unbiased estimator of $\theta^k$ could be $T_1$, taking value $1$, if $X_1=...=X_k=1$, and $0$ otherwise. Also $T=\sum X_i$ is a sufficient statistic. Also, based on Lehmann-Scheffe theorem and Rao-Blackwell, if i find $E(T_1|T)$, this should be the UMVUE of $\theta^k$. However, i have problem calculating $E(T_1|T)$. Can anyone proceed? I have already done the following: $E(T_1|T)=0*(T_1=0|T=t) +1*P(T_1=1|T=t) $ = $P(T_1=1|T=t) = P(X_1=...=X_k=1|\sum X_i=t)$. From there, Bayes theorem can be applied, and this is where i am stuck on how to proceed.

Answer 1

If $k=n$ and the sample consists just of $X_1,\dots,X_k$, then $T_1={\bf 1}_{\{T=k\}}$ is a function of $T$, so $E(T_1 | T)=T_1$, and $T_1$ itself is the desired UMVUE for $\theta^k$.

Next, consider the more interesting case where $n>k$ and and $T=\sum_{i=1}^n X_i$. For every set $S \subset \{1,\dots, n\}$ with $|S|=k$, define $T_S:=\prod_{i \in S} X_i$. By symmetry, $E(T_S|T)=E(T_1|T)=\theta^k$ for every such set $S$. Thus $$T_*:={n \choose k}^{-1}\sum_{|S|=k} T_S $$ satisfies $E(T_*|T)=E(T_1|T)$ as well. Finally, observe that ${n \choose k} T_*$ counts the number of subsets of size $k$ of the set $\{i \le n: X_i=1\},$ so $$T_*:={n \choose k}^{-1} {T \choose k} $$ is a function of $T$. Therefore, $$E(T_1|T)=E(T_*|T)=T_* \,,$$ and $T_*$ is the desired UMVUE for $\theta^k$.