Let we have random variables $\left(X_1, X_2, .., X_n \right)$ with pmf $p(X)=\theta(1-\theta)^x; x=0,1,2,\ldots\quad \mathrm{and}\quad 0<\theta<1$
I want to estimate $\mathbb{E}\Big[ I[X_1=0] \Big| \sum_i X_i = t \Big]$
In my textbook direct solution given as $\frac{n-1}{t+n-1} \qquad (n\geq2)$.
Can you please suggest some approach how to estimate this. I have general idea on how to estimate expectation, but could not get how can I proceed with this expectation.
$\sum_iX_i$ is a sufficient statistic for $\theta$, that is, conditioned on this sum the distribution of the data doesn’t depend on $\theta$. In the present case, all $n$-tuples that sum to $t$ are equiprobable. By stars and bars, there are $\binom{t+n-1}t$ such tuples. If $X_1=0$, the remaining $n-1$ values have to sum to $t$, and there are $\binom{t+n-2}t$ such tuples. Since all tuples that sum to $t$ are equiprobable, the desired probability is just the ratio
$$ \frac{\binom{t+n-2}t}{\binom{t+n-1}t}=\frac{n-1}{t+n-1}\;. $$