Suppose that $X_1,\ldots,X_n$ are $n$ i.i.d. random variables from the Poisson distribution truncated on the left at $0.$
Find the UMVUE of $P(X_1 =1).$
I am trying to do it by computing the expectation of a function $h(T)$ of my statistic, which is just the sum of $X_i$, and this expectation is equal to $P(X_1 =1),$ so I want to find $h(T),$ which will be the UMVUE, since $T$ is minimal complete sufficient statistic.
The problem is that I obtain a big double sum in this expectation and I don't know how to deal with it.
I have also tried to compute the MLE, but I don't know how to obtain it in this case.
Anyone could help me please?
thanks in advance
Best Answer
To add to the final line in Michael Hardy's answer, we have that
$$ E(T \mid X_1 + \cdots + X_n) = \Pr(X_1 = 1) \cdot \frac{\Pr(X_2 + \cdots + X_n = x-1)}{\Pr(X_1 + \cdots + X_n = x)}$$
Now given $X_1 + X_2 \cdots + X_n = x$ in the denominator, we can say that $X_2 + \cdots + X_n = x-X_1$.
So in the numerator, we obtain $X_2 + \cdots + X_n = x - X_1 = x -1$ , giving $X_1 = 1$.
Hence $$ E(T \mid X_1 + \cdots + X_n) = \Pr(X_1 = 1) \cdot \Pr(X_1 = 1)$$
and this is easily obtainable by substituting in $x = 1$ into $\Pr(X = x) = \frac{\lambda^x}{x!(e^\lambda - 1)}$