UMVUE of Poisson distribution truncated at zero

Question

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

Answer 1

I am surmising that by "Poisson distribution truncated on the left at $0,$" you mean the conditional distribution given that it's not $0.$ Thus one has $$ \Pr(X_1 = x) = \frac{\lambda^x e^{-\lambda}/(x!)}{\Pr(X_1\ge 1)} = \frac{\lambda^x e^{-\lambda}/(x!)}{1 - e^{-\lambda}} = \frac{\lambda^x}{x!(e^\lambda - 1)}. $$ Thus the likelihood is $$ L(\lambda) = \frac{\lambda^{x_1+\cdots+x_n}}{(e^\lambda -1)^n} \times \text{constant} $$ (where "constant" means not depending on $\lambda$), and then we have $$ \ell(\lambda) = \log L(\lambda) = (x_1+\cdots+x_n)\log\lambda - n\log(e^\lambda - 1) + \text{constant} $$ $$ \ell\,'(\lambda) = \frac {x_1+\cdots+x_n} \lambda - \frac{ne^\lambda}{e^\lambda - 1}. $$ I'm not sure whether finding a zero of that function can be done by using Lambert's W, but you can use numerical methods when you know $(x_1+\cdots+x_n)/n.$

I'm going to ask for some clarification of the other parts of your question before posting more here.

Answer 2

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 $\Pr(X = x) = \frac{\lambda^x}{x!(e^\lambda - 1)}$