# UMVUE of the probability a Poisson R.V is odd?

Problem: Let $$X_i \sim Pois(\lambda)$$. Find the UMVUE of the probability that $$X_1$$ is odd.

My attempt:

I don't think there's any obvious unbiased estimator to use conditioning. So instead I write

$$P(X_1 \text{ is odd}) = \sum_{k=0}^\infty \frac{\lambda^{2k+1}}{(2k+1)!}e^{-\lambda}$$

The complete sufficient statistic is $$T(X)= \sum_{i=1}^n X_i \sim Pois(n\lambda)$$, so an unbiased estimator must satisfy

$$\sum_{k=0}^\infty \delta(k)\frac{(n\lambda)^k}{k!}e^{-n\lambda} =\sum_{k=0}^\infty \frac{\lambda^{2k+1}}{(2k+1)!}e^{-\lambda}$$

But here I am stuck. I know I need to write out the power series of each exponential, but then I get left with the product of two sums which does not seem very helpful.

$$\sum_{k=0}^\infty \delta(k)\frac{(n\lambda)^k}{k!}\sum_{j=0}^\infty \frac{\lambda^j}{j!}=\sum_{j=0}^\infty \frac{(n\lambda)^j}{j!}\sum_{k=0}^\infty \frac{\lambda^{2k+1}}{(2k+1)!}$$

Using Fubinis I thiink I can write this as

$$\sum_{k=0}^\infty \sum_{j=0}^\infty \delta(k)\frac{(n\lambda)^k}{k!} \frac{\lambda^j}{j!}=\sum_{k=0}^\infty\sum_{j=0}^\infty \frac{(n\lambda)^k}{k!} \frac{\lambda^{2j+1}}{(2j+1)!}$$

This seems to suggest $$\delta(t)$$ is of the form $$\delta(t) = \frac{t!}{(2t+1)!}$$ when $$t$$ is odd, and 0 otherwise?

I am not confident with my attempt at all.

We have the exact expression, verifiable using the power series expansion of $$e^{\lambda}$$: $$\sum_{k=0}^\infty \frac{\lambda^{2k+1}}{(2k+1)!}=\frac{1}{2}(e^{\lambda}-e^{-\lambda})$$
So that reduces the probability to $$P(X_1=\text{odd})=\frac{1}{2}(1-e^{-2\lambda})=g(\lambda)\,,\text{ say }$$
Since $$T\sim \mathsf{Poi}(n\lambda)$$ we have $$E(a^T)=e^{n\lambda(a-1)}$$. This equals $$e^{-2\lambda}$$ for $$a=1-\frac{2}{n}$$.
So UMVUE of $$e^{-2\lambda}$$ based on a sample of $$n$$ observations is $$h(T)=\left(1-\frac{2}{n}\right)^T$$
This means UMVUE of $$g(\lambda)$$ is $$\frac{1}{2}\left[1-h(T)\right]=\frac{1}{2}\left[1-\left(1-\frac{2}{n}\right)^T\right]$$
• $T=\sum X_i$ to clarify. – StubbornAtom May 24 at 9:53