# UMVUE of the probability of a conditional poisson probability $P_{\lambda} (X=r )$ [duplicate]

Consider a $$X_1, ... X_n \sim~ Poisson(\lambda)$$, I want to obtain the UMVUE of $$P_{\lambda} (X=r)$$.

This is my approach: $$\operatorname{\mathbb{E}}_{\theta}[h(t)] = P_{\lambda} (X=r)$$. The probability is dependent on $$\lambda$$ so I am inclined to find the conditional probability and proceed as $$\operatorname{\mathbb{E}}_{\theta}[h(t)] = P(X=r, \lambda)/ P(\lambda)$$. I am not sure if this is necessary but would appreciate some advice.

Incorporating provided suggestions I get: $$\sum_{t=0}^{\infty} h(t) \lambda^t\exp(-t)/ t! = \lambda^r\exp(-r)/ r!$$ I will expand the RHS: $$\lambda^r\exp(-r)/ r! = \lambda^r\sum_{n=0}^{\infty} (-r)^{n}/ n!r!$$

But I am stuck.

• The probability is supposed to depend on $\lambda$; that is not a problem. You need to write down the equation $E[h(T)]=P(X=r)$ for some appropriate statistic $T$ and solve for $h$. Commented Sep 18, 2020 at 5:22
• Use Rao-Blackwell: start with an arbitrary unbiased estimator of $P_\lambda(X=r)$ and condition by a minimal sufficient complete statistic. Commented Sep 18, 2020 at 7:20
• This is my approach: let h(t) = 1(X= R) . Then E[h(t)] = P(X=R). h(x) is therefore the UMVUE for P(X=R). Commented Sep 19, 2020 at 2:38