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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.

user1916067
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    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 $h$. – StubbornAtom Sep 18 '20 at 05:22
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    Use Rao-Blackwell: start with an arbitrary unbiased estimator of $P_\lambda(X=r)$ and condition by a minimal sufficient complete statistic. – Xi'an Sep 18 '20 at 07:20
  • 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). – user1916067 Sep 19 '20 at 02:38

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