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Let $X_1,...X_n$ be $\text{Poi}(\lambda)$ distributed random variables. I want to construct a minimal variance unbiased estimator (MVUE) for $\lambda$.

By the Neyman Lemma, I know that $T:=\sum_{i=1}^nX_i$ is a minimal sufficient statistic for $\lambda$.

Now I want to use the Lehmann-Scheffe Theorem, which states that if we have a sufficient and complete statistic $T$ for a parameter $\lambda$ and find an unbiased function $\phi(T)$, then $\phi(T)$ is a MVUE for $\lambda$

I want to show that $T$ is complete. Given a measurable function $g$ I want to show that $\mathbb{E}_\lambda[g(T)]=0\Rightarrow g(T)=0$ almost surely.

Since $T\sim \text{Poi}(n\lambda)$ we have that $\mathbb{E}_\lambda[g(T)]=\sum_{k=0}^\infty g(k) e^{-\lambda n}\frac{(n\lambda)^k}{k!}=e^{-n\lambda}\sum_{k=0}^\infty \frac{g(k)}{k!}z^k$ where $z:=n\lambda$. Now by the theory of power series we must have that $\frac{g(k)}{k!}=0\Rightarrow g(k)=0$ which is what we wanted to show.

If I can construct an unbiased estimator which is a function of $T$, I am done. So lets check $\mathbb{E}[T]=n\lambda$. I correct it by $S:=T\frac{1}{n}$ and it follows by Lehmann-Scheffe Lemma that $S(T)$ is a MVUE of $\lambda$.

Is this correct?

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    $\begingroup$ In the exponential family, I would think something like this stats.stackexchange.com/questions/322381/… would be easier when showing completeness $\endgroup$ – pedernv Nov 20 '19 at 10:17
  • $\begingroup$ @Xi'an If (g(k)=0) I know that (g(T)=0) almost surely. $\endgroup$ – EpsilonDelta Nov 20 '19 at 14:07
  • $\begingroup$ This is not what is required by completeness: if $\mathbb E[g(T)]=0$ then prove that $g\equiv 0$. $\endgroup$ – Xi'an Nov 20 '19 at 15:09
  • $\begingroup$ @Xi'an See en.wikipedia.org/wiki/Completeness_(statistics) $\endgroup$ – EpsilonDelta Nov 20 '19 at 15:25
  • $\begingroup$ @Xi'an Ok, but how is $g\equiv 0$ the same as $P_\lambda[g(T)=0]=1,\ \forall \lambda\in[0,\infty)$? Your condition is stronger, isn't it? My reasoning in the proof is, that I can conclude from $g(k)=0,\ \forall k\in\mathbb{N}$ that $g=0, \text{ a.s.}$. I did not mean to be rude, please excuse that I just sent the link! $\endgroup$ – EpsilonDelta Nov 20 '19 at 20:11

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