I just wanted to make sure what I am doing is correct.

$X_1, ..., X_n$ is a random sample from a Poisson($\theta$) distribution

$T = \sum X_i$ is a minimal sufficient statistic since

$$\prod^n_{i=1} \frac{e^{-\theta}\theta^{x_i}}{x_i !} = e^{-n\theta}\theta^{\sum x_i}\frac{1}{x_1!x_2!***x_n!}$$

Can be factorized into two functions, one which does not depend on $\theta$.

Now, to use the Lehmann-Scheffe theorem, I need to check for completeness of T.

$E[h(t)] = 0$ $\iff$ $\sum h(t) \frac{e^{-n\theta}(n\theta)^t}{t!}$ = 0

Can only be possible if h(t) = 0 for t = 0, 1, 2, ... $\rightarrow$ T is complete.

Using L-S, $$\sum_{t=0}^\infty h(t)(n\theta)^t\frac{e^{-n\theta}}{t!} = g(\theta) = \theta$$

$$\iff \sum_{t=0}^\infty h(t)(n\theta)^t\frac{1}{t!} = e^{n\theta}\theta$$

$$\iff \sum_{t=0}^\infty h(t)(n\theta)^t\frac{1}{t!} = \sum_{t=0}^\infty (n\theta)^t \frac{1}{t!}\theta$$

Therefore h(t) = $\theta$ is the UMVUE? Not quite sure at this point.

Thanks for any help

  • 1
    $\begingroup$ I do not understand why $0$ becomes $g(\theta)=\theta$. Stick to $\mathbb{E}_\theta[h(T)]=0$ for all $\theta$'s and apply first to $\theta=0$, this implies $h(0)=0$. Then proceed by recursion. $\endgroup$
    – Xi'an
    Commented Dec 7, 2014 at 21:21
  • $\begingroup$ That is just the formula for the Lehmann-Scheff Theorem where $g(\theta)$ = $E_\theta(h(T))$. $\endgroup$
    – tim.mof
    Commented Dec 7, 2014 at 22:05
  • 1
    $\begingroup$ I guess since T ~ Poisson($n\theta$) then E[T] = $n\theta$ so T/n is the UMVUE of $\theta$ $\endgroup$
    – tim.mof
    Commented Dec 7, 2014 at 22:37

1 Answer 1


I think you complicated the computation when you went through the expected value. You already got a complete sufficient statistic (you could have avoid the whole check of minimal sufficiency and completeness simply using the fact that the Poisson distribution belongs to the exponential family of parameter one which is also full rank, so $\sum_{i=1}^n X_i $ is such a statistic).

Then if you consider properly Lehmann-Scheffè theorem, it allows you to choose between two different but equivalent methods to find an UMVUE:

-Using Rao-Blackwell; ingredients: an unbiased estimator $U$ (even trivial) for your parameter $g(\theta)$; a sufficient and complete statistic $T$ for $\theta$. Then the UMVUE is $W=\mathrm E[U|T]$;

-Direct application of Lehman-Scheffè: ingredients: a statistic $W$ that is both unbiased for $g(\theta)$ and a function of $T$, a sufficient and complete statistic for $\theta$.

I'd suggest to use this second method to find an UMVUE for $\theta$, indeed you already have $T$, so the first thing to do is check if it is unbiased for $\theta$ or not. In this case, using the well known distribution of the sum of Poisson random variables, you get that $\mathrm E[T]=n\theta$. So you can correct your estimator, exploiting the fact that the expected value is a linear functional. Hence $W=T/n$ is your UMVUE, indeed if we check its properties: it is a function of $T$; it is unbiased for $\theta$, indeed $\mathrm E[W]=\mathrm E[T/n]=\mathrm E[T]/n=\theta$.

  • 1
    $\begingroup$ If T is the sum of n i.i.d. Poisson r.v then shouldn't $E[T] = n \theta$ and not $\frac{\theta}{n}$ ? $\endgroup$ Commented Sep 19, 2021 at 20:38
  • $\begingroup$ That's indeed correct, thanks! $\endgroup$
    – wonderer
    Commented Feb 8, 2023 at 7:28

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