# Complete statistic: Poisson Distribution

Context

I am having difficulty trying to understand a step of a proof which relies on a property of series.

Proof

Suppose that $X_1, X_2, \ldots , X_n$ is a random sample of size $n$ from a Poisson distribution with parameter $\lambda > 0$. The goal is to show that $T = \sum_{i=1}^n X_i$ is a complete statistic.

Since we know that $T = \sum_{i=1}^n X_i \sim \mathrm{Poisson}(n\lambda)$:

$$\mathbb{E}(h(T)) = \sum_{k=0}^{\infty} h(k) \, e^{-n\lambda} \, \frac{(n\lambda)^k}{k!} = 0\Longrightarrow \sum_{k=0}^{\infty} h(k) \, \frac{(n\lambda)^k}{k!} = 0$$

The textbook I am using and some others sources I've found argue that:

$$\boxed{\displaystyle\sum_{k=0}^{\infty} h(k) \, \frac{(n\lambda)^k}{k!} = 0 \Longrightarrow h(k) \, \frac{(n\lambda)^k}{k!} = 0 \qquad \forall k}$$

It probably is an obvious result from calculus, but I am unable to prove it.

If $h(k) \, (n\lambda)^k/k! = 0$ for all $k$ then $T$ is a complete statistic because $\lambda$ is nonnegative and then $h(k) = 0$ for all $k$ .

• The result is evident. Given that >> $\sum_{k=0}^\infty h(t) \frac{(n\lambda)^k}{k!}=0$. Since $\frac{(n\lambda)^k}{k!}\not= 0$ for $(n,k) \epsilon \mathbb{N}$ and $\lambda >0$, we must have $h(t)=0$. You have also written all this youself except for the typo where you say that $\frac{(n\lambda)^k}{k!}=0$. That is false. In point of fact, you have replied to your question yourself within the question itself :)
– nb1
Commented Jan 4, 2012 at 12:17
• Sorry, there were even more typos. I wrote $h(t)$ but it should read $h(k)$ instead. Of course, $h(t)$ would not depend on $k$ and the result would be evident. Commented Jan 4, 2012 at 12:30
• Thank you again. I am failing to express myself correctly. The result that I want to prove is "an infinite summation is zero iff each term in it is identically zero", but I have writen the implication for a particular series I am dealing with. Commented Jan 4, 2012 at 12:58
• Is the hypothesis that $$\sum_{k=0}^{\infty} h(k) \, \frac{(n\lambda)^k}{k!} = 0$$ for a particular $\lambda>0$, or is it that this is true for all $\lambda>0$? Commented Jan 4, 2012 at 15:05
• I think the hypothesis is that $$\sum_{k=0}^\infty h(k)\frac{(n\lambda)^k}{k!} = 0$$ for all $\lambda > 0$, since we are trying to prove that $\sum_{i=1}^n X_i$ is a complete statistic for the Poisson "family" of distributions. Commented Jan 4, 2012 at 15:16

If $s(\lambda) = \sum_{k=0}^\infty h(k)\frac{(n\lambda)^k}{k!}$ and $s(\lambda) =0$ for all $\lambda$, then clearly $h(0)=0$ since $s(0)=h(0)$.

Similarly if you find the $m$th derivative of $s(\lambda)$ at $\lambda=0$, which must also be $0$, you will have $h(m)=0$ for all $m$.

• Just a thing: the hypothesis is that $s(\lambda) = 0$ for all $\lambda >0$ and we are setting $\lambda = 0$ to conclude that $h(k) = 0$ for all $k$, is this proof still valid? Commented Jan 4, 2012 at 16:46
• I think my point is that I require $s(\lambda)$ to be "flat" in every sense at zero. Commented Jan 4, 2012 at 23:34
• Hi @Henry can you maybe help me with this question: math.stackexchange.com/questions/2598781/… Commented Jan 9, 2018 at 23:35
• you cannot put $s(0)$ as 0 lies outside the domain of a Poisson distribution. I do not understand your further comment about "flat"-ness. Commented Oct 15, 2020 at 13:44
• @MaverickMeerkat I would have thought you can regard $s(0)$ as representing $s(\lambda)$ as $\lambda \to 0^+$, or you can say that "a Poisson distributed random variable with mean $0$" can be taken to mean a random variable which is $0$ with probability $e^{-0}\frac{0^0}{0!}=1$. Commented Oct 15, 2020 at 15:09

$\sum_{k=0}^{\infty}h(k)\frac{(n\lambda)^{k}}{k!}=0$ is a polynomial function of $\lambda$ and only have specific roots. Therefore the equation cannot hold for all $k$ unless $h(k)=0$.

• It's not a polynomial function though, it's an analytic function. Commented Sep 19, 2017 at 22:11