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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 $k$ .

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    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.2012-01-04

2 Answers 2

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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 $m$.

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    Hi @Henry can you maybe help me with this question: https://math.stackexchange.com/questions/2598781/poisson-probability-per-week2018-01-09
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$\sum_{k=0}^{\infty}h(k)\frac{(n\lambda)^{k}}{k!}=0$ is a polynomial function of $\lambda$ and only have specific roots. Therefore the equation cannot hold for all $k$ unless $h(k)=0$.

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    It's not a polynomial function though, it's an analytic function.2017-09-19