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


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 $k$ .

  • 1
    $\begingroup$ 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 :) $\endgroup$ – nb1 Jan 4 '12 at 12:17
  • $\begingroup$ 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. $\endgroup$ – Robert T. Jan 4 '12 at 12:30
  • 1
    $\begingroup$ 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. $\endgroup$ – Robert T. Jan 4 '12 at 12:58
  • 1
    $\begingroup$ 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 $\lambda>0$? $\endgroup$ – Brian M. Scott Jan 4 '12 at 15:05
  • 1
    $\begingroup$ 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. $\endgroup$ – Robert T. Jan 4 '12 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 $m$.

  • $\begingroup$ @ Henry: Thanks for your answer! $\endgroup$ – Robert T. Jan 4 '12 at 16:27
  • 1
    $\begingroup$ 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? $\endgroup$ – Robert T. Jan 4 '12 at 16:46
  • $\begingroup$ I think my point is that I require $s(\lambda)$ to be "flat" in every sense at zero. $\endgroup$ – Henry Jan 4 '12 at 23:34
  • $\begingroup$ Hi @Henry can you maybe help me with this question: math.stackexchange.com/questions/2598781/… $\endgroup$ – John Smith Jan 9 '18 at 23:35

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

  • 1
    $\begingroup$ It's not a polynomial function though, it's an analytic function. $\endgroup$ – Chill2Macht Sep 19 '17 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.