Poisson complete statistic

Question

I have the same question as this thread, but I cannot understand the proof.

The problem is, given $f(\lambda)=\sum_{k=0}^\infty g(k)\frac{(n\lambda)^k}{k!}=0,\forall\lambda>0$. How to show $g(k)\equiv0$?

The accepted answer in that thread claims that $f(0)=g(0)$. Why? I believe $f(0)=0$. But I cannot see why $g(0)=0=f(0)$.

One comment claims "an infinite summation is zero iff each term in it is identically zero". But there is no proof.

So, can anyone prove $g(k)\equiv0$ or the claim "an infinite summation is zero iff each term in it is identically zero"?

Thanks!

Answer 1

You have $$ h(\mu) = \sum_{k=0}^\infty g(k)\frac{\mu^k}{k!} = 0 \text{ for all values of }\mu\ge0. $$ This is $$ h(\mu) = g(0) + g(1)\mu + g(2)\frac{\mu^2}2+ g(3)\frac{\mu^3}6 + g(4)\frac{\mu^4}{24} + g(5)\frac{\mu^5}{120}+\cdots. $$ If $\mu=0$ then all terms of this series except the first one are $0$, so you're left with $g(0)$. Thus $h(0)=g(0)$.

Then you have $$ h'(\mu) = g(1) + g(2)\mu + g(3)\frac{\mu^2}2 + g(4)\frac{\mu^3}6 + g(5)\frac{\mu^4}{24} + \cdots. $$ If $\mu=0$ the all terms of this series then all terms of this series except the first one are $0$, so you're left with $g(1)$. Thus $h'(0)=g(1)$.

Then you have $$ h''(\mu) = g(2) + g(3)\mu + g(4)\frac{\mu^2}2 + g(5)\frac{\mu^3}6 + \cdots. $$ If $\mu=0$ the all terms of this series then all terms of this series except the first one are $0$, so you're left with $g(2)$. Thus $h''(0)=g(1)$.

Continuing in this way, you see that $h^{(n)}(0)=g(n)$ for $n=0,1,2,3\ldots$.

If $h(\mu)=0$ for all values of $\mu\ge0$ then $h^{(n)}(0)=0$ for all $n$, so all values of $g$ are $0$.

Answer 2

1) $f(0)=g(0)$ from the description of $f(\lambda)=\sum_{k=0}^{\infty}g(k)(n\lambda)^k/k!$. When you plug in $\lambda=0$, you get the term $g(0)$, and so $f(0)=g(0).$

2) To prove the claim that an infinite power series is $0$ if and only if every term is $0$, differentiate, and put $\lambda=0$ repeatedly.

3) If you are worried about the validity of "plugging in $\lambda=0$" or differentiating the power series term by term with respect to $\lambda$, rest assured- they are valid mathematical operations, due to uniform convergence of the power series.