0

Assume that $h(S)$ is a realvalued function of a random variable with distribution $Erlang(n,\theta)$ (h(S) not depending on $\theta$). Then conclude from $\mathbb{E}(h(S))=0$ almost surely that $h(S)=0$ a.s..

Clearly it would suffice to show the conclusion from $\int h(S) s^{n-1}e^{-\theta s}ds = 0$. However I don't see how to prove this.

My idea was to use two different arbitrary values of the parameter $\theta$ to get a contradiction for $h(S)\not = 0$ since $h(S)$ cannot depend on $\theta$. However I wasn't successful so far with this approach. Does anybody have hint?

GottlobtFrege
  • 478
  • 2
  • 11
  • The correct version of your title is: a nonnegative function is almost surely 0 if its expectation is 0. So you need to check that in your case the function is nonnegative. – GEdgar Jan 22 '21 at 11:30
  • no, the function h(s) doesn't have to be nonnegative (however, if it was the conclusion would follow easily) – GottlobtFrege Jan 22 '21 at 11:33
  • Pretty much the same: https://math.stackexchange.com/q/2686724/321264. – StubbornAtom Feb 27 '21 at 06:52

1 Answers1

2

This is essentially the completeness property of the Erlang/Gamma family of distributions under consideration, which follows from the fact that this distribution is a member of a regular (full rank) exponential family with parameter $\theta$.

Or you can directly say that $\int_0^\infty h(s) s^{n-1}e^{-\theta s}\,ds$ is the Mellin transform of $h(s)e^{-\theta s}$, so that by uniqueness of integral transforms

$$\int_0^\infty h(s) s^{n-1}e^{-\theta s}\,ds =0 \quad,\forall\,\theta>0\implies h(s)e^{-\theta s}=0 \,,\,\text{a.e.}\quad\forall\,\theta$$

And hence $$h(s)=0\,,\,\text{a.e.}$$

StubbornAtom
  • 17,052