# Moments of function of Poisson process

(I'm new to Poisson processes, so please edit if my terminology is incorrect.)

This is a special case of a problem I'm working on; hoping for intuition that will generalize to a multi-dimensional Poisson process. The context is approximating a distribution defined by the $\arg\min$ of a (non-Poisson) process in Theorem 1 of this paper.

Setup: consider a Poisson process on $[0,\infty)$, with points $\{ \Gamma_k\}_{k=1}^\infty$, $\Gamma_k=E_1+\cdots+E_k$, $E_i\stackrel{iid}{\sim}\rm{Exponential}(1)$.

Consider $u=\inf\{t:1\le\sum_{k=1}^\infty \mathbb 1(\Gamma_k\le t)\}$, where $1(\cdot)$ is the indicator function (one if true, zero if false). In this special case, $u=\Gamma_1=E_1$, and thus $u\sim\rm{Exp}(1)$. We have a closed form solution for $u$, and the first two (central) moments are $\mathbb E(u)=1$ and $\rm{Var}(u)=1$.

QUESTION: can we derive $\mathbb E(u)=1$ and $\rm{Var}(u)=1$ without using the closed form of $u$? (In my general problem, there is no closed form.) So we know $u=\inf\{t:1\le\sum_{k=1}^\infty \mathbb 1(\Gamma_k\le t)\}$, but we can't just use $u=\Gamma_1$ to calculate moments.

Notes/thoughts:

As a reminder, the mean measure of the considered Poisson process is $m(A)=\lambda(A)$, where $\lambda$ is Lebesgue measure (e.g., $\lambda([a,b])=b-a$; just the total length of set $A\subset [0,\infty)$). This means that the expected number of points in any interval is equal to the length of the interval.

Also, the probability of one event occurring in $[t,t+dt]$ is $dt$ since the rate is one here.

Also potentially helpful: define the random counting measure $\hat N(t)=\sum_{k=1}^\infty 1(\Gamma_k\le t)$. Then, $\mathbb E(\hat N(t))=\lambda([0,t])=t$, and $u=\inf\{t:1\le \hat N(t)\}$.

For the first moment, I noticed that in this case, $\inf\{t:1\le \mathbb E(\hat N(t))\}=\inf\{t:1\le t\}=1=\mathbb E(u)$. That is, solving for $u$ after plugging in the mean measure happens to yield the mean of $u$. This struck me as not true generally: $\mathbb E(u(X))\ne u(\mathbb E(X))$. But maybe some property of the Poisson process and/or $u$'s characterization means that $u$ is a linear'' function of the process, so this is generally true here?

For the second moment, I haven't made any progress. Tried thinking about the expectation $\mathbb E\left[(u-\mathbb E(u))^2\right]=\mathbb E[(u-1)^2]$, but don't know how I could get that without cheating. Also wondered if there was a variance of the process,'' if the plug-in approach for the first moment turns out to be valid. But basically stuck.

Last, recall either $\rm{Var}(u)$ or $\mathbb E(u^2)$ is sufficient, since $\rm{Var}(u)=\mathbb E(u^2)-(\mathbb E(u))^2$.

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While MO is not SE I'll let it stay, but please do not cross-post unless you lose hope in getting an answer on the original site. –  mbq Sep 3 '11 at 11:30
Ok, sorry; I haven't yet figured out which site is better for questions like this--would have guessed MO, but got a great response here (SE) on my first question. But feel free to close this if that's more appropriate; I can ask to reopen once I "lose hope" –  David M Kaplan Sep 7 '11 at 15:27