# Moments of function of Poisson process

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

Edit: per comments, here is a (more) general version of the originally posted problem (which is now at the bottom, below the line); I hope it's more clear/helpful.

The context is approximating a distribution defined by the $\arg\min$ of a (non-Poisson) process in Theorem 1 of this paper.

Setup: I want to know the first two moments of $\Delta_m \stackrel{d}{=}\arg\min_u Z(u)$, where $u=(u_1,\ldots,u_d)'\in\mathbb R^d$. The $d$ elements correspond to regressors, where the first element is one (the constant term) and the other elements are regressors distributed according to measure $\mu$. Centering the non-constants at zero, $$Z(u)\equiv -(mu_1,0,0,\ldots,0)' + D + \sum_{k\ne0} \int_0^{u'X_k}\\!\left[1(\Gamma_k\le s)- 1(\Gamma_k<0)\right]\\,\mathrm ds,$$ where $m$ is a positive integer, index $k$ takes integer values, $X_k$ are iid with distribution $\mu$, $\Gamma_k=E_1+\cdots+E_k$ for $k>0$ with $E_i\stackrel{iid}{\sim}\rm{Exp}(1)$ independent of all $X_k$ also, $\Gamma_k=-(E_{-1}+\cdots+E_{-k})$ for $k<0$, and $D$ is a random (independent) vector with mean zero and known (if a bit complicated) distribution that I hope I can lay aside for now and add back later. The original paper says the points $\{(\Gamma_k,X_k')':k\ne0\}$ are points of a Poisson process with mean measure $$m(d\epsilon,dx)\equiv \lambda(d\epsilon)\mu(dx),$$ where $\lambda$ is Lebesgue measure.

With $d=2$, we can graph the points $(X_k,\Gamma_k)$ in two dimensions, using only the non-constant element of $X_k$. Picking $(u_1,u_2)'$ determines a line in this plane, $\Gamma=u_1+u_2X$. The first first-order condition (FOC) says that the number of points in the green region minus the number in the red is equal to $m$: http://econ.ucsd.edu/~dkaplan/images/FOC1_v4_small.png

The second FOC is similar, but weighting each point by the $X_k$ value, and should equal zero. Or if you weight the points by $|X_k|$, the red and green for $X_k<0$ switch: http://econ.ucsd.edu/~dkaplan/images/FOC2_v1_small.png

Note: these are not exact equalities, because the integral term in $Z(u)$ is discontinuous; the FOCs are more like "the smallest $u$ such that these are $\ge m$ and $\ge0$."

Question: what are the first two moments of $\Delta_m\stackrel{d}{=}\arg\min_u Z(u)$?

I think the mean is $(\Gamma_m,0)$, but can't prove it; I have no ideas for the variance.

Original post, for special case $d=1$.

Original 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 $\rm{Var}(u)=1$.

Original 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.

Original 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 $\rm{Var}(u)=\mathbb E(u^2)-(\mathbb E(u))^2$.

• I think it would help if you made your question more concrete by telling us what you really want to calculate. For example, maybe you want to replace $1(\lambda_k\le t)$ with some other function? – Ori Gurel-Gurevich Sep 3 '11 at 3:30
• It also would help if you can let us know the more general problem you are trying to solve, for which there might be a common underlying pattern as the Poisson case. For example, my first attempt at guessing the general problem is, for a non-exploding counting process N (increments are all 1), get the distributional information about its first jump time (or at least 1st and 2nd moments). Is this correct or not? – Poldavian Sep 3 '11 at 13:29
• @Poldavian and Ori: thanks for the comments--I have added to my question accordingly. I hope the general problem (general $d$ instead of $d=1$ or $d=2$) is more clear now. Please (anyone) let me know of any further improvements I can make. – David M Kaplan Sep 7 '11 at 17:49