Asymptotic behaviour of absolute different of two independent Poisson processes

Question

Suppose we have $X_1,X_2\sim Po(\lambda)$ ($X_1$ and $X_2$ are independent). Consider the interval $[0,1]$ with 100000 subintervals of length $\Delta=\frac{1}{100000}$. I can calculate:

$$E|X_1-X_2|=2e^{-2\lambda}\sum_{i,k=1}^{\infty}k\frac{\lambda^{2i+k}}{(i+k)!i!}$$

Therefore, $E|\Delta X_1-\Delta X_2|=2e^{-2\lambda\Delta}\sum_{i,k=1}^{\infty}k\frac{(\lambda\Delta)^{2i+k}}{(i+k)!i!}$. If now I walk along vector$X=X_1-X_2$ with step of size $m$, we then have: $E|\Delta X_1-\Delta X_2|=2e^{-2\lambda m\Delta}\sum_{i,k=1}^{\infty}k\frac{(\lambda m\Delta)^{2i+k}}{(i+k)!i!}$. I ran some simulation and find out that the probability $|\Delta X_1-\Delta X_2|$ have more than 2 jumps is very small. Thus,$E|\Delta X_1-\Delta X_2|=2e^{-2\lambda m\Delta}[\lambda m\Delta+(\lambda m\Delta)^2+\frac{1}{2}(\lambda m\Delta)^3]$. I would like to examine the asymptotic behaviour of $\sum_1^n|\Delta X_1-\Delta X_2|$. We have: $$\frac{1}{n}\sum_1^n|\Delta X_1-\Delta X_2|\to E|\Delta X_1-\Delta X_2|$$ or $$\frac{1}{n}\sum_1^n|\Delta X_1-\Delta X_2|\sim 2e^{-2\lambda m\Delta}[\lambda m\Delta+(\lambda m\Delta)^2+\frac{1}{2}(\lambda m\Delta)^3]$$ Note that $n=\frac{1}{m\Delta}$, then $$\sum_1^n|\Delta X_1-\Delta X_2|\sim \frac{1}{m\Delta}\times 2e^{-2\lambda m\Delta}[\lambda m\Delta+(\lambda m\Delta)^2+\frac{1}{2}(\lambda m\Delta)^3]$$ I got stuck at this step since I cant say anything about asymptotic behaviour of the LHS using the RHS when $m$ getting smaller. Does it behave as order 1 of $m$? Can some one please tell me if I am on the right track or I made a mistake somewhere?

Answer 1

For any $m\in\mathbb{N}$, we have: $$\mathbb{P}[X_1-X_2=m]=\sum_{n\in\mathbb{Z}}\mathbb{P}[X_1=m+n]\mathbb{P}[X_2=n]=\lambda^m e^{-2\lambda}\sum_{n=0}^{+\infty}\frac{\lambda^{2n}}{(m+n)!n!}$$ or: $$\mathbb{P}[X_1-X_2=m]=e^{-2\lambda}I_m(2\lambda)$$ where $I_m$ is a hyperbolic Bessel function.

By symmetry, the probability mass function of $X_1-X_2$ is given by: $$\mathbb{P}[X_1-X_2=m]=e^{-2\lambda}I_{|m|}(2\lambda)$$ for any $m\in\mathbb{Z}$, and: $$\mathbb{E}[|X_1-X_2|]=e^{-2\lambda}\sum_{m=1}^{+\infty}2m\, I_m(2\lambda)=\frac{e^{-2\lambda}}{\pi}\int_{0}^{\pi}\exp(2\lambda\cos\theta)\sum_{m=1}^{+\infty}2m\cos(m\theta)\,d\theta.$$ Since: $$ I_m(2\lambda)\leq\frac{\lambda^m}{m!}I_0(2\lambda) $$ we have the upper bound:

$$\mathbb{E}[|X_1-X_2|]\leq e^{-2\lambda}I_0(2\lambda)\sum_{m=1}^{+\infty}\frac{2m\lambda^m}{m!}=\color{red}{2\lambda e^{-\lambda}I_0(2\lambda)}\approx\color{blue}{e^\lambda\sqrt{\frac{\lambda}{\pi}}}.$$