# Independent Poisson processes and the geometric distribution

In the last post we saw that a geometric sum of iid exponential random variables is the interarrival distribution of a split Poisson process. Now, by “magic” geometric random variables will come from considering two Poisson processes.

Let ${\{N(t):t\geqslant0\}}$ and ${\{M(t):t\geqslant0\}}$ be independent Poisson processes with intensities ${\nu}$ and ${\lambda}$. Let

$\displaystyle \tau = \inf\{t>0:N(t)=1 \}$

be the time of first arrival in ${N(t)}$. What are the distributions of ${M(\tau)}$ and ${N(2\tau)-N(\tau)}$?

It is clear that ${\tau\sim\mathrm{Exp}(\nu)}$, i.e. ${\tau}$ has density ${f_\tau(t)=\nu e^{\nu t}\mathsf 1_{(0,\infty)}(t)}$. By independence, we can compute the distribution of ${M(\tau)}$ by conditioning on ${\tau}$; for ${m\in\{0,1,2,\ldots\}}$ we have

\displaystyle \begin{aligned} \mathbb P(M(\tau) = m) &= \int_0^\infty f_{M(\tau),\tau}(m,t)\ \mathsf dt\\ &= \int_0^\infty f_{M(\tau)\mid \tau=t}(m\mid\tau)f_\tau(t)\ \mathsf dt\\ &= \int_0^\infty e^{-\mu t}\frac{(\mu t)^m}{m!} \nu e^{-\nu t}\ \mathsf dt\\ &= \mu^m\nu \int_0^\infty \frac{t^m}{m!} e^{-(\mu+\nu)t}\ \mathsf dt\\ &= \frac{\mu^m \nu}{(\mu+\nu)^{m+1}}\\ &= \left(\frac\mu{\mu+\nu} \right)^m \left(\frac\nu{\mu+\nu} \right). \end{aligned}

It follows that ${M(\tau)\sim\mathrm{Geo}\left(\frac\nu{\mu+\nu} \right)}$.

By the independent increment property of Poisson processes, for ${n\in\{0,1,2\ldots\}}$ we have

\displaystyle \begin{aligned} \mathbb P(N(2\tau)-N(\tau)=n) &= \int_0^\infty f_{N(2\tau)-N(\tau),\tau}(n,t)\ \mathsf dt\\ &= \int_0^\infty f_{N(2\tau)-N(\tau)\mid \tau=t}(n\mid \tau)f_\tau(t)\ \mathsf dt\\ &= \int_0^\infty e^{-\nu t}\frac{(\nu t)^n}{n!}\nu e^{-\nu t}\ \mathsf dt\\ &= \nu^{n+1}\int_0^\infty \frac{t^n}{n!} e^{-2\nu t}\ \mathsf dt\\ &= \frac{\nu^{n+1}}{(2\nu)^{n+1}}\\ &= \left(\frac12 \right)^{n+1}. \end{aligned}

It follows that ${N(2\tau)-N(\tau)\sim\mathrm{Geo}\left(\frac12 \right)}$.

Now consider the first arrival time of ${M(t)}$,

$\displaystyle \sigma = \inf\{t>0:M(t)=1 \}.$

Since ${\sigma\sim\mathrm{Exp}(\mu)}$, we have ${\tau\wedge\sigma\sim\mathrm{Exp}(\mu+\nu)}$, and by symmetry, ${N(\sigma)\sim\mathrm{Geo}\left(\frac\mu{\mu+\nu} \right)}$. What about the distribution of ${M(\tau)\min N(\sigma)}$?

In general, if ${X\sim\mathrm{Geo}(p_1)}$ and ${Y\sim\mathrm{Geo}(p_2)}$ are independent, then for ${k\in\{0,1,2,\ldots\}}$

\displaystyle \begin{aligned} \mathbb P(X\wedge Y=k) &= \mathbb P(X=k)\mathbb P(Y>k) + \mathbb P(Y=k)\mathbb P(X>k) + \mathbb P(X=k)\mathbb P(Y=k)\\ &= (1-p_1)^kp_1(1-p_2)^{k+1} + (1-p_2)^kp_2(1-p_1)^{k+1}+(1-p_1)^kp_1(1-p_2)^kp_2\\ &= (1-p_1)^k(1-p_2)^k(p_1(1-p_2) + p_2(1-p_1) + p_1p_2)\\ &= ((1-p_1)(1-p_2))^k(1-(1-p_1)(1-p_2)) \end{aligned}

so that ${X\wedge Y\sim\mathrm{Geo}(1-(1-p_1)(1-p_2))}$.

Therefore

$\displaystyle M(\tau)\wedge N(\sigma)\sim\mathrm{Geo}\left(1-\frac{\mu\nu}{(\mu+\nu)^2} \right).$