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).

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