# Poisson Process timed on other arrivals

I've been struggling on this question for a while. Particularly, part b. My thought process was that it uses 2C1 competing exponentials to end up with that result, although it does not make intuitive sense to me why thats the case or how I'd even arrive at that result. Any intuition provided would be greatly appreciated! Another approach I was considering was to use Geometric.

Consider three independent Poisson processes $$(N_i(t), t\geqslant 0)$$ with rates $$\lambda_i$$, for $$i=1,2,3$$. Let $$\tau =\inf\{t>0: N_3(t)=1\}$$.

(a) Describe the distribution of $$N_1(\tau)$$.

(b) Find $$\mathbb E[N_2(\tau)\mid N_1(\tau)]$$. Verify that this is equal to $$2(2/3)^{N_1(\tau)}$$ in the case that all $$\lambda_i=\lambda$$.

• I am not sure, but I think you can write $P(N_1(\tau) = k) = \int_{t=0}^{\infty} P(N_1(t) = k) P(t=\tau)dt = \int_{\tau=0}^{\infty} e^{-\lambda_1\tau}\frac{(\lambda_1\tau)^k}{k!} \lambda_3 e^{-\lambda_3 \tau}d\tau$ and t denotes first arrival time in $N_3(t)$ (interarrival time $X_1$), so $t \sim Exp(\lambda_3)$ May 12, 2021 at 10:55
• See if this problem helps: math.stackexchange.com/questions/1450955/… May 12, 2021 at 11:10

For part (a), merge the first and third process. Each arrival to the merged process arrives to process $$i \in \{1,3\}$$ with probability $$\frac{\lambda_i}{\lambda_1 + \lambda_3}$$ and so $$\Pr[N_1(\tau) = k] = \left(\frac{\lambda_1}{\lambda_1 + \lambda_3}\right)^k \left(\frac{\lambda_3}{\lambda_1 + \lambda_3}\right).$$
For part (b), merge all three processes. By the same logic, $$\Pr[N_1(\tau) = k \text{ and } N_2(\tau) = \ell]$$ is $$\binom{k+\ell}{k} \left(\frac{\lambda_1}{\lambda_1 + \lambda_2 + \lambda_3}\right)^k \left(\frac{\lambda_2}{\lambda_1 + \lambda_2 + \lambda_3}\right)^\ell \left(\frac{\lambda_3}{\lambda_1 + \lambda_2 + \lambda_3}\right).$$ Dividing, we get the conditional distribution of $$N_2(\tau)$$ given $$N_1(\tau)$$, and it is a negative binomial distribution.
I don't get the given value for $$\mathbb E[N_2(\tau)\mid N_1(\tau)]$$; I get $$\frac{N_1(\tau)+1}{2}$$ in the case that all the rates are equal. The value $$2(2/3)^{N_1(\tau)}$$ makes no sense for $$\mathbb E[N_2(\tau)\mid N_1(\tau)]$$; when $$N_1(\tau)$$ is large, we expect $$N_2(\tau)$$ to be large as well, but $$2(2/3)^{N_1(\tau)}$$ decreases as $$N_1(\tau)$$ increases. Instead, I get $$2(2/3)^k$$ for $$\Pr[N_2(\tau)=0 \mid N_1(\tau)=k]$$.
Recall that the distribution of $$\tau$$ is simply exponential with parameter $$\lambda_3$$. Hence we may find the distribution of $$N_1(\tau)$$ by computing: \begin{align} \mathbb P(N_1(\tau) = j) &= \int_0^\infty \mathbb P(N_1(\tau)=j\mid \tau = t) \lambda_3 e^{-\lambda_3 t}\ \mathsf dt\\ &= \int_0^\infty \mathbb P(N_1(t) = j)\lambda_3 e^{-\lambda_3 t}\ \mathsf dt\\ &= \int_0^\infty e^{-\lambda_1 t} \frac{(\lambda_1 t)^j}{j!}\lambda_3 e^{-\lambda_3 t}\ \mathsf dt\\ &= \frac{\lambda_1^j\lambda_3}{(\lambda_1+\lambda_3)^{j+1}} \end{align}
• Intuitively, if $N_1(\tau)$ is small, that makes it more likely that $\tau$ is small, so $N_2(\tau)$ is also likely to be small. May 20, 2021 at 4:02