# Probability in 2-D Poisson process

I saw this problem in Durrett.

Consider two independent Poisson processes $N_1(t)$ and $N_2(t)$ with rates 1 and 2. What is the probability that the two-dimensional process $(N_1(t),N_2(t))$ ever visits the point $(i, j)$?

I wrote this as $P(N_1(t)=i,N_2(t)=j)=P(N_1(t)=i)P(N_2(t)=j)=\frac{e^{-\lambda_1t}(\lambda_1t)^i}{i!}\frac{e^{-\lambda_2t}(\lambda_2t)^j}{j!}$

Is this all that is there to the problem or is there something I am missing?

The process $(N_1(t),N_2(t))_{t\geqslant0}$ starts from $(0,0)$ and jumps from $(i,j)$ to $(i+1,j)$ or to $(i,j+1)$ with respective probabilities $p=\lambda_1/(\lambda_1+\lambda_2)$ and $q=1-p=\lambda_2/(\lambda_1+\lambda_2)$.

Let $h(i,j)$ denote the probability to visit $(i,j)$, then $h(0,0)=1$ and, for every $i\geqslant1$ and $j\geqslant1$, $h(i,0)=p^i$, $h(0,j)=q^j$ and $h(i,j)=ph(i-1,j)+qh(i,j-1)$.

Thus, the reduced variables $c(i,j)=h(i,j)/(p^iq^j)$ are such that $c(i,0)=c(0,j)=1$ and $c(i,j)=c(i-1,j)+c(i,j-1)$ hence one can show recursively that $c(i,j)=(i+j)!/(i!\,j!)$. Finally, for every $(i,j)$, $$h(i,j)=\frac{(i+j)!}{i!\,j!}\,p^iq^j=\frac{(i+j)!}{i!\,j!}\,\frac{\lambda_1^i\,\lambda_2^j}{(\lambda_1+\lambda_2)^{i+j}}.$$ Sanity check: For every $n\geqslant0$, $\sum\limits_{i=0}^nh(i,n-i)=1$. Do you see why?

I never would thought this as a Markov chain. Nice thought! I get the same answer from a different approach. The joint process must reach i+j total points sometimes. The only concern then is when the joint process reach i+j points, there shall be i points from N1 and j points from N2. So it reduces down to a very simple binomial and that's the answer.