Let $\{N(t) \}_{t\geq0}$ be a Poisson process with rate $\lambda$. Suppose that each "arrival" is independently of type "$i$" with probability $p_i$. Then I know that the $\{N_i(t) \}$ are independent Poisson processes with rates $\lambda p_i$. The problem I'm having difficulty with is stated as follows:

Suppose cars arrive according to a Poisson process with rate $\lambda=3$ cars per minute, and each car is independently either Blue with probability 1/2, or Green with probability 1/3, or Red with probability 1/6.

Let $N(t)$ be the total number of cars that arrive by time t, and $N_R(t)$ the number of Red cars that arrive by time $t$.

*I wish to find $P(N_R(3)=2 \, | \, N(4) =4)$.

I would know how to solve this using independence if $t$ was equal in the condition $-$ i.e., $$ \begin{aligned} P(N_R(4)=2 \, | \, N(4) =4) &= \frac{P(N_R(4)=2 \, , \, N_B (4) + N_G(4) =2)}{P(N(4)=4)} \\ &= \frac{P(N_R(4)=2)P(N_B(4)+N_G(4)=2)}{P(N(4)=4)} \end{aligned} $$ however, I have not been able to figure out how to handle the different time. I've only been able to confidently get that, $$ P(N_R(3)=2 \, | \, N(4) =4) = P(N_R(3)=2 \, | \, N_R(4) + N_G(4)+N_B(4) =4) $$


$$ P(N_R(3)=2\mid N_R(4) + N_G(4) + N_B(4) =4) = \frac{P(N_R(3)=2, N_R(4) + N_G(4) + N_B(4) =4 )}{P(N_R(4) + N_G(4) + N_B(4) =4)}$$

By definition $P(N_R(4) + N_G(4) + N_B(4) =4) = P(N=4)$.

On the other hand, $$P(N_R(3)=2, N_R(4) + N_B(4) + N_G(4) =4 ) \\= \sum_{i=2}^4 P(N_R(3)=2, N_R(4)=i, N_G(4)+N_B(4)=4-i) \\ = \sum_{i=2}^4 P(N_R(3)=2, N_R(4)=i) P( N_G(4)+N_B(4)=4-i)$$ by independence $N_R, N_G$ and $N_B$. Note that $N_G(4)+ N_B(4) \sim Poi(10) $ as $N_G(4) \sim Poi(4)$ and $N_B(4) \sim Poi(6)$ are independent.

For $2\leq i \leq 4$, $$P(N_R(3)=2, N_R(4)=i) = P(N_R(3)=2, N_R(4)-N_R(3)=i-2) = P(N_R(3)=2)P(N_R(1)=i-2)$$ by independence and stationarity of increments of Poisson processes. So all you now have to calculate are the probabilities of the events of the form $P(M_t=n)$ where $M$ is a PP with rate $\lambda$. And you know $M_t \sim Poi(\lambda t)$.

  • $\begingroup$ Thanks for answer! The only part that remains confusing to me is the need for a double summation. You've indexed something by j, but I cannot see it within the summand. Thanks. $\endgroup$ – Mr. Wayne Apr 11 at 13:11
  • $\begingroup$ Thanks for pointing it out. There is no $j$. I fixed it. The summation is just used to calculate the total probability after decomposing the bigger event into all possible cases. $\endgroup$ – Sayantan Apr 11 at 13:26

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