# Computing the conditional probability of a Poisson process with thinning.

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 $$M_t \sim Poi(\lambda t)$$.
• 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. – Sayantan Apr 11 at 13:26