Two poisson processes - no independence Say I have two poisson processes which are not independent,
$N_t$: The number of cars that go over a bridge, with rate 10 per minute
There is a 7/10 chance that cars going over this bridge are red. Therefore,
$M_t$: The number of red cars that go over a bridge, with rate 7 per minute
How would I calculate probabilities between these dependent events, $P(N_1=5,M_{0.5}=4)$? (i.e. probability of 4 red cars in 0.5 minutes and 5 total cars in a minute)
 A: If I understand your question correctly, we want the probability of $4$ red cars in the first half-minute, and a total of $5$ cars in the first minute. Given that there are $4$ reds in the first half-minute, this event can happen in $2$ ways: (i) $0$ non-reds in the first half-minute, and $1$ car (any colour) in the next half minute or (ii) $1$ non-red in the first half minute, and $0$ cars in the next half-minute.
We calculate the probability of (i) and leave the computation of (ii), and the addition, to you.
The number of non-reds in a half-minute has Poisson distribution, parameter $3/2$. So the probability of $0$ non-reds in the first half-minute is $e^{-3/2}$. The probability of $1$ car in the next half minute is $e^{-5}\frac{5^1}{1!}$. Because of independence, we multiply.
Added: Putting things together, we get that the probability of $4$ reds in the first half-minute, and a total of $5$ cars in the first minute, is
$$e^{-7/2}\frac{(7/2)^4}{4!}\left[e^{-3/2}\frac{(3/2)^0}{0!}e^{-5}\frac{5^1}{1!}+ e^{-3/2}\frac{(3/2)^1}{1!}e^{-5}\frac{5^0}{0!}                        \right].$$
