Let $\lbrace N(t)\rbrace_{t\geq 0}$ be a Poisson process with intensity $\lambda = 3$. Compute $$P\left[N(6) = 2 \,|\, N(8) = 4, N(3) = 1\right].$$
I understand when there is only one condition i.e. $P\left[N(6) = 2 | N(8) = 4\right]$. Since that is $$P\left[N(6) = 2 , N(8) = 4\right]/ P\left[N(8) = 4\right] = P\left[N(6) = 2 , N(8)-N(6) = 2\right]/ P\left[N(8) = 4\right].$$ But I am unsure on how to go about doing this one
Any tips on where to begin? Thanks
You can handle this using the definition of conditional probability, stationary increments and independent increments. $$ \begin{align*} P\left[N(6) = 2 \,|\, N(8) = 4, N(3) = 1\right] &= \frac{P[N(6) = 2,N(8) = 4, N(3) = 1]}{P[N(8) = 4, N(3) = 1]}\\ &=\frac{P[N(3) = 1, N(6) - N(3) = 1, N(8) - N(6) = 2]}{P[N(8) = 4, N(3) = 1]}\\ &=\frac{P[N(3) = 1]P[N(6) - N(3) = 1]P[N(8) - N(6) = 2]}{P[N(8) = 4, N(3) = 1]}\\ &=\frac{P[N(3) = 1]P[N(3)= 1]P[N(2)= 2]}{P[N(8) = 4, N(3) = 1]}\\ \end{align*} $$
I'll let you finish it up from there.