Customers arrive at a bank according to a Poisson process with
rate $\lambda$ per minute. Given that two customers arrived in the first 4
minutes, what is the probability that one arrived in the first minute and the other arrived in the last minute.
The setup for the problem is the following
$$P[N_1=1,N_3=1,N_3=1|N_4=2]=\frac{P[N_1=1,N_3=1,N_4=2]}{P[N_4=2]}$$
by bayes' theorem and conditionality which is equal to
$$\frac{P[N_1=1]P[N_2=0]P[N_1=1]}{P[N_4=2]}$$
I understand the inclusion of the term $P[N_2=0]$ due to the stationarity of increments of a poisson process but I don't understand how we have removed $P[N_4=2]$ and replaced the term with $P[N_1=1]$.
Best Answer
By the definition of conditional probability, $$ p:=\mathbb P(N(1)=1, N(4)-N(3)=1\mid N(4)=2) = \frac{\mathbb P(N(1)=0, N(4)-N(3)=1,N(4)=2)}{\mathbb P(N(4)=2)}. $$ Conditioning on $\{N(1)=0, N(4)-N(3)=0\}$, we have \begin{align} p &= \frac{\mathbb P(N(4)=2\mid N(1)=1, N(4)-N(3)=1)\mathbb P(N(1)=1,N(4)-N(3)=1)}{\mathbb P(N(4)=2)}\\ &= \frac{\mathbb P(N(3)-N(1)=0)\mathbb P(N(1)=1)\mathbb P(N(4)-N(3)=1)}{\mathbb P(N(4)=2)}\\ &= \frac{e^{-2\lambda}\cdot e^{-\lambda}\lambda\cdot e^{-\lambda}\lambda}{e^{-4\lambda}(4\lambda)^2/2!}\\ &= \frac{2\lambda^2}{16\lambda^2}\\ &= \frac18. \end{align}