Let be $X(t)$ be a Poisson process with parameter $λ = 1/3$. Find $P[X(6) -X(3)=7]$

Question

Please can someone help me with this problem in my exam tomorrow in Stochastic Processes

Let be $X(t)$ be a Poisson process with parameter $λ = 1/3$. Find $P[X(6) -X(3)=7]$

Thanks in advance!

Answer 1

You should know by now that for independent blocks of time, $$X(6)-X(3)\sim \text{Poisson}(3\lambda).\tag 1$$

In other words, if $a\geq 0$ and $b>0$, then number of arrivals in $(a,a+b)$ follows $$X(a+b)-X(a) \sim\text{Poisson}(b\lambda).\tag 2$$

Then using $(1)$, we have $$P\{X(6)-X(3)) = 7\} = e^{3(1/3)}\frac{(3(1/3))^7}{7!} =\frac{e}{7!}.$$