A Poisson Process has lambda = 0.9 What is the probability that the fourth event occurs between time 2 and time 5?
The source I'm reading has the answer pegged as 54.9%.
I got the same answer but my work is totally different, I just want to make sure my method makes sense and I didn't get the same answer by coincidence.
I figured I wanted Pr[N(2) < 4 and N(5) < 4] aka less than 4 events at time 2 and less than 4 events at time 5. I calculated that P(N(2)<4)=.8913 and P(N(5)<4)=.3423 and by subtracting .8913-.3423=.549 Honestly the subtraction was a guess here, I'm not sure if that makes sense or not.
Let $\{N(t):t\geqslant 0\}$ be a Poisson process with rate $\lambda>0$, $n$ a fixed positive integer, and $0<s<t$ fixed real numbers. Let $T_n$ denote the $n^{\mathrm{th}}$ arrival time, then $T_n$ has density $f_n(x) = \lambda\frac{(\lambda x)^{n-1}}{(n-1)!}e^{-\lambda x}\mathsf 1_{(0,\infty)}(x)$ (it is a good exercise to prove this). It follows that \begin{align} \mathbb P(s<T_n<t) &= \int_{(s,t)} f_n(x)\ \mathsf dx\\ &=\int_s^t \lambda\frac{(\lambda x)^{n-1}}{(n-1)!}e^{-\lambda x}\ \mathsf dx\\ &= \frac{\Gamma (n,s \lambda )-\Gamma (n,t \lambda )}{(n-1)!}. \end{align} Substituting $\lambda=\frac9{10}$, $n=4$, $s=2$, and $t=5$, we have $$ \frac{674}{125}e^{-\frac95}-\frac{493}{16} e^{-9/2}\approx0.548996. $$