Waiting time of Poisson process follows Gamma distribution

Question

Suppose that events occur according to a Poisson process with rate $\lambda$, so that for every $t > 0$, the number of occurrences $N(t)$ in the time interval $[0,t]$ has a Poisson distribution with parameter $\lambda t$. Let $T_n$ be the waiting time to the occurrence of the $n$th event. Show that $T_n$ has a gamma distribution with parameters $(n, \lambda)$.

$F(t)=1-P(T_n>t)=1-P(N(t)\ge n-1)=1-\sum_{i=0}^{n-1}\frac{e^{-\lambda t}(\lambda t)^i}{i!}$. I have to prove that the derivative of this expression is equal to $\frac{\lambda e^{-\lambda t}(\lambda t)^{n-1}}{\Gamma(n)}$. How to do it?

Answer 1

Well, $$ \frac{\mathrm{d}}{\mathrm{d} t} \frac{\mathrm{e}^{-\lambda t}(\lambda t)^i}{i!} = -\lambda \frac{\mathrm{e}^{-\lambda t}(\lambda t)^i}{i!} + \lambda i \frac{\mathrm{e}^{-\lambda t}(\lambda t)^{i-1}}{i!} = \lambda\left(- \frac{\mathrm{e}^{-\lambda t}(\lambda t)^i}{i!} + \frac{\mathrm{e}^{-\lambda t}(\lambda t)^{i-1}}{(i-1)!} \right) $$ Therefore: $$ \frac{\mathrm{d}}{\mathrm{d} t} \sum_{i=0}^{n-1} \frac{\mathrm{e}^{-\lambda t}(\lambda t)^i}{i!} = -\lambda \sum_{i=0}^{n-1} \left(- \frac{\mathrm{e}^{-\lambda t}(\lambda t)^i}{i!} + \frac{\mathrm{e}^{-\lambda t}(\lambda t)^{i-1}}{(i-1)!} \right) = -\lambda \left( - \frac{\mathrm{e}^{-\lambda t}(\lambda t)^{n-1}}{(n-1)!} \right) $$ Because $\sum_{i=0}^{n-1} (f(i)-f(i-1)) = f(n-1) - f(-1)$ for any $f$.