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?

  • $\begingroup$ Maybe you have to prove that, but if you just want to show that the distribution is gamma, observe that it is the sum of n interarrival times which are i.i.d. exponential. Show that such a fctn is gamma by using moment generating functions. $\endgroup$ – mike Apr 17 '12 at 11:28
  • $\begingroup$ One of the inequalities must must change. The expression $F(t)$, the probability that the waiting time $T_n$ was below or equal to $t$ is equal to the probability that at time $t$ the number of occurrences is $n$ or larger (such that the $n$-th event must have occurred before $t$). Hence $$P(N(t)>n) = 1-P(N(t) \leq n-1)$$ and not $$1-P(N(t)\geq n-1)$$ @amWhy $\endgroup$ – Martijn Weterings Aug 31 '18 at 12:03

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 $f$.


A different approach that also uses a Poisson distribution with parameter $\lambda t$ would be:

$$P(t < T_n < t+dt) = \underbrace{ P(N(t) = n-1)}_{\substack{\text{the probability for }\\\text{$(n-1)$ arrivals at time $t$}}} \qquad \qquad \times \qquad \underbrace{\vphantom{P(N(t) = n-1)}\lambda dt}_{\substack{\text{the probability for }\\\text{arrival between time $t$ and $t+dt$}}} $$

leading to

$$P(t < T_n < t+dt)= \text{Poisson}(n-1,\lambda t) \times \lambda dt = \lambda\frac{(\lambda t)^{n-1} e^{-\lambda t}}{(n-1)!} dt = \frac{\lambda e^{-\lambda t}(\lambda t)^{n-1} }{\Gamma(n)} dt$$

The connection with your approach is as following:

You could view the derivative as $$\frac{d}{dt} P(N(t)<n) = \frac{d}{dt} P(N(t)=0) + \frac{d}{dt} P(N(t)=1) + \cdots +\frac{d}{dt} P(N(t)=n-1)$$


$$\underbrace{{\frac{d}{dt} P(N(t)=k)}}_{\text{change of k}} = \underbrace{\lambda P(N(t)=k-1)\vphantom{\frac{d}{dt} P(N(t)=k)}}_{\text{gain from k-1 to k}} -\underbrace{\lambda P(N(t)=k)\vphantom{\frac{d}{dt} P(N(t)=k)}}_{\text{loss from k to k+1}}$$

and all those terms cancel (much like the answer of Sasha):

$$\frac{d}{dt} P(N(t)<n) = -\lambda P(N(t)=n-1)$$

So the the derivative of $P(N(t) < n)$ the rate at which you are surpassing the 'level' $n-1$ is equal to how many there is currently at the 'level' $n-1$ and how fast those will rise to the 'level' $n$. What happens at lower 'levels' does not matter for the change of $P(N(t) < n)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.