# A relationship between Poisson distribution and gamma distribution

We define $$N(t)$$ to be number of events in the interval $$[0,t]$$. We assume that $$N(t) \sim P(\lambda t)$$ for $$\lambda > 0$$. Let $$X$$ be the waiting time until the $$n$$-th event, we need to prove that $$X \sim \Gamma(n , \lambda)$$.

In order to proof that I thought about splitting the interval to $$n$$ smaller intervals, each interval will indicate the $$i$$-t'h event $$i = 1,2,\dots,n$$ and say that each smaller interval $$T_i \sim Exp(\lambda)$$. then say that $$X = T_1 + T_2 +\dots +T_n$$ and then prove it by using the moment-generating function of the gamma distribution and the MGF of the exponential distribution.

Is it correct?

Times between two successive events follow exponential distribution. Time to $$n$$-th event is then a sum of exponential r.v.-s, which is a gamma distribution. You have the right idea to use MGF-s. The proof can be found here: Gamma Distribution out of sum of exponential random variables
• Intervals are random and governed by $\lambda$ so I don't think you need to split anything here. Please see the reference. – dnqxt Apr 18 at 15:40