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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?

Jean-Claude Arbaut
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Yarinnahum4
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1 Answers1

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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

dnqxt
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  • Thank you! I just wanted to know if I can say that each smaller interval is indeed an exponential distribution. – Yarinnahum4 Apr 18 '19 at 15:35
  • 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 '19 at 15:40