Question

Suppose the number of events that happen in time t follows a Poisson distribution with parameter λ.

Show that the time when the third even occurs follows a gamma distribution with α = 3,β = 1/λ.

Answer 1

I we'll ti ane arrival foll Let Ne=no. Well first show that, the time taken for -arrival follows a exponential (A), "Neano. of arrivales during time period to The taken for aon arrival assuming Xt: time taken for aon arri someone arrive at arrive at time t. thes (x+²) = No Me = Nexn) by definition. how applying the probability on L.H.S. we have P(X=> )= 1-PCX+ sn) using equivalence relation we have P(X+50)=1-P(Neth N==0) But, PCNEEA-N4 =o)= P(NMFO) this by poisson pmf for rate & and time n. We have P(Nein Ne=0)= ane te =) P(X+ Sn)= 1-enn - the distribution of teft e? A exponential (')). - Now, we know, arrivals are undependent, thus arrivals of time taken for three arrivals is sum of 3 exponential (^_^) distribution, which is a gamma (2=3,'') • -24 on Thuas, the distribution of awrence of 3rd arrival follows a gamma (a= 3, B??). Hence proved,