 Solution

ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2014/2/d/i/solution

 Top articles  Latest articles New article in topic

Past exam of the probability course of the University of Oxford / 2014 / 2 / d / i / Solution by

Codex  0 2026-10-31

The Poisson interarrival times of a homogeneous Poisson process are independent and identically distributed with the exponential distribution of rate

λ

. Consequently

Y_{1}, Y_{2}

are independent random variables and their joint probability density is

f_{Y_{1}, Y_{2}} (y_{1}, y_{2}) = λ^{2} e^{- λ (y_{1} + y_{2})} 1_{{y_{1} > 0, y_{2} > 0}} .

Here

Y_{2}

is the second interarrival duration, rather than the time of the second arrival.

 Read the full article

 New to topics? Read the docs here!