Thread: Conditional Distribution of arrival times

1. Conditional Distribution of arrival times

Just got my final back, and this was the only problem I couldn't figure out. Could someone show me how to do this? Thanks!

A Poisson process N(t) has rate $\displaystyle \lambda$= 1. Let $\displaystyle X_2$ be the time between the first and second arrivals of the process. Suppose there are exactly 2 arrivals in the interval $\displaystyle 0\leq t\leq 1$, so N(1) = 2. Find the conditional pdf of $\displaystyle X_2$ given that N(1) = 2. Use your result to calculate the mean and standard deviation of $\displaystyle X_2$ given that N(1) = 2.

We were given the hint to use the theorem of conditional arrival times, namely that given N(t) = n, the n arrival times $\displaystyle S_1,...,S_n$ have the same distribution as the order statistics corresponding to n independent random variables uniformly distributed on (0,t).

The only headway I was able to make was that $\displaystyle X_2 = S_2 - S_1$, so the conditional pdf of $\displaystyle X_2$ would be the pdf of $\displaystyle S_2$ minus the pdf of $\displaystyle S_1$. But when I used that to find the expected value, I got a ln(t), which is undefined at 0 so I figured I was wrong.

2. Re: Conditional Distribution of arrival times

The answer to that question is strongly related to the following theorem in...

http://www.nas.its.tudelft.nl/people...UP_Poisson.pdf

Theorem 7.3.3 Given that exactly one event of a Poisson process $\displaystyle \{X(t);\ t \ge 0\}$ has occurred during the interval [0, t], the time of occurrence of this event is uniformly distributed over [0, t].

Naw if $\displaystyle T_{1}$ and $\displaystyle T_{2}$ are two two times of occurrence in [0,1], then the random variable $\displaystyle \Delta T= |T_{2}-T_{1}|$ has p.d.f. given by...

$\displaystyle p(\tau)=\begin{cases}2\ (1-\tau)} &\text{if}\ 0<\tau<1\\ 0 &\text{elsewhere}\end{cases}$ (1)

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$