Question

Consider a Poisson process with rate between the 5rd and 7th arrivals? = 3. What is the distribution of the time

Name the distribution and any parameters. Show mathematically, the distribution between the 5th and 7th arrival.

Answer 1

Note that poisson probability distribution and exponential distribution follows memoryless property. Therefore the amount of waiting time between 5th and 7th arrival is same as the waiting time for 2 arrivals. Let the number of arrivals be given by random variable X and waiting time by random variable Y.

Then the distribution of waiting time is given as:

P(Y > y) = Probability that waiting time is more than y

which should be equal to probability that there is no or 1 arrival in time y

P(Y > y) = P(X = 0) + P(X = 1)

Now for time y, the distribution for X is given as:

$X \sim Poisson(\lambda y)$

$X \sim Poisson(3y)$

Therefore the probability now is computed as:

$P(Y > y) = P(X = 0) + P(X = 1)$

$P(Y > y) = e^{-3y} + 3ye^{-3y}$

Therefore, we have the cumulative distribution of Y given as:

$F_y(y) = P(Y \leq y) = 1 - P(Y > y) = 1 - e^{-3y} - 3ye^{-3y}$

Therefore the distribution now can be obtained by differentiating the cumulative distribution with respect to y as:

$f(y) = 3e^{-3y} - 3e^{-3y} + 9ye^{-3y} = 9ye^{-3y}$

This is the required distribution of the time here.