Question

A Geiger counter beeps according to a Poisson process with rate of 1 beep per minute.

(a). Let N_1 be the number of beeps in the first minute, N_2 the number of beeps in the second minute, N_3 the number of beeps in the third minute, etc. Let K be the smallest number for which N_K = 0, so minute # K is the first minute without a beep. Find E(K).

(b). What is the probability that the time until the next beep is more than 1 minute but less than 2 minutes?

(c), For any numbers 0 < a < b, find a formula for the probability that the time T in minutes until the next beep is between a and b.

(d). For the T in part (b), find P( T > 3 | T> 1 ).

Answer 1

(a)

We know that the interarrival time intervals between Poisson process follows exponential distribution. Let T be the interarrival time intervals between Poisson process, then T ~ Exp(1)

Probability that there is no beep in 1 minute period = P(T > 1) = exp(-1) = 0.3679

Given, K be the smallest number for which N_K = 0. This can be modeled as Gemetric distribution, where the probability of success, p = 0.3679. Thus, K ~ Geom(0.3679)

By Gemetric distribution, E(K) = 1/p = 1/0.3679 = 2.718

(b)

P(1 < T < 2) = P(T > 1) - P(T > 2)

= exp(-1) - exp(-2)

= 0.2325

(c)

P(a < T < b) = P(T > a) - P(T > b)

= exp(-a) - exp(-b)

(d)

P( T > 3 | T> 1 ) = P(T > 3 and T> 1) / P(T> 1)

=  P(T > 3) / P(T> 1)

= exp(-3) / exp(-1)

= 0.1353