Exponential distribution and Poisson process, Schemes and Mind Maps of Elementary Mathematics

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Exponential distribution and Poisson process

Anastasiia Kim

March 2, 2020

Exponential Distribution

In practice, the exponential distribution often arises as the distribution of the amount of time until some specific event occurs. I (^) the amount of time (starting from now) until an earthquake occurs I (^) until a new war breaks out I (^) waiting time between phone calls

Probability density function of exponential random variables

Example

Suppose that the length of a phone call in minutes is an exponential random variable with parameter λ = 1 / 10. If someone arrives immediately ahead of you at a public 10 telephone booth, find the probability that you will have to wait more than 10 minutes Let X denote the length of the call made by the person in the booth. I (^) wait more than 10 minutes

P ( X > 10) = 1 − P ( X < 10) = 1 − F (10) = 1 − (1 − e −^1 /^10 ·^10 ) =. 368

I (^) the probability that you will wait between 10 and 20 minutes

P (10 < X < 20) = F (20) − F (10) = e −^1 − e −^2 =. 233

Example

Suppose that the number of miles that a car can run before its battery wears out is exponentially distributed with an average value of 10,000 miles. If a person desires to take a 5000-mile trip, what is the probability that he or she will be able to complete the trip without having to replace the car battery? What can be said when the distribution is not exponential? I (^) if the lifetime distribution is Exponential with λ = 1 / 10:

P (remaining time > 5) = 1 − F (5) = e −^1 /^2 =. 604

I (^) If the lifetime distribution not Exponential, let’s denote by t the number of miles that the battery had been in use prior to the start of the trip, then

P (lifetime > t + 5|lifetime > t ) =

1 − F ( t + 5) 1 − F ( t )

where F ( t ) is the cdf for the corresponding distribution.

Example

The time between the arrival of electronic messages at your computer is exponentially distributed with a mean of two hours. I (^) What is the probability that you do not receive a message during a two-hour period? I (^) If you have not had a message in the last four hours, what is the probability that you do not receive a message in the next two hours? I (^) What is the expected time between your fifth and sixth messages?

Poisson process

Suppose we can model the number of calls arriving during an t-minute time window with a Poisson distribution. Assume that the calls arrive completely at random in time during the t-minutes I (^) Let the expected number of calls during a 1-minute interval be λ = 2 (a rate) I (^) For example, the expected number of calls in 2-minutes is 4 calls I (^) Then the expected number of calls in t-minutes is λt = 2 t calls Let N (r.v.) denote the number of calls in an t-minute time interval in a Poisson process with a rate parameter of λ events per minute. Then,

N ∼ Poisson ( λt )

Exponential distribution and Poisson process

How long you have to wait for an event depends on how often events occur. N is the number of calls in an t-minute time interval

N ∼ Poisson ( λt )

Let X be the wait time (continuous) until the first call.

P ( X > t ) = P (you wait at least t minutes for first call) =

= P (there were no calls in the first t minutes) = P ( N = 0) = e − λt^ ( λt )^0 0!

= e − λt

F ( t ) = P ( X ≤ t ) = 1 − P ( X > t ) = 1 − e − λt is the cumulative distribution function of X. Wait time in a Poisson process is modeled with the Exponential distribution.