MA 320-001: Introductory Probabilitydmu228/ma320/lectures/sec5.2.pdfThe Poisson Distribution Deﬁnition (Poisson Process) Let the number of changes that occur in a given continuous

MA 320-001: Introductory Probability

David Murrugarra

Department of Mathematics,University of Kentucky

http://www.math.uky.edu/~dmu228/ma320/

Spring 2017

David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 1 / 17

mailto:[email protected]

http://www.math.uky.edu/

http://www.uky.edu/

http://www.math.uky.edu/~dmu228/ma320/


Waiting Time in a Poisson Process

When observing a process of Poisson Type, we counted thechanges occurring in a given interval. This number was a discreterandom variable with Poisson distribution.

But not only is the number of changes a random variable, thewaiting times between successive changes are also randomvariables.

We are going to consider the waiting time X until the first changein a Poisson process.

Then X is a continuous-type random variable.
























The Poisson Distribution

Definition (Poisson Process)Let the number of changes that occur in a given continuous interval becounted. Then we have an approximate poisson process withparameter λ > 0 if the following conditions are satisfied:

1 The number of changes occurring in nonoverlapping intervals areindependent.

2 The probability of exactly one change occurring in a sufficientlyshort interval of length h is approximately λh.

3 The probability of two or more changes occurring in a sufficientlyshort interval is essentially zero.




Definition (Poisson Distribution)The random variable X has a Poisson distribution if its densityfunction is of the form

f (x) =λxe−λ

x!, x = 0,1,2, ...,

where λ > 0.

In this case, µ = σ2 = λ.




If events in a poisson process occur at a mean rate of λ per unit, theexpected number of occurrences in an interval of length t is λt .

For example, if phone calls arrive at a switchboard following a Poissonprocess at a mean rate of three per minute, then the expected numberof phone calls in a 5-minute period is (3)(5) = 15.

Moreover, the number of occurrences say, x , in the interval of length thas the Poisson density function,

f (x) =(λt)xe−λt

x!, x = 0,1,2, ...



Section 5.2: Exponential Distribution

The exponential distribution describes the waiting time until the firstchange in a Poisson process

Let X denote waiting time until the first change occurs in a Poissonprocess in which the mean number of changes in the unit interval is λ.Then F (x) = 0, for x < 0.

For x ≥ 0,

F (x) = P(X ≤ x) = 1− P(X > x)= 1− P(no changes in [0, x ])= 1− exp−λx

Thus, when x > 0, the probability density function (pdf) of X is

F ′(x) = f (x) = λexp−λx






For x ≥ 0,









For x ≥ 0,







Definition (Exponential Distribution)We often let λ = 1/θ and say that the random variable X has anexponential distribution if its p.d.f. is defined by

f (x) =1θ

e−x/θ, 0 ≤ x <∞,

In this case, µ = θ and σ2 = θ2.

The probability density function (pdf) of an exponential distribution is

f (x) =

{λe−λx x ≥ 0,0 x < 0.




Let X have an exponential distribution with mean µ = θ. then thecumulative distribution function of X is

F (x) ={

0, −∞ < x < 0,1− e−x/θ, 0 ≤ x <∞.



Exponential Distribution: PDF

Figure: Exponential Distribution pdf



Exponential Distribution: CDF

Figure: Exponential Distribution cdf




ExampleCustomers arrive in a certain shop according to an approximatePoisson process at a mean rate of 20 per hour. What is the probabilitythat the shopkeeper will have to wait more than 5 minutes for thearrival of the first customer?

Let X denote the waiting time in minutes until the first customer arrives.

Note that λ = 1/3 is the expected number of arrivals per minute. Thus

θ =1λ= 3 and

f (x) = ?







θ =1λ= 3 and

f (x) = ?







θ =1λ= 3 and

f (x) =13

e−(1/3)x , 0 ≤ x <∞.






θ =1λ= 3 and

f (x) =13

e−(1/3)x , 0 ≤ x <∞.

Hence,

P(X > 5) =∫ ∞

5

13

e−(1/3)xdx = e−5/3 = 0.1889.




Rule (Memoryless Property)An exponentially distributed random variable T obeys the relation

Pr (T > s + t |T > s) = Pr(T > t), ∀s, t ≥ 0.

When T is interpreted as the waiting time for an event to occur relativeto some initial time, this relation implies that, if “T ” is conditioned on afailure to observe the event over some initial period of time “s”, thedistribution of the remaining waiting time is the same as the originalunconditional distribution.

The exponential distributions and the geometric distributions are theonly memoryless probability distributions.




Rule (Memoryless Property)An exponentially distributed random variable T obeys the relation

Pr (T > s + t |T > s) = Pr(T > t), ∀s, t ≥ 0.

When T is interpreted as the waiting time for an event to occur relativeto some initial time, this relation implies that, if “T ” is conditioned on afailure to observe the event over some initial period of time “s”, thedistribution of the remaining waiting time is the same as the originalunconditional distribution.

The exponential distributions and the geometric distributions are theonly memoryless probability distributions.



Memorylessness

ExampleSuppose that a certain type of electronic component has anexponential distribution with a mean life of 500 hours. If X denotes thelife of this component (or the time to failure of this component), then

Note thatθ =

1λ= 500

andf (x) =

1500

e−t/500, 0 ≤ x <∞.



Memorylessness


Note thatθ =

1λ= 500

andf (x) =

1500

e−t/500, 0 ≤ x <∞.



Memorylessness


P(X > x) =∫ ∞

x

1500

e−t/500dt = e−x/500.



Memorylessness


What is the probability the the component will last an additional 600hours, given that it has operated for 300 hours?

P(X > 900|X > 300) = P(X > 600) = e−6/5.

That is, the probability the the component will last an additional 600hours, given that it has operated for 300 hours, is the same as theprobability that it will last 600 hours when first put into operations.



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MA 320-001: Introductory Probabilitydmu228/ma320/lectures/sec5.2.pdfThe Poisson Distribution Deﬁnition (Poisson Process) Let the number of changes that occur in a given continuous