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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
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.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 2 / 17
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.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 2 / 17
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.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 2 / 17
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.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 2 / 17
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.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 3 / 17
The Poisson Distribution
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.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 4 / 17
The Poisson Distribution
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, ...
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 5 / 17
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
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 6 / 17
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
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 6 / 17
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
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 6 / 17
Section 5.2: Exponential Distribution
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.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 7 / 17
Section 5.2: Exponential Distribution
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 <∞.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 8 / 17
Exponential Distribution: PDF
Figure: Exponential Distribution pdf
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 9 / 17
Exponential Distribution: CDF
Figure: Exponential Distribution cdf
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 10 / 17
Section 5.2: Exponential Distribution
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) = ?
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 11 / 17
Section 5.2: Exponential Distribution
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) = ?
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 11 / 17
Section 5.2: Exponential Distribution
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) =13
e−(1/3)x , 0 ≤ x <∞.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 12 / 17
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) =13
e−(1/3)x , 0 ≤ x <∞.
Hence,
P(X > 5) =∫ ∞
5
13
e−(1/3)xdx = e−5/3 = 0.1889.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 13 / 17
Section 5.2: Exponential Distribution
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.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 14 / 17
Section 5.2: Exponential Distribution
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.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 14 / 17
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 <∞.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 15 / 17
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 <∞.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 15 / 17
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
P(X > x) =∫ ∞
x
1500
e−t/500dt = e−x/500.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 16 / 17
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
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.
David Murrugarra (University of Kentucky) MA 320: Section 5.2 Spring 2017 17 / 17