Lesson 1 Negative Exponential

6.7 EXPONENTIAL DISTRIBUTION

• The exponential distribution is the third (and last) continuous probability distribution we will consider.

• It has useful applications in the reliability of components which may fail suddenly and in the theory of queuing.

• It also has an important connection with the (discrete) Poisson distribution as follows.

• Suppose events occur randomly in time.

• Then we know (from Section 6.3) that the number of events per unit time has a Poisson distribution.

• It can be shown that the time between successive random events has an exponential distribution.

Exponential Random Variables: Pdf

• Let λ be a positive real number. We write X~exponential(λ) and say that Xis an exponential random variable with parameter λ if the pdf of X is

, if 0( )

0, otherwise

xe xf x

λλ − ≥=


• A look at the graph of the pdf is informative. Here is a graph for λ=0.5. Note that it has the same shape as every exponential graph with negative exponent (exponential decay). The tail shrinks to 0 quickly enough to make the area under the curve equal 1. Later we will see that the expected value of an exponential random variable is 1/λ (in this case 2). That is the balance point of the lamina with shape defined by the pdf. (graph on next slide)



• A simple integration shows that the area under f is 1:

( )00

0

lim lim

lim 0 1 1

t tx x x

t t

t

t

e dx e dx e

e e

λ λ λ

λ λ

λ λ∞ − − −

−∞ →∞ →∞

− −

→∞

= = −

= − − − = − + =

∫ ∫

Exponential Random Variables: Cdf

P(X<x)= dxex x∫

−0

λλ

[ ] xxe 0

λ−−=

xe λ−−1=


• Essentially the same computation as above shows that . Here is the graph of the cdf for X~exponential(0.5).

Exponential Random Variables:Expectation

• Finding the expected value of X~exponential(λ) is not hard as long as we remember how to integrate by parts. Recall that you can derive the integration by parts formula by differentiating a product of functions by the product rule, integrating both sides of the equation, and then rearranging algebraically. Thus integration by parts is the product rule used backwards and sideways.


Given functions u and v, .

Integrate both sides wrt x to get .

Rearrange to get .

In practice we look for u and dv in our original integral

and convert to the RHS of

x x xD uv uD v vD u

uv udv vdu

udv uv vdu

= +

= +

= −

∫ ∫∫ ∫

the equation to see if that is easier.


The distribution function, F(x)

Probabilities for the exponential distribution

Exponential Random Variables:(i) Pdf

, if 0( )

0, otherwise

xe xf x

λλ − ≥=

(ii) Cdfotherwise0

0,1)(

=

≥−= − xexF xλ

(iii) mean =λ1

(iv) Variance2

1

λ=

(v) median =λ2ln

Exponential Random Variables:Applications

• Exponential distributions are sometimes used to model waiting times or lifetimes. That is, they model the time until some event happens or something quits working. Of course mathematics cannot tell us that exponentials are right to describe such situation. That conclusion depends on finding data from such real-world situations and fitting it to an exponential distribution.

.


• As the coming examples will show, this formula greatly facilitates finding exponential probabilities.


• Suppose the wait time X for service at the post office has an exponential distribution with mean 3 minutes. If you enter the post office immediately behind another customer, what is the probability you wait over 5 minutes? Since E(X)=1/λ=3 minutes, then λ=1/3, so X~exponential(1/3). We want

.

1 55

3 3

( 5) 1 ( 5) 1 (5)

1 1 0.189

P X P X F

e e− ⋅ −

> = − ≤ = −

= − − = ≈


• Under the same conditions, what is the probability of waiting between 2 and 4 minutes? Here we calculate . 4 2

3 3

2 4

3 3

(2 4) (4) (2) 1 1

0.250

P X F F e e

e e

− −

− −

≤ ≤ = − = − − −

= − ≈


• The trick in the previous example of calculating

is quite common. It is the reason the cdf is so useful in computing probabilities of continuous random variables.

.

( ) ( ) ( )P a X b F b F a≤ ≤ = −

Example 6.16

The mean time between successive telephone calls follow a negative

exponential distribution, arriving randomly at a switchboard is 30

seconds.

(a) What is the probability that the time between successive

telephone calls will

be:

(i) less than 15 seconds;

(ii) between 15 and 30 seconds;

(iii) between 30 and 60 seconds;

(iv) more than 60 seconds?

(b) What is the median time between successive telephone calls?

successive telephone calls are arriving randomly with a mean time

between calls of 30 seconds, X, the time between calls, has an

exponential distribution with a parameter X such that 1/A = 30

seconds

(Note that X is the mean number of calls per second, whereas 1/λ is the mean time, in seconds, between successive calls.)

0.6931= 20.8 secondsm =

λλλλ

(a) (i) P(X < 15) = F(15) = 15

30

1

1

−

− e

(ii) P(15 < X < 30)

)1(30

30

1

−

− e )1(15

30

1

−

− e−=

= 0.2387

(b)

= F(30)−F(15)

P(30 < X < 60)

)1(60

30

1

−

− e )1(30

30

1

−

− e=

= 0.2325

= F(60)−F(30)

−

(iii)

(iv) P(X>60) = 1 − F(60)

)1(160

30

1

−

−− e=

= 0.1353

= 0.3935

Documents

Lesson 1 Negative Exponential