<< Chapter < Page Chapter >> Page >

Memorylessness of the exponential distribution

In [link] recall that the amount of time between customers is exponentially distributed with a mean of two minutes ( X ~ Exp (0.5)). Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a customer to arrive within the next minute. With the exponential distribution, this is not the case–the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the memoryless property . Specifically, the memoryless property says that

P ( X > r + t | X > r ) = P ( X > t ) for all r ≥ 0 and t ≥ 0

For example, if five minutes has elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using r = 5 and t = 1 in the foregoing equation.

P ( X >5 + 1 | X >5) = P ( X >1) = e ( 0.5 ) ( 1 ) ≈ 0.6065.

This is the same probability as that of waiting more than one minute for a customer to arrive after the previous arrival.

The exponential distribution is often used to model the longevity of an electrical or mechanical device. In [link] , the lifetime of a certain computer part has the exponential distribution with a mean of ten years ( X ~ Exp (0.1)). The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is P ( X >17| X >10) = P ( X >7) = 0.4966.

Refer to [link] where the time a postal clerk spends with his or her customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that he or she will spend at least an additional three minutes with the postal clerk?

The decay parameter of X is m = 1 4 = 0.25, so X Exp (0.25).

The cumulative distribution function is P ( X < x ) = 1 – e –0.25 x .

We want to find P ( X >7| X >4). The memoryless property says that P ( X >7| X >4) = P ( X >3), so we just need to find the probability that a customer spends more than three minutes with a postal clerk.

This is P ( X >3) = 1 – P ( X <3) = 1 – (1 – e –0.25⋅3 ) = e –0.75 ≈ 0.4724.

This graph shows an exponential distribution. The graph slopes downward. It begins at the point (0, 0.25) on the y-axis and approaches the x-axis at the right edge of the graph. The region under the graph to the right of x = 3 is shaded to represent P(x > 3) = 0.4724.

1–(1–e^(–0.25*2)) = e^(–0.25*2).

Try it

Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. If a bulb has already lasted 12 years, find the probability that it will last a total of over 19 years.

Let T = the lifetime of the light bulb. Then T Exp ( 1 8 ) .

The cumulative distribution function is P ( T < t ) = 1 − e - t 8

We need to find P ( T >19| T = 12). By the memoryless property ,

P ( T >19| T = 12) = P ( T >7) = 1 – P ( T <7) = 1 – (1 – e –7/8 )= e -7/8 ≈ 0.4169.

1 – (1 – e^(–7/8)) = e^(–7/8).

Relationship between the poisson and the exponential distribution

There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of μ units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean λ = 1/μ. Recall from the chapter on Discrete Random Variables that if X has the Poisson distribution with mean λ , then P ( X = k ) = λ k e λ k ! . Conversely, if the number of events per unit time follows a Poisson distribution, then the amount of time between events follows the exponential distribution. ( k ! = k *( k -1*)( k –2)*( k -3)…3*2*1)

Questions & Answers

how do you translate this in Algebraic Expressions
linda Reply
why surface tension is zero at critical temperature
Shanjida
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!
QuizOver.com Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Introduction to statistics i - stat 213 - university of calgary - ver2015revb. OpenStax CNX. Oct 21, 2015 Download for free at http://legacy.cnx.org/content/col11874/1.3
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Introduction to statistics i - stat 213 - university of calgary - ver2015revb' conversation and receive update notifications?

Ask