Exponential Distribution | Memoryless Random Variable

The exponential distribution is one of the widely used continuous distributions. It is often used to model the time elapsed between events. HOMEVIDEOSCALCULATORCOMMENTSCOURSESFORINSTRUCTORLOGIN FORINSTRUCTORSSignInEmail:Password:Forgotpassword? ←previous next→ VideoAvailable4.2.2ExponentialDistribution Theexponentialdistributionisoneofthewidelyusedcontinuousdistributions.Itisoftenusedto modelthetimeelapsedbetweenevents.Wewillnowmathematicallydefinetheexponentialdistribution, andderiveitsmeanandexpectedvalue.Thenwewilldeveloptheintuitionforthedistributionand discussseveralinterestingpropertiesthatithas. Acontinuousrandomvariable$X$issaidtohaveanexponentialdistributionwithparameter $\lambda>0$,shownas$X\simExponential(\lambda)$,ifitsPDFisgivenby \begin{equation} \nonumberf_X(x)=\left\{ \begin{array}{ll} \lambdae^{-\lambdax}&\quadx>0\\ 0&\quad\textrm{otherwise} \end{array}\right. \end{equation} Figure4.5showsthePDFofexponentialdistributionforseveralvaluesof$\lambda$. Fig.4.5-PDFoftheexponentialrandomvariable. Itisconvenienttousetheunitstepfunctiondefinedas \begin{equation} \nonumberu(x)=\left\{ \begin{array}{ll} 1&\quadx\geq0\\ 0&\quad\textrm{otherwise} \end{array}\right. \end{equation} sowecanwritethePDFofan$Exponential(\lambda)$randomvariableas $$f_X(x)=\lambdae^{-\lambdax}u(x).$$ LetusfinditsCDF,meanandvariance.For$x>0$,wehave $$F_X(x)=\int_{0}^{x}\lambdae^{-\lambdat}dt=1-e^{-\lambdax}.$$ SowecanexpresstheCDFas $$F_X(x)=\big(1-e^{-\lambdax}\big)u(x).$$ Let$X\simExponential(\lambda)$.Wecanfinditsexpectedvalueasfollows,usingintegrationbyparts: $EX$ $=\int_{0}^{\infty}x\lambdae^{-\lambdax}dx$ $=\frac{1}{\lambda}\int_{0}^{\infty}ye^{-y}dy$ $\textrm{choosing$y=\lambdax$}$ $=\frac{1}{\lambda}\bigg[-e^{-y}-ye^{-y}\bigg]_{0}^{\infty}$ $=\frac{1}{\lambda}.$ Nowlet'sfindVar$(X)$.Wehave $EX^2$ $=\int_{0}^{\infty}x^2\lambdae^{-\lambdax}dx$ $=\frac{1}{\lambda^2}\int_{0}^{\infty}y^2e^{-y}dy$ $=\frac{1}{\lambda^2}\bigg[-2e^{-y}-2ye^{-y}-y^2e^{-y}\bigg]_{0}^{\infty}$ $=\frac{2}{\lambda^2}.$ Thus,weobtain $$\textrm{Var}(X)=EX^2-(EX)^2=\frac{2}{\lambda^2}-\frac{1}{\lambda^2}=\frac{1}{\lambda^2}.$$ If$X\simExponential(\lambda)$,then$EX=\frac{1}{\lambda}$andVar$(X)=\frac{1}{\lambda^2}$. Aninterestingpropertyoftheexponentialdistributionisthatitcanbeviewedasacontinuousanalogue ofthegeometricdistribution.Toseethis,recalltherandomexperimentbehindthegeometricdistribution: youtossacoin(repeataBernoulliexperiment)untilyouobservethefirstheads(success).Now,suppose thatthecointossesare$\Delta$secondsapartandineachtosstheprobabilityofsuccessis$p=\Delta\lambda$. Alsosupposethat$\Delta$isverysmall,sothecointossesareveryclosetogetherintimeandtheprobability ofsuccessineachtrialisverylow.Let$X$bethetimeyouobservethefirstsuccess.Wewillshowinthe SolvedProblemssectionthatthedistributionof$X$convergesto$Exponential(\lambda)$as$\Delta$ approacheszero. Togetsomeintuitionforthisinterpretationoftheexponentialdistribution,supposeyouarewaiting foraneventtohappen.Forexample,youareatastoreandarewaitingforthenextcustomer.Ineach millisecond,theprobabilitythatanewcustomerentersthestoreisverysmall.Youcanimaginethat, ineachmillisecond,acoin(withaverysmall$P(H)$)istossed,andifitlandsheadsanewcustomers enters.Ifyoutossacoineverymillisecond,thetimeuntilanewcustomerarrivesapproximatelyfollows anexponentialdistribution. Theaboveinterpretationoftheexponentialisusefulinbetterunderstandingthepropertiesofthe exponentialdistribution.Themostimportantofthesepropertiesisthattheexponentialdistribution ismemoryless.Toseethis,thinkofanexponentialrandomvariableinthesenseoftossingalot ofcoinsuntilobservingthefirstheads.Ifwetossthecoinseveraltimesanddonotobserveaheads, fromnowonitislikewestartalloveragain.Inotherwords,thefailedcointossesdonotimpact thedistributionofwaitingtimefromnowon.Thereasonforthisisthatthecointossesareindependent. Wecanstatethisformallyasfollows: $$P(X>x+a|X>a)=P(X>x).$$ If$X$isexponentialwithparameter$\lambda>0$,then$X$isamemorylessrandomvariable,thatis $$P(X>x+a\hspace{5pt}|\hspace{5pt}X>a)=P(X>x),\hspace{10pt}\textrm{for}a,x\geq0.$$ Fromthepointofviewofwaitingtimeuntilarrivalofacustomer,thememorylesspropertymeansthat itdoesnotmatterhowlongyouhavewaitedsofar.Ifyouhavenotobservedacustomeruntiltime$a$, thedistributionofwaitingtime(fromtime$a$)untilthenextcustomeristhesameaswhenyoustartedattimezero. Letusprovethememorylesspropertyoftheexponentialdistribution. \begin{align}%\label{} \nonumberP(X>x+a|X>a)&=\frac{P\big(X>x+a,X>a\big)}{P(X>a)}\\ \nonumber&=\frac{P(X>x+a)}{P(X>a)}\\ \nonumber&=\frac{1-F_X(x+a)}{1-F_X(a)}\\ \nonumber&=\frac{e^{-\lambda(x+a)}}{e^{-\lambdaa}}\\ \nonumber&=e^{-\lambdax}\\ \nonumber&=P(X>x).\\ \end{align} ←previous next→ TheprintversionofthebookisavailablethroughAmazonhere.