Continuous Random Variables: The ExponentialRandom VariableJuana [email protected] Department of StatisticsFeb 11, 2013J. Sanchez Continuous Random VariablesAnnouncementsProject is due on Wednesday, beginning of class. Please, read theinstructions first. You will get mistakes if you don’t.Today’s lecture has some examples already seen on Friday but nowin the context of the exponential. We will go over them fast for thatreason.Your homework 4 was returned today in lecture.J. Sanchez Continuous Random VariablesTodayRoss Chapter 5, Section 5.5. (not 5.5.1)I. Introduction. Phenomena that are exponentialII. Exponential Random Variable: pdf and CDFIII. Exponential random variable:Moment Generating FunctionIV. Expected Value and Variance of Exponential Random Variablesusing the definition.V. Any kth moment of the exponential random variable.VI. Percentiles of an exponential random variable and theinterquartile rangeVII. Expectation and Variance of linear combinations of exponentialrandom variablesVIII. The memoryless property of the exponential random variable.IX. Applications of the Exponential Distribution to queuingJ. Sanchez Continuous Random VariablesI. Introduction. Phenomena that are exponentialThe amount of time between events that occur at a Poisson rate λcan be modeled as an exponential random variable. For example,the time between earthquakes, or the time between wars, or the timebetween phone calls.The amount of time until an event occurs for the first time can alsobe modeled as an exponential random variable.Other variables that may not fit that description can also be foundto be properly fit by an exponential model.J. Sanchez Continuous Random VariablesNotesNotesNotesNotesII. Exponential Random Variable: pdf and CDFA random variable X is exponential with parameter λ if :Probability density function (pdf)f (x) =λe−λxx ≥ 0, λ > 00 otherwiseλ is the rate parameter.Cumulative Distribution Function (CDF)F (x ) = Prob(X ≤ x) =Zxt=0λe−λtdt= −e−λt|x0= 1 − e−λxx ≥ 0, λ > 0and F0(x) = λe−λX= f (x)We can use the cdf to compute probabilities of exponential randomvariables: P(a < X < b) = F (b) − F (a)J. Sanchez Continuous Random VariablesFinding probabilities for exponential random variablesExampleSuppose that the length of a phone call in minutes is an exponential r.v.with parameter λ =110. If someone arrives immediately ahead of you at apublic telephone booth, find the probability that you will have to wait(a) more than 10 minutesP(X > 10) =Z∞10110e−110xdx = −e110x |∞10= e−1= 0.368 = 1−F (10)(b) between 10 and 20 minutesP(10 < X < 20) = P(X < 20) − P(X < 10) = e−1− e−2= 0.233J. Sanchez Continuous Random VariablesFigure: The density of an exponential random variable with λ = 1/10. Theshaded area represents the Probability that X is larger than 10J. Sanchez Continuous Random VariablesFigure: The density of an exponential random variable with λ = 1/10. Theshaded area represents the Probability that X is between 10 and 20J. Sanchez Continuous Random VariablesNotesNotesNotesNotesExampleThe amount of time, in hours, that a computer functions before breakingdown is a continuous r.v. with pdff (x) =1100e−1100xx ≥ 0This is the exponential random variable.What is the cumulative distribution function for this continuous r.v.?F (x ) =Zx01100e−1100tdt = 1 − e−1100xCheck that we did the right thing by taking the derivative of yourcumulative distribution function and seeing whether it is equal tof(x).Compute P(4 ≤ X ≤ 10) using the cumulative distribution function.Interpret what your computation means in the context of thisproblem.P(4 ≤ X ≤ 10) = F (10)−F (4) = 1−e−110010−1 − e−11004= 0.0559J. Sanchez Continuous Random VariablesIII. Exponential random variable:Moment GeneratingFunctionMgfxMx(t) = E (etx)=R∞0etxλe−λxdx= λR∞0etxe−λxdx= λR∞0e−(λ−t)xdx=−1(λ−t)λe−(λ−t)x|∞0=λλ−tJ. Sanchez Continuous Random VariablesExerciseLet X be an exponential random variable. Using the Moment GeneratingFunction (Mgf) of X, find E(X) and Variance of X. The Mgf isMx(t) =λλ − tJ. Sanchez Continuous Random VariablesIn Class ExerciseMx(t) =λλ − tdMdt=λ(λ − t)2d2Mdt2=2(λ(λ − t)(λ − t)4=2λ(λ − t)3Evaluate both derivatives at t = 0.dMdt|t=0=λ(λ)2=1λd2Mdt2|t=0=2λ2Var(X ) =2λ2−1λ2=1λ2J. Sanchez Continuous Random VariablesNotesNotesNotesNotesµx= M0(t) |t=0=λ(λ−t)2|t=0=1λTo find the σ2xwe first find the E (X2).E (X2) = M00(t) |t=0=2λ(λ−t)3|t=0=2λ2So we can see that the variance of an exponential random variable is thesquare of the expected value.σ2x=2λ2− (1λ)2=1λ2J. Sanchez Continuous Random VariablesIV. Expected Value and Variance of Exponential RandomVariables using the definitionSummaryThe parameter λ is the reciprocal of the expected value. Soµx= E (X) =1λσ2x=1λ2The variance is the expected value squared.J. Sanchez Continuous Random VariablesProof of Expected ValueRecall integration by parts from calculusRudv = uv | −Rvduµx=Z∞0xλe−λxdx, make λe−λxdx = dv u = x=Z∞0udv= −xe−λx|∞0+Z∞0e−λxdx= 0 −e−λxλ|∞0=1λJ. Sanchez Continuous Random VariablesProof of VarianceRecall integration by parts from calculusRudv = uv | −Rvduσ2x= E (x2) − [µx]2You may integrate E (x2) by parts by making λe−λxdx = dv and u = x2.E (X2) =Z∞0x2λe−λxdx= −x2e−λx|∞0+Z∞02xe−λxdx= 0 +2λµx=2λ2σ2xσ2x=2λ2−1λ2=1λ2J. Sanchez Continuous Random VariablesNotesNotesNotesNotesV. Any kth moment of the exponential random variableusing the definitionNotice thatExk=k!λkWhy?Exk= λZ∞0xke−λxdx= λΓ(k + 1)1λk+1=1λkk!J. Sanchez Continuous Random VariablesThe gamma function appeared and will appear again.DefinitionΓ(α) =Z∞0e−yyα−1dy= e−yyα−1|∞0+Z∞0e−y(α − 1)yα−2dy= (α − 1)Z∞0e−yyα−2dy= (α − 1)Γ(α − 1)Recall integration by parts:Rudv = uv | +Rvdu We used integration byparts to do the integral for the gamma function above.J. Sanchez Continuous Random VariablesFigure: The density of an exponential random variable with λ = 1. Theexpected value is1λ= 1. The color line indicates the value of X where theexpected value is.J. Sanchez Continuous Random VariablesVI. Percentiles of an exponential random variable and theinterquartile rangeExampleThe amount of time, in hours, that a computer functions before breakingdown is a
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