Exponential distribution moment generating function to find ...

For the exponential distributed random variable Y, one can show that the moments E(Yn) are E(Yn)=n!λn where E(Y)=1λ. MathematicsStackExchangeisaquestionandanswersiteforpeoplestudyingmathatanylevelandprofessionalsinrelatedfields.Itonlytakesaminutetosignup. Signuptojointhiscommunity Anybodycanaskaquestion Anybodycananswer Thebestanswersarevotedupandrisetothetop Home Public Questions Tags Users Unanswered Teams StackOverflowforTeams –Startcollaboratingandsharingorganizationalknowledge. CreateafreeTeam WhyTeams? Teams CreatefreeTeam Teams Q&Aforwork Connectandshareknowledgewithinasinglelocationthatisstructuredandeasytosearch. Learnmore Exponentialdistributionmomentgeneratingfunctiontofindthemean AskQuestion Asked 8years,4monthsago Modified 8years,4monthsago Viewed 28ktimes 3 3 $\begingroup$ Withmean=2withexponentialdistribution Calculate $E(200+5Y^2+4Y^3)=432$ $E(200)=200$ $E(5Y^2)=5E(Y^2)=5(8)=40$ $E(4Y^3)=4E(Y^3)=4(48)=192$ $E(Y^2)=V(Y)+[E(Y)]^2=2^2+2^2$ $E(Y^3)=m_Y^3(0)=48(1-2(0))^{-4}=48$ isthisright? probabilityprobability-distributions Share Cite Follow askedDec7,2013at16:59 afsdfdfsafafsdfdfsaf 1,53733goldbadges2525silverbadges4949bronzebadges $\endgroup$ 1 $\begingroup$ Pleasehelpwiththis $\endgroup$ – afsdfdfsaf Dec8,2013at3:28 Addacomment  |  2Answers 2 Sortedby: Resettodefault Highestscore(default) Datemodified(newestfirst) Datecreated(oldestfirst) 10 +50 $\begingroup$ Let$Y\sim\operatorname{Exp}(\lambda)$thatiswithpdf $$ f(y)= \begin{cases} \lambda\operatorname{e}^{-\lambday}&y\ge0\\ 0&y<0 \end{cases} $$ withparameter$\lambda>0$. Themomentscanbeevaluateddirectlyorusingthemomentgeneratingfunction. DirectEvaluation $$ \begin{align} \Bbb{E}(Y^n)&=\int_{-\infty}^{\infty}y^nf(y)\operatorname{d}y=\int_{0}^{\infty}y^n\lambda\operatorname{e}^{-\lambday}\operatorname{d}y\\ &=\frac{1}{\lambda^n}\int_{0}^{\infty}t^n\operatorname{e}^{-t}\operatorname{d}t\qquad\qquad\qquad\qquad\qquad(\text{putting}\lambday=t)\\ &=\frac{1}{\lambda^n}\Gamma(n+1)=\frac{n!}{\lambda^n}\tag1 \end{align} $$ usingtheGammaFunction$\Gamma(x)=\int_0^\inftyt^{x-1}{\rme}^{-t}\,{\rmd}t$andthat$\Gamma(n+1)=n!$foranypositiveinteger$n$. UsingtheMomentGeneratingFunction Themomentgeneratingfunction(mgf)is $$ \begin{align} M_Y(t)=\Bbb{E}\left(\operatorname{e}^{tY}\right)&=\int_{-\infty}^{\infty}\operatorname{e}^{ty}f(y)\operatorname{d}y=\int_{0}^{\infty}\operatorname{e}^{ty}\lambda\operatorname{e}^{-\lambday}\operatorname{d}y\\ &=\lambda\int_{0}^{\infty}\operatorname{e}^{(t-\lambda)y}\operatorname{d}y=\frac{\lambda}{t-\lambda}\left[\operatorname{e}^{(t-\lambda)y}\right]_{0}^{\infty}=\frac{\lambda}{\lambda-t}\hspace{1cm}\text{for}t0$,thenintegratingbypartsgives$E(Y^n)=[-x^ne^{-\lambdax}]_0^\infty+\int_0^\inftynx^{n-1}e^{-\lambdax}=\frac{n}{\lambda}E(Y^{n-1})=\frac{n}{\lambda}\cdot\frac{(n-1)!}{\lambda^{n-1}}=\frac{n!}{\lambda^n}$. $\endgroup$ – universalset Dec10,2013at17:06 Addacomment  |  YourAnswer ThanksforcontributingananswertoMathematicsStackExchange!Pleasebesuretoanswerthequestion.Providedetailsandshareyourresearch!Butavoid…Askingforhelp,clarification,orrespondingtootheranswers.Makingstatementsbasedonopinion;backthemupwithreferencesorpersonalexperience.UseMathJaxtoformatequations.MathJaxreference.Tolearnmore,seeourtipsonwritinggreatanswers. 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