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Show that the result given in Exercise 3.158 also holds for continuous random variables. That is, show that, if Y is a random variable with moment-generating function m(1) and U is given by U = aY + b, the moment-generating function of U is em(at). If Y has mean u and variance o?, use the moment-generating function of U to derive the mean and variance of U. Recall: my(t) = E(etY) and whent = 0 then my(0) = E(e°Y) = E(1) = 0 and d*m(t)] = tk- dik and M) = F(*) = E (*) %3D = E (X et×). Similarly, %3D diz M(t) () = E (X² c²X). = E %3D

Show that the result given in Exercise 3.158 also holds for continuous random variables. That is, show that, if Y is a random variable with moment-generating function m(1) and U is given by U = aY + b, the moment-generating function of U is em(at). If Y has mean u and variance o?, use the moment-generating function of U to derive the mean and variance of U. Recall: my(t) = E(etY) and whent = 0 then my(0) = E(e°Y) = E(1) = 0 and d*m(t)] = tk- dik and M) = F(*) = E (*) %3D = E (X et×). Similarly, %3D diz M(t) () = E (X² c²X). = E %3D

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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4.137 Show that the result given in Exercise 3.158 also holds for continuous random variables. That is, show that, if Y is a random variable with moment-generating function m(t) and U is given by U = aY +b, the moment-generating function of U is em(at). If Y has mean u and variance o?, use the moment-generating function of U to derive the mean and variance of U. Recall: my(t) = E(etY) and when t = 0 then my(0) = E(eoY) = E(1) = 0 and d*m(t)] dık and d dt - E () = E (X e² *). Similarly, M(1) = E (**) = E E (X²e+x). = 1 Hence, in general we get d" M(t) = E (e* *) %3D dt" dt" uP dt" = E - E (X" e² ×). If we set t = 0 in the nth derivative, we get d" = E (X" e²×)\,-o = E (X"). It=0
Transcribed Image Text:4.137 Show that the result given in Exercise 3.158 also holds for continuous random variables. That is, show that, if Y is a random variable with moment-generating function m(t) and U is given by U = aY +b, the moment-generating function of U is em(at). If Y has mean u and variance o?, use the moment-generating function of U to derive the mean and variance of U. Recall: my(t) = E(etY) and when t = 0 then my(0) = E(eoY) = E(1) = 0 and d*m(t)] dık and d dt - E () = E (X e² *). Similarly, M(1) = E (**) = E E (X²e+x). = 1 Hence, in general we get d" M(t) = E (e* *) %3D dt" dt" uP dt" = E - E (X" e² ×). If we set t = 0 in the nth derivative, we get d" = E (X" e²×)\,-o = E (X"). It=0
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