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 =...

Extracted 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