In: Statistics and Probability
here for exponential distribution mgf =M(t)=/(-t)
1st derivative of mgf =M1(t) =(d/dt)*/(-t) =/(-t)2
2nd derivative of mgf =M2(t) =(d/dt)*/(-t)2 =*2!/(-t)3
therefore kth derivative of mgf Mk(t) =(dk/dt)(/(-t)) =k!* /(-t)k+1
hence E(Yk) = Mk(0) =k!* /(-0)k+1 =k!*/()k+1 =k!/k