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