Question

Let X and Y be independent random variables with moment-generating function MX(t) and MY (t) respectively. If a and b are constants and U = aX +bY , show that the moment-generating function of U is MU (t) = MX(at).MY (bt).

Answer 1

$M_X(t) = E(e^{tX}) \\ M_Y(t) = E(e^{tY})$

X and Y are independent

$U = aX +bY \\ M_U(t) = E(e^{tU}) = E[e^{t(aX+bY)}] \\ = E[e^{(at)X} \times e^{(bt)Y}] \\ = E[e^{(at)X}] \times E[e^{(bt)Y}] \\ M_U(t) = M_X(at) \times M_Y(bt)$