moment generating function basics

Question

if X,Y discrete random variables such that Y=aX+b , how do I prove $M_y(t)=e^{tb}M_X(at)$.

My attempt:

$M_y(t)$=$\sum$$e^{tax}.e^{tb}p(y)$. but then I notice this works iff p(x)=p(y) why is this the case? Thanks

Answer 1

Note that $$ M_Y(t)=Ee^{tY}=Ee^{t(aX+b)}=Ee^{tb}e^{taX}=e^{tb}Ee^{(ta)X}=e^{tb}M_{X}(at) $$

Answer 2

I think this is correct using the properties of expectation but I admit it has been a while. If I've made an error I will correct.

$\mathbb{M}_Y(t) =\mathbb{E} (e^{Yt} ) = \mathbb{E}(e^{(aX+b)t } ) =\mathbb{E}(e^{(aXt+bt) }) =\mathbb{E}(e^{aX }e^{bt})= e^{bt} \mathbb{E}(e^{(aXt) }) =e^{bt}\mathbb{M}_X(at) $