# Let $X$ and $Y$ be independent and identically distributed random variables with moment generating function then $E(\dfrac{e^{tX}}{e^{tY}})$

Let $X$ and $Y$ be independent and identically distributed random variables with moment generating function $M(t)=E(e^{tX});\ \ -\infty<t<\infty$ then $E(\dfrac{e^{tX}}{e^{tY}})$ equals ?

$(A)=M(t)M(-t)$

$(B)=1$

$(C)=(M(t))^2$

$(D)=\frac{M(t)}{M(-t)}$

My input: Since its given that random variables are i.i.d so $E(\dfrac{\require{\cancel}\cancel{e^{tX}}}{\cancel{e^{tY}}})=E(1)=1$

can I do that?

• What happens e.g. for $X:=1$ (constant random variable) and $Y:=2$ (also constant random variable)? Does the expression equal $1$? – saz Feb 4 '18 at 16:38
• You assumed $X=Y$ with probability $1$ when you did the cancellation. That is not necessarily true. – StubbornAtom Feb 4 '18 at 16:42
• @saz Oh yes i was thinking totally wrong. I took a case when they both take same probability. Silly me. – Daman Feb 4 '18 at 16:42
• @StubbornAtom i am showing new work wait. – Daman Feb 4 '18 at 16:42
• Another example: $X,Y \sim {\cal N}(0,1)$ iid. – GNUSupporter 8964民主女神 地下教會 Feb 4 '18 at 17:28

$X,Y$ are independent of each other, so $e^{tX}$ and $e^{tY}$ are independent of each other. This gives $M(t) = E[e^{tX}] = E[e^{tY}]$
Moreover, $M(t) = E[e^{tX}]$ and $M(-t) = E[e^{-tY}]$ are independent of each other, this gives $$E\left[\frac{e^{tX}}{e^{tY}}\right] = E[e^{tX} \cdot e^{-tY}] = E[e^{tX}] \cdot E[e^{-tY}] = M(t) M(-t).$$
No you cannot do that. What you can do: $$E\left(\frac{e^{t X}}{e^{t Y}}\right)=E\left(e^{t X} e^{-t Y}\right) = E(e^{t X}) E(e^{-t Y}) = M(t) M(-t)$$ where the second step is due to independence. As the random variables are identically distributed $M_X(t) = M_Y(t) =M(t)$.
Observe that since $X$ and $Y$ are independent, it follows that $e^{tX}$ and $e^{-tY}$ are independent for $t\in\mathbb{R}$. Thus $$Ee^{tX}e^{-tY}=Ee^{tX}E{e^{-tY}}=M_{X}(t)M_{Y}(-t)=M_{X}(t)M_X(-t)$$ since $X, Y$ are equal in distribution.