# Moment generating function (MGF) of the ratio distribution $\displaystyle\frac{X}{Y}$

If we know the moment generating functions (MGFs) of the random variables $X$ and $Y$ to be $M_{X}(s)$ and $M_{Y}(s)$, respectively. The MGF of the sum $X+Y$ will $M_{X}(s) \cdot M_{Y}(s)$.

So what is the MGF of the ratio distribution $\displaystyle\frac{X}{Y}$?

• To be pedantic, the formula you gave holds when $X$ and $Y$ are independent, but not in general. As for your actual question, I don't think there are any simple rules in general. If $X$ and $Y$ are independent you can try to use some sort of convolution formula and evaluate some complicated integral, but in general it's one of those problems that in practice you have to tackle on a case-by-case basis unfortunately. Jun 21 '16 at 1:28

## 1 Answer

It is nowhere near as neat. \begin{align}\mathsf M_{X/\, Y}(t) =&~ \mathsf E(e^{tX/\, Y}) \\[1ex]=&~\mathsf E(\mathsf E(e^{tX /\, Y}\mid Y)) \\[1ex]=&~ \mathsf E(\mathsf M_{X\mid Y}(t/Y))\end{align}

Now, if $X$ and $Y$ are independent (as required for the summation formula you gave) then this becomes: \begin{align}\mathsf M_{X/\, Y}(t) =&~ \mathsf E(\mathsf M_X(t/Y)) \end{align}

But it won't get any simpler than that except perhaps for special cases.