Doubt in product of MGF(moment generating functions) of two random variables

Question

We are given two independent standard normal random variables X and Y. We need to find out the M.G.F of XY.

Let $u=XY$

$M_{XY}(t)=M_{u}(t)=E(e^{xyt}) =E(e^{ut})= \int_{u=-\infty}^{+\infty}e^{ut} f_u(u) du, u =xy $

Now how to split $f_{xy}(xy)$

I saw the solution in Finding the M.G.F of product of two random variables.

i recently started probability and statistics. I did not understand how $f_{XY}(XY)$ is split (here it is actually not joint distribution of 2 random variables right? it is pdf of single random variable XY). Kindly elaborate

Answer 1

First, don't switch between cases. Mathematics' symbols are case sensitive.

Further, it is preferred standard to use upper-case for random variables, and lower-case for scalar terms (constants, parameters, et cetera).

Now, by the definition, a moment generating function of random variable $U$ with parameter $t$ is the expectation of random variable $\mathrm e^{tU}$: $$\mathsf M_U(t)=\mathsf E(\mathrm e^{tU})$$

When $U$ is defined as the product of random variables $X$ and $Y$, we therefore have: $$\mathsf M_{XY}(t)=\mathsf E(\mathrm e^{tXY})$$

By the Law of Total Expectation, and the fact that the random variables are independent.

$$\begin{align}\mathsf M_{XY}(t)&=\mathsf E(\mathsf E(\mathrm e^{(tX)Y}\mid X))\\&=(2\pi)^{-1}\int_\Bbb R \mathrm e^{-x^2/2}\int_\Bbb R \mathrm e^{-y^2/2}\cdot\mathrm e^{txy}~\mathrm d y~\mathrm d x\\&=\sqrt{2\pi~}^{-1}\int_\Bbb R \mathrm e^{-x^2/2}\cdot \mathrm e^{t^2x^2/2}~\mathrm d x\\&=\sqrt{2\pi~}^{-1}\int_\Bbb R\mathrm e^{-(1-t^2)x^2/2}~\mathrm d x\\&=\sqrt{1-t^2~~}^{-1}\quad\big[\Re(t^2)<1\big] \end{align}$$

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Taking that a few smaller steps at a time.

$$\mathsf M_{XY}(t)=\mathsf E(\mathsf E(\mathrm e^{(tX)Y}\mid X))\\=\mathsf E(\mathsf M_Y(tX))$$

Now, $Y\sim\mathcal{N}(0,1^2)$ (standard normal), so $\mathsf M_Y(s)=\mathrm e^{s^2/2}$. If you do not already have that available$$\begin{align}\mathsf M_Y(s)&=\mathsf E(\mathrm e^{sY})\\&=\int_\Bbb R\mathrm e^{sy}\cdot\sqrt{2\pi~}^{-1}\mathrm e^{-y^2/2}\mathrm d y\\&=\sqrt{2\pi}^{-1}\int_\Bbb R\mathrm e^{sy-y^2/2}\mathrm d y\\&=\mathrm e^{s^2/2}\end{align}$$

... Okay. Anyway, using that:

$$\begin{align}\mathsf M_{XY}(t)&=\mathsf E(\mathrm e^{(tX)^2/2})\\&=\int_\Bbb R\mathrm e^{t^2x^2/2}\cdot\sqrt{ 2\pi~}^{-1}~\mathrm e^{-x^2/2}~\mathrm d x\\&=\sqrt{ 2\pi~}^{-1}\int_\Bbb R\mathrm e^{-(1-t^2)x^2/2}\mathrm d x\\&=\sqrt{1-t^2~}^{-1}\quad\big[\Re(t^2)<1\big]\end{align}$$