# PDF, mean and variance of product of two dependent random variables

Consider two dependent random variables

$$X \sim \text{Ga}(\alpha,\beta) \quad \quad \quad \quad Y|X \sim \text{N} \Big( 0,\frac{1}{X} \Big).$$

How do I compute the PDF, mean and variance of the random variable $$Z = X \times Y$$?

## 2 Answers

The moments are obtained using iterated moment formulae. Using the law of iterated expectation you have:

\begin{aligned} \mathbb{E}(Z) &= \mathbb{E}(\mathbb{E}(Z|X)) \\[6pt] &= \mathbb{E}(\mathbb{E}(X \cdot Y|X)) \\[6pt] &= \mathbb{E}(X \cdot \mathbb{E}(Y|X)) \\[6pt] &= \mathbb{E}(X \cdot 0) \\[6pt] &= \mathbb{E}(0) = 0. \\[6pt] \end{aligned}

Using the law of iterated variance you have:

\begin{aligned} \mathbb{V}(Z) &= \mathbb{V}(\mathbb{E}(Z|X)) + \mathbb{E}(\mathbb{V}(Z|X)) \\[6pt] &= \mathbb{V}(\mathbb{E}(X \cdot Y|X)) + \mathbb{E}(\mathbb{V}(X \cdot Y|X)) \\[6pt] &= \mathbb{V}(X \cdot \mathbb{E}(Y|X)) + \mathbb{E}(X^2 \cdot \mathbb{V}(Y|X)) \\[6pt] &= \mathbb{V}(X \cdot 0) + \mathbb{E} \Big( X^2 \cdot \frac{1}{X} \Big) \\[6pt] &= \mathbb{V}(0) + \mathbb{E}(X) \\[6pt] &= \frac{\alpha}{\beta}. \\[6pt] \end{aligned}

To find the PDF of $$Z$$ we can start by observing that this distribution is symmetric around the point $$z=0$$, so we can proceed for the positive case and then use symmetry for the rest. To find the distribution function you can use the law of total probability for any $$z \geqslant 0$$ to get:

\begin{aligned} F_Z(z) = \mathbb{P}(Z \leqslant z) &= \mathbb{P}(X \cdot Y \leqslant z) \\[6pt] &= \int \limits_0^\infty \mathbb{P}(X \cdot Y \leqslant z|X=x) \cdot f_X(x) \ dx \\[6pt] &= \int \limits_0^\infty \mathbb{P} \Big( Y \leqslant \frac{z}{x} \Big| X=x \Big) \cdot f_X(x) \ dx \\[6pt] &= \int \limits_0^\infty \Phi \Big( \frac{z}{\sqrt{x}} \Big) \cdot \text{Ga}(x|\alpha, \beta) \ dx. \\[6pt] \end{aligned}

For all $$z \geqslant 0$$ the density corresponding to this distribution function is:

\begin{aligned} f_Z(z) = \frac{dF_Z}{dz}(z) &= \frac{d}{dz} \int \limits_0^\infty \Phi \Big( \frac{z}{\sqrt{x}} \Big) \cdot \text{Ga}(x|\alpha, \beta) \ dx \\[6pt] &= \int \limits_0^\infty \frac{\partial}{\partial z} \Phi \Big( \frac{z}{\sqrt{x}} \Big) \cdot \text{Ga}(x|\alpha, \beta) \ dx \\[6pt] &= \int \limits_0^\infty \frac{1}{\sqrt{x}} \cdot \phi \Big( \frac{z}{\sqrt{x}} \Big) \cdot \text{Ga}(x|\alpha, \beta) \ dx \\[6pt] &= \int \limits_0^\infty x^{-1/2} \cdot \frac{1}{\sqrt{2 \pi}} \exp \Big( -\frac{1}{2} \frac{z^2}{x} \Big) \cdot \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} \exp(- \beta x) \ dx \\[6pt] &= \frac{1}{\sqrt{2 \pi}} \frac{\beta^\alpha}{\Gamma(\alpha)} \int \limits_0^\infty x^{\alpha -3/2} \exp \Big( - \beta x -\frac{1}{2} \frac{z^2}{x} \Big) \ dx. \\[6pt] \end{aligned}

Using symmetry, we then have:

$$f_Z(z) = \frac{1}{\sqrt{2 \pi}} \frac{\beta^\alpha}{\Gamma(\alpha)} \int \limits_0^\infty x^{\alpha -3/2} \exp \Big( - \beta x -\frac{1}{2} \frac{z^2}{x} \Big) \ dx \quad \quad \quad \text{for } z \in \mathbb{R}.$$

There is no closed-form expression for this integral, so the density cannot be simplified any further. The density can be computed numerically using standard methods of numerical computation of integrals.

Use the multivariate change of variable formula.

Let $Z = XY$ and $V = Y$ so that $Z/V = X$.

Then we have the multivariate change of variable formula, as shown here:

$f_{Z,V}(z,v) = f_{X,Y}(x = \frac{Z}{V}, y = V) * |J|$ where $|J|$ is the Jacobian.

Now, before we do that, we need to figure out what $f_{X,Y}(x,y)$ is.

First, note that the density of Y is actually a conditional density, that is $Y|X \sim N(0, \frac{1}{X})$

Since we know $f_{Y|X}(y) = \frac{f_{X,Y}(x,y)}{f_{X}(x)}$ we can simply multiply $f_{Y|X}(y)$ (which is given to us) by $f_{X}(x)$ (which is also given as a Gamma($\alpha, \beta$)) to obtain the joint density. Then, using the above steps (which I am assuming you probably know how to do) you can obtain the joint density of Z and V.

Then, integrate the resulting joint density over all values of V to get the marginal of Z, $f_Z(z)$, from which you can then obtain the desired moments needed for the mean and variance of Z.

Hope this helps.