# Even probability density

Here’s another interesting problem I came across on math.stackexchange: Suppose ${X}$ is a continuous random variable with density ${f_X}$ and ${Y=XZ}$ where ${X}$ and ${Z}$ are independent and

$\displaystyle \mathbb P(Z=1) = \mathbb P(Z=-1) = \frac12.$

Show that ${Y}$ is a continuous random variable, and that the density of ${Y}$ is even, i.e. ${f_Y(y) = f_Y(-y)}$ for each ${y\in\mathbb R}$.

Solution: If ${y\in\mathbb R}$ then

\displaystyle \begin{aligned} F_Y(y) &= \mathbb P(Y \leqslant y)\\ &= \mathbb P(XZ\leqslant y)\\ &= \mathbb P(XZ\leqslant y, Z=1) + \mathbb P(XZ\leqslant y, Z=-1)\\ &= \mathbb P(XZ\leqslant y\mid Z=1)\mathbb P(Z=1) + \mathbb P(XZ\leqslant y\mid Z=1)\mathbb P(Z=-1)\\ &= \frac12\left(\mathbb P(X\leqslant y) + \mathbb P(-X\leqslant y) \right)\\ &= \frac12\left(F_X(y) + 1 - F_X(-y) \right). \end{aligned}

As the distribution function of ${Y}$ is differentiable, ${Y}$ is continuous, so we may compute the density by differentiation:

\displaystyle \begin{aligned} f_Y(y) &= \frac{\mathsf d}{\mathsf dy} F_Y(y)\\ &= \frac{\mathsf d}{\mathsf dy}\left[\frac12\left(F_X(y) + 1 - F_X(-y) \right)\right]\\ &= \frac12\left(f_X(y) + f_X(-y)\right). \end{aligned}

By symmetry it is evident that ${f_Y(y) = f_Y(-y)}$.

Note that this does not depend on the distribution of ${X}$!