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}!

2 thoughts on “Even probability density

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