Function of Random Variables

A random variabel \(\mathbf{X}\) with a function \(g(\cdot)\) applied to it results in a new random variable \(\mathbf{Y} = g(\mathbf{X})\). The probability density function (PDF) of the distribution cannot be applied directly to \(g(\cdot)\), instead, it must be derived using the cumulative distribution function (CDF), which results in a new CDF.

Given random variable \(\mathbf{X}\) with normal distribution \(\mathbf{X} \sim \mathcal{N}(\mu, \sigma^2)\), and a function \(g(\mathbf{X}) = a\mathbf{X}+b\). We have:

\[ \mathbf{Y} = a\mathbf{X}+b \\ \]

The CDF of \(\mathbf{Y}\), donated by \(F_\mathbf{Y}\):

\[ \begin{align*} F_\mathbf{Y}(y) = P(\mathbf{Y} < y) &= P(a\mathbf{X}+b)\\ &= P(\mathbf{X} < \frac{y-b}{a})\\ &= F_\mathbf{X}(\frac{y-b}{a}) \end{align*} \]

To obtain the PDF, take the derivative of the CDF:

\[ \begin{align*} f_\mathbf{y}(y) &= \frac{d}{dy} F_\mathbf{X}\left(\frac{y-b}{a}\right)\\ &= \frac{1}{a} f_\mathbf{X}\left(\frac{y-b}{a}\right) \end{align*} \]

Thus,

\[ \mathbf{Y} \sim \mathcal{N}(a\mu+b, a^2\sigma^2) \]

Example :

Given a random variable \(\mathbf{X}\) with standart normal distribution \(\mathbf{X} \sim \mathcal{N}(0, 1)\), and the function \(g(\mathbf{X}) = \mathbf{X}^2\). The CDF of \(\mathbf{Y}\) is:

\[ \begin{align*} F_\mathbf{Y}(y)=P(\mathbf{Y} < y)&=P(\mathbf{X}^2 < y)\\ &= P(-\sqrt{y} < \mathbf{x} < \sqrt{y})\\ &= \Phi(\sqrt{y}) - \Phi(-\sqrt{y}) \\ \end{align*} \]

Then, the PDF of \(\mathbf{Y}\) is:

\[ \begin{align*} f_\mathbf{Y}(y)&=\frac{d}{dy} F_\mathbf{Y}(y)\\ &= \Phi'(\sqrt{y})\cdot \frac{y^{-\frac{1}{2}}}{2} - \Phi'(-\sqrt{y})\cdot \frac{y^{-\frac{1}{2}}}{2}\\ &= y^{-\frac{1}{2}} \Phi'(\sqrt{y})\\ &= y^{-\frac{1}{2}} f_\mathbf{X}(\sqrt{y}) \end{align*} \]

where PDF of \(f_\mathbf{X}\) is :

\[ f_\mathbf{X}(x)=\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}} \]

Subtitute \(f_\mathbf{X}\) to \(f_\mathbf{Y}\),

\[ f_\mathbf{Y}(y) = \frac{y^{-\frac{1}{2}}e^{-\frac{x^2}{2}}}{\sqrt{2\pi}} \]