Distribution Of The Negative of A Normally Distributed Random Variable

This post derives the distribution for an random variable that appears in the derivation of a result in another post.

In this post, I derive the distribution of the negative of a normally distributed random variable. Let’s say we have \(X \sim N(\mu, \sigma^2)\), and we want to find the distribution of \(Y=-X\).

I start with the identity: \begin{align} P(Y \le y) = P(-X \le y) = P(X \ge -y) = 1 - P(X \le -y) \end{align}

Since \(X \sim N(\mu, \sigma^2)\), we have \begin{align} P(X \le -y) = \int^{-y}_{-\infty} \frac{1}{\sqrt{2\pi\sigma^2}} \exp^{-\frac{(x-\mu)^2}{2\sigma^2}} dx \end{align}

Which gives us \begin{align} P(Y \le y) = 1 - P(X \le -y) = 1 - \int^{-y}_{-\infty} \frac{1}{\sqrt{2\pi\sigma^2}} \exp^{-\frac{(x-\mu)^2}{2\sigma^2}} dx \end{align}

Using Leibniz integral rule, we have

\begin{align} f_{Y}(y) = \frac{\partial P(Y \le y)}{\partial y} = & - \frac{1}{\sqrt{2\pi\sigma^2}} \exp^{-\frac{(-y-\mu)^2}{2\sigma^2}} \left(\frac{\partial (-y)}{\partial y}\right) \\ = & - \frac{1}{\sqrt{2\pi\sigma^2}} \exp^{-\frac{(-y-\mu)^2}{2\sigma^2}} (-1) \\ = & \frac{1}{\sqrt{2\pi\sigma^2}} \exp^{-\frac{(-y-\mu)^2}{2\sigma^2}} \\ = & \frac{1}{\sqrt{2\pi\sigma^2}} \exp^{-\frac{(y+\mu)^2}{2\sigma^2}} \\ = & \frac{1}{\sqrt{2\pi\sigma^2}} \exp^{-\frac{(y-(-\mu))^2}{2\sigma^2}} \end{align}

This is nothing but the density function for a Normally distributed random variable with mean \(-\mu\) and variance \(\sigma^2\), hence \(Y = N(-\mu, \sigma^2)\)