Compute the distribution of $f(X)$ if $X\sim\mathcal{N}(\mu,\sigma^2)$ and $f$ a function

Question

Let $X\sim\mathcal{N}(\mu, \sigma^2)$. We know its pdf being $$ \frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac12\left(\frac{x-\mu}{\sigma}\right)^2\right). $$

Given a function $f$ (possibly non linear, can be as smooth as you wish), is there a way to know the pdf of $f(X)$ as well ?

Answer 1

$X\sim N(\mu, \sigma^2)$. Then,

• p.d.f. $f_X(x) = \frac{1}{\sigma}\varphi\left(\frac{x-\mu}{\sigma}\right)$;
• c.d.f. $F_X(x) = \Phi\left(\frac{x-\mu}{\sigma}\right)$,

where $\varphi(\cdot)$ and $\Phi(\cdot)$ are the p.d.f and c.d.f. of a standard normal distribution $N(0, 1)$ respectively.

Let $Y=f(X)$. Then, the c.d.f. of $Y$ is

$$ \begin{aligned} F_Y(y) &= \text{Pr}\left(Y \leq y\right) = \text{Pr}\left(f(X) \leq y\right) = \\ &= \text{Pr}\left(X \leq f^{-1}(y)\right) = F_X\left(f^{-1}(y)\right) = \\ &= \Phi\left(\frac{f^{-1}(y)-\mu}{\sigma}\right). \end{aligned} $$

Now, it is possible to calculate the p.d.f. of $Y$ (its existence depends on the properties of $f(\cdot)$ function).

$$ \begin{aligned} f_Y(y) &= F'_Y(y) = \Phi'\left(\frac{f^{-1}(y)-\mu}{\sigma}\right) = \\ &= \frac{1}{\sigma}\varphi\left(\frac{f^{-1}(y)-\mu}{\sigma}\right)\left(f^{-1}(y)\right)' = \\ &= \frac{1}{\sigma}\varphi\left(\frac{f^{-1}(y)-\mu}{\sigma}\right)\frac{1}{f'\left(f^{-1}(y)\right)}. \end{aligned} $$