Example 5. The random variable $X$ is given by the probability density function

$$
p(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-x^{2} / 2 \sigma^{2}}
$$

Find the probability density function of $Y=X^{3}$.
Solution. Since the function $y=x^{3}$ is differentiable and strictly monotonic, we can apply formula (2.7.4).

We find the function $x=\varphi(y)$, which is the inverse of the function $y=x^{3}$: $\varphi(y)=x=y^{1 / 3}$ and $p[\varphi(y)]$. According to the condition, $p(x)$ is defined by formula (I), so

$$
p[\varphi(y)]=p\left[y^{1 / 3}\right]=\frac{1}{\sigma \sqrt{2 \pi}} e^{-y^{2 / 3} / 2 \sigma^{2}}
$$

We find the derivative of the inverse function with respect to $y$:

$$
\varphi^{\prime}(y)=\left(y^{1 / 3}\right)^{\prime}=\frac{1}{3 y^{2 / 3}}
$$

Substituting expressions (II) and (III) into formula (2.7.4), we obtain the desired probability density function of the function $Y=X^{3}$:

$$
g(y)=\frac{1}{3 \sigma y^{2 / 3} \sqrt{2 \pi}} e^{-y^{2 / 3} / 2 \sigma^{2}}
$$