Probability Density Function: Small question Let $X \sim N(\mu, \sigma^2)$ and $f_{X} {(x)} = \frac{1}{\sigma\sqrt{2\pi}} \exp \left( -\frac{(x-\mu)^2}{2\sigma^2} \right)$, where $-\infty < x < \infty.$
Find the PDF of $X^3$.
Do you literally just sub $x^3$ into the function given?
 A: Let $Y=X^3$. Find first an expression for the probability that $Y\le y$, that is, for the cumulative distribution function $F_Y(y)$ of $Y$.  Since $Y \le y$ iff $X^3\le y$ iff $X \le y^{1/3}$, we want the probability that $X\le y^{1/3}$. Thus
$$F_Y(y)=\Pr(X\le y^{1/3})=\int_{-\infty}^{y^{1/3}} f_X(x)\,dx.$$
To find the density function $f_Y(y)$ of $Y$, differentiate $F_Y(y)$. To do this, use the Fundamental Theorem of Calculus. By the Chain Rule, you will get 
$$f_Y(y)=\frac{1}{3}y^{-2/3}f_X(y^{1/3}).$$
If differentiating under the integral sign is no longer a familiar process to you, see the Wikipedia article.
Remark: Note that substituting worked, sort of. Except that it was $y^{1/3}$ that we substituted, and there is an extra term needed for the derivative of $y^{1/3}$. What made the calculation relatively straightforward is that $t^3$ is an increasing function of $t$. Things are less nice if you want the density function of $g(X)$, where $g(t)$ is not a monotone function. A similar strategy works, but the details get more complicated.  
