If I have a random variable X with a given Probability Density Function, How do I find the PDF of the area of a circle with radius X?

Question

To find the PDF of the area of the circle, do I just substitute the PDF of the random variable X in for the radius in the circle area equation?

Answer 1

If we are careful, we can use cumulative distribution functions to solve this question.

Let $X$ be a nonnegative (I assume $X$ to be nonnegative because a negative radius is a bit senseless) random variable whose cumulative distribution function is given by $F_X:[0,\infty)\to[0,1]$.

If we define $Y=\pi X^2$, then $Y$ is the area of a circle of radius $X$, which is what we want. If we let $F_Y:[0,\infty)\to[0,1]$ denote the cumulative distribution function of $Y$, then for every $x\in[0,\infty)$, we see that \begin{align*} F_Y(x)&=\Pr[Y\leq x]\\ &=\Pr[\pi X^2\leq x]\\ &=\Pr\left[X^2\leq \frac x{\pi}\right]\\ &=\Pr\left[X\leq \sqrt{\frac x{\pi}}\right]&\text{(since $X$ is nonnegative)}\\ &=F_X\left(\sqrt{\frac x{\pi}}\right). \end{align*}

Now comes the point where we relate the distributions to the densities, which is where we have to be careful. If we assume that $X$ has a density $f_X$, then we may assume that $F_X$ is absolutely continuous and that $$\frac{d}{dx}F_X(x)=f_X(x)$$ almost everywhere. Furthermore, given that $x\mapsto\sqrt{\frac{x}{\pi}}$ is absolutely continuous (see here) and nondecreasing, then the composition $$F_Y(x)=F_X\left(\sqrt{\frac x{\pi}}\right)$$ is absolutely continuous (the fact that $x\mapsto\sqrt{\frac{x}{\pi}}$ is nondecreasing is of utmost importance, since compositions of absolutely continuous functions are not absolutely continuous in general). Now, since $F_Y$ is absolutely continuous, we conclude that it has a density $f_Y$ which is given by $$\frac{d}{dx}F_Y(x)=f_Y(x)$$ almost everywhere. The above derivative can easily be computed using the chain rule.