Here's an interesting question:
The area of a circle is exponentially distributed with parameter
$\lambda$. Find the distribution of PDF of the radius of the circle.
Then assume that the radius is exponentially distributed (with the
same rate $\lambda$) and find the PDF of the circle’s area.
So an exponential distribution means that the density function would be along the lines of:
PDF(x) = f(x) = $\lambda e^{-\lambda x}dx$
then I think one would do something along these lines:
$P${$R<=r$} = $ \frac{1}{2\pi} \int_{B(0)}\int r^\frac{r^2}{2}dr$ = $\int^0_r r^{\frac{r^2}{2}} dr$
to find the PDF of the radius, right?
How would I apply this to find the PDF of the circle's area?
Best Answer
Let the random variable $X$ denote the area, and let the random variable $R$ denote the radius. We have been told that the density function of the random variable $X$ is $\lambda e^{-\lambda x}$ for $x\ge 0$, and $0$ elsewhere. That is what is meant by saying that area has exponential distribution with parameter $\lambda$.
We want the pdf of $R$. To get it, we first find an expression for the cumulative distribution function of $R$, that is, the probability that $R\le r$. This is $0$ for $r\lt 0$, so assume $r\ge 0$. We have $X=\pi R^2$. So $R \le r$ iff $X \le \pi r^2$, and therefore $$P(R\le r)=P(X\le \pi r^2)=\int_0^{\pi r^2} \lambda e^{-\lambda x}\,dx.$$ We could integrate, it is not hard, and then differentiate to get the density function of $R$. But it is easier (at least for me) to differentiate under the integral sign, remembering to use the Chain Rule. We get that (for $r\ge 0$) $$f_R(r)= (2\pi r)\lambda e^{-\lambda \pi r^2}.$$
Now for the other problem. Here it is $R$ that has the exponential distribution. We want to find $P(X\le x)$. Since $X=\pi R^2$, this is true iff $R\le \sqrt{x/\pi}$. Express this as an integral, in the style of the first part. Then evaluate the integral and differentiate, or differentiate without evaluating.