# Exponentially Distributed Random Variable - P.D.F

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?

• You would not apply this, they are two separate problems. The integral is not right. For one thing, where is the exponential distribution? – André Nicolas May 1 '12 at 2:56
• I think I see what you were saying, and I think I fixed my integral... What else am I doing wrong here? I've been breaking my head open with this one for a while. – gfppaste May 1 '12 at 3:00
• I posted a solution for the first problem, and a start on the second, since you seem to have real difficulties. There may be places in the first problem where you need extra help. I will not give extra help on the second, except to confirm whether your answer is right. – André Nicolas May 1 '12 at 3:15
• Thank you very much for your help on this one... you've made the problem make a lot more sense. – gfppaste May 1 '12 at 3:20

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.