# Why Cant We Directly Find the PDF of the Transformation of Random?

## How To Convert PDF Online?

Sign & Make It Legally Binding

## Easy-to-use PDF software     ## Why can't we directly find the PDF of the transformation of random variables, say g(X) from the random variable X. Why do we have to first convert PDF into CDF and then have to differentiate to get to the PDF of the transformed random variable?

I see that you asked for the cumulative distribution, but I’ll answer a slightly different (and somewhat more difficult) question since this kind of question at this time of the year seems suspiciously like a question I might give my own students on a take home exam. If you remind me via a comment on this answer in a week or so, I’ll add a section to the answer that gives the CDF in addition to the density. If X and Y are independent, then so are X^2 and Y^2. Similarly, if X and Y instead followed a \text{Unifrom}(0,1) distribution, then X^2 and Y^2 would have the same distribution as t do with X and Y following a \text{Unifrom}(-1,1) distribution. So one approach to this problem is to first find the distribution of the square of a \text{Unifrom}(0,1) random variable. Then, convolve that density with itself to get the density for the sum of the squares of two independent random variables with that newfound distribution. The first step is easily done with the method of transformations. The density function for X^2 is given by f_{X^2}(x) = \begin{cases} \frac 1{2{\sqrt x}} & x\in(0,1)\\ 0 & \text{otherwise}\end{cases} Notice that this implies that X^2 follows a \text{Beta}\left(\frac 12, 1\right) distribution. So now we see that we need to find the distribution of the sum of two independent random variables with this Beta distribution. I’ll omit the Calculus here, but I will give you a link to the WolframAlpha command that gives you the result. The density function of the result you seek is given by f_{Z}(z)=\begin{cases}\frac \pi 4 & 0

4.5

satisfied