pdf of a simple random variable calculated two different ways.. with two different answers so we have a random variable Y with a uniform pdf on the interval [0,1]. the question is, what is the pdf of W=Y^2.
method 1 using the transformation of variables formula:
$f_w=f_y(g^{-1}(y))\dfrac{d}{dy}g^{-1}(y)=\dfrac{1}{2 \sqrt{w}}$
method 2 treating W like a product distriubtion of $Y*Y$ (
http://en.wikipedia.org/wiki/Product_distribution)
$f_w = \int_{-\infty}^{\infty} \dfrac{1}{|y|}f_y(w/y)f_y(y)dy = \int_w^1 \dfrac{1}{|y|}dy = -ln(w)$
both have similar cdfs. i don't know which is right or why i got different answers.

Let's do it another way. We find the cdf of $W$. For $0\lt w\lt 1$, we have $W\le w$ if and only if $Y^2\le w$ if and only if $Y\le \sqrt{w}$. But $\Pr(Y\le \sqrt{w})=\sqrt{w}$.
For the density, differentiate. We get $\frac{1}{2\sqrt{w}}$ on $(0,1)$, and $0$ elsewhere. 
Remark: A problem with second approach is that it applies to a product of statistically independent random variables, which $Y$ and $Y$ are not.
