Density of Random Variables - Probability Let X be a random variable with X ~ R(0,1). Density of X is given by:
$f_{X}(x) = \begin{cases}
1  &  x \in ]0,1[  \\
0 & \text{otherwise}  \\
\end{cases}$
1) Find the density for $Y = X^2$
The answer should be $ f_{Y}(y)=\frac{1}{2\sqrt{y}}, y\in]0,1[ $ from what i'm told.
Please explain how i make this conclusion. Also if you could explain the part with ~R(0,1). Any help is appreciated.
 A: The $R(0,1)$ is a little strange, the usual notation is something like $U(0,1)$, for uniform.
We will find the density function of $Y$. We choose the slightly long way to do it. (There is a short more mechanical way.) 
First we find the cumulative distribution function $F_Y(y)$ of $Y$, then we differentiate to find the density function $f_Y(y)$.
First we do the easy part. Recall that $F_Y(y)=\Pr(Y\le y)$. This is $0$ for $y\le 0$, and $1$ when $y\ge 1$. Next we deal with the interesting part, when $0\lt y\lt 1$. We have
$$\Pr(Y\le y)=\Pr(X^2\le y)=\Pr(0\lt X\lt \sqrt{y})=\sqrt{y}.$$
Thus $F_Y(y)=\sqrt{y}$ when $0\lt y\lt 1$.
Finally, differentiate $F_Y(y)$ to get the density. We get $f_Y(y)=0$ if $y\le 0$ or if $y\ge 1$, and $f_Y(y)=\frac{1}{2\sqrt{y}}$ when $0\lt y\lt 1$.
A: One way to do this is to start with the distribution function of X, given by F_X(x) = 0 for x<0, F_X(x) = x for 0 <=x <= 1 and F_X(x) = 1 for x > 1.
Then realise that F_X(x) = Prob(X<=x).
Now we want to know F_Y (y) so that we can work out its density function.
This means we need to work out, for each y, Prob(Y<=y) = Prob (X^2<=y) = Prob (X <= sqrt(y))
We need not worry about signs ONLY because we know X is non-negative. Do you see this?
...Thus we get that F_Y(y) = 0 for y<0, sqrt(y) for 0<=y<=1, and 1 for y> 1.
Differentiate this and you should get your answer.
