I was doing a question in probability 1st year, the question is:
$X$ is has a uniform distribution on $[0,1]$. What's the probability distribution and the density of $\sqrt{X}$?
What I have so far is $P(\sqrt(X) < x)=P(X<x^2)$.
Now, as we know that X follows the uniform distribution then
$ =((x^2-0)/(1-0))= x^2 $
then I don't really know how to calculate the domain of the function, because when P($X <x ^2)= 1$ then $x=1$ or $x=-1$
I don't know which of both values should I take
There is such a theorem for function of random variable: Let $X$ be a continuous random variable and $Y=g(x)$ where $g(X)$ is continuous in $X$ and strictly monotonous, then $$f_Y(y)=f_X[g^{-1}(y)]\left|\cfrac {\mathrm{d}g^{-1}(y)}{\mathrm{d}y}\right| $$
So $f_Y(y)=f_X(x^2)\cfrac{\mathrm{d}y^2}{\mathrm{d}y}=2y \qquad y\in[0,1]$
$\displaystyle F_Y(y)=\begin{cases}{0 \qquad y<0\\\int_0^y 2u\mathrm{d}u=y^2 \qquad y\in[0,1]\\1 \qquad y>1} \end{cases}$