How to determine cumulative distribution function of squared random variable?

Question

I don't understand how can I get the

$F(x)$(cumulative distribution funciton)

given only

$X = RND^2$

where RND means (continuous) random variable, the paper tells the answer is $F(x) = \sqrt x$ , if x $\epsilon (0,1)$ it seems that I need to take square root but why?

I know $F(x) = p(X < x)$ but from here I can't proceed.

Answer 1

You have $X=Y^2$, where $Y \sim U(0,1)$. Now we want the cdf of $X$.

$P(X<x)=P(Y^2<x)=P(Y<\sqrt x)=F_Y(\sqrt x)$

Now you plug in the value $\sqrt x$ into the cdf of $Y$:

$$F_Y(y)=\begin{cases} 0 & \text{for } y < 0 \\ y & \text{for } y \in [0,1) \\ 1 & \text{for } x \ge 1 \end{cases}$$

To obtain the pdf you just differentiate the cdf of Y.