Random variable X is uniformly distributed over the interval [0,1]. $Y = X^{2}$.I need to find the distribution function for $Y$. So i did this:
$$P(Y \leq x)= P(X^{2} \leq x) = P(X \leq \sqrt{x}) = F(\sqrt{x}) = \sqrt{x}$$
So distribution function is $\sqrt{x}$. But what the intervals for it? How to find them?
More accurately the CDF of $X$ is given by $$ F_X(x) = \begin{cases} 0 & x \leq 0, \\ x & x \in [0, 1], \\ 1 & x \geq 1. \end{cases} $$ The rest of your working is fine for when $x \geq 0$, so the CDF of $Y$ is given by $$ F_Y(x) = F_X(\sqrt{x}) = \begin{cases} \sqrt{x} & x \in [0, 1], \\ 1 & x \geq 1. \end{cases}$$ For $x \leq 0$ we must have $F_Y(x) = 0$. So putting it together gives $$ F_Y(x) = \begin{cases} 0 & x \leq 0, \\ \sqrt{x} & x \in [0, 1], \\ 1 & x \geq 1. \end{cases}$$