CDF of $Y$ given the CDF of $X$ and $Y=X^2$

Question

Let $X$ be a continuous random variable with CDF $F_X$, and $Y=X^2$. Find the CDF of $Y$.

My solution: $$F_Y(t)=P(Y \leq t) = P(X^2 \leq t) = P( -\sqrt{t} \leq X \leq \sqrt{t}) = $$ $$P(X \leq \sqrt{t}) - P(X \leq -\sqrt{t}) = F_X(\sqrt{t}) - F_X(-\sqrt{t})$$ (assuming $t \geq 0$, otherwise $F_Y(t)=0$).

Is this correct?

Answer 1

$$F_Y(t)=P(Y \leq t) = P(X^2 \leq t) = P( -\sqrt{t} \leq X \leq \sqrt{t}) = $$ $$P(X \leq \sqrt{t}) - P(X < -\sqrt{t}) = F_X(\sqrt{t}) - F_X((-\sqrt{t})-)$$ where $F(s-)$ denotes the left-hand limit of $F(s)$ at $s$. Since $X$ is continuous it follows that we can replace $F(s-)$ by $F(s)$ so we get $$F_Y(t)=F_X(\sqrt{t}) - F_X((-\sqrt{t})).$$