How to show that the cdf of Y is $F(\sqrt{Y})$ if cdf of $X$ is $F(x)$, and $Y=X^2$?

Question

I was trying to work through how to find the CDF of $Y$ if we know CDF of $X$ is $F_X(x)$, and $Y=X^2$.

So if I plug in $y=x^2$ then I can intuitively get

$$F_Y(y)=+\sqrt y$$

I solved this simply by solving $Y=X^2$ for $X$ and get $X=\sqrt Y$. Is it true in general that I can just solve the relationship between to Random Variables to get the cdf of one from the other?

Related: Probability. Find the CDF of $Y = X^2 $

Answer 1

You have $$ F_Y(y) = \mathbb{P}[Y < y] = \mathbb{P}[X^2<y] = \mathbb{P}[|X| < \sqrt{y}] = F_X(\sqrt{y}) - F_X(-\sqrt{y}) $$ Assuming $X \ge 0$, you have $F_X(-\sqrt{y}) = 0$...