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

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?

## 1 Answer

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

• Thanks. Now I am wondering - if we know a relationship between $X$ and $Y$ like that, when is it not possible to merely solve the equation for $X$ in terms of $Y$ and substitute into $F_X(x)$? Or is that always possible? – nundo Oct 4 '18 at 16:05
• @nundo what if the function is not 1-1? or does not have an easy inverse... – gt6989b Oct 4 '18 at 16:14
• I see so we are relying on the fact that there is an inverse function in order to make these statements. – nundo Oct 10 '18 at 14:04