1
$\begingroup$

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 $

$\endgroup$
2
$\begingroup$

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 $F_X(-\sqrt{y}) = 0$...

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – nundo Oct 4 '18 at 16:05
  • 1
    $\begingroup$ @nundo what if the function is not 1-1? or does not have an easy inverse... $\endgroup$ – gt6989b Oct 4 '18 at 16:14
  • $\begingroup$ I see so we are relying on the fact that there is an inverse function in order to make these statements. $\endgroup$ – nundo Oct 10 '18 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.