MHB Cumulative distribution function

[Julio1]

Hi !, my problem is the following:

Let $F_X (x)$ an distribution function strictly monotone for the random variable $X$ and it's defined the new random variable $Y=F_X (X).$ Find the cumulative distribution function of $Y$.

Spoiler

In this case, $F_X (x)$ is an cdf, but I don't how does for find the cdf of $Y.$ I think that need have the density function, but I don't have is information.

 

[Evgeny.Makarov]

Let $F_X (x)$ an distribution function strictly monotone for the random variable $X$ and it's defined the new random variable $Y=F_X (x).$

I don't understand the phrase "it's defined the new random variable $Y=F_X (x)$". $F_X(x)$, for each given number $x$, is a number. There is no random variable here. Recall that a random variable is a function from the sample space to $\Bbb R$. Do you mean $Y=F_X (X)$, i.e., $Y=F_X\circ X$?

In mathematics, uppercase and lowercase letters often denote different objects.

 

[Julio1]

I don't understand the phrase "it's defined the new random variable $Y=F_X (x)$". $F_X(x)$, for each given number $x$, is a number. There is no random variable here. Recall that a random variable is a function from the sample space to $\Bbb R$. Do you mean $Y=F_X (X)$, i.e., $Y=F_X\circ X$?

In mathematics, uppercase and lowercase letters often denote different objects.

Thanks!, you're right, the correct is $Y=F_X (X).$ In this case, $Y=F_X(X)$ is an transformation?

 

[Evgeny.Makarov]

In this case, $Y=F_X(X)$ is an transformation?

I am not sure what you mean by this.

Since $F_X$ is strictly monotonic, it has an inverse $F_X^{-1}$. So
\[
F_Y(y)=\text{Pr}(Y<y)=\text{Pr}(F_X(X)<y)=\text{Pr}(F_X^{-1}(F_X(X))<F_X^{-1}(y))=\text{Pr}(X<F_X^{-1}(y))=\ldots
\]
Can you simplify this further?