Finding individual PDF of two conjoined dependent variables

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Given two random variables distributed using a conjoined PDF $f_{X,Y}(x,y)$, with $0 < x < y < \infty$, I should find the individual PDF for each $X$ and $Y$.

I do this by evaluating the integrals:

\begin{equation*} f_X(x) = \int_0^x f_{X,Y}(x,y) dy \end{equation*}

\begin{equation*} f_Y(y) = \int_0^y f_{X,Y}(x,y) dx \end{equation*}

This yields two single-variable functions, but this seems strange to me, since apparently $X$ and $Y$ are dependent – the domain of $X$ is clearly bounded by $Y$'s value. Perhaps should the $f_X(x)$ be integrated from 0 to $y$?

Are these integrals correct or did I miss something?

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You should do it by evaluating the integrals:

\begin{equation*} f_X(x) = \int_0^{\infty} f_{X,Y}(x,y) dy \end{equation*}

\begin{equation*} f_Y(y) = \int_{0}^{\infty} f_{X,Y}(x,y) dx\ \end{equation*}

The first for any fixed $x>0$ and the second for any fixed $y>0$.

Because the PDF only takes values $\neq0$ if $0<x<y<\infty$ this comes to the same as evaluating: \begin{equation*} f_X(x) = \int_x^{\infty} f_{X,Y}(x,y) dy \end{equation*}

\begin{equation*} f_Y(y) = \int_0^{y} f_{X,Y}(x,y) dx\ \end{equation*}