Given $Y$, the conditional cdf of $X$ is $F_{X|Y}(x|y)=\left(\frac{x}{y}\right)^2$, where $x\leq y$. If $Y$ is exponentially distributed with intensity $\lambda$, what is the marginal cdf of $X$?
My approach is the followings: $$F_{X}(x) =\int^{\infty}_{x} \left(\frac{x}{y}\right)^2 \cdot f_Y{\left(y\right)}dy$$ $$=\int^{\infty}_{x} \left(\frac{x}{y}\right)^2 \lambda \exp \{ -\lambda y \} dy$$
However, it seems to be false. How do i obtain the marginal cdf of $X$?
============================[ Update ]===============================
From the references, I know that
$$f_X(x)= \int^{\infty}_{-\infty} f_{X,Y}\left( x,y\right) dy,$$ $$f_{X|Y}\left(x|y\right)=\frac{f_{X,Y}(x,y)}{f_Y(y)}.$$
Therefore, in my opinion, $f_{X|Y}\left(x|y\right){f_Y(y)}= f_{X,Y}(x,y)$.
Now, I further suppose that, $$f_{X|Y}\left(x|y\right)= \left\{ \begin{array}{c} 0 && x<0 \\ \frac{2x}{y^2} && 0 \leq x \leq y \\ 0 && y < x \end{array}, \right. $$ and
$$f_Y \left( y\right)= \left\{ \begin{array}{c} 0 && y \leq 0 \\ \lambda \exp \{-\lambda y \} && 0 \leq y \\ \end{array}. \right. $$ Therefore, I assume that $$ \begin{array}{c} f_X(x) && = && \int^{\infty}_{-\infty} f_{X,Y}(x,y)~dy \\ && = && \int^{\infty}_{-\infty}f_{X|Y}\left(x|y\right){f_Y(y)}~dy\\ && = && \int^{\infty}_{x} \frac{2\lambda x}{y^2} \exp \{-\lambda y \}~dy . \end{array} $$ $$ \begin{array}{c} F_X(c) && = && \int^{c}_{0} \int^{\infty}_{x} \frac{2\lambda x}{y^2} \exp \{-\lambda y \}~dy~dx . \end{array} $$ This is the correct answer.
You obtain the marginal CDF of $X$ by first computing the marginal pdf of $X$ and then integrating to get the marginal CDF. So, the conditional CDF of $X$ given that $Y = y$ where $y \geq 0$ is $$F_{X\mid Y}(x \mid y) = \begin{cases}0, &x < 0, \\ \displaystyle\left(\frac xy\right)^2, &0 \leq x < y, \\1, &x\geq y.\end{cases}$$ Can you determine $f_{X\mid Y}(x \mid y)$ from this? How about $f_{X,Y}(x,y)$?