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.
Best 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 $f_{X,Y}(x,y)$?