Marginal distribution from joint distribution Let X be a random variable $X$ with distribution
$$f_X(x) = \begin{cases} 2x & x \in [0, 1] \\ 0 & \text{else}  \end{cases}$$. Given $X = x$, suppose the conditional distribution of $Y$ is uniform on $[0, 1/x]$. We wish to find the marginal distribution of $Y$. My thinking is the following. Find the joint distribution as
$$ f_{X, Y}(x, y) = f_{Y | X}(y | x) f_X(x) $$
We have
$$ f_{Y | X} ( y | x) = \begin{cases} x & y \in [0, 1/x] \\ 0 & \text{else} \end{cases} $$
So
$$ f_{X, Y}(x, y) = \begin{cases} 2x^2 & 0 \leq y \leq 1 /x, 0 \leq x \leq 1 \\ 0 & \text{else} \end{cases} $$
Then integrate this with respect to $x$
$$ f_Y(y) = \int_{-\infty}^{\infty} f_{X, Y}(x, y) \: dx = \int_0^{1/y} 2x^2 \: dx = \frac{2}{3y^3}$$
However, this isn't a valid probability distribution as it doesn't integrate to 1. What's wrong with this reasoning?

                              
Since
$$f_{X, Y}(x, y) = 0\ \text{ unless }\ 0 \leq y \leq 1 /x,\, 0 \leq x \leq 1$$
we can invert the order of the constraints into
$$f_{X, Y}(x, y) = 0\ \text{ unless }\ 0 \leq y,\, 0 \leq x \leq \min\{1,y^{-1}\}$$
Thus
$$f_Y(y)=\int f_{X,Y}(x,y)\,\text dx=\int_0^{\min\{1,y^{-1}\}} 2x^1\,\text dx=\begin{cases} 2/3 &\text{if }y<1\\
2y^{-3}/3 &\text{if }y>1\end{cases}$$
with
$$\int_1^\infty 2y^{-3}/3\,\text dy= 1^{-2}/3=1/3$$
