I have a quick question. Perhaps the answer is obvious but I'm stuck on this one either way.
The problem is:
"Let $X$ have a uniform distribution on (0,1), and given that $X=x$, let the conditional distribution of $Y$ be uniform on $(0,1/x)$. (b) Find $f_y(y)$, the marginal pdf of $Y$."
So, I figure the solution is really straight forward. By a proposition in my text book, the marginal pdf of $Y$ is simply
$f_y(y)=\int_{-\infty}^{\infty} f(x,y)dx$. And because we know $x \in (0,1)$ and the joint pdf (which is $f(x,y) = x$) is zero everywhere else, we simply get $f_y(y)=\int_{-\infty}^{\infty} f(x,y)dx = \int_{0}^{1} xdx = 1/2$.
Now, by text book says that $f_y(y) = 1/2$ for $y\in (0,1)$ and $f_y(y) = 1/(2y^2)$ for $y \geqq 1$.
Im thinkning that because $y \in (0,1/x) $ and $x\in (0,1)$ then obviously this $\implies y \in (0,\infty) $ and therefore we absolutley need to check how the marginal pdf of $Y$ behaves for $y \geqq 1$. I cant however seem to figure out how to express the integrals in the different intervals.
Some help would be appreciated, thanks!
You have $(Y\mid X=x)\sim\mathcal{U}(0,1/x)$ where $x>0$.
Now, $0<y<1/x\,,0<x<1\implies 0<x<\min(1,1/y)=m$ (say).
By total probability theorem, marginal pdf of $Y$ is
$$f_Y(y)=\int_0^m f_{Y\mid X}(y\mid x)f_X(x)\,\mathrm{d}x$$
$$=\int_0^mx\,\mathrm{d}x=\frac{m^2}{2}$$
$$=\frac{1}{2}\mathbf1_{0<y<1}+\frac{1/y^2}{2}\mathbf1_{y\geqslant1}$$