Finding the unconditional distribution I found two similar questions. One has a good answer What is the distribution of an unconditioned random variable knowing the conditional distribution? . I have a similar problem that I think should be **_much_** simpler. We have: > $ X \in U(0,1) $. > > $ Y|X=x \in U(0,x) $. From this I know how to calculate $ E[Y]=1/4 $ and $ Var(Y) $ but I want to find the distribution of $ Y $. $$ f_Y(y)= \int_{-\infty}^{\infty} f_{X,Y}(x,y) \,dx = \int_{-\infty}^{\infty} f_Y(y|x) f_X(x) \,dx $$ Now, $ f_X(x)=1 $ and $ f_Y(y|x)=1/x $. But I am completely lost at how to reason about the limits of integration here. Clearly the integral diverges unless I plug in something different, but what? Or is this the wrong way to tackle this problem?

> But I am completely lost at how to reason about the limits of integration here.

Yes, and this is because the densities are **not** what you write. Rather, $f_X$ and $f(\ \mid x)$ are defined on the whole real line $\mathbb R$ by $$ f_X(x)=\mathbf 1_{0\lt x\lt 1},\qquad f(y\mid x)=\frac1x\cdot\mathbf 1_{0\lt y\lt x}. $$ Note that $f_X(x)$ exists for every $x$ and that, if $x$ is in $(0,1)$, $f(y\mid x)$ exists for every $y$. One usually extends $f(\ \mid x)$ by $f(y\mid x)=0$ for every $y$ if $x$ is not in $(0,1)$. Then, as you explained, for every $y$, $$ f_Y(y)=\int_{-\infty}^{\infty}f_Y(y\mid x)f_X(x)\mathrm dx, $$ that is, $$ f_Y(y)=\int_0^1\frac1x\mathbf 1_{0\lt y\lt x}\mathrm dx=\mathbf 1_{0\lt y\lt 1}\int_y^1\frac1x\mathrm dx=\text{____}\cdot\mathbf 1_{0\lt y\lt 1}. $$