If $X \sim U(0,1)$ and $Y \sim U(0,X)$, what is the joint distribution of $(X,Y)$?

Question

Problem: Let $X \sim U(0,1)$, $Y \sim U(0,X)$. Which is the joint distribution $f(x,y)$?

Attempt: $$\mathbb P(X \leq a, Y \leq b) = \mathbb E(\mathbb P (Y \leq b \mid X=x \leq a)) = \int_0^a \frac b x \, dx $$ which is of course a problem!

Thanks!

Answer 1

For the two random variables the joint distribution is obtained by integrating the joint probability density over the rectangle $[0,a] \times [0,b]$. Note that the density is zero when $y > x$ and outside the rectangle $[0,1]\times [0,1]$.

If $b \geqslant a$, then

$$F(a,b) = \int_0^a \left( \int_0^x \frac{1}{x} \, dy\right) \, dx = a$$

otherwise, if $b < a$, then

$$F(a,b) = \int_0^b \left( \int_0^x \frac{1}{x} \, dy\right) \, dx + \int_b^a \left( \int_0^b \frac{1}{x} \, dy\right) \, dx = b + b \log(a/b)$$

Answer 2

Since $X\sim U(0,1)$ and $X = x$, let $Y$ have uniform distribution on the interval $(0,x)$. So we have $$f_{X}(x) = 1 \ \ \ 0\leq x \leq 1$$ and $$f_{Y|X}(y|x) = \frac{1}{x} \ \ \ 0\leq y \leq x$$ Hence the joint density is

$$f(x,y) = f_{Y|X}(y|x)f_{X}(x) = \frac{1}{x}\times 1 = \frac{1}{x} \ \ \ 0\leq x \leq 1, 0\leq y \leq x$$