Uniform random variables and conditional distributions.

Let $$U \sim U([0, 1])$$, be uniformly distributed on the interval $$[0, 1]$$. Let $$X \sim U([0, U])$$ and $$Y \sim U([0, 1-U])$$.

(a) Find the conditional density of $$U$$ given $$Y$$.

(b) Find the joint density of $$X$$ and $$Y$$.

For part (a), I tried to use Bayes rule in the following way:

$$f_{U | Y}\left(u | y\right) = \frac{f_{Y | U}\left(y | u\right) f_{U}\left(u\right)}{f_{Y}\left(y\right)}$$

Now, we know that $$f_U(u) = 1$$ for $$0 \leq u \leq 1$$ and 0 otherwise. The conditional density for $$Y$$ given that $$U = u$$ is $$f_{Y | U}\left(y | u\right) = \frac{1}{1-u}$$ for $$0 \leq y \leq 1-u$$. Additionally, the density of $$Y$$ can be given by:

$$f_Y(y) = \int_{u = 0}^{u = 1}{f_{Y | U}\left(y | u\right) f_U(u) du} = \int_{0}^{1}{ \frac{1}{1-u} du}$$

However, this integral is divergent. So I am confused as to what could be done for this problem. I find that I run into similar divergence of integrals in part (b) as well.

You correctly state that $$f_{Y\mid U}(y\mid u)=\frac1{1-u}$$ for $$0\le y\le1-u$$, but then you use it as if it were $$\frac1{1-u}$$ everywhere. If you take into account that it’s $$0$$ outside that range, you get
$$f_Y(y)=\int_0^1f_{Y\mid U}(y\mid u)f_U(u)\,\mathrm du=\int_0^{1-y}\frac1{1-u}\,\mathrm du=-\log y\;.$$
• Oh, I see my mistake. And so this would yield $f_{U | Y}(u|y) = -\frac{1}{ln(y)} \frac{1}{1-u}$. But, take the case for finding $f_X(x)$. If X is conditioned on U, then the integral for the marginal density of X still gives a divergent term of $ln(0)$ since $0 \leq x \leq u$, so $0 \leq u \leq x$, and the integral would be $\int_{0}^{x}{\frac{1}{u} du}$? Apr 3, 2020 at 22:46
• @EoinS: Take another look at "$0\le x\le u$, so $0\le u\le x$" :-) Apr 3, 2020 at 22:52