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.


1 Answer 1


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

$$ f_Y(y)=\int_0^1f_{Y\mid U}(y\mid u)f_U(u)\,\mathrm du=\int_0^{1-y}\frac1{1-u}\,\mathrm du=-\log y\;. $$

  • $\begingroup$ 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 $\int_{0}^{x}{\frac{1}{u} du}$? $\endgroup$
    – Eoin S
    Apr 3, 2020 at 22:46
  • 1
    $\begingroup$ @EoinS: Take another look at "$0\le x\le u$, so $0\le u\le x$" :-) $\endgroup$
    – joriki
    Apr 3, 2020 at 22:52

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