# Suppose that U has a uniform distribution on [0, 1] and that, conditional on U = u, the distribution of V is uniform on [0, u]. What is the probability density function of V?

Feb 26, 2016

Probability density of random variable $V$ is
$f \left(x\right) = {\int}_{x}^{1} \frac{\mathrm{dy}}{y} = - \ln \left(x\right)$

#### Explanation:

The probability density $f \left(x\right)$ of random variable $V$ is a result of a combination of two factors:
(a) random variable $U$ should take some value $y$ greater than $x$ (with probability density $1$) and, for each such value $y$,
(b) random variable $V$ should take a value $x$ with probability density $\frac{1}{y}$.

These two above factors are two independent random variable, so the probabilities of combined events must be multiplied.

Now the probability density of $U$ (which is $1$) should be multiplied by probability density of $V$ (which is $\frac{1}{y}$) and integrate by $y$ from $x$ to $1$:
$f \left(x\right) = {\int}_{x}^{1} \frac{\mathrm{dy}}{y} = - \ln \left(x\right)$

Just to check, integral of this probability density from $0$ to $1$ should be equal to $1$:
${\int}_{0}^{1} \left[- \ln \left(x\right)\right] \mathrm{dx} = {\left[x - x \cdot \ln \left(x\right)\right]}_{0}^{1} = 1$

Graphically, the probability density of random variable $V$ looks like this:

$f \left(x\right) = - \ln \left(x\right)$
graph{-ln(x) [-.1, 1, -5, 5]}