Question

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?

Answer 1

Probability density of random variable `V` is
`f(x) = int_x^1 dy/y = -ln(x)`

Explanation:

The probability density `f(x)` 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 `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 `1/y`) and integrate by `y` from `x` to `1`:
`f(x) = int_x^1 dy/y = -ln(x)`

Just to check, integral of this probability density from `0` to `1` should be equal to `1`:
`int_0^1 [-ln(x)]dx = [x-x*ln(x)]_0^1 = 1`

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

`f(x)=-ln(x)`
graph{-ln(x) [-.1, 1, -5, 5]}

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