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?

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vrangett · Accepted Answer

Probability density of random variable V is  f(x)=∫x1dyy=−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 1y.  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 1y) and integrate by y from x to 1:  f(x)=∫x1dyy=−ln⁡(x)  Just to check, integral of this probability density from 0 to 1 should be equal to 1:  ∫01[−ln⁡(x)]dx=[x−x⋅ln⁡(x)]01=1  Graphically, the probability density of random variable V looks like this:  f(x)=−ln⁡(x)  graph {−ln⁡(x)[−.1,1,−5,5]}

Suppose that U has a uniform distribution on [0, 1]

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