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?

1 Answer
Feb 26, 2016

Answer:

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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