 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2012/m/m2/b/iv/solution

For

0 < x < 1

, the conditional density is obtained by dividing the joint density by the positive marginal density:

f_{Y ∣ X = x} (y) = \frac{24 x y}{12 x ( 1 - x ) ^{2}} = \frac{2 y}{( 1 - x ) ^{2}}, 0 \leq y \leq 1 - x .

Consequently

f_{Y ∣ X = 1/2} (y) = {8 y, 0, 0 \leq y \leq 1/2, otherwise,

which integrates to one. Its conditional expectation is

E (Y ∣ X = 1/2) = \int_{0}^{1/2} y (8 y) d y = \frac{1}{3} .

This uses the usual continuous conditional-density version, defined by the density ratio at a point where

f_{X}

is positive; conditioning does not require the event

X = 1/2

itself to have positive probability.

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