 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/discrete-mathematics-continuous-mathematics-and-probability/8/b/ii/solution

E Y = E [X /2] = q / (2 p)

. The law of total variance separates the two sources of randomness:

Var (Y) = E [Var (Y ∣ X)] + Var (E [Y ∣ X]) = \frac{1}{4} E X + \frac{1}{4} Var (X) .

Substitution gives

E Y = \frac{1 - p}{2 p}, Var (Y) = \frac{q ( p + 1 )}{4 p ^{2}} = \frac{1 - p ^{2}}{4 p ^{2}} .

Y

2 p / (1 + p)

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