 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/i/solution

Given

X = k

, the score

Y

1/2

. Its conditional expected value and variance are

k /2

and

k /4

. Thus

E [Y ∣ X = k] = \frac{k}{2}, E [Y^{2} ∣ X = k] = \frac{k}{4} + \frac{k ^{2}}{4} .

This is binomial thinning of the first-round count; the second formula uses the identity

E [Y^{2} ∣ X] = Var (Y ∣ X) + E [Y ∣ X]^{2}

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