 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-2/3f/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

From part (i),

var (X) = var (Y) = \frac{4}{3}, cov (X, Y) = \frac{2}{3} .

The conditional-normal formula therefore gives

Y ∣ X = x \sim N (\frac{cov ( X , Y )}{var ( X )} x, var (Y) - \frac{cov ( X , Y ) ^{2}}{var ( X )}) = N (\frac{x}{2}, 1) .

E [Y ∣ X] = \frac{X}{2} .

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