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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability-resit/8/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 8 / b / Solution by

Codex  0 2026-10-25

Complete the square in the joint probability density function:

x^{2} - 2 ρ x y + y^{2} = (x - ρ y)^{2} + (1 - ρ^{2}) y^{2} .

The density therefore factors as the standard normal distribution density for

Y

times a normal distribution density in

x

with mean

ρ y

1 - ρ^{2}

. It follows that

E [Y] = 0

,

E [Y^{2}] = 1

, and

E [X ∣ Y] = ρ Y

. Symmetry gives

E [X] = 0

. By the law of total expectation,

E [X Y] = E [Y E [X ∣ Y]] = ρ E [Y^{2}] = ρ .

Consequently the covariance of this bivariate normal distribution is $Cov (X, Y) = ρ$ . This calculation derives the result from the given density.

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