 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-2/12f/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

σ_{1}, σ_{2} > 0

and

∣ ρ ∣ < 1

. Put

a = (x_{1} - μ_{1}) / σ_{1}

and

b = (x_{2} - μ_{2}) / σ_{2}

a^{2} - 2 ρ ab + b^{2} = (b - ρ a)^{2} + (1 - ρ^{2}) a^{2} .

Consequently the joint probability density function factors as

f_{X_{1}, X_{2}} (x_{1}, x_{2}) = \frac{e ^{- a^{2} /2}}{2 π σ _{1}} \frac{e x p [ - ( b - ρ a ) ^{2} / ( 2 ( 1 - ρ ^{2} ))]}{2 π σ _{2} 1 - ρ ^{2}} .

For fixed

x_{1}

, the second factor is a normalized normal distribution density in

x_{2}

, with mean

μ_{2} + ρ σ_{2} a

σ_{2} 1 - ρ^{2}

. Its integral is one; explicitly substitute

w = (b - ρ a) / 1 - ρ^{2}

in the Gaussian integral. Thus the marginal density is

f_{X_{1}} (x_{1}) = \frac{1}{2 π σ _{1}} exp [- \frac{( x _{1} - μ _{1} ) ^{2}}{2 σ _{1}^{2}}] .

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