 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ia/paper-2/3f/a/solution

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

With

σ_{1}, σ_{2} > 0

and

∣ ρ ∣ < 1

f (x_{1}, x_{2}) = \frac{1}{2 π σ _{1} σ _{2} 1 - ρ ^{2}} exp [- \frac{1}{2 ( 1 - ρ ^{2} )} (\frac{( x _{1} - μ _{1} ) ^{2}}{σ _{1}^{2}} - \frac{2 ρ ( x _{1} - μ _{1} ) ( x _{2} - μ _{2} )}{σ _{1} σ _{2}} + \frac{( x _{2} - μ _{2} ) ^{2}}{σ _{2}^{2}})] .

The strict correlation bound makes its covariance matrix positive definite, so this is an ordinary two-dimensional density rather than a distribution supported on a line.

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