 Solution

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

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

Now complete the square in the other order:

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

f_{X_{2}} (x_{2}) = \frac{1}{2 π σ _{2}} e^{- b^{2} /2} .

The factor

e^{- b^{2} /2}

f_{X_{1} ∣ X_{2} = x_{2}} (x_{1}) = \frac{1}{2 π σ _{1} 1 - ρ ^{2}} exp [- \frac{( a - ρ b ) ^{2}}{2 ( 1 - ρ ^{2} )}] .

Since

a - ρ b = \frac{x _{1} - ( μ _{1} + ρ σ _{1} ( x _{2} - μ _{2} ) / σ _{2} )}{σ _{1}},

this is a normalized normal distribution density. Hence

X_{1} ∣ X_{2} = x_{2} \sim N (μ_{1} + ρ \frac{σ _{1}}{σ _{2}} (x_{2} - μ_{2}), σ_{1}^{2} (1 - ρ^{2})) .

Conditioning on an exact observation is interpreted through this conditional probability density, rather than conditioning by a ratio of event probabilities.

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