 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-221/1/ii/solution

Codex  0 Created 2026-09-28 Updated 2026-10-14

p (x_{1}, u, x_{2}, x_{3}, x_{4}) = p (x_{1}) p (u) p (x_{2} ∣ x_{1}, u) p (x_{3} ∣ x_{1}, x_{2}) p (x_{4} ∣ x_{3}, u) .

The graph gives

U ⊥ X_{1}

and

U ⊥ X_{3} ∣ (X_{1}, X_{2})

. Therefore Bayes' theorem gives

p (u ∣ x_{1}, x_{2}, x_{3}) = p (u ∣ x_{1}, x_{2}) = \frac{p ( u ) p ( x _{2} ∣ x _{1} , u )}{p ( x _{2} ∣ x _{1} )} .

Consequently

x_{2} \sum p (x_{4} ∣ x_{1}, x_{2}, x_{3}) p (x_{2} ∣ x_{1}) = x_{2}, u \sum p (x_{4} ∣ x_{3}, u) p (u) p (x_{2} ∣ x_{1}, u) = u \sum p (x_{4} ∣ x_{3}, u) p (u),

which contains no

x_{1}

. This observed equality is a Verma constraint: it is implied by the latent-variable causal graph even though it is not a conditional independence.

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