Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 221 / 2 / ii

Let

K = Σ^{- 1}

R = {1, \dots, p} ∖ {j, k}

, and condition on

V_{R} = v_{R}

. In the Gaussian exponent, all terms depending jointly on

v_{j}

and

v_{k}

are contained in

- \frac{1}{2} (K_{jj} v_{j}^{2} + 2 K_{jk} v_{j} v_{k} + K_{kk} v_{k}^{2} + 2 b_{j} (v_{R}) v_{j} + 2 b_{k} (v_{R}) v_{k}) .

K_{jk} = 0

, this conditional density is a product of one function of

v_{j}

and one function of

v_{k}

V_{j}

and

V_{k}

given

V_{R}

holds.

Conversely, conditional independence makes this everywhere-positive conditional density factorize. Its mixed second derivative must therefore vanish:

\frac{\partial ^{2}}{\partial v _{j} \partial v _{k}} lo g f (v_{j}, v_{k} ∣ v_{R}) = - K_{jk} = 0.

Hence the conditional independence holds exactly when

(Σ^{- 1})_{jk} = 0

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