= Solution
Let $K=\Sigma^{-1}$ be the <precision matrix>, let $R=\{1,\ldots,p\}\setminus\{j,k\}$, and condition on $V_R=v_R$. In the Gaussian exponent, all terms depending jointly on $v_j$ and $v_k$ are contained in
$$
-\frac12\left(K_{jj}v_j^2+2K_{jk}v_jv_k+K_{kk}v_k^2
+2b_j(v_R)v_j+2b_k(v_R)v_k\right).
$$
If $K_{jk}=0$, this conditional density is a product of one function of $v_j$ and one function of $v_k$, so the <conditional independence> of $V_j$ and $V_k$ given $V_R$ holds.
Conversely, conditional independence makes this everywhere-positive conditional density factorize. Its mixed second derivative must therefore vanish:
$$
\frac{\partial^2}{\partial v_j\partial v_k}
\log f(v_j,v_k\mid v_R)=-K_{jk}=0.
$$
Hence the conditional independence holds exactly when $(\Sigma^{-1})_{jk}=0$.
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