A Conditional independence

Lemma A.1

✓

If \(X \perp \! \! \! \! \perp Y \mid Z\), then \(X \perp \! \! \! \! \perp (Y, Z) \mid Z\).

Proof ▶

Lemma A.2 Contraction

If \(X \perp \! \! \! \! \perp Y \mid Z\) and \(X \perp \! \! \! \! \perp Z\), then \(X \perp \! \! \! \! \perp (Y, Z)\).

Proof ▶

It suffices to show that \(\mathcal{L}(X \mid Y, Z) = \mathcal{L}(X)\). By conditional independence, \(\mathcal{L}(X \mid Y, Z) = \mathcal{L}(X \mid Z)\). By independence, \(\mathcal{L}(X \mid Z) = \mathcal{L}(X)\).

Lemma A.3

If \(X \perp \! \! \! \! \perp Y \mid Z, W\) and \(X \perp \! \! \! \! \perp Z \mid W\), then \(X \perp \! \! \! \! \perp (Y, Z) \mid W\).

Proof ▶

It suffices to show that \(\mathcal{L}(X \mid Y, Z, W) = \mathcal{L}(X \mid W)\). By the first hypothesis, \(\mathcal{L}(X \mid Y, Z, W) = \mathcal{L}(X \mid Z, W)\). By the second hypothesis, \(\mathcal{L}(X \mid Z, W) = \mathcal{L}(X \mid W)\).

Lemma A.4

✓

A family of random variables \((X_i)_{i \in \mathbb {N}}\) is independent if and only if for all \(n \in \mathbb {N}\), \(X_{n+1}\) is independent of \((X_0, \ldots , X_n)\).

Proof ▶