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Undirected graphical models and sparse precision estimation — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Visible RBM units are conditionally independent but dependent marginally

Example

Assume AC. In a binary restricted Boltzmann machine with two visible units, one hidden unit, $a = (0, 0)$ , $b = 0$ and $W_{11} = W_{21} = \log 3$ , the visible units are independent conditional on the hidden unit. Each conditional success probability is $1 / 2$ when $H = 0$ and $3 / 4$ when $H = 1$ . Marginally, $P (V_{1} = 1) = P (V_{2} = 1) = 7 / 10$ and $P (V_{1} = 1, V_{2} = 1) = 1 / 2 \neq 49 / 100$ , so $V_{1}$ and $V_{2}$ are dependent.

Example references

The Elements of Statistical Learning, §17.4.4

Hinton, A Practical Guide to Training Restricted Boltzmann Machines, §2

Facts & Assumptions

Given: The RBM parameters and binary state spaces in the Example.

[F1]

The binary RBM conditionals are products of Bernoulli laws with success probabilities $σ (b_{j} + (W^{T} v)_{j})$ and $σ (a_{i} + (W h)_{i})$ . Binary restricted Boltzmann machine, exact likelihood gradient, and finite contrastive divergence.

[F2]

Conditional independence given a sigma-algebra is the bounded-function conditional-expectation factorization. Conditional independence given a sigma-algebra.

[A1]

AC is the explicit hypothesis of the conditional-independence convention, and here means every family of nonempty sets has a choice function. Conditional independence given a sigma-algebra The Axiom of Choice.

Verification

technique · direct

1.1A1F1F2algebra

With $a_{i} = 0$ , $b = 0$ and $W_{i 1} = \log 3$ , [F1] gives $P (V_{i} = 1 ∣ H = h) = σ (h \log 3)$ , which is $1 / 2$ at $h = 0$ and $3 / 4$ at $h = 1$ . The conditional visible mass factors as the product of these two Bernoulli masses, so [A1] and [F2] give conditional independence.

2.1givenstep 1.1algebra

Summing the unnormalized joint mass over the two visible bits gives weight $(1 + 1)^{2} = 4$ to $h = 0$ and $(1 + 3)^{2} = 16$ to $h = 1$ . Thus $Z = 20$ and $P (H = 0) = 1 / 5$ , $P (H = 1) = 4 / 5$ .

3.1step 1.1step 2.1algebra∎

Averaging the conditional success probabilities from step 1.1 yields $P (V_{i} = 1) = \frac{1}{5} \cdot \frac{1}{2} + \frac{4}{5} \cdot \frac{3}{4} = \frac{7}{10}$ for each i. Likewise, conditional independence gives $P (V_{1} = 1, V_{2} = 1) = \frac{1}{5} \cdot \frac{1}{4} + \frac{4}{5} \cdot \frac{9}{16} = \frac{1}{2}$ , whereas the product of the marginals is $\frac{49}{100}$ . Since these differ, the visible variables are marginally dependent.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A three-node Gaussian chain has conditionally independent endpoints

Example

Assume AC. Let $Θ = (\begin{array}{ccc} 2 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 2 \end{array})$ , $Σ = Θ^{- 1}$ , and $X \sim N_{3} (0, Σ)$ . Put an edge between $i$ and $j$ exactly when $Θ_{i j} \neq 0$ . Then the graph is the path $1 - 2 - 3$ , and $X_{1}$ and $X_{3}$ are conditionally independent given $X_{2}$ .

Example references

The Elements of Statistical Learning, §17.3.1

Evans, Graphical Models, §5.1

Graph diagram

The vertices are the variables $X_{1}, X_{2}, X_{3}$ ; an edge is drawn exactly when the corresponding off-diagonal precision entry is nonzero. The missing $1 - 3$ edge records $Θ_{13} = 0$ . The verification uses the quadratic form and the proved Gaussian theorem, not the diagram.

diagram failed to render:
\begin{tikzpicture}[every node/.style={draw,circle,minimum size=8mm}]
  \node (x1) at (0,0) {$X_1$};
  \node (x2) at (1.6,0) {$X_2$};
  \node (x3) at (3.2,0) {$X_3$};
  \draw (x1)--(x2);
  \draw (x2)--(x3);
\end{tikzpicture}

Facts & Assumptions

Given: The Gaussian vector, precision matrix and graph convention in the Example.

[F1]

The Gaussian precision-zero theorem says that $Θ_{i j} = 0$ exactly when $X_{i}$ and $X_{j}$ are conditionally independent given all other coordinates, for positive-definite covariance. Gaussian precision zeros characterize pairwise conditional independence.

[F2]

In a finite undirected graph, an edge is the unordered pair of its endpoints. Finite undirected graphs, Markov properties, and clique factorization.

[A1]

AC is the explicit hypothesis of the Gaussian theorem; here it means every family of nonempty sets has a choice function. Gaussian precision zeros characterize pairwise conditional independence The Axiom of Choice.

Verification

technique · direct

1.1givenalgebra

For $x = (x_{1}, x_{2}, x_{3})^{T}$ , $x^{T} Θ x = x_{1}^{2} + x_{3}^{2} + (x_{1} - x_{2})^{2} + (x_{2} - x_{3})^{2}$ . This is positive for every nonzero $x$ : if the sum vanished, each square would vanish and successively $x_{1} = x_{3} = 0$ , $x_{2} = x_{1} = x_{3} = 0$ . Hence $Θ$ is positive definite and $Σ = Θ^{- 1}$ is a valid positive-definite covariance.

1.2F2givenalgebra

The only nonzero off-diagonal entries are $Θ_{12} = Θ_{21} = - 1$ and $Θ_{23} = Θ_{32} = - 1$ , while $Θ_{13} = Θ_{31} = 0$ . By [F2], the graph is exactly the path $1 - 2 - 3$ .

2.1A1F1step 1.1step 1.2∎

Theorem [F1] applies with $p = 3$ , covariance $Σ ≻ 0$ from step 1.1 and precision $Θ = Σ^{- 1}$ . Since $Θ_{13} = 0$ , it gives $X_{1} ⊥ ⁣ ⁣ ⁣ ⊥ X_{3} ∣ σ (X_{2})$ .

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Increasing λ creates an exact zero edge in a two-variable graphical lasso

Example

Take observations $x_{1} = (1, 1)^{T}$ and $x_{2} = (- 1, - 1)^{T}$ , so their centered sample covariance is $S = (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ . Under the all-entry penalty objective, every $λ \geq 1$ gives the unique maximizer $Θ = (1 + λ)^{- 1} I_{2}$ with $Θ_{12} = 0$ , while for $0 < λ < 1$ every optimizer has $Θ_{12} \neq 0$ . In particular, $λ = 2$ yields a zero edge and $λ = \frac{1}{2}$ yields a nonzero edge.

Example references

Friedman, Hastie and Tibshirani, §2

The Elements of Statistical Learning, §17.3.2

Facts & Assumptions

Given: The two observations and the all-entry graphical-lasso convention in the Example.

[F1]

The centered sample covariance and objective use $∥ Θ ∥_{1} = \sum_{i, j} ∣ Θ_{i j} ∣$ , including diagonal entries. Gaussian sparse-precision objective with the all-entry penalty.

[F2]

For $S ⪰ 0$ and $λ > 0$ , the objective has a unique maximizer characterized by the symmetric entrywise KKT condition; at a zero off-diagonal entry the subgradient lies in $[- 1, 1]$ . The all-entry graphical-lasso objective has a unique KKT optimizer.

Verification

technique · direct

1.1givenF1algebra

The sample mean is zero, and $S = \frac{1}{2} (x_{1} x_{1}^{T} + x_{2} x_{2}^{T}) = (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ . This matrix is positive semidefinite because $u^{T} S u = (u_{1} + u_{2})^{2} \geq 0$ .

2.1F2step 1.1algebra

Let $λ \geq 1$ , $a = (1 + λ)^{- 1}$ , $Θ = a I_{2}$ , and let $Γ$ have diagonal entries $1$ and off-diagonal entries $- 1 / λ$ . Then $Θ ≻ 0$ , $Γ_{12} \in [- 1, 1]$ , and $Θ^{- 1} - S - λ Γ = 0$ : the diagonal entries are $(1 + λ) - 1 - λ = 0$ and the off-diagonal entries are $0 - 1 - λ (- 1 / λ) = 0$ . By [F2], this is the unique maximizer, with zero off-diagonal edge.

3.1givenF2step 1.1step 2.1algebra∎

Let $0 < λ < 1$ . If an optimizer had $Θ_{12} = 0$ , its off-diagonal KKT equation from [F2] would require $- 1 - λ Γ_{12} = 0$ for some $Γ_{12} \in [- 1, 1]$ , hence $Γ_{12} = - 1 / λ < - 1$ , a contradiction. Existence in [F2] therefore forces every optimizer to have $Θ_{12} \neq 0$ . Taking $λ = 2$ and $λ = \frac{1}{2}$ gives the stated two cases.

Undirected graphical models and sparse precision estimation — Examples

2 · Logical flowchart

3 · Definitions, theorems and proofs

4 · Examples, counterexamples and false statements

Visible RBM units are conditionally independent but dependent marginally

Example

Example references

Facts & Assumptions

Verification

A three-node Gaussian chain has conditionally independent endpoints

Example

Example references

Graph diagram

Facts & Assumptions

Verification

Increasing λ creates an exact zero edge in a two-variable graphical lasso

Example

Example references

Facts & Assumptions

Verification

Sources