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Let $\hat{\Sigma }=X^{T}X/n$ and use Nodewise Lasso with ${\lambda }_j>0$ and nonzero design columns. Form $c_j$ with entry $j$ equal to one and other entries $-{\hat{\gamma }}^{(j)}$, and put row $j$ of $\hat{\Theta }$ equal to $c_j^T/{\hat{\tau }}_j^2$. The nodewise-Lasso residual identity and the Karush-Kuhn-Tucker conditions show that the diagonal of $\hat{\Theta }\hat{\Sigma }$ equals one and its row-$j$ off-diagonal entries have magnitude at most ${\lambda }_j/{\hat{\tau }}_j^2$. For the Debiased Lasso,The displayed bound follows row by row from Holder inequality. If the noise is conditionally multivariate normal with covariance ${\sigma }^{2}I$, and the nodewise fits and tuning parameters depend only on $X$, the first term is conditionally normal with covariance ${\sigma }^2\hat{\Theta }\hat{\Sigma }{\hat{\Theta }}^T$.

In the normal linear model, assume $\epsilon \sim N_n(0,{\sigma }^{2}I)$ is independent of $X$. Independent rows of $X$ with multivariate normal distribution $N_p(0,I_p)$ give a simple random design for Debiased Lasso. Let $p\to \infty$, $\log p=o(n)$ and $s\log p=o(\sqrt{n})$, with fixed positive noise scale. Choose the Lasso penalty and all Nodewise Lasso penalties as sufficiently large constant multiples of $\sqrt{\log p/n}$, including the noise scale in the former. Uniform sample cross-correlations are bounded by this scale, so zero is the nodewise solution with probability tending to one. Column squared norms divided by $n$ converge uniformly to one. To check the Compatibility condition for the Lasso directly, put $\hat{\Sigma }=X^{T}X/n$ and $\eta ={\max }_{j,k}|{\hat{\Sigma }}_{jk}-1_{\{j=k\}}|$. On the Lasso cone condition, $\Vert \delta {\Vert }_1\le 4\Vert {\delta }_S{\Vert }_1$, and the Cauchy-Schwarz inequality givesFor $s\ge 1$, entrywise concentration of a Gaussian Gram matrix gives $\eta \le D_0\sqrt{\log p/n}$ with probability tending to one for sufficiently large fixed $D_0$. The stated sparsity condition implies $s\eta \to 0$ on this event, so one can take ${\varphi }^2\ge 1/2$ eventually. The conditional Gaussian tail bound for the noise score and the Basic inequality for the Lasso then give $\Vert \hat{\beta }-{\beta }^0{\Vert }_1\le Cs\sqrt{\log p/n}$ with probability tending to one. Consequently the debiased-Lasso remainder bound is at most $As\log p/\sqrt{n}$ with probability tending to one, for a fixed constant $A$. These assumptions allow both $p$ and $s$ to grow, for example $p=n^2$, $s=\lfloor n^{1/4}\rfloor$.

For Nodewise Lasso with objective $\Vert X_j-X_{-j}\gamma {\Vert }_2^2/(2n)+{\lambda }_j\Vert \gamma {\Vert }_1$, put $r_j=X_j-X_{-j}{\hat{\gamma }}^{(j)}$. Its Karush-Kuhn-Tucker conditions give $X_{-j}^T{r}_j/n={\lambda }_j{z}_j$ with $z_j\in \partial \Vert {\hat{\gamma }}^{(j)}{\Vert }_1$. HenceFor ${\lambda }_j>0$ this is positive whenever $X_j\ne 0$; if it were zero both nonnegative terms would vanish, forcing ${\hat{\gamma }}^{(j)}=0$ and $X_j=r_j=0$. Nonzero columns are therefore an essential hypothesis for division by this quantity.

Fit the conditional law of each variable given all others and use its predictor support to estimate its neighbours. In a Gaussian graphical model, this is linear regression with coefficients given by conditional regression coefficients from a Gaussian precision matrix. The Nodewise Lasso supplies sparse fits; the OR rule for nodewise graph selection and AND rule for nodewise graph selection combine possibly disagreeing directional supports into an undirected graph.

Put $\Omega ={\Sigma }^{-1}$, the precision matrix. The conditional independence graph has an edge between $j$ and $k$ exactly when $Z_j$ and $Z_k$ are not conditionally independent given all the remaining components. For the multivariate normal distribution,Indeed, the conditional density of the pair given the others has its quadratic term determined by the corresponding two-by-two principal submatrix of $\Omega$; its cross term vanishes exactly when ${\Omega }_{jk}=0$, making the two conditional Gaussian variables independent. This is a Gaussian graphical model.

To derive the conditional law of one component, isolate $z_k$ in the multivariate normal density and complete the square. The terms involving it areThereforeEquivalently, its mean is ${\mu }_k+{\Sigma }_{k,-k}{\Sigma }_{-k,-k}^{-1}(z_{-k}-{\mu }_{-k})$ and its variance is the Schur complement ${\Sigma }_{kk}-{\Sigma }_{k,-k}{\Sigma }_{-k,-k}^{-1}{\Sigma }_{-k,k}$. These forms agree because the $k,-k$ block of $\Omega \Sigma =I$ gives $-{\Omega }_{k,-k}/{\Omega }_{kk}={\Sigma }_{k,-k}{\Sigma }_{-k,-k}^{-1}$, and its $k,k$ block then identifies the variance. This proves the relevant conditional multivariate normal distribution formula.

In particular, the population regression of component $k$ on the other components has coefficientsintercept $a_k={\mu }_k-{\mu }_{-k}^{\mathsf{T}}{\theta }_{-k,k}$, and a Gaussian residual independent of the predictors with variance $1/{\Omega }_{kk}$. These are the conditional regression coefficients from a Gaussian precision matrix. The coefficient is zero exactly when the corresponding edge is absent. Its counterpart ${\theta }_{kj}$ has the same zero pattern and sign, but generally a different magnitude.

This motivates nodewise regression: for each $k$, regress $X_k$ on $X_{-k}$, estimating an intercept as well. For sparse graph estimation, use the Nodewise Lasso criterionThe Lasso penalty encourages many fitted coefficients to be zero and remains usable when the number of predictors exceeds the sample size. Ordinary unpenalized sample regressions would usually give a dense estimated graph. Separate regressions can disagree about a pair, so an undirected graph can use the OR rule for nodewise graph selection, retaining an edge if either fitted direction is nonzero, or the more conservative AND rule for nodewise graph selection, retaining it only if both are nonzero. These are choices for combining estimated neighbourhoods, rather than identities forced by finite-sample fits.

The printed joint objective uses a Group Lasso penalty on the unordered pairsThis is paired group Lasso for Gaussian graph estimation. Each group represents a single undirected edge, coupling its two directional regressions. A sum of group norms encourages entire pairs to be zero; a squared group norm would give smooth ridge-type shrinkage instead of edge deletion. The constraint ${\Theta }_{kk}=0$ removes self-prediction, and the intercepts are not penalized.

The free symbols called ${\mu }_k$ in the objective are regression intercepts, not generally the marginal means of the original Gaussian vector. Profiling them out givesand replaces all columns by their centred versions in the squared-error terms. This is profiling intercepts in nodewise regression. Scaling predictors to comparable units can also help interpret a common penalty, although the displayed objective itself is well defined on the given data.

The grouped KKT conditions make the sparsity mechanism precise. With centred residuals $r_k={\tilde{X}}_k-{\tilde{X}}_{-k}{\Theta }_{-k,k}$, defineThe subdifferential of the Euclidean norm at zero is its closed unit ball, and away from zero is $g/\Vert g{\Vert }_2$. Hence a minimizer satisfiesThese are the paired group Lasso optimality conditions. They jointly threshold the two directional scores, matching the population fact that both directions represent the same edge.

Estimate the graph by including $\{j,k\}$ exactly when $\Vert {\hat{g}}_{jk}{\Vert }_2>0$. This graph is undirected by construction. The penalty does not impose ${\Theta }_{jk}={\Theta }_{kj}$, nor does it mathematically require both coordinates of every selected pair to be nonzero. That distinction is appropriate because the population regression coefficients themselves need not have equal magnitudes. The fitted coefficient matrix is not automatically a symmetric positive definite precision matrix; its pair supports, rather than its entries as a precision estimate, supply the requested conditional independence graph.

Put $\hat{\Sigma }=X^{T}X/n$. The Debiased Lasso correction isFor the Nodewise Lasso, let $r_j=X_j-X_{-j}{\hat{\gamma }}^{(j)}$. Its Karush-Kuhn-Tucker conditions areso $z_{jk}=\mathrm{sgn}({\hat{\gamma }}_k^{(j)})$ at a nonzero coefficient and $|z_{jk}|\le 1$ at a zero coefficient. Multiplying by ${\hat{\gamma }}^{(j)}$ gives the nodewise-Lasso residual identityWe require nonzero design columns, a necessary qualification missing from the unrestricted printed formulation. Because ${\lambda }_j>0$, ${\hat{\tau }}_j^2=0$ would force both $r_j=0$ and ${\hat{\gamma }}^{(j)}=0$, and hence $X_j=0$. Thus nonzero columns suffice for ${\hat{\tau }}_j^2>0$ even when $p>n$ or the full Gram matrix is singular. A zero column instead gives ${\hat{\tau }}_j^2=0$ and its coefficient is unidentifiable from $Y$; the proposed normalization and confidence interval do not exist there.

For each $j$, define $c_j\in {\mathbb{R}}^p$ by $(c_j{)}_j=1$ and $(c_j{)}_{-j}=-{\hat{\gamma }}^{(j)}$. Set row $j$ of $\hat{\Theta }$ to $c_j^T/{\hat{\tau }}_j^2$. Since $X{c}_j=r_j$,This row construction is the nodewise normalization used by van de Geer and coauthors. The matrix $\hat{\Theta }$ need not be symmetric or be an actual inverse.

Substitute the linear model into the correction. The exact decomposition isIn the fixed-design normal linear model, or with $\epsilon \mid X\sim N_n(0,{\sigma }^{2}I)$ and nodewise estimates and tuning chosen from $X$ alone, a linear transformation of a multivariate normal distribution gives $W\mid X\sim N_p(0,{\sigma }^2\hat{\Omega })$. Marginal normality of $\epsilon$ without this conditional assumption would not suffice for a response-dependent random design. No independence between $W$ and $\Delta$ is asserted. The diagonal cancellation and the Holder inequality give the debiased-Lasso remainder bound

For a coordinate variance, ${\hat{\Theta }}_j{X}^T=r_j^T/{\hat{\tau }}_j^2$, whenceWriting $z_{1-\alpha /2}$ for the corresponding standard normal quantile, an approximate confidence interval isEquivalently its half-width is $z_{1-\alpha /2}\sigma \Vert X_j-X_{-j}{\hat{\gamma }}^{(j)}{\Vert }_2/(n{\hat{\tau }}_j^2)$. We assume $0<\alpha <1$ and $\sigma >0$. For asymptotic coverage, the standardized remainder ${\Delta }_j/(\sigma \sqrt{{\hat{\Omega }}_{jj}})$ must have convergence in probability to zero and the coordinate variance must be nondegenerate. Exact conditional normality of $W$ alone does not give finite-sample exact coverage for $\hat{b}$.

One sufficient random-design regime is the Gaussian identity design for debiased Lasso: take independent rows $X_i\sim N_p(0,I_p)$, independently of $\epsilon$, fixed $\sigma >0$, $p=p_n\to \infty$, $s=s_n\ge 1$, andChoose $\lambda =C\sigma \sqrt{\log p/n}$ and all ${\lambda }_j=D\sqrt{\log p/n}$ for sufficiently large fixed $C,D$. Here is why this regime has the requested remainder bound. Uniformly over columns, their squared norms divided by $n$ tend to one, while all off-diagonal sample cross-correlations have magnitude at most a constant times $\sqrt{\log p/n}$ with probability tending to one. Taking $D$ larger makes the zero vector satisfy all nodewise Karush-Kuhn-Tucker conditions strictly, so ${\hat{\gamma }}^{(j)}=0$ for every $j$ on that event, and ${\hat{\tau }}_j^2=\Vert X_j{\Vert }_2^2/n\ge 1/2$ eventually. This also gives ${\hat{\Omega }}_{jj}=1/{\hat{\tau }}_j^2\to 1$ uniformly.

In this regime, the entrywise entrywise concentration of a Gaussian Gram matrix and Lasso cone condition argument in Gaussian identity design for debiased Lasso gives a Compatibility condition for the Lasso bounded away from zero with probability tending to one. Together with the noise-score event and the Basic inequality for the Lasso, this yields, for a fixed $C_1$,For example, the cone inequality and compatibility bound give $\Vert \hat{\beta }-{\beta }^0{\Vert }_1\le 12\lambda s/{\varphi }^2$ on their joint event. Therefore, with a fixed $A\ge 2D{C}_1$,The stated sparsity condition makes this remainder vanish and supplies the preceding confidence-interval requirement. Both dimensions may grow: $p=n^2$ and $s=\lfloor n^{1/4}\rfloor$ satisfy these assumptions. The concentration and Gaussian-design compatibility results are sufficient standard conditions here; the question does not require their proofs. The displayed rate is a high-probability fixed-constant bound, rather than only a claim of boundedness in probability.

The moral graph of a Directed acyclic graph is formed by joining any two parents of a common child and then removing the directions from all edges. For D-separation, a path in the underlying undirected graph is active given $S$ when every noncollider on it is outside $S$, and every collider has itself or a descendant in $S$. Disjoint vertex sets $A$ and $B$ are D-separated by $S$ if no such active path connects them.

The adjacency characterization isA one-edge path cannot be blocked by conditioning on other vertices. For a nonadjacent pair, choose the later vertex $v$ in a topological ordering. The earlier vertex $u$ is a nondescendant and is not a parent of $v$, so the parents of $v$ separate the pair by the local Markov property of a directed acyclic graph.

Use faithfulness of a directed acyclic graph to mean that graphical and probabilistic independences agree, for all disjoint sets:This includes the global Markov property of a directed acyclic graph as well as the absence of extra independences. The conditional independence graph of $P$ is the undirected graph in which $u-v$ is absent exactly when $Z_u\perp Z_v\mid Z_{V\setminus \{u,v\}}$.

To identify this graph, condition on every vertex other than $u,v$. Any noncollider in the interior of a path is then conditioned on, so an active path must have only colliders internally. Two adjacent interior vertices cannot both be colliders: the arrow joining them has an arrowhead at only one of its ends. Thus an active path can have no interior vertex, giving a direct edge, or just one, giving $u\to w\leftarrow v$. In the latter case the conditioned common child $w$ activates the path. These possibilities are exactly the edges added or retained by moralization. Faithfulness therefore provesThis is why the moral graph equals the conditional independence graph under faithfulness.

For Gaussian graph estimation, collect the data into an $n\times p$ matrix $D$ and let $S_n=n^{-1}{\sum }_i{x}_i{x}_i^T$. The precision matrix $\Theta ={\Sigma }^{-1}$ has ${\Theta }_{jk}=0$ precisely for absent graph edges, by Gaussian conditional independence. In neighbourhood selection, fit a Lasso regression of each column on the others:Select the nonzero coefficients as neighbours of $j$, then make the graph undirected using either the OR rule, retaining an edge selected in at least one regression, or the AND rule, requiring both. These are instances of Nodewise Lasso.

The Graphical Lasso estimates the whole precision matrix jointly, for example byRead off its nonzero off-diagonal entries as graph edges. This expression uses an off-diagonal penalty; an all-entry penalty is another convention.

To reduce computation in the PC algorithm, start its skeleton search from an estimate of the conditional independence graph rather than the complete graph. The exact graph contains the true DAG skeleton, so oracle tests can remove its extra edges while retaining the true adjacencies. Fewer initial edges and smaller neighbourhoods reduce both candidate pairs and candidate conditioning sets.

For correctness of the orientation stage, preserve valid separating-set records. With an exact conditional independence graph, an initially absent pair $u,v$ has the full complement $V\setminus \{u,v\}$ as a separating set. Record that set, and record the usual discovered sets for edges removed later. Leaving an initially absent pair with an empty default could falsely mark an unshielded noncollider as a collider. This describes PC initialization from a conditional independence graph; with an estimated initial graph, screening errors can propagate into the output.

One standard construction uses the Debiased Lasso. Starting from the Square-root Lasso estimate $\hat{\beta }$, estimate a vector ${\hat{\Theta }}_j$ that approximately inverts the $j$th column of the empirical Gram matrix $\hat{\Sigma }=X^{T}X/n$, for example by a Nodewise Lasso. Defineand estimate $\sigma$ by $\hat{\sigma }=\Vert Y-X\hat{\beta }{\Vert }_2/\sqrt{n}$. The approximate two-sided level-$\alpha$ test rejects $H_j^{\beta }$ whenwhere $z_{1-\alpha /2}$ is a standard normal quantile.

Sufficient high-dimensional conditions include a Compatibility condition for the Lasso bounded away from zero, $\gamma \asymp \sqrt{\log p/n}$, $\log p=o(n)$, andtogether with the corresponding sparsity and consistency conditions for the nodewise inverse-Gram estimate. Under these assumptions the Debiased-Lasso asymptotic normality makes the rejection probability under $H_j^{\beta }$ tend to $\alpha$.