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For three-dimensional Brownian motion, every radius $r>0$ and every initial point satisfyIf the initial point is on one of the spheres, it is already a hit at time zero. If it is inside a lattice-centered open ball, its eventual exit hits that sphere. Otherwise choose at each integer time a nearest lattice center. The current displacement from it has norm at most $\sqrt{3}/2$, so the multivariate normal distribution of the next unit increment gives a uniform positive chance of entering its radius-$r/2$ open ball. The geometric tail bound from a uniform escape probability makes eventual entry certain, and continuity forces a boundary crossing. This periodic-family recurrence does not contradict the Brownian sphere-hitting probability in dimension three for one fixed sphere.

For a random vector indexed by vertices $V$, its conditional independence graph is the undirected graph with $u-v$ absent exactly when $Z_u\perp Z_v\mid Z_{V\setminus \{u,v\}}$. In a nonsingular multivariate normal distribution, its edges are the nonzero off-diagonal entries of the precision matrix.

An ellipsoidal confidence region defined by a positive-definite matrix $A$ and a calibrated threshold $c$. For a known-covariance multivariate normal distribution sample, $A=n{\Sigma }^{-1}$ and the $1-\alpha$ quantile of the chi-squared distribution with $d$ degrees of freedom give exact coverage for the $d$-dimensional mean.

For a centered Gaussian random variable in a Banach space, its first Gaussian chaos is the closed linear subspace of L2 space generated by its continuous linear observations. Every element is a centered Gaussian random variable, and every finite collection has a multivariate normal distribution: approximate in L2 space and pass to the characteristic function. The first Gaussian chaos supplies the scalar coordinates of its Cameron-Martin space of a Gaussian random variable in a Banach space.

For a stationary unit-variance normal distribution autoregression with coefficient $a\in (-1,1)$, its vector of length $T\ge 2$ has precision matrix $J$ that is tridiagonal: the two endpoint diagonal entries are $1/(1-a^2)$, interior entries are $(1+a^2)/(1-a^2)$, and adjacent off-diagonal entries are $-a/(1-a^2)$. An independent observation $r=\lambda Z+\eta$, with covariance $vI$, updates precision and information vector toThe conditional law is multivariate normal distribution $N(Q^{-1}h,Q^{-1})$.

Let $0<\beta \le 1$, let $X\in {\mathbb{R}}^p$, and proposeThe proposal distribution is $N(\sqrt{1-{\beta }^2}X,{\beta }^2{I}_p)$. If $X\sim N(0,I_p)$ independently of $Z$, then $(X,Y)$ is a jointly multivariate normal distribution invariant under exchanging $X$ and $Y$, because both random vectors have covariance matrix $I_p$ and their cross-covariance matrices are both $\sqrt{1-{\beta }^2}{I}_p$. Consequently its density $q$ satisfieswhere $\varphi$ is the standard-normal density. Thus the proposal is reversible with respect to the standard normal distribution.

A Gaussian Bayesian network specifies a linear normal conditional model at each node of a Bayesian network: $X_j\mid X_{\mathrm{pa}(j)}\sim N({\alpha }_j+{\gamma }_j^T{X}_{\mathrm{pa}(j)},{\sigma }_j^2)$. Independent local errors and an acyclic ordering generate a joint multivariate normal distribution. Positive conditional variances give a nonsingular joint multivariate normal distribution. The graph constrains regression coefficients and hence conditional independence.

For a nondegenerate multivariate normal distribution, the conditional independence graph is read from zeros in its precision matrix. Conditional independence of two components given all others is equivalent to a zero cross term in their conditional Gaussian quadratic density, hence to ${\Omega }_{jk}=0$. It does not generally correspond to a zero in the covariance matrix.

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 $m=Au+\eta$, with independent $u\sim \mathcal{N}(0,I_d)$ and $\eta \sim \mathcal{N}(0,{\delta }^2{I}_k)$, the multivariate normal distribution of the Bayesian posterior has posterior mean $b_{\delta }=(A^{T}A+{\delta }^2{I}_d{)}^{-1}{A}^{T}m$ and covariance matrix $C_{\delta }={\delta }^2(A^{T}A+{\delta }^2{I}_d{)}^{-1}$. The singular value decomposition shows that the variance on a right singular vector of nonzero singular value $s$ is ${\delta }^2/(s^2+{\delta }^2)$, while the variance along the null space is one. Thus, for fixed data, the mean tends to the minimum-norm least-squares solution and the covariance tends to the orthogonal projection onto the null space. Unobserved components retain their prior distribution even as the noise vanishes.

A Gaussian process is a stochastic process whose every finite vector of values has a multivariate normal distribution. Its law is determined by its mean and covariance functions.

A Gaussian random vector has a multivariate normal distribution; equivalently, every real linear combination of its coordinates is a normal random variable, allowing zero variance. Its expected value vector and covariance matrix determine its law. Subvectors are independent when their cross-covariances vanish, a feature particular to joint Gaussianity.

For independent and identically distributed random variables $X,Y$ with a centered multivariate normal distribution, a continuously differentiable function $f$ with bounded gradient, and a convex function $\Psi$ for which the expressions are integrable, the displayed inequality holds. First apply Jensen inequality conditionally to $f(X)-f(Y)$. For the Gaussian rotation of independent copies, the chain rule gives $f(X)-f(Y)={\int }_0^{\pi /2}⟨\nabla f(U_{\theta }),V_{\theta }⟩d\theta$. A second application of Jensen inequality to the uniform measure on $[0,\pi /2]$, followed by the Gaussian rotation of independent copies, proves the inequality. The useful choice $\Psi (v)=e^{\lambda |v|}$ yields a Gaussian concentration inequality through an exponential moment of an absolute standard normal variable.

If $X,Y$ are independent and identically distributed random variables with a centered multivariate normal distribution and covariance matrix $\Sigma$, then $(U_{\theta },V_{\theta })$ has the same probability law as $(X,Y)$. Its two marginal covariance matrices are $\Sigma$ and its cross-covariance is zero. The pair has a multivariate normal distribution, so the two components are independent. This remains valid when $\Sigma$ is singular.

If a positive-definite matrix $Q=L{L}^T$ is the precision matrix of a multivariate normal distribution, draw independent $\xi \sim N(0,I)$ and solve $L^T\eta =\xi$. Then $m+\eta$ has covariance $L^{-T}{L}^{-1}=Q^{-1}$. This avoids forming the inverse and exploits a banded Cholesky decomposition when $Q$ is banded.

Two subvectors of a jointly multivariate normal distribution are independent exactly when their cross-covariance matrix is zero.

With a full-column-rank $n\times p$ design and isotropic normal errors, ordinary least squares gives $\hat{\beta }\sim N_p(\beta ,{\sigma }^2(X^{T}X{)}^{-1})$, independent of the residual squared length. Orthogonal design and residual projections of the multivariate normal distribution are independent, and $\mathrm{RSS}/{\sigma }^2\sim {\chi }_{n-p}^2$. The maximum likelihood variance estimate is $\mathrm{RSS}/n$, not the unbiased estimate $\mathrm{RSS}/(n-p)$.

A continuous vector-valued local martingale $M$ starting at zero is standard multidimensional Brownian motion if its quadratic covariations satisfy $[M^i,M^j{]}_t={\delta }_{ij}t$. For each deterministic vector $\xi$, the Itô formula makes $\exp (i\xi \cdot M_t+|\xi {|}^{2}t/2)$ a complex local martingale. Its absolute value is bounded on each finite horizon, so it is a true martingale. Its conditional expectation identity givesThis conditional characteristic function identifies a centered multivariate normal distribution with covariance $(t-s)I$, independent of the past. Thus the coordinates are independent Brownian motions. The converse follows immediately from the standard coordinate quadratic covariations.