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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.

The correlation coefficient computed under a conditional distribution, using its conditional covariance and positive conditional variances. It can depend on the observed conditioning value. For a multivariate normal distribution, conditioning on other coordinates gives the corresponding partial correlation.

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 real-valued stochastic process $(X_t)$ with finite second moments, write $m(t)=\mathbb{E}{X}_t$. Its covariance function isThe diagonal gives the variance at each time; off-diagonal values record linear dependence between times. For the independent standard normal random variables above, $C_X(m,n)=1_{\{m=n\}}$, whereas the shared Gaussian process $Y_n=Z$ has $C_Y(m,n)=1$ for all $m,n$. For a fair simple random walk on the integer line, $C(m,n)=\min (m,n)$; Brownian motion has the continuous-time counterpart $C(s,t)=\min (s,t)$. These formulas quantify how much random history the observations share.

Every covariance function is a positive-semidefinite kernel. Indeed, for any real coefficients $a_1,\dots ,a_k$,Thus a proposed covariance function must make every finite covariance matrix positive semidefinite. For a Gaussian process, the mean function and the covariance function determine every finite-dimensional distribution, including degenerate ones such as the shared process. Conversely, any real positive-semidefinite kernel satisfying $C(s,t)=C(t,s)$, together with a specified mean function, defines consistent multivariate normal distributions; the Kolmogorov extension theorem then constructs a Gaussian process. For a general stochastic process, covariance alone does not determine the full law. The exceptional completeness of this description is what makes Gaussian process regression possible with a mean and a covariance model.

For fixed $x_1,\dots ,x_d$, the vector $({\alpha }_n(x_j){)}_{j=1}^d$ converges in distribution to $(B(G(x_j)){)}_{j=1}^d$, where $B$ is a standard Brownian bridge. To prove it, write the vector as $n^{-1/2}$ times the sum of independent centred bounded indicator vectors. Their covariance matrix is $G(\min (x_i,x_j))-G(x_i)G(x_j)$. The multivariate central limit theorem gives a centred multivariate normal distribution with this covariance, exactly the bridge vector. Singular covariance matrices, including endpoint times or repeated values of $G$, cause no problem. This is a finite-dimensional assertion; a full function-space limit additionally requires tightness.

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 nonsingular multivariate normal distribution with precision matrix $\Omega$, fixing all coordinates except $X_i$ leaves a density proportional to $\exp [-{\Omega }_{ii}(x_i-m_i{)}^2/2]$ after completing the square. Thus its conditional variance is the reciprocal diagonal precision entry. This is conditioning on every other coordinate, not the marginal variance $({\Omega }^{-1}{)}_{ii}$.

For independent observations from a nonsingular multivariate normal distribution with unknown mean, let $S=n^{-1}{\sum }_i(y_i-\bar{y})(y_i-\bar{y}{)}^T$. The unrestricted maximum-likelihood estimator of covariance is $S$, while the diagonal restriction gives $D=\mathrm{diag}(S_{11},\dots ,S_{pp})$. The fitted trace terms both equal $p$, so twice the log-likelihood difference is $W=n\log (\det D/\det S)=-n\log \det R$, where $R$ is the sample correlation matrix. Under the diagonal null hypothesis and fixed dimension, the Wilks theorem gives $W\Rightarrow {\chi }_{p(p-1)/2}^2$.

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 centered Gaussian process with almost surely continuous sample paths and covariance $\mathbb{E}[B_s{B}_t]=\min (s,t)$ is standard Brownian motion. Here Gaussian process means that every finite-dimensional distribution is a multivariate normal distribution; having separate normal distributions at each time would not suffice. This characterization is stated in Pitman, Basic Properties of Brownian Motion, Definition 15.2.

To compute the Brownian covariance kernel directly, suppose $0\le s\le t$. Since $B_t=B_s+(B_t-B_s)$ and the added Brownian increment is centered and independent of $B_s$,Symmetry gives $\min (s,t)$ for arbitrary nonnegative times. For instance, $\mathrm{Cov}(B_1,B_4)=1$ and their correlation is $1/\sqrt{1\cdot 4}=1/2$. Positions are therefore related even though disjoint Brownian increments are independent: both positions contain the displacement accumulated during the first unit of time. Conversely, taking differences of $\min (s,t)$ gives zero covariance for disjoint increments; their joint multivariate normal distribution turns this into independence.

If $S_1$ has a multivariate normal distribution with mean $\mu$ and invertible covariance $V$, and $g$ has bounded gradient, Gaussian integration by parts gives $\mathrm{Cov}(S_1,g(S_1))=V\mathbb{E}[\nabla g(S_1)]$. Consequently the risky holdings of its one-period quadratic hedge equal the expected payoff gradient. The initial cost is $R^{-1}\mathbb{E}[g(S_1)]+(S_0-R^{-1}\mu )\cdot \mathbb{E}[\nabla g(S_1)]$. Bounded gradient ensures linear growth and square integrability, even if the payoff is not bounded.

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.

Take the original synthetic model $T\sim N(0,0.6)$ and ${\epsilon }_1,{\epsilon }_2\sim N(0,0.4)$, mutually independent random variables, with the second argument denoting variance. The pair $(X,Y)$ has a multivariate normal distribution with mean zero, unit marginal variances, and Pearson correlation coefficient $0.6$. Its conditional multivariate normal distribution isFor a first reading $x=2$, the second reading has conditional expectation $1.2$ and conditional variance $0.64$, hence standard deviation $0.8$. There is still conditional probability $1-\Phi (1)\approx 0.1587$ of $Y>2$, a movement farther from zero, where $\Phi$ is the standard normal distribution function. The line $0.6x$ describes a conditional mean, not an individual trajectory.

A short derivation isolates $R=Y-0.6X$. Its covariance with $X$ is zero and its variance is $0.64$. Because $(X,R)$ has a multivariate normal distribution, zero covariance implies independence: $R\sim N(0,0.64)$ independently of $X$. This also explains why reversing the two measurements gives $\mathbb{E}[X\mid Y=y]=0.6y$. The statistical association does not identify a direction of causal effect.