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After integrating out the multivariate normal distribution $\beta$, the marginal distribution isUp to terms independent of ${\sigma }^2$, the log-likelihood isDifferentiating and setting the result to zero gives the maximum marginal likelihood estimatorThis is an Empirical Bayes method because the estimated hyperparameter is then inserted into the prior and posterior distributions.

Fix $0\le r\le t$. Since $M$ is a martingale,Every finite vector consisting of $M_{t+s}-M_t$ and past values $M_{r_1},\dots ,M_{r_n}$ has a multivariate normal distribution. Therefore uncorrelated jointly normal variables are independent, so the increment is independent of every finite vector of past values. A Monotone class theorem then extends this to independence from ${\mathcal{F}}_t=\sigma (M_r:0\le r\le t)$. This is the independent increments of a Gaussian martingale.

True. The Dambis-Dubins-Schwarz theorem, with an independent continuation of the Brownian motion if $f$ is bounded, representsBecause $f$ is deterministic, every finite vector $(M_{t_1},\dots ,M_{t_n})$ is a finite vector of a Brownian motion at deterministic times and therefore has a multivariate normal distribution. Hence $M$ is a Gaussian process. This is the deterministic quadratic variation characterizes a Gaussian continuous local martingale result.

First integrate out ${\eta }_i$: conditional on ${\xi }_i$, $y_i\sim N(\alpha +\beta {\xi }_i,s^2)$ with $s^2={\sigma }^2+{\sigma }_y^2$. Since $(x_i,y_i)$ is an affine transformation of independent normal variables, it is bivariate normal with meanand covarianceIts determinant simplifies toFor $r_i=(x_i-\mu ,y_i-\alpha -\beta \mu {)}^T$, the observed-data likelihood is therefore

Put $\rho =e^{-\Delta t/\tau }=e^{-x}$ and retain $R=k(0,0)$, which equals one for the stated exponential covariance function. The covariance matrix of $(y_1,y_2,y_3)$ isConditioning a multivariate normal distribution and simplifying givesThe absence of $y_1$ from both conditional moments shows that $y_3$ and $y_1$ are conditionally independent given $y_2$. Equivalently, the exponential-kernel Gaussian process is the stationary Ornstein-Uhlenbeck process, which is Markov.

By Brownian scaling,where $Z$ has the standard three-dimensional multivariate normal distribution. The right-hand side tends to zero in probability, since $Z$ has no atom at the origin. Part (d) gives almost-sure convergence to $Y$, which also implies convergence in probability to $Y$. Uniqueness of a limit in probability therefore gives $Y=0$ almost surely. Hence $|B_t|\to \infty$ almost surely, proving the transience of Brownian motion in dimension at least three in dimension three.

The family $(Y_x)$ is a Gaussian random field. For distinct $x,x^{\prime }$ its covariance isTwo distinct vertices of ${\mathbb{Z}}^d$ have a common neighbour exactly when their graph distance, equivalently their ${\ell }^1$ distance, is two. Since jointly Gaussian variables are independent exactly when they are uncorrelated,Thus the field has finite-range dependence, even though nearest-neighbour values are independent.

The precision matrix is the inverse of a nonsingular covariance matrix. For a multivariate normal distribution, a zero off-diagonal precision entry encodes conditional independence of the corresponding coordinates given all remaining coordinates.

Quadratic discriminant analysis uses class-conditional Gaussian distributions with class-dependent covariance matrices. Their log density ratio is a quadratic polynomial in the observation.

For known zero mean, invertible observation covariance $\Sigma$, target covariance vector $c$ and target variance $v_0$, simple kriging predicts $c^T{\Sigma }^{-1}z$ with error variance $v_0-c^T{\Sigma }^{-1}c$. Completing the covariance quadratic form proves optimality. A multivariate normal distribution makes this the exact conditional mean and variance.

For centered multivariate normal distributions represented by vectors $U,V$ with equal corresponding variances and $\mathbb{E}[U_i{U}_j]\ge \mathbb{E}[V_i{V}_j]$ for all distinct indices, $\mathbb{P}(U_i\le u_i\text{ for all }i)\ge \mathbb{P}(V_i\le u_i\text{ for all }i)$ for every threshold vector. In particular the supremum of $U$ is stochastically smaller than the supremum of $V$. Equal corresponding variances are essential to this formulation. The related Sudakov-Fernique inequality compares expected values of maxima using variances of increments and does not require equal pointwise variances.

For centered multivariate normal distributions, if $\mathbb{E}(U_i-U_j{)}^2\le \mathbb{E}(V_i-V_j{)}^2$ for every pair, then the displayed comparison holds. For a quick proof, make $U,V$ independent and set $Z_r=\sqrt{1-r}U+\sqrt{r}V$. Apply Gaussian integration by parts to $F_{\tau }(z)={\tau }^{-1}\log {\sum }_i{e}^{\tau z_i}$. With softmax function weights $p_i$ and $D=\mathrm{Cov}(V)-\mathrm{Cov}(U)$, the derivative of $\mathbb{E}{F}_{\tau }(Z_r)$ is $\frac{\tau }{4}\mathbb{E}{\sum }_{i,j}{p}_i{p}_j(D_{ii}+D_{jj}-2D_{ij})\ge 0$. Since $\max z_i\le F_{\tau }(z)\le \max z_i+{\tau }^{-1}\log n$, let $\tau \to \infty$. Singular covariance matrices follow by adding identical small independent normal noise to both vectors and taking a limit.