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The head of the Brownian snake driven by $g$ is the centered Gaussian process $(Z_t{)}_{0\le t\le 1}$ with covariance function $\mathbb{E}[Z_s{Z}_t]=m_g(s,t)$. Part i shows that these finite-dimensional distributions exist consistently. Moreover,If $g$ has Hölder constant $L$, then $d_g(s,t)\le 2L|t-s{|}^{\alpha }$. The absolute moment formula for a centered normal distribution consequently gives, for every $p\ge 2$,The Kolmogorov continuity theorem, with $p$ arbitrarily large, produces a modification that is $\gamma$-Hölder continuous for every $\gamma <\alpha /2$. Taking $\gamma =\alpha /2-\epsilon$ proves the claim whenever $0<\epsilon <\alpha /2$; for larger $\epsilon$ the assertion is vacuous.

Every linear combination of $X_{t_1},\dots ,X_{t_n}$ is a deterministic constant plus one stochastic integral of a deterministic $L^2$ function against $B$. Part a makes every such combination Gaussian. By the linear-combination characterization of a multivariate normal distribution, $(X_{t_1},\dots ,X_{t_n})$ is jointly Gaussian, so $X$ is a Gaussian process.

For every finite collection of positive times, $(B_{t_1},\dots ,B_{t_m})$ is a linear transformation of the corresponding values of $W$, so $B$ is a centered Gaussian process. If $0<s\le t$, thenThus $B$ has the covariance of Brownian motion.

Its paths are continuous for $t>0$. As $t\downarrow 0$, put $u=1/t$. The given almost-sure limit givesThe Gaussian-process characterization of Brownian motion now proves the time inversion of Brownian motion:

False. Let $Z\sim N(0,1)$ and define $X_t=tZ$. This is a centered continuous Gaussian process. Its natural filtration satisfies $Z=X_s/s\in {\mathcal{F}}_s^X$ for every $s>0$, and hence, for $0<s<t$,with positive probability. Thus $X$ is not a martingale and does not belong to the stated martingale class.

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.

The Zero-boundary Gaussian free field on the unit disc is the centered Gaussian process indexed by finite Borel measures of finite Green energy, with covarianceThis covariance determines all its finite-dimensional distributions.

The Domain Markov property of the Gaussian free field says that for every suitable open $U\subset \mathbb{D}$ one can writewhere $h_U$ is a zero-boundary Gaussian free field on $U$, independent of $h^{\mathbb{D}\setminus U}$, while $h^{\mathbb{D}\setminus U}$ is harmonic on $U$ and carries the information from the field outside $U$.

Part (c), iterated over disjoint nested annuli, gives independent increments, and the law $X_s-X_t\stackrel{d}{=}{X}_{s-t}$ gives stationary increments. Every finite vector is jointly Gaussian by the definition of the Zero-boundary Gaussian free field, and a continuous version was assumed. Moreover $X_0=0$, because the field has zero boundary values.

Thus $X$ is a continuous centered Gaussian process with stationary increments and independent increments. Its variance is a continuous additive function on the nonnegative real numbers, so $\mathrm{Var}(X_t)={\sigma }^{2}t$ for some ${\sigma }^2\ge 0$. The Gaussian-process characterization of Brownian motion now givesfor standard Brownian motion $B$. This is the Circle-average process of the Gaussian free field is Brownian motion.

Let $K\in {\mathbb{R}}^{n\times n}$ have entries $K_{ij}=\kappa (x_i,x_j)$, let $k_{\ast }\in {\mathbb{R}}^n$ have entries $(k_{\ast }{)}_i=\kappa (x_i,x_{\ast })$, and let $k_{\ast \ast }=\kappa (x_{\ast },x_{\ast })$. The Gaussian process prior and independent Gaussian noise implyApplying the conditional multivariate normal distribution gives the Gaussian process regression posteriorwhere

A Dirichlet process mixture model avoids fixing the number of occupied functions. Let $G_0=N_n(0,K)$ be the finite-dimensional Gaussian process law on the common input grid and specifyA draw from a Dirichlet process is almost surely discrete, so several $f_i$ coincide and thereby form clusters. The number of occupied clusters is random and can grow with the data.

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.

The variable $Z={\int }_0^1{B}_{t}dt$ is the $L^2$ limit of Riemann sums of the Gaussian process $B$, so it is a Gaussian random variable. Its mean is zero, and Fubini's theorem with $\mathbb{E}[B_s{B}_t]=\min (s,t)$ givesTherefore $Z\sim N(0,1/3)$.

Let $e_n(t)=\sqrt{2}\sin (n\pi t)$, let $(g_n)$ be independent standard normal variables, and choose $a_n=e^{-n^2}$. DefineFinite collections of values are limits of centered Gaussian vectors, so this is a centered Gaussian process.

The functions $(e_n)$ are an orthonormal basis of $H=L^2(0,1)$. Hencewhich proves ${\mu }_X(H)=1$. For every integer $k\ge 0$,Thus, almost surely and simultaneously for all $k$, the differentiated series converges uniformly on $(0,1)$. Termwise differentiation gives an almost surely infinitely differentiable version.

Finally, let $F\subset H$ be finite-dimensional. Choose a nonzero $h\in F^{\perp }$. Thenis a centered normal variable of variancebecause every $a_n$ is positive and $(e_n)$ is complete. The event $\{X\in F\}$ is contained in $\{⟨X,h{⟩}_H=0\}$, which has probability zero because a nondegenerate normal distribution has no atoms. Therefore ${\mu }_X(F)=0$ for every finite-dimensional linear subspace $F$.

This covariance function is periodic with period $P>0$. Map time to the circle $u(t)=(\cos (2\pi t/P),\sin (2\pi t/P))$ and pull back a Gaussian kernel; since $\Vert u(t)-u(t^{\prime }){\Vert }^2=4{\sin }^2(\pi (t-t^{\prime })/P)$, this proves positive semidefiniteness. A Gaussian process with this kernel has $\mathrm{Var}(f(t+P)-f(t))=0$ and therefore repeats at every fixed phase. Replacing the inner $\pi$ by $2\pi$ changes the actual period to $P/2$.

A zero-mean Gaussian process need not have zero average in each realization. For the periodic covariance function, the constant Fourier component iswhere $I_0$ is a modified Bessel function. Replacing $f(t)$ by $f(t)-P^{-1}{\int }_0^{P}f(u)du$ gives a Gaussian process with kernel $k_P(t,t^{\prime })-c_0$. This is a positive-semidefinite kernel because it is a covariance after a linear transformation, and each sample has zero period average. The real integral representation of the modified Bessel function I0 yields the displayed constant.

If a stationary Gaussian process has covariance spectral density $\hat{K}$, its reproducing-kernel Hilbert space has normon the functions for which this integral is finite.