Proof of Gaussian Process Characterization of Standard Brownian Motion

Throughout, clauses (i)–(iv) refer to Standard Brownian Motion, and all expectations, variances, and covariances of jointly Gaussian random variables below are defined and finite by Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector. We use that the covariance is symmetric, $\operatorname{Cov}(X,Y)=\operatorname{Cov}(Y,X)$, directly from its defining formula, since multiplication of real numbers commutes. We also use repeatedly: (monotonicity) if an event $A$ contains an event $A'$ with $P(A')=1$, then $P(A)=1$, since $A=A'\cup(A\setminus A')$ is a disjoint union, so $P(A)\ge P(A')=1$ by additivity and nonnegativity of the measure $P$, while $P(A)\le P(\Omega)=1$ by the same argument applied to $\Omega=A\cup(\Omega\setminus A)$; and (modification) Almost Sure Modifications of Gaussian Random Vectors are Gaussian: random variables almost surely equal, componentwise, to the components of a Gaussian random vector form a Gaussian random vector with the same mean vector and covariance matrix.

Part 1: a standard Brownian motion satisfies (a) and (b).

Assume $B$ is a standard Brownian motion. Condition (b) is clause (ii) verbatim. It remains to prove (a). By clause (i), $P(B_0=0)=1$.

Marginals. Fix $t>0$. By clause (iii) with $s=0$, the increment $B_t-B_0$ is Gaussian with $\mathbb{E}[B_t-B_0]=0$ and $\operatorname{Var}(B_t-B_0)=t$. The event $\{B_t=B_t-B_0\}$ contains $\{B_0=0\}$ (the equality holds pointwise wherever $B_0=0$), hence has probability $1$ by monotonicity; by Almost Sure Modifications of Gaussian Random Vectors are Gaussian, $B_t$ is Gaussian with $\mathbb{E}[B_t]=0$ and $\operatorname{Var}(B_t)=t$. Similarly, the constant $0$ is Gaussian by Gaussian Random Vectors and Jointly Gaussian Random Variables with the representation having $m=0$ and $\mu_1=0$, and $P(B_0=0)=1$; hence $B_0$ is Gaussian with $\mathbb{E}[B_0]=0$ and $\operatorname{Var}(B_0)=0$, by Almost Sure Modifications of Gaussian Random Vectors are Gaussian and claim 2 of Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector. Thus $\mathbb{E}[B_t]=0$ and $\operatorname{Var}(B_t)=t$ for every $t\ge0$.

Joint Gaussianity at increasing positive times. Let $d$ be a natural number and let $0<s_1<\dots<s_d$. Put $s_0=0$ and $D_k=B_{s_k}-B_{s_{k-1}}$ for $1\le k\le d$. By clause (iii), each $D_k$ is Gaussian; by clause (iv) applied to the sequence $0=s_0<s_1<\dots<s_d$, the increments $D_1,\dots,D_d$ are independent. By Independent Gaussian Random Variables are Jointly Gaussian, $(D_1,\dots,D_d)$ is a Gaussian random vector. Define $Y_k=\sum_{l=1}^{k}D_l$ for $1\le k\le d$; by Affine Transformations of Gaussian Random Vectors are Gaussian (affine image with constants $0$ and matrix entries $M_{kl}=1$ for $l\le k$, $M_{kl}=0$ for $l>k$), the tuple $(Y_1,\dots,Y_d)$ is a Gaussian random vector. By telescoping, pointwise on all of $\Omega$,

so each event $\{B_{s_k}=Y_k\}$ contains $\{B_0=0\}$ and has probability $1$ by monotonicity. By Almost Sure Modifications of Gaussian Random Vectors are Gaussian, $(B_{s_1},\dots,B_{s_d})$ is a Gaussian random vector.

Joint Gaussianity at arbitrary distinct times. Let $d$ be a natural number and let $t_1,\dots,t_d\ge0$ be distinct; at most one of them is $0$. If $d=1$ and $t_1=0$, then $B_{t_1}=B_0$ is Gaussian as shown in the marginals paragraph. Otherwise let $s_1<\dots<s_e$ be the positive values among $t_1,\dots,t_d$ arranged increasingly, so that $e\ge1$ and $e\in\{d-1,d\}$; by the previous paragraph $(B_{s_1},\dots,B_{s_e})$ is a Gaussian random vector. Define a tuple $(W_1,\dots,W_d)$ as an affine image of it: for $1\le i\le d$, if $t_i>0$ let $j(i)\in\{1,\dots,e\}$ be the unique index with $s_{j(i)}=t_i$ and let $W_i$ be the $j(i)$-th component (constant $0$, coefficient $1$ on that component and $0$ elsewhere); if $t_i=0$ let $W_i$ be the constant $0$ (all coefficients $0$). By Affine Transformations of Gaussian Random Vectors are Gaussian, $(W_1,\dots,W_d)$ is a Gaussian random vector. For $t_i>0$ the equality $B_{t_i}=W_i$ holds pointwise on $\Omega$, and for $t_i=0$ the event $\{B_{t_i}=W_i\}=\{B_0=0\}$ has probability $1$; hence $P(B_{t_i}=W_i)=1$ for every $i$, and by Almost Sure Modifications of Gaussian Random Vectors are Gaussian the tuple $(B_{t_1},\dots,B_{t_d})$ is a Gaussian random vector. Hence $B$ is a Gaussian process.

Covariances. By symmetry of the covariance it suffices to treat $0\le s\le t$. If $s=t$, then $\operatorname{Cov}(B_s,B_t)=\operatorname{Var}(B_t)=t=\min(s,t)$ by the marginals paragraph. Next let $0<s<t$. By clause (iv) applied to $0<s<t$, the increments $D_1=B_s-B_0$ and $D_2=B_t-B_s$ are independent, and each is Gaussian by clause (iii), with $\mathbb{E}[D_1]=\mathbb{E}[D_2]=0$, $\operatorname{Var}(D_1)=s$, $\operatorname{Var}(D_2)=t-s$. By Independent Gaussian Random Variables are Jointly Gaussian, $(D_1,D_2)$ is a Gaussian random vector with $\operatorname{Cov}(D_1,D_2)=0$. By Affine Transformations of Gaussian Random Vectors are Gaussian, the pair $(Y_1,Y_2)=(D_1,\ D_1+D_2)$ is a Gaussian random vector with, by the mean and covariance formulas of that lemma,

Pointwise on $\Omega$, $Y_1=B_s-B_0$ and $Y_2=B_t-B_0$, so the events $\{B_s=Y_1\}$ and $\{B_t=Y_2\}$ contain $\{B_0=0\}$ and have probability $1$; by Almost Sure Modifications of Gaussian Random Vectors are Gaussian,

Finally let $s=0<t$. Let $W_2=B_t-B_0$, Gaussian by clause (iii), and let $(W_1,W_2)$ be the affine image of the one-term Gaussian random vector $(W_2)$ whose first coordinate is the constant $0$ (all coefficients $0$) and whose second coordinate is $W_2$ itself; by Affine Transformations of Gaussian Random Vectors are Gaussian this pair is a Gaussian random vector, and its covariance formula gives $\operatorname{Cov}(W_1,W_2)=0$, all coefficients in the first row being $0$. The events $\{B_0=W_1\}=\{B_0=0\}$ and $\{B_t=W_2\}\supseteq\{B_0=0\}$ have probability $1$, so by Almost Sure Modifications of Gaussian Random Vectors are Gaussian, $\operatorname{Cov}(B_0,B_t)=0=\min(0,t)$. Together with the expectations from the marginals paragraph, condition (a) holds.

Part 2: conditions (a) and (b) imply that $B$ is a standard Brownian motion.

Assume (a) and (b). Clause (ii) is condition (b) verbatim.

Clause (i). By (a) and Jointly Gaussian Families of Random Variables and Gaussian Processes, the one-term tuple $(B_0)$ is a Gaussian random vector with $\mathbb{E}[B_0]=0$ and $\operatorname{Var}(B_0)=\min(0,0)=0$. By the degenerate case (claim 1) of Standardization and Cumulative Distribution Function of a Gaussian Random Variable, $P(B_0=0)=1$; the set $\{B_0=0\}$ is an event as noted in Standard Brownian Motion, so $B_0=0$ almost surely, which is clause (i).

Clause (iii). Fix real $0\le s<t$. The pair $(B_s,B_t)$ is a Gaussian random vector by (a) and Jointly Gaussian Families of Random Variables and Gaussian Processes, the times being distinct. By Affine Transformations of Gaussian Random Vectors are Gaussian (affine image with constant $0$ and coefficients $-1,1$), the increment $B_t-B_s$ is a Gaussian random variable with

and, by the covariance formula of the same lemma together with (a) and the symmetry of the covariance,

using $\min(s,t)=s$ and $\min(s,s)=s$, $\min(t,t)=t$. Hence clause (iii) holds.

Clause (iv). Let $p$ be a natural number and let real numbers $0\le t_0<t_1<\dots<t_p$ be given. The times are distinct, so $(B_{t_0},B_{t_1},\dots,B_{t_p})$ is a Gaussian random vector by (a). Define $D_k=B_{t_k}-B_{t_{k-1}}$ for $1\le k\le p$. The tuple $(D_1,\dots,D_p)$ is the image of $(B_{t_0},\dots,B_{t_p})$ under the affine map with constants $0$ whose $k$-th row has coefficient $-1$ on the component with time index $t_{k-1}$, coefficient $1$ on the component with time index $t_k$, and $0$ elsewhere; hence it is a Gaussian random vector by Affine Transformations of Gaussian Random Vectors are Gaussian, and the covariance formula of that lemma together with (a) gives, for $1\le k<l\le p$,

where the minima were evaluated using $t_{k-1}<t_k\le t_{l-1}<t_l$, which holds since $k\le l-1$. By Pairwise Uncorrelated Jointly Gaussian Random Variables are Independent, the increments $D_1,\dots,D_p$ are independent. Hence $B$ has independent increments, which is clause (iv).

All four clauses of Standard Brownian Motion hold, so $B$ is a standard Brownian motion. $\blacksquare$