Brownian Motions: I

June 19, 2020

This will be part of (hopefully) a series of notes I will write down whilst preparing for some informal talks in a study group, in the process of understanding Brownian motions. I am reading mainly from Robert Haslhofer’s course notes available here, although I might occasionally use other resources, most notably perhaps Durrett’s “Probability: Theory and Examples”. Be warned that I have a less than adequate understanding of probability and I might make many mistakes and have fundamental misunderstandings; these notes are written mostly for personal bookkeeping purposes. Acknowledgement goes to Ritvik Radhakrishnan and Bhaswar Bhattacharya for going through this with me and pointing out several errors and typos.

A real-valued continuous time stochastic process is a measurable function $X : \Omega \times [0, \infty) \to \Bbb R$ for some probability space $(\Omega, \mathcal{A}, \Bbb {P})$, which we think of as a family of random variables $\{X_t\}_{t \geq 0}$ defined on this probability space. Such a process $B : \Omega \times [0, \infty) \to \Bbb R$ is a Brownian motion starting at $x_0 \in \Bbb R$ if $B_0 = x_0$, the process has independent increments, i.e., for $0 \leq t_1 < \cdots < t_k$ the increments $B_{t_2} - B_{t_1}, \cdots, B_{t_k} - B_{t_{k-1}}$ are independent random variables and moreover the increments have a Gaussian law: $B_{t + h} - B_t \sim N(0, h)$ for any $t \geq 0$, $h > 0$. Finally, we demand that the process is sample-continuous, that is, the event $\{\omega \in \Omega : t \mapsto B_t(\omega) \text{ is continuous}\} \subset \Omega$ contains an $\mathcal{A}$-measurable subset of probability $1$.

I will devote the post to establishing existence of Brownian motions. A particularly appealing approach is to try to construct a Brownian motion as a random ray on $\Bbb R$. To set it up, let us denote $C[0, \infty)$ as the space of continuous maps $\gamma : [0, \infty) \to \Bbb R$, and $C_0$ be the subspace of all such maps $\gamma$ such that $\gamma(0) = 0$. We equip this space with the compact-open topology, and let $\mathcal{B}(C_0)$ denote the Borel $\sigma$-algebra. We would like the Brownian motion to be a random variable $B : (\Omega, \mathcal{A}, \Bbb P) \to (C_0, \mathcal{B}(C_0))$ valued in this space. We can switch perspectives by trying to construct the pushforward measure $\mu_B := B_* \Bbb P$ instead, at which point the Brownian motion will be the random variable defined on the probability space $(C_0, \mathcal{B}(C_0), \mu_B)$ tautologically by $B : (C_0, \mathcal{B}(C_0), \mu_B) \to (C_0, \mathcal{B}(C_0))$ with $B(\gamma) = \gamma$.

We start by considering open subsets of $C_0$ of the following form: Let $0 < t_1 < \cdots < t_k$ be a finite sequence of times and $U_1, \cdots, U_k \subset \Bbb R$ be open subsets. Define

\[V(t_1, \cdots, t_k, U_1, \cdots, U_k) = \{\gamma \in C_0 : \gamma(t_i) \in U_i \; \forall\, 1 \leq i \leq k\}\]

Let $p_t(x, y) := \exp(-(x-y)^2/2t)/\sqrt{2\pi t}$ denote the density of the normal distribution $N(x, t)$ at $y$. We define:

\[\displaystyle \mu_B(V(t_1, \cdots, t_k, U_1, \cdots, U_k)) = \int_{U_1 \times \cdots \times U_k} p_{t_1}(0, x_1) \cdots p_{t_k - t_{k-1}}(x_{k-1}, x_k) dx_1 \cdots dx_k\]

One would expect this to be sufficient information to define a probability measure $\mu_B$ on the full measure space $(C_0, \mathcal{B}(C_0))$. We shall quote the following theorem in order to proceed:

Theorem. (Kolmogorov extension theorem) Let $(X_i, \mathcal{B}_i)$ be a family of Borel measure spaces on Polish spaces indexed by $i \in I$ an arbitrary indexing set. For any subset $J \subseteq I$ let us denote $X_J = \prod_{j \in J} X_j$ and $\mathcal{B}_J = \bigotimes_{j \in J} \mathcal{B}_j$ be the corresponding product $\sigma$-algebra. Let $\mu_J$ be a family of probability measures on the measure spaces $(X_J, \mathcal{B}_J)$ for every finite subset $J \subset I$ which are Kolmogorov consistent, i.e., for any pair of finite subsets $J_1 \subset J_2$ of $I$, and any measurable set $A \in \mathcal{B}_{J_1}$, $\mu_{J_2}(\pi_{J_2, J_1}^{-1}(A)) = \mu_{J_1}(A)$. Then there exists a unique probability measure on $(X_I, \mathcal{B}_I)$ which is consistent with the family of measures $\{\mu_J\}_{J \subset I}$, i.e., for any finite subset $J \subset I$, and any measurable set $A \in \mathcal{B}_I$, $\mu(\pi_J^{-1}(A)) = \mu_J(A)$.

Let us thus enlarge our probability space to $\Bbb R^{[0, \infty)}$ equipped with the product $\sigma$-algebra $\mathcal{B}_{\Bbb R}^{[0, \infty)}$. For any finite subset $J = \{t_1 < t_2 < \cdots < t_k\} \subset [0, \infty)$, our definition for $\mu_B$ provides us with a Kolmogorov consistent family of measures $\mu_J$ on $(\Bbb R^J, \mathcal{B}^{\otimes J})$. Thus, we obtain a measure $\mu_B$ on $(\Bbb R^{[0, \infty)}, \mathcal{B}_{\Bbb R}^{[0, \infty)})$.

Unfortunately, we run into a problem. $C_0 \subset \Bbb R^{[0, \infty)}$ does not happen to be a measurable set in the product $\sigma$-algebra. This is simple enough to see: the product $\sigma$-algebra $\mathcal{B}_{\Bbb R}^{[0, \infty)}$ is generated by finite cylinder events

\[\prod_{f \in F} U_f \times \prod_{i \notin F} \Bbb R \subset \Bbb R^{[0, \infty)}\]

where $F \subset [0, \infty)$ is finite and $U_F \subset \Bbb R$ are Borel. Under countable union of such events, we can form the countable cylinder events $\prod_{d \in D} U_d \times \prod_{i \notin D} \Bbb R$ where $D \subset [0, \infty)$ is countable. These are clearly closed under further countable unions and complementation, and thus generates a $\sigma$-algebra. Hence, these must be all the possible events appearing in the $\sigma$-algebra $\mathcal{B}_{\Bbb R}^{[0, \infty)}$. Thus, such events must depend only on countably many time coordinates, which $C_0$ clearly does not — if continuity of a path was definable by simply looking at countably many points on the path, we would be in trouble!

There is a fairly straightforward analytic fix for this. We switch our point of view to random variables instead of measures once again: So far we have successfully defined a stochastic process

\[\begin{gather*}B : (\Bbb R^{[0, \infty)}, \mathcal{B}_{\Bbb R}^{[0, \infty)}, \mu_B) \times [0, \infty) \to \Bbb R \\ B(\gamma, t) = \gamma(t)\end{gather*}\]

It is moreover clear from construction that $B$ satisfies all the axioms of a Brownian motion except sample-continuity, by the fact that $\mu_B$ agrees with an independent vector of normal distributions with variance equal to time-increment on finite cylinder events, which is equivalent to the finite-dimensional distributions of $B$ to be equal in distribution to the Gaussian distribution:

\[(B_{t_1}, \cdots, B_{t_k}) \sim N(\mathbf{0}_k, \text{diag}(t_1, t_2 - t_1, \cdots, t_k - t_{k-1}))\]

We shall modify this process to be sample-continuous by appealing once again to Kolmogorov:

Theorem. (Kolmogorov continuity theorem) Let $X : (\Omega, \mathcal{A}, \Bbb P) \times [0, \infty) \to \Bbb R$ be a stochastic process. Assume there exists constant $\alpha, \beta, C > 0$ such that

\[\displaystyle \Bbb E\vert X_s - X_t\vert ^\beta \leq C \vert s - t\vert ^{1 + \alpha}\;\; \forall \; s, t > 0\]

Then there exists a stochastic process $Y : (\Omega, \mathcal{A}, \Bbb P) \times [0, \infty) \to \Bbb R$ such that $\Bbb P(X_t = Y_t) = 1$ for all $t \geq 0$ such that for any $\gamma < \alpha/\beta$, the sample paths of $Y$ are locally $\gamma$-Holder continuous with probability $1$. In particular, $Y$ is sample-continuous.

Proof. Let us define the sets $A_n = \{\vert X_{i/2^n} - X_{(i-1)/2^n}\vert \leq 1/2^{n \gamma} \; \forall\; 0 < i \leq 2^n \}$. Further, define $B_N = \bigcap_{n \geq N} A_n$; it is an easy exercise to check that for any pair of diadic rationals $s, t$ such that $\vert s - t\vert \leq 1/2^N$, there exists a constant $K = K(\gamma)$ depending only on $\gamma$ such that on $B_N$,

\[\vert X_s - X_t\vert \leq K \vert s - t\vert ^\gamma\]

Notice that $\Bbb P(\vert X_{i/2^n} - X_{(i-1)/2^n}\vert > 1/2^{n\gamma}) \leq \Bbb E \vert X_{i/2^n} - X_{(i-1)}/2^n\vert ^{\beta} \cdot 2^{n \beta \gamma}$ by Markov inequality applied to $\beta$-th exponent. Thus,

\[\displaystyle \begin{aligned}\Bbb P(A_n^c) \leq \sum_{1 \leq i \leq 2^n} P(\vert X_{i/2^n} - X_{(i-1)/2^n}\vert > 1/2^{n\gamma}) & \leq 2^{n(1+\beta\gamma)} \Bbb E\vert X_{i/2^n} - X_{(i-1)/2^n}\vert ^\beta \\ &\leq C 2^{n(1+\beta \gamma)} 2^{-n(1+\alpha)} = C 2^{-n\lambda}\end{aligned}\]

where $\lambda = \alpha - \beta \gamma > 0$, using the hypothesis on the $\beta$-th moments of $X$ and choice of $\gamma < \alpha/\beta$. Thus, $\Bbb P(B_N^c) \leq \sum_{n \geq N} P(A_n^c) = C 2^{-N \gamma}/(1 - 2^{-\gamma})$. Since $\sum_{N \geq 1} \Bbb P(B_N^c) < \infty$, by the Borel-Cantelli lemma the probability that infinitely many events $B_N^c, N = N(\omega) \geq 1$ occur is zero. In particular, for almost every $\omega \in \Omega$, there exists $N = N(\omega) \geq 1$ such that $\omega \in B_N$. Thus, for almost every sample path of $X$, there exists $\delta = \delta(\omega) > 0$ such that $\vert X_t - X_s\vert \leq C \vert t - s\vert ^\gamma$ holds true whenever $s, t$ are diadic rationals such that $\vert s - t\vert \leq \delta$: take $\delta = 2^{-N}$.

Finally define the modification $Y$ of $X$ by setting $Y_t = X_t$ if $t > 0$ is a diadic rational, otherwise find a sequence $\{t_n\}$ of diadic rationals and define $Y_t = \lim_n X_{t_n}$ pointwise. This is well-defined since the sample paths $t \mapsto X_t$ are almost surely continuous on the diadic rationals, which is a dense subset of $[0, \infty)$, hence has a continuous extension to $[0, \infty)$. It remains to be seen that $\Bbb P(X_t = Y_t) = 1$ for a non-diadic rational $t > 0$. For any $\varepsilon > 0$, observe

\[\Bbb P(\vert X_t - X_{t_n}\vert > \varepsilon) \leq \Bbb E\vert X_t - X_{t_n}\vert ^\beta/\varepsilon^\beta \leq C/\varepsilon^\beta \cdot \vert t - t_n\vert ^{1+\alpha}\]

Thus, $\{X_{t_n}\}$ converges in probability to $X_t$. Since the pointwise limit of this sequence in $Y_t$, we are forced to have $X_t = Y_t$ almost everywhere. This fully establishes the theorem.

Finally, observe that the process $B : \Omega \times [0, \infty) \to \Bbb R$ we constructed earlier has the property in the hypothesis with $\alpha = 1, \beta = 4$ since $B_t - B_s \sim N(0, \vert t - s\vert )$ for all $t, s > 0$, hence has fourth moment proportional to $\vert t - s\vert ^2$. Thus, using Kolmogorov’s continuity theorem, we can modify $B$ to be a sample-continuous process. This concludes the construction of a 1-dimensional Brownian motion.