My background is functional analysis rather than probability, but I would like to understand what is a Brownian motion. Below I'm giving my current understanding, can anyone verify whether I'm correct. Thank you.

Canonical Brownian motion $\mathbf{B}$ is a space $C([0, \infty), \mathbb{R}^n)$ equipped with a certain probability measure. We can write it as $\mathbf{B}= (B_t)_{t \geq 0}$, where $B_t ( f ) = f(t)$, for all $f \in C([0, \infty), \mathbb{R}^n)$. Measure is given by $$P \big( \bigcap_{j=1}^{k} B_{t_j}^{-1}(F_j) \big) = \int_{F_1 \times \ldots \times F_k} p(t_1, x, x_1) \ldots p(t_k - t_{k-1}, x_{k-1}, x_k) \mathrm{d}x_1\ldots \mathrm{d} x_k,$$ where $$ p(t,x,y) = (2\pi t)^{-\frac{n}{2}}\exp\big( - \frac{| x- y|^2}{2t}\big)$$ for $y \in \mathbb{R}^n$ and it is defined on a $\sigma-$field $\sigma( \{ B_t \colon t \geq 0\})$. From this definition we can conclude that $B_t$ is a Gaussian process, that is, that for all $0 \leq t_1 \leq \ldots \leq t_k$ the random variable $(B_{t_1}, \ldots, B_{t_k}) \in \mathbb{R}^{nk}$ has a normal distribution. $B_t$ has independent increments, and $t \mapsto B_t(f) = f(t)$ is continuous and hence show that canonical Brownian motion has properties which define usual Brownian motion - as a Wiener process.

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    $\begingroup$ Sounds like you understand pretty well! Sometimes we specify that the initial point is $x=0$, and call the resulting measure the Wiener measure. $\endgroup$
    – user940
    Jun 6, 2014 at 13:08

1 Answer 1


The realization of Brownian motion as the evaluation maps $(B_t)_{t \geq 0}$ on $C([0,\infty) \to \Bbb{R}^n)$ is certainly popular, and the development follows along the lines of what you wrote using Kolmogorov extension type arguments to create the Brownian process from the finite dimensional distributions (basically the formula you gave for $P(\cap_1^k B_{t_j}^{-1}(F_j))$ are the finite dimensional distributions).

As you are coming from a background in functional analysis, I think it is likely meaningful to point out here the semigroup associated with Brownian motion. You might have noticed that the density $p(t,x,y)$ you wrote for Brownian motion is also the fundamental solution to the heat equation on $\Bbb{R}^n$: $$\partial_t u(t,x) = \frac{1}{2} \Delta_x u(t,x)$$ where $\Delta_x$ is the Laplacian acting on the $x$ variable. In particular, for sufficiently regular $f$, you can solve $$ \begin{cases} \partial_t u(t,x) = \frac{1}{2}\Delta_x u(t,x) & (t, x) \in [0,\infty) \times \Bbb{R}^n \\ u(0,x) = f(x) & x \in \Bbb{R}^n \end{cases} $$ by convolution $$u(t,x) = \int_{\Bbb{R}^n} f(y) p(t,x,y) \, dy$$ Written in terms of semigroups, we have that $u(t,x) = e^{\frac{t}{2} \Delta_x}f (x)$ where $e^{\frac{t}{2} \Delta_x}$ is the heat semigroup.

At the same time, since $p(t,x,y)$ is the density of $B_t$ you have $$ \Bbb{E}_x[f(B_t)] = \int_{\Bbb{R}^n} f(y) p(t,x,y) \,dy $$ where $\Bbb{E}_x$ is the expected value where the process starts at the point $x$; $B_0 = x$. Putting these pieces together, you have that $$ \Bbb{E}_x[f(B_t)] = e^{\frac{1}{2}t \Delta_x} f(x)$$ For this reason, we say that that semigroup associated to Brownian motion is the heat semigroup and that the generator of Brownian motion is $\frac{1}{2} \Delta_x$. What is particularly nice from your standpoint is that you can now investigate many of the properties of Brownian motion from investigating the (densely defined, essentially self-adjoint) Laplacian. More generally, you can follow this paradigm to look at Markov processes and their associated Markov semigroups.


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