Definition of Brownian motion
A Brownian motion $B_t$ is a stochastic process with the following properties:
-
Stationary increments.
If $s<t$, then $B_t - B_s$ has the same distribution as $B_{t-s} - B_0$. -
Independent increments.
If $s<t$, then $B_t - B_s$ is independent of all $B_r$ for $r \le s$. -
Continuous paths.
The function $t \mapsto B_t$ is almost surely continuous.
Note: The definition does not assume Gaussian increments, but from these three properties it follows that
$$
B_t \sim \mathcal{N}(0,t).
$$
We now ask the natural question:
If a process satisfies $X_t \sim \mathcal{N}(0,t)$ for all $t$, must it be Brownian motion?
The answer is no, as we demonstrate below.
A Gaussian process from an SDE
Consider the stochastic differential equation $$ \begin{align} dX_t &= -\lambda X_t\, dt + \sigma(t)\, dB_t, \\ X_0 &= 0. \end{align} $$
Rewriting, $$ dX_t + \lambda X_t, dt = \sigma(t), dB_t. $$
Multiply both sides by the integrating factor $e^{\lambda t}$: \begin{align} e^{\lambda t} dX_t + \lambda X_t e^{\lambda t} dt &= \sigma(t) e^{\lambda t} dB_t, \\ d(X_t e^{\lambda t}) &= \sigma(t) e^{\lambda t} dB_t. \end{align}
Integrating from $0$ to $t$: $$ X_t e^{\lambda t} - X_0 = \int_0^t \sigma(s) e^{\lambda s} dB_s. $$
Thus $$ X_t = \int_0^t \sigma(s) e^{-\lambda (t-s)}, dB_s. $$
Since this is a stochastic integral with deterministic integrand, $X_t$ is Gaussian with
variance
$$
\int_0^t |\sigma(s) e^{-\lambda (t-s)}|^2 ds.
$$
Forcing the variance to equal $t$
We ask whether we can choose $\sigma(t)$ such that $$ \mathrm{Var}(X_t) = t. $$
So impose $$ \int_0^t \sigma(s)^2 e^{-2\lambda (t-s)} ds = t. $$
Multiply both sides by $e^{2\lambda t}$: $$ \int_0^t \sigma(s)^2 e^{2\lambda s} ds = t e^{2\lambda t}. $$
Differentiate with respect to $t$: $$ \sigma(t)^2 e^{2\lambda t} = e^{2\lambda t} + 2\lambda t e^{2\lambda t}. $$
Therefore $$ \sigma(t) = \sqrt{,1 + 2\lambda t,}. $$
This choice ensures $$ X_t \sim \mathcal{N}(0,t). $$
But $X_t$ is not Brownian motion
Even though $X_t$ has the correct marginal distribution, its increment structure is wrong:
- increments of $X_t$ depend on the past, because of the exponential factor $e^{-\lambda (t-s)}$
- increments are not independent
- increments are not stationary
These can be verified directly from the covariance function: $$ \mathbb{E}[X_t X_s] = \int_0^{\min(s,t)} \sigma(u)^2 e^{-\lambda(t-u)} e^{-\lambda(s-u)} du, $$ which is incompatible with Brownian motion.
Thus, $X_t$ is a Gaussian process with the correct variance, but not a Brownian motion.