Brownian motion is Markov
Brownian motion
is the archetypal Markov process, and the reason is its independent
increments. Let us prove it carefully, one line at a time, so that nothing is
smuggled in. Fix a present time t, a horizon
s > 0, and write \mathcal{F}_t for the
history of the path up to t (the
filtration).
We want the future value W_{t+s}.
Step 1 — split the future into present plus increment. Add and subtract the
present value. This is just algebra and is always true:
W_{t+s} = \underbrace{W_t}_{\text{present}} + \underbrace{(W_{t+s} - W_t)}_{\text{the increment}}.
Step 2 — name what each piece knows. The present value
W_t is \mathcal{F}_t-measurable — it is
part of the history, so conditioning on \mathcal{F}_t treats it as
a known constant. The increment W_{t+s} - W_t, by the
independent-increments property of Brownian motion, is independent of
\mathcal{F}_t:
(W_{t+s} - W_t) \perp \mathcal{F}_t.
Step 3 — condition on the whole history. Take the conditional law of
W_{t+s} given \mathcal{F}_t. The first
term is a known constant W_t; the second is independent of
\mathcal{F}_t, so conditioning on
\mathcal{F}_t does nothing to it — its conditional law equals its
unconditional law, which is N(0, s) (a Brownian increment over a
gap of length s is centred at 0 with
variance s). A constant plus an independent
N(0,s) is Gaussian, recentred at the constant:
\mathcal{L}\big(W_{t+s} \mid \mathcal{F}_t\big) = N\big(W_t,\; s\big).
Step 4 — read off the only input that survived. The right-hand side mentions
the past only through the single number W_t; the rest of
the history \mathcal{F}_t has dropped out entirely. So conditioning
on the full history gives the same answer as conditioning on the present value alone:
\mathbb{P}\big(W_{t+s} \in A \mid \mathcal{F}_t\big) = \mathbb{P}\big(W_{t+s} \in A \mid W_t\big).
That is exactly the Markov property. Independent increments did all the work: they are
precisely what makes the route taken irrelevant once you know where you are now.
Brownian motion (W_t)_{t\ge 0} is a Markov process. For every
t \ge 0 and horizon s > 0, the
conditional law of W_{t+s} given the history
\mathcal{F}_t depends on that history only through the present
value, and is Gaussian:
\mathcal{L}\big(W_{t+s} \mid \mathcal{F}_t\big) = N\big(W_t,\, s\big), \qquad\text{hence}\qquad \mathbb{P}\big(W_{t+s} \in A \mid \mathcal{F}_t\big) = \mathbb{P}\big(W_{t+s} \in A \mid W_t\big).
For a Markov process the future law is captured by a transition
probability p(s, x, t, y) — the probability density (or
kernel) of being at y at time t given
you were at x at the earlier time s.
Because the present screens off the past, these kernels are all you need; they chain
together through the Chapman–Kolmogorov relation
p(s, x, u, z) = \int p(s, x, t, y)\, p(t, y, u, z)\, dy,
which is just the Markov property written for densities. For Brownian motion the kernel is
the Gaussian p(s,x,t,y) = \tfrac{1}{\sqrt{2\pi(t-s)}}\, e^{-(y-x)^2 / 2(t-s)}.
The property we proved fixes the present time t in advance. The
strong Markov property upgrades this: it lets you restart the clock not at
a deterministic instant but at a
stopping time
\tau — a random time whose arrival you can recognise without
peeking into the future. The claim is that, conditioned on
\mathcal{F}_\tau (everything known up to
\tau), the shifted process
\widetilde W_s := W_{\tau + s} - W_\tau, \qquad s \ge 0,
is again a Brownian motion, started afresh at the origin and independent of the
past \mathcal{F}_\tau. In other words the process forgets its
history not only at fixed clock times but at any rule-based moment — the instant it first
hits a level, say. This is not automatic: it holds because Brownian paths are continuous
and a stopping time can be approximated from above by discrete times, at each of which the
ordinary (fixed-time) Markov property applies. The strong version is what makes hitting-time
and reflection arguments work, and it is the engine behind much of the pricing of
path-dependent options.