Two key examples
Two facts carry an astonishing amount of the theory. Throughout, fix times
s \le t and use two properties of
Brownian motion:
the increment W_t - W_s is independent of the
history \mathcal{F}_s, and it is distributed
N(0,\, t-s) — so it has mean
0 and variance
t-s.
(a) Brownian motion is a martingale
Step 1 — split the future into present plus increment. Pure algebra:
\mathbb{E}[W_t \mid \mathcal{F}_s] = \mathbb{E}\big[\, W_s + (W_t - W_s) \,\big|\, \mathcal{F}_s \,\big].
Step 2 — split the conditional expectation. Conditional expectation is
linear, so the two terms separate:
= \mathbb{E}[W_s \mid \mathcal{F}_s] + \mathbb{E}[\,W_t - W_s \mid \mathcal{F}_s\,].
Step 3 — pull out the known term. The present value
W_s is \mathcal{F}_s-measurable (it is
part of the history), so conditioning leaves it untouched:
\mathbb{E}[W_s \mid \mathcal{F}_s] = W_s.
= W_s + \mathbb{E}[\,W_t - W_s \mid \mathcal{F}_s\,].
Step 4 — drop the conditioning on the increment. Because
W_t - W_s is independent of
\mathcal{F}_s, conditioning on the history changes nothing — its
conditional mean equals its plain mean, which is 0:
\mathbb{E}[\,W_t - W_s \mid \mathcal{F}_s\,] = \mathbb{E}[\,W_t - W_s\,] = 0.
Step 5 — collect the terms.
\mathbb{E}[W_t \mid \mathcal{F}_s] = W_s + 0 = W_s.
So W_t is a martingale: the best forecast of any future value is
today's value.
(b) W_t^2 - t is a martingale
On its own W_t^2 drifts upward — it is a submartingale, since
\mathbb{E}[W_t^2] = t grows with t. The
claim is that subtracting t compensates that drift
exactly. We compute \mathbb{E}[W_t^2 \mid \mathcal{F}_s].
Step 1 — write the future as present plus increment, then square. Using
(p+q)^2 = p^2 + 2pq + q^2 with
p = W_s and q = W_t - W_s:
\mathbb{E}[W_t^2 \mid \mathcal{F}_s] = \mathbb{E}\big[\, (W_s + (W_t - W_s))^2 \,\big|\, \mathcal{F}_s \,\big] = \mathbb{E}\big[\, W_s^2 + 2 W_s (W_t - W_s) + (W_t - W_s)^2 \,\big|\, \mathcal{F}_s \,\big].
Step 2 — split into three conditional expectations (linearity again):
= \mathbb{E}[W_s^2 \mid \mathcal{F}_s] + 2\,\mathbb{E}[\,W_s (W_t - W_s) \mid \mathcal{F}_s\,] + \mathbb{E}[\,(W_t - W_s)^2 \mid \mathcal{F}_s\,].
Step 3 — the first term. W_s^2 is
\mathcal{F}_s-measurable, so it passes through untouched:
\mathbb{E}[W_s^2 \mid \mathcal{F}_s] = W_s^2.
Step 4 — the cross term. The factor W_s is known
given \mathcal{F}_s, so it pulls out of the conditional
expectation; what remains is the mean-zero increment:
2\,\mathbb{E}[\,W_s (W_t - W_s) \mid \mathcal{F}_s\,] = 2 W_s \, \mathbb{E}[\,W_t - W_s \mid \mathcal{F}_s\,] = 2 W_s \cdot 0 = 0.
Step 5 — the squared term. The increment is independent of
\mathcal{F}_s, so its conditional second moment equals its plain
second moment — which, since the increment has mean 0, is its
variance t-s:
\mathbb{E}[\,(W_t - W_s)^2 \mid \mathcal{F}_s\,] = \mathbb{E}[\,(W_t - W_s)^2\,] = \operatorname{Var}(W_t - W_s) = t - s.
Step 6 — collect the three pieces.
\mathbb{E}[W_t^2 \mid \mathcal{F}_s] = W_s^2 + 0 + (t - s).
Step 7 — subtract t from both sides. Since
t is a constant, \mathbb{E}[t \mid \mathcal{F}_s] = t,
so
\mathbb{E}[\,W_t^2 - t \mid \mathcal{F}_s\,] = W_s^2 + (t - s) - t = W_s^2 - s.
The right-hand side is the process W_t^2 - t evaluated at the
present time s — so W_t^2 - t is a
martingale. The compensator that restores the fair-game balance is precisely
t, your first glimpse of quadratic variation,
\langle W\rangle_t = t. Refresh the figure to draw a fresh
Brownian path against its mean, with W_t^2 - t overlaid — it too
hovers around zero.
Let (W_t) be a Brownian motion adapted to
(\mathcal{F}_t). Then both
W_t \qquad\text{and}\qquad W_t^2 - t
are martingales: \mathbb{E}[W_t \mid \mathcal{F}_s] = W_s and
\mathbb{E}[\,W_t^2 - t \mid \mathcal{F}_s\,] = W_s^2 - s for all
s \le t. The compensator in the second is exactly the quadratic
variation, \langle W\rangle_t = t.
Example (b) is more than a curiosity — it is the germ of stochastic calculus. Rearranging
Step 6 gives, in increment form,
\mathbb{E}[\,(W_t - W_s)^2 \mid \mathcal{F}_s\,] = t - s,
which says the expected squared increment of Brownian motion equals the
length of the time step. Summing such squared increments over a fine partition,
the fluctuations cancel and the total converges to the elapsed time — that limit is the
quadratic variation
[W]_t = \langle W \rangle_t = t.
This is exactly the new term that ordinary calculus lacks. In a Taylor expansion of
f(W_t) the second-order piece is normally negligible, but here
(dW_t)^2 behaves like dt rather than
vanishing, leaving a surviving correction. That is the content of
Itô's lemma,
df(W_t) = f'(W_t)\, dW_t + \tfrac{1}{2} f''(W_t)\, dt,
whose \tfrac12 f''\,dt term is the compensator we first met as
the -t in W_t^2 - t. (Take
f(x) = x^2: then f'' = 2, and the
correction is exactly dt.)
A martingale stays fair even if you quit at a random,
non-clairvoyant
time. For a continuous martingale (M_t) and a stopping time
\tau satisfying a regularity condition (e.g.
\tau bounded, or
(M_{t \wedge \tau}) uniformly integrable),
\mathbb{E}[M_\tau] = \mathbb{E}[M_0].
Combined with the two martingales above this is a workhorse. Stopping
W_t at the first exit time
\tau from an interval (-b, a) gives
the exit probabilities, and stopping W_t^2 - t at the same
\tau yields \mathbb{E}[\tau] = \mathbb{E}[W_\tau^2]
— the expected time to escape — straight from "a fair game stays fair when you quit on a
rule". As always the regularity condition is essential: an unbounded stopping time can
break the equality.