Fix an adapted square-integrable integrand H and let the upper limit
run, turning the
Itô integral
into a process:
M_t = \int_0^t H_s\, dW_s, \qquad 0 \le t \le T.
This object is the workhorse of mathematical finance — a self-financing trading gain, a
risk-neutral price increment — and it inherits a beautiful list of properties from the
construction. The headline is that M_t is a
continuous martingale:
a fair bet on a fair game stays fair. We collect the properties, then prove the martingale one
in full.
The martingale property, line by line
We show \mathbb{E}[M_t \mid \mathcal{F}_s] = M_s for all
s \le t. The whole proof rests on one structural fact: the Itô
integral, like an ordinary integral, is additive over adjacent intervals.
Step 1 — split the integral at the present time s.
The integral from 0 to t is the integral
up to s plus the integral over the remaining slice
(s, t]:
M_t = \int_0^t H\, dW = \int_0^s H\, dW + \int_s^t H\, dW = M_s + \int_s^t H\, dW.
Step 2 — take the conditional expectation and use linearity. Condition on
everything known at s and split the two pieces:
\mathbb{E}[M_t \mid \mathcal{F}_s] = \mathbb{E}[M_s \mid \mathcal{F}_s] + \mathbb{E}\!\left[\int_s^t H\, dW \;\middle|\; \mathcal{F}_s\right].
Step 3 — the past term is known. The running integral
M_s is \mathcal{F}_s-measurable — it is
built only from increments up to time s — so it passes through the
conditioning untouched:
\mathbb{E}[M_s \mid \mathcal{F}_s] = M_s.
Step 4 — the future slice has conditional mean zero. The increment
\int_s^t H\, dW is itself an Itô integral, built from left-endpoint
positions H_{t_i} (each known at t_i \ge s)
times future increments that are independent of \mathcal{F}_s and
mean-zero. The very argument that gave the integral mean zero, applied conditionally on
\mathcal{F}_s, gives
\mathbb{E}\!\left[\int_s^t H\, dW \;\middle|\; \mathcal{F}_s\right] = 0.
Step 5 — collect. Adding Steps 3 and 4,
\mathbb{E}[M_t \mid \mathcal{F}_s] = M_s + 0 = M_s.
So M_t = \int_0^t H\, dW is a martingale: the best forecast of any
future value of the integral, given the present, is its current value. "Martingale in,
martingale out" — integrating an adapted bet against Brownian motion never creates drift.
Let H be adapted with
\mathbb{E}\big[\int_0^T H_s^2\, ds\big] < \infty, and set
M_t = \int_0^t H_s\, dW_s. Then:
-
Linearity:
\int_0^t (aH + bK)\, dW = a\int_0^t H\, dW + b\int_0^t K\, dW.
-
Zero mean: \mathbb{E}[M_t] = 0 for every
t.
-
Martingale:
\mathbb{E}[M_t \mid \mathcal{F}_s] = M_s for all
s \le t.
-
Variance (Itô isometry):
\mathbb{E}[M_t^2] = \mathbb{E}\big[\int_0^t H_s^2\, ds\big].
-
Continuous paths: t \mapsto M_t has continuous
sample paths (with probability one).
-
Quadratic variation:
[M]_t = \int_0^t H_s^2\, ds.
The martingale M_t has zero mean, so all of its "size" is variance —
and the quadratic variation
records exactly where that variance comes from along the path. For the Itô integral it is
[M]_t = \int_0^t H_s^2\, ds.
Read it as a rate: over an instant dt the martingale accumulates
squared fluctuation d[M]_t = H_t^2\, dt — the local "energy" injected
by the noise, scaled by how aggressively the strategy H_t is
positioned right then. (The differential shorthand is
(H\, dW)^2 = H^2 (dW)^2 = H^2\, dt, the
(dW)^2 = dt rule once more.) Taking the expectation recovers the
isometry, \mathbb{E}[M_t^2] = \mathbb{E}[\int_0^t H^2\, ds] = \mathbb{E}[[M]_t]
— so the quadratic variation is the pathwise refinement of the variance.
A caveat — local martingales. Everything above assumed the
L^2 condition \mathbb{E}[\int_0^T H^2\, ds] < \infty.
When H is only locally square-integrable (the integral of
H^2 is finite up to a sequence of stopping times growing to
T, but its expectation may blow up), the construction still works and
M_t is a local martingale — a martingale "up to a
localising sequence". A local martingale need not have constant mean, so the genuine martingale
property (and the clean \mathbb{E}[M_t] = 0) can fail without the
integrability guard. This is a real subtlety in pricing: a strategy that is only a local
martingale can manufacture an apparent free lunch, which integrability conditions exist to rule
out.