We would like to make sense of a stochastic integral
\int_0^T H_s \, dW_s,
the running gain of a strategy H traded against a Brownian source
of risk W. The obvious move is to reuse the classical
Riemann–Stieltjes integral
\int H \, dg path by path — fix an outcome
\omega, freeze the Brownian path
t \mapsto W_t(\omega), and integrate against it like any other
function. This page shows, in detail, why that fails — and why the failure is
structural, not a technicality.
Two obstructions appear. First, the Stieltjes integral
\int H \, dg only exists when the integrator
g has finite total variation, and a Brownian path
has infinite total variation on every interval. Second — and more revealingly — the
Riemann sum we would write down depends on where inside each sub-interval we sample
the integrand, precisely because the
quadratic variation
is non-zero. There is no single number the sum converges to; there is a family of
them, one per sampling rule.
The total-variation obstruction
The Riemann–Stieltjes integral \int_0^T H \, dg is built as a limit
of sums \sum_i H(\xi_i)\,\big(g(t_{i+1}) - g(t_i)\big). For that
limit to exist independently of the partition and the sample points
\xi_i, classical analysis demands that g
have finite total variation,
V(g; [0,T]) = \sup_{\text{partitions}} \sum_{i=0}^{n-1} \big|\, g(t_{i+1}) - g(t_i) \,\big| < \infty.
A Brownian path fails this badly. We saw on the
quadratic-variation
page that \sum_i (\Delta W_i)^2 \to T, a finite positive number. But
the absolute increments dominate their squares once they are small,
|\Delta W_i| \ge (\Delta W_i)^2 / \max_j |\Delta W_j|, and as the
mesh shrinks \max_j |\Delta W_j| \to 0 (paths are continuous). So
\sum_i |\Delta W_i| \;\ge\; \frac{\sum_i (\Delta W_i)^2}{\max_j |\Delta W_j|} \;\xrightarrow{\;\text{mesh}\to 0\;}\; \frac{T}{0^{+}} = +\infty.
The total variation of a Brownian path is infinite on every interval. The classical theorem
that would have handed us \int H\, dW simply does not apply: its one
hypothesis is the one thing Brownian motion does not have.
The centrepiece: \int_0^T W\, dW depends on the sample point
The total-variation argument says the classical integral cannot be defined. The next
computation says something sharper: even the most natural Riemann sum gives
different answers depending on where we sample — so there is no integral to rescue.
We compute the simplest interesting case, H_t = W_t, line by line.
Partition [0, T] by
0 = t_0 < t_1 < \cdots < t_n = T and abbreviate
\Delta W_i = W_{t_{i+1}} - W_{t_i}. Sample the integrand at the
left endpoint of each sub-interval:
S^{\mathrm{L}}_n = \sum_{i=0}^{n-1} W_{t_i}\,\big(W_{t_{i+1}} - W_{t_i}\big) = \sum_{i=0}^{n-1} W_{t_i}\,\Delta W_i.
Step 1 — the algebraic identity that linearises the product. The trick is to
write each term W_{t_i}\,\Delta W_i using the difference of two
squares. From the elementary identity
b^2 - a^2 = (a + b)(b - a) with
a = W_{t_i}, b = W_{t_{i+1}}, and the
observation that a + b = 2a + (b - a),
W_{t_{i+1}}^2 - W_{t_i}^2 = \big(W_{t_{i+1}} + W_{t_i}\big)\,\Delta W_i = \big(2 W_{t_i} + \Delta W_i\big)\,\Delta W_i = 2\,W_{t_i}\,\Delta W_i + (\Delta W_i)^2.
Step 2 — solve for the term we want. Rearranging the line above isolates the
left-endpoint product:
W_{t_i}\,\Delta W_i = \tfrac{1}{2}\big(W_{t_{i+1}}^2 - W_{t_i}^2\big) - \tfrac{1}{2}(\Delta W_i)^2.
Step 3 — sum over the partition. Add the identity over
i = 0, \dots, n-1:
S^{\mathrm{L}}_n = \tfrac{1}{2}\sum_{i=0}^{n-1}\big(W_{t_{i+1}}^2 - W_{t_i}^2\big) \;-\; \tfrac{1}{2}\sum_{i=0}^{n-1}(\Delta W_i)^2.
Step 4 — the first sum telescopes. Consecutive squares cancel, leaving only
the endpoints, and W_0 = 0:
\sum_{i=0}^{n-1}\big(W_{t_{i+1}}^2 - W_{t_i}^2\big) = W_{t_n}^2 - W_{t_0}^2 = W_T^2 - W_0^2 = W_T^2.
Step 5 — the second sum is the quadratic variation. As the mesh shrinks, the
sum of squared increments converges (in L^2) to the quadratic
variation [W]_T = T:
\sum_{i=0}^{n-1}(\Delta W_i)^2 \;\xrightarrow{\;L^2\;}\; [W]_T = T.
Step 6 — pass to the limit. Putting Steps 4 and 5 into Step 3,
\int_0^T W\, dW \;:=\; \lim_{\text{mesh}\to 0} S^{\mathrm{L}}_n = \tfrac{1}{2}\,W_T^2 - \tfrac{1}{2}\,T.
There is the surprise. Ordinary calculus, treating W like a smooth
variable, would have predicted \int W\, dW = \tfrac12 W_T^2 — the
familiar "\int x\, dx = \tfrac12 x^2". The
-\tfrac12 T is a brand-new correction, and it is exactly half the
quadratic variation.
Now sample on the right instead
Repeat the computation sampling the integrand at the right endpoint,
W_{t_{i+1}}. The only change is the sign of the
(\Delta W_i)^2 bookkeeping. From the same Step 1 identity,
W_{t_{i+1}}\,\Delta W_i = \tfrac12(W_{t_{i+1}}^2 - W_{t_i}^2) + \tfrac12(\Delta W_i)^2
(the cross-term flips because we are pairing the larger factor with the increment), so summing
and taking the limit,
S^{\mathrm{R}}_n = \tfrac12\big(W_T^2 - W_0^2\big) + \tfrac12 \sum_i (\Delta W_i)^2 \;\longrightarrow\; \tfrac{1}{2}\,W_T^2 + \tfrac{1}{2}\,T.
The two answers differ:
S^{\mathrm{R}}_n - S^{\mathrm{L}}_n = \sum_{i=0}^{n-1}(\Delta W_i)^2 \;\longrightarrow\; T.
Left and right sampling disagree by exactly the quadratic variation T
— and they disagree in the limit, not just before it. For a smooth integrator this gap
would vanish (squared increments are second order and die); for Brownian motion it survives.
That is precisely why "the" Riemann–Stieltjes integral of W does not
exist: the limit is not well defined until we commit to a sampling rule.
Let (W_t) be a standard Brownian motion. Then, on every interval
[0, T]:
-
Infinite total variation:
\sum_i |W_{t_{i+1}} - W_{t_i}| \to \infty as the mesh shrinks,
so the classical Riemann–Stieltjes integral \int H\, dW is
undefined path by path — its sole hypothesis fails.
-
Sampling-point dependence: the left- and right-endpoint Riemann sums of
\int_0^T W\, dW converge to different limits,
\tfrac12 W_T^2 - \tfrac12 T \qquad\text{vs.}\qquad \tfrac12 W_T^2 + \tfrac12 T,
differing by the quadratic variation [W]_T = T.
-
The choice of sampling point is therefore a genuine definitional choice, not a
convergence detail — the integral only exists once that choice is fixed.
Since the answer depends on the sampling point, the construction of the stochastic integral
names its convention. Two are standard:
-
Itô samples at the left endpoint
W_{t_i}, giving
\int_0^T W\, dW = \tfrac12 W_T^2 - \tfrac12 T.
-
Stratonovich samples at the midpoint
\tfrac12(W_{t_i} + W_{t_{i+1}}), giving the calculus-flavoured
\int_0^T W \circ dW = \tfrac12 W_T^2 with no correction — at the
cost of being anticipating.
Finance chooses the left point — the
Itô integral — for a financial reason, not a
mathematical one: at time t_i a trader can only know the value of
the position W_{t_i} they hold before the next move
\Delta W_i happens. The left point is the only choice that is
non-anticipating — you bet, then the market moves — and that is exactly what
makes the Itô integral a fair game (a martingale). Stratonovich's midpoint peeks half a step
into the future, which is unphysical for a trading strategy. The two are related by a clean
dictionary, so nothing is lost; finance simply lives on the left endpoint.