The Stratonovich Integral

The Itô integral evaluates the integrand at the start of each tiny time step — a deliberate choice, forced by the requirement that a trading strategy can only use information available before the price moves. But that choice has a cost: Itô calculus needs its own special chain rule (Itô's lemma) because ordinary calculus's chain rule silently breaks. There's a second, equally valid way to build the integral that keeps ordinary calculus intact — you just have to give up the ability to trade on it.

The analogy: pricing a fill at the open versus at the midpoint

Imagine averaging the fill price of many small trades executed continuously over a jagged, jumpy price series. One convention says: for each tiny slice of time, use the price at the very start of the slice — this is what an executable trading strategy is forced to do, since you commit to a position before you see where the price goes next. A second convention says: use the price at the midpoint between the start and end of the slice, effectively peeking slightly into the future of each tiny step to average the two endpoints. Both are legitimate ways to define "the integral," but they give numerically different answers whenever the underlying path is as jagged as Brownian motion — and only the first one describes something an actual trader could execute.

Writing it down

Both integrals approximate $\int_0^T f(W_t)\, dW_t$ as a sum over small time steps $[t_i, t_{i+1}]$, but they differ in where $f$ is evaluated within each step. The Itô integral uses the left endpoint:

$$\int_0^T f(W_t)\, dW_t \approx \sum_i f(W_{t_i}) \left(W_{t_{i+1}} - W_{t_i}\right).$$

The Stratonovich integral, written with a small circle, uses the midpoint value instead:

$$\int_0^T f(W_t) \circ dW_t \approx \sum_i f\!\left(\frac{W_{t_i}+W_{t_{i+1}}}{2}\right) \left(W_{t_{i+1}} - W_{t_i}\right).$$

In words: same jagged path, same tiny increments — the only difference is whether you evaluate $f$ using only information you had at the start of the step (Itô) or using a value that blends in where the path ends up (Stratonovich). Because Brownian motion is so jagged (its quadratic variation is nonzero — see Quadratic Variation), this seemingly small difference in evaluation point does not wash out in the limit; the two integrals converge to genuinely different random variables. The relationship between them, for $f(W_t)$, is

$$\int_0^T f(W_t) \circ dW_t = \int_0^T f(W_t)\, dW_t + \frac{1}{2}\int_0^T f'(W_t)\, dt.$$

In words: Stratonovich equals Itô plus a specific correction term built from the derivative of $f$ — and that correction is exactly what allows the Stratonovich integral to obey the ordinary chain rule, since it's designed to absorb the extra term Itô's lemma would otherwise need to add separately.

Worked example 1: the two integrals of $W_t\, dW_t$, computed both ways

Take $f(W_t) = W_t$ itself, so you're computing $\int_0^T W_t\, dW_t$ both ways. By Itô's lemma applied to $W_t^2$: $d(W_t^2) = 2W_t\, dW_t + dt$, so integrating both sides from 0 to $T$ gives $W_T^2 - W_0^2 = 2\int_0^T W_t\, dW_t + T$, and since $W_0=0$,

$$\int_0^T W_t\, dW_t = \frac{W_T^2 - T}{2} \quad \text{(Itô)}.$$

The extra $-T/2$ is the signature "correction term" that never appears in ordinary calculus. Now apply the conversion formula with $f(x)=x$, so $f'(x)=1$: $\int_0^T W_t \circ dW_t = \int_0^T W_t\, dW_t + \frac12 \int_0^T 1\, dt = \frac{W_T^2-T}{2} + \frac{T}{2} = \frac{W_T^2}{2}$. The Stratonovich version comes out exactly $W_T^2/2$ — precisely the answer ordinary calculus would give if $W_t$ were an ordinary smooth function. That's the whole point of Stratonovich: it's built to reproduce the ordinary calculus answer.

Each path sampled above is jagged at every scale — no matter how finely you slice it into steps, the path still wiggles within each slice. That's precisely the property that keeps the left-endpoint sum (Itô) and the midpoint sum (Stratonovich) from converging to the same number: with a smooth path the two conventions would agree in the limit, but Brownian motion's jaggedness never smooths out, however small the steps get.

Worked example 2: plugging in numbers

Suppose after simulating a path, $W_T = 1.2$ and $T = 1$ (in some consistent time units). The Itô integral evaluates to $\frac{1.2^2 - 1}{2} = \frac{1.44-1}{2} = 0.22$. The Stratonovich integral evaluates to $\frac{1.2^2}{2} = 0.72$. Same path, same endpoints, a difference of exactly $0.5 = T/2$ between the two — matching the correction term computed symbolically above, and a concrete illustration that "the integral of a Brownian path" is not a single unambiguous number until you've specified which convention you mean.

What this means in practice

• Trading strategies and pricing models use Itô, because the left-endpoint evaluation is the only one consistent with "you commit to a position using only information available at that moment" — using Stratonovich for a P&L calculation would silently assume access to future information within each infinitesimal step.
• Physics and engineering SDEs often use Stratonovich, because physical noise processes are frequently better modeled as a limit of smooth, correlated noise (where Stratonovich is the natural limit) rather than the idealized instantaneous, uncorrelated noise of Itô calculus.
• The conversion formula is the practical bridge: a model specified in Stratonovich form (common in some numerical schemes and physics literature) can always be rewritten in Itô form by adding the correction drift term, so a quant reading a Stratonovich SDE can convert it to the tradable Itô version before touching it.