The Itô Integral

Here is a question every quant eventually has to answer precisely: if you hold a changing number of shares in something whose price is wiggling continuously, what is your total profit at the end of the day? The answer feels obvious — add up "shares held" times "price change" over every instant. The trouble is that a Brownian path has no slope at any point and wanders infinitely far up and down in any interval, however short. The ordinary machinery for adding up things like that quietly breaks. The Itô integral is the repair, and the repair is more interesting than the problem.

The picture before any symbols

Forget continuity for a moment. Imagine a trading day chopped into one-minute bars. At the start of each minute you decide how many shares to hold; you then sit still while the price does whatever it does; at the top of the next minute the bar's move lands in your P&L, and you get to re-decide.

Your profit for the day is the sum, over minutes, of

(shares you were holding at the start of the minute) × (price change during that minute)

Two things about that sentence carry the entire theory. First, your position is fixed before you see the move — you are never allowed to size a trade using the print it is about to earn. Second, you can only decide with what you know so far. That is the whole of it. The Itô integral is that sum, with the minutes shrunk to nothing.

Building it

Write $W_t$ for a standard Brownian path — the price driver — and $H_t$ for the number of units you hold at time $t$. We want to give meaning to

$$I_T \;=\; \int_0^T H_t \, dW_t .$$

In words: the total P&L over $[0,T]$ from holding $H_t$ units of something whose price moves like $W$.

Start with the easy case, a simple position: $H$ is constant on each of finitely many intervals, changing only at the times $0 = t_0 < t_1 < \dots < t_n = T$. Then there is nothing to argue about — the integral is literally the trading sum,

$$I_T \;=\; \sum_{i=0}^{n-1} H_{t_i}\,\bigl(W_{t_{i+1}} - W_{t_i}\bigr).$$

In words: on each interval, take the position you held at the start of it and multiply by the move that followed; add those up. Note the subscript on $H$: it is $t_i$, the left end of the interval. That is the no-peeking rule written as an index.

For a general position, approximate $H$ by simple positions and take a limit. The limit exists — and this is the whole technical content — because of one identity, the Itô isometry:

$$E\!\left[\left(\int_0^T H_t\,dW_t\right)^{\!2}\right] \;=\; \int_0^T E\!\left[H_t^2\right] dt .$$

In words: the variance of your P&L is the time-average of your squared position. Doubling your size doubles your P&L's standard deviation; holding a position for twice as long multiplies its variance by two. The isometry says the map from "position schedule" to "P&L" preserves size, so a sequence of positions that converges gives a sequence of P&Ls that converges too. That is what makes the definition legitimate.

Worked example 1: a step position, priced out

Take $T = 1$ year cut into four quarters, $\Delta t = 0.25$. You hold 2 units for the first half of the year and 5 units for the second half:

$$H_t = 2 \ \text{ on } [0, 0.5), \qquad H_t = 5 \ \text{ on } [0.5, 1].$$

First, what does the isometry predict before we see any prices? $H$ is a fixed schedule, so $E[H_t^2] = H_t^2$ and

$$\int_0^1 H_t^2\,dt \;=\; 2^2 \times 0.5 \;+\; 5^2 \times 0.5 \;=\; 2 + 12.5 \;=\; 14.5 .$$

So the year's P&L has mean 0 and variance 14.5, a standard deviation of $\sqrt{14.5} = 3.81$. Because the schedule is deterministic, the P&L is exactly normal with those two numbers — you can quote a confidence interval on the strategy before running it.

Now a realised path. Suppose the four quarterly moves come in as

$$\Delta W = (+0.5,\; -0.3,\; +0.7,\; -0.4).$$

Multiply each by the position that was standing when it happened:

$$I_1 = 2(0.5) + 2(-0.3) + 5(0.7) + 5(-0.4) = 1.0 - 0.6 + 3.5 - 2.0 = 1.9 .$$

A profit of 1.9 against a predicted standard deviation of 3.81 — an unremarkable half-sigma year. The point is that both numbers came out of the same definition, one before the fact and one after.

Worked example 2: the left endpoint is not a detail

Now let the position be the driver: $H_t = W_t$. This is the classic $\int_0^T W_t\,dW_t$, and it is where ordinary calculus fails loudly.

Use the same four moves, so the path visits

$$W = (0,\; 0.5,\; 0.2,\; 0.9,\; 0.5) \quad \text{at} \quad t = (0,\; 0.25,\; 0.5,\; 0.75,\; 1).$$

Left endpoints (Itô). Multiply each move by the level standing before it:

$$0(0.5) + 0.5(-0.3) + 0.2(0.7) + 0.9(-0.4) = 0 - 0.15 + 0.14 - 0.36 = -0.37 .$$

Right endpoints. Same moves, but using the level after each one:

$$0.5(0.5) + 0.2(-0.3) + 0.9(0.7) + 0.5(-0.4) = 0.25 - 0.06 + 0.63 - 0.20 = +0.62 .$$

Two answers, from the same path, differing by 0.99. In ordinary calculus that gap would shrink to zero as you refine the grid. Here it does not, and you can see exactly why. Subtract the two sums term by term and each term is $(W_{t_{i+1}} - W_{t_i})^2$, so the gap is the sum of squared moves:

$$0.5^2 + 0.3^2 + 0.7^2 + 0.4^2 = 0.25 + 0.09 + 0.49 + 0.16 = 0.99 .$$

That is the quadratic variation of the path, and for Brownian motion it converges to $T = 1$ no matter how finely you chop. The gap is not an approximation error. It is a real, permanent quantity of size $T$.

There is a tidy closed form hiding in the arithmetic. For the left sum,

$$\int_0^T W_t\,dW_t \;=\; \tfrac{1}{2}W_T^2 \;-\; \tfrac{1}{2}T .$$

Check it: $\tfrac12 (0.5)^2 - \tfrac12(0.99) = 0.125 - 0.495 = -0.37$, matching the sum above exactly (using the path's realised quadratic variation 0.99 in place of its limit 1). In words: the answer school calculus would give you, $\tfrac12 W_T^2$, minus a correction of half the elapsed time. That $-\tfrac12 T$ is the reason Itô's lemma has a second-derivative term at all.

What the left endpoint buys you

Because the position is fixed before each move, and Brownian increments have mean zero and are independent of the past, every term in the sum has expectation zero given what you knew when you placed it. That gives the property everything else rests on: $\int_0^t H_s\,dW_s$ is a martingale in $t$. Its expected value is always zero, and its best forecast tomorrow is its value today. In trading language, a strategy that only bets on a fair coin, using no future information, cannot have positive expected P&L — and the Itô integral is built so that this is a theorem rather than a hope.

Take the explorer below and set it to brownian. Pick any path and pick a moment on it. Everything to the left is what your position may depend on; everything to the right is what your position gets multiplied by. Turn the volatility up and watch how much of the path's total movement is invisible at any coarse resolution — that hidden movement is the quadratic variation that made the two sums above differ.

What this means in practice

• Every SDE is shorthand for one of these. When you write $dS_t = \mu S_t\,dt + \sigma S_t\,dW_t$, the honest object is the integral equation $S_t = S_0 + \int_0^t \mu S_u\,du + \int_0^t \sigma S_u\,dW_u$. See stochastic differential equations.
• Hedging P&L. The error left over from a discretely rebalanced delta hedge is exactly the difference between an Itô integral and its finite-step approximation. The residual scales like $\sqrt{\Delta t}$, which is why halving your rebalancing interval only cuts hedge slippage by about 30 percent.
• Simulation. Euler–Maruyama is the left-endpoint sum with the grid left finite. Coding it with the end-of-step state by accident introduces a bias of order $\Delta t$ that does not shrink relative to the answer — a classic and silent bug.
• Pricing theory. The martingale representation theorem says every martingale driven by $W$ is an Itô integral for some $H$. That $H$ is the hedge ratio, which is why every derivative in a complete market can be replicated.
• Backtests. The left-endpoint rule is the mathematical form of "trade on the next bar's open, not this bar's close". Break it and your integral is not a martingale any more, which is a formal way of saying your backtest prints money that does not exist.