The Itô Integral

3.1Where You Evaluate Matters

In ordinary calculus, an integral is built from thin rectangles: cut time into $n$ pieces, and in each piece multiply the height of a curve by the width. It doesn't matter where inside a piece you read off the height (left end, middle, right end). As the pieces shrink, every choice gives the same answer.

Try the same with Brownian motion. In each piece, multiply the path's value $W$ by the piece's rise $\Delta W$, and add everything up. The slider $\theta$ ("theta") chooses where inside each piece $W$ is read: $\theta = 0$ is the left end, $\theta = 1$ the right end. Drag it, then make the pieces finer.

$\theta$

Where inside each piece the path is read: 0 for the left end, ½ for the middle, 1 for the right end.

$t_i^{\theta}$

The reading point in piece $i$: a fraction $\theta$ of the way from its left end to its right end. The orange dots mark them.

$W_{t_i^{\theta}}$

The path's value at that reading point: the "height" of the rectangle.

$\Delta W_i$

The rise of the path across piece $i$ (section 2.1): the "width" of the rectangle, measured in rises of $W$ instead of in time.

$\tfrac12 W_t^2$

The answer ordinary calculus would give: the rule $\int x\,dx = \tfrac12 x^2$, with $W$ in place of $x$.

$(\theta - \tfrac12)\,t$

An extra amount that depends on the reading point. For a smooth curve it would be 0. Here it isn't.

In plain words: reading $W$ later inside a piece means reading it after part of that piece's rise has already happened. So the height and the width are no longer independent: both include the same bit of movement. Each piece then contributes a little extra, on average $\theta$ times the piece's squared rise. By section 2.1, squared rises add up to $t$ instead of vanishing, so the total extra is $\theta\,t$. Different reading points give genuinely different answers, and we must pick a rule.

Itô's rule is $\theta = 0$: always read at the left end. It is the only choice that never peeks at the future. If $W$ is a price and the height is how much you hold, you must decide your holding before the price moves. With that rule, $\int_0^t W\,dW = \tfrac12 W_t^2 - \tfrac12 t$. That extra $-\tfrac12 t$ is the first glimpse of Itô calculus. (The middle point, $\theta = \tfrac12$, is also used. It is called the Stratonovich integral, and it keeps the ordinary-calculus answer. We return to it in Chapter 4.)

import numpy as np
rng = np.random.default_rng(7)
N = 2 ** 16
W = np.concatenate([[0], np.cumsum(rng.normal(0, np.sqrt(1 / N), N))])
n = 2 ** 12
k = N // n                                           # fine steps per piece
for theta in (0, 0.5, 1):
    read = W[np.arange(n) * k + round(theta * k)]    # W at the reading point of each piece
    total = np.sum(read * np.diff(W[::k]))
    print(f"theta = {theta}: sum = {total:.3f}   theory = {0.5 * W[-1] ** 2 + theta - 0.5:.3f}")

theta = 0: sum = -0.396   theory = -0.385
theta = 0.5: sum = 0.120   theory = 0.115
theta = 1: sum = 0.627   theory = 0.615

3.2Building the Integral

The clearest way to think about the Itô integral is as trading gains. $W$ is the price of something. Time is split into 8 periods, and at the start of each period you choose how many units to hold (the bars): positive means you bet on a rise, negative means you bet on a fall. Your gain in a period is the number of units held × the price's rise.

Drag across the bars to set your bets. The orange line shows your running gains along one price path. The histogram shows your final gains across 2,000 different price paths.

$\int_0^1 \cdots\, dW_t$

The Itô integral from time 0 to time 1: the total of "holding × rise", added up over all the time.

$H_t$

Your holding at time $t$: the height of the bar for the period that contains $t$. It is chosen at the start of the period, before the price moves (Itô's left-end rule from section 3.1).

$H_j$

The holding during period $j$.

$t_{j-1},\ t_j$

The start and end of period $j$. Here, $t_j = j/8$.

$W_{t_j} - W_{t_{j-1}}$

How much the price rises during period $j$.

In plain words: for bets that stay fixed within each period, the integral is just the sum above: bet × rise, period by period. The full Itô integral allows the holding to change continuously, and even to depend on what the price has done so far (the next section does exactly that). It is defined as what these sums settle down to as the periods get shorter and shorter. The one rule that never changes: each holding is fixed before the move it bets on.

Keyboard: Tab to the bars, ← → to pick a period, ↑ ↓ to change its bet. The histogram's curve is the bell curve that theory predicts, centred at 0.

Whatever bets you choose, your average gain is 0. Betting on Brownian motion is a fair game (section 2.2), and the Itô integral inherits that: it is a martingale. The spread of your gains depends only on how big your bets are: the variance equals "bet² × time", added up. The next section shows this rule holds even for bets that react to the price.

import numpy as np
rng = np.random.default_rng(8)
bets = np.array([1, 2, 0.5, -1, -1, 0, 1.5, 1])       # holding in each of 8 periods
rises = rng.normal(0, np.sqrt(1 / 8), (100_000, 8))   # price rise per period, per path
gains = rises @ bets                                   # sum of bet x rise: the Ito integral
print(f"mean {gains.mean():+.4f} (0)   variance {gains.var():.3f} (sum bet^2 / 8 = {np.sum(bets ** 2) / 8:.3f})")

mean +0.0022 (0)   variance 1.322 (sum bet^2 / 8 = 1.312)

3.3Itô Isometry

Now let the bet depend on the price itself. With $H = W$ you hold more the higher the price has gone. With $H = \operatorname{sign}(W)$ you hold +1 when the price is above its start and −1 when below. For each rule, the computer trades along 10,000 price paths. The chart shows three averages as more and more paths are included.

$\int_0^t H_s\,dW_s$

Your total trading gains up to time $t$ (section 3.2). Here $t = 1$.

$\mathbb{E}[\;\cdot\;]$

The average over all possible price paths (section 2.2). The chart estimates it with up to 10,000 paths.

$\big(\int H\,dW\big)^2$

Your gains squared: a measure of how big your gain or loss is, whichever way it went.

$\int_0^t H_s^2\,ds$

An ordinary integral over time: the size of your holding, squared, added up over time. It measures how much risk you carried.

$\operatorname{sign}(W)$

+1 if $W$ is above 0, −1 if below.

isometry

"Same size". The size of the random gains (left side) equals the size of the holdings (right side).

In plain words: the random, hard-to-predict quantity on the left (your squared gains) has the same average as a plain accumulation of squared holdings over time on the right. The reason is sections 1.2 and 2.1 again: rises in different periods don't interact, so only each period's own squared rise counts, and a squared rise averages the elapsed time. That is why $(dW)^2 = dt$ reappears here as "$H^2 dt$". Meanwhile the gains themselves average 0, whatever rule you use.

Horizontal axis: number of paths averaged (log scale). Dashed line: theory for the two orange and blue averages.

The squared gains and the accumulated squared holdings converge to the same number, and the gains converge to 0. For $H = W$ that number is $\int_0^1 s\,ds = \tfrac12$, because $W_s^2$ averages $s$. These two facts, "mean zero" and "the isometry", are what make the Itô integral usable: they turn questions about a random integral into ordinary calculus.

import numpy as np
rng = np.random.default_rng(9)
paths, steps = 20_000, 200
dt = 1 / steps
dW = rng.normal(0, np.sqrt(dt), (paths, steps))
W = np.concatenate([np.zeros((paths, 1)), np.cumsum(dW, axis=1)[:, :-1]], axis=1)   # W at left ends
for name, H in (("1", np.ones_like(W)), ("W", W), ("sign(W)", np.sign(W))):
    I = np.sum(H * dW, axis=1)                        # Ito integral, left-end rule
    risk = np.sum(H ** 2 * dt, axis=1)
    print(f"H = {name:8} E[I] = {I.mean():+.3f}   E[I^2] = {np.mean(I ** 2):.3f}   E[int H^2 dt] = {risk.mean():.3f}")

H = 1        E[I] = +0.001   E[I^2] = 1.006   E[int H^2 dt] = 1.000
H = W        E[I] = +0.003   E[I^2] = 0.505   E[int H^2 dt] = 0.499
H = sign(W)  E[I] = +0.009   E[I^2] = 0.990   E[int H^2 dt] = 0.995