Every quant desk in the world runs on one question: how risky is this position? A trading strategy that holds H_s units of a Brownian-driven asset accumulates the random total \int_0^T H_s\, dW_s — and the risk of that position is its variance. But how do you compute the variance of a random integral? Path by path it is hopeless: every realisation of W gives a different jagged answer, and there are uncountably many of them.

The Itô isometry is the conversion rate between two worlds. It says the hard object — the mean-square size of a stochastic integral, living in probability space — equals an easy object: an ordinary time-integral of the squared integrand.

On the left is the L^2(\Omega) size of a random variable. On the right is a plain L^2(dt \times d\mathbb{P}) size of the integrand — integrate H^2 in time, then average. The isometry says these are equal, and that single equality plays two roles at once. It is the construction tool that lets the Itô integral be extended from simple step processes to every square-integrable adapted integrand, and it is the daily workhorse for computing the variance of anything driven by Brownian noise — a hedging error, a discretisation error, the width of a confidence band around a simulated price. One identity, both jobs.

Warm-up: one step, verified from scratch

Before the general proof, check the isometry by hand on the simplest possible integrand: a process that takes a single value on a single interval. Fix times s < t, let \varphi be \mathcal{F}_s-measurable (a value you commit to at time s, using only information available then), and set

Left-hand side. The increment W_t - W_s is independent of \mathcal{F}_s, and \varphi^2 is \mathcal{F}_s-measurable — so the expectation of the product factorises:

using that W_t - W_s \sim N(0,\, t-s), whose mean square is its variance, t - s.

Right-hand side. The squared integrand is H_u^2 = \varphi^2 on (s, t] and zero elsewhere, so

The two sides match exactly. Notice what did the work: independence let the expectation factorise, and variance = elapsed time evaluated the increment. The general proof below is nothing but this calculation repeated across every interval — plus one extra observation about why the intervals don't interfere with each other.

The derivation, line by line (simple H)

We prove it for a simple adapted process H_s = \sum_{i=0}^{n-1} H_{t_i}\mathbf{1}_{(t_i, t_{i+1}]}(s); the general case then follows by the very approximation the isometry makes possible. Write \Delta W_i = W_{t_{i+1}} - W_{t_i} and \Delta t_i = t_{i+1} - t_i, so the integral is \int_0^T H\, dW = \sum_i H_{t_i}\Delta W_i.

Step 1 — expand the square into a double sum. A finite sum squared is the double sum of all pairwise products:

Taking expectations and using linearity,

Split the double sum into off-diagonal terms (i \neq j) and diagonal terms (i = j). We show every off-diagonal term is zero, then evaluate the diagonal.

Step 2 — the off-diagonal terms vanish. Take i < j (the case i > j is symmetric). Then the three factors H_{t_i}, H_{t_j}, \Delta W_i are all \mathcal{F}_{t_j}-measurable: H_{t_i} and \Delta W_i happened before t_j, and H_{t_j} is adapted, set at t_j. Only the last increment \Delta W_j reaches into the future. Condition on \mathcal{F}_{t_j} and pull out everything known:

The future increment \Delta W_j is independent of \mathcal{F}_{t_j} and mean-zero, so the inner conditional expectation is 0, and the whole term collapses:

Every cross term is killed by the same "future increment has no correlation with the past" mechanism that made the integral mean-zero. This is the extra observation promised in the warm-up: the one-step calculations on different intervals never interfere, because independent increments annihilate every mixed product. Only the diagonal survives.

Step 3 — the diagonal terms. On the diagonal i = j the term is \mathbb{E}\big[H_{t_i}^2\,(\Delta W_i)^2\big] — exactly the warm-up calculation. Condition on \mathcal{F}_{t_i} and pull out the known coefficient H_{t_i}^2:

The squared increment is independent of \mathcal{F}_{t_i}, so its conditional mean is its plain mean — the variance of a mean-zero N(0, \Delta t_i):

Therefore each diagonal term is

Step 4 — sum the diagonal and recognise the time-integral. Adding the surviving terms,

But \sum_i H_{t_i}^2\,\Delta t_i is exactly the (deterministic-in-time) Riemann sum of \int_0^T H_s^2\, ds for the step integrand H^2 — and since H is simple, it equals that integral on the nose. Hence

Two ingredients did everything: independent increments killed the off-diagonal, and variance = elapsed time ((dW)^2 = dt again) evaluated the diagonal.

Only the diagonal survives — the L² Pythagoras picture

The picture of the proof: lay out the n \times n grid of pairwise terms \mathbb{E}[H_{t_i} H_{t_j}\,\Delta W_i\,\Delta W_j]. Every off-diagonal cell (i \neq j) is killed by the independent future increment and vanishes; only the diagonal cells survive, each contributing \mathbb{E}[H_{t_i}^2]\,\Delta t_i — and their sum is \mathbb{E}[\int_0^T H^2\, ds].

There is a geometric way to read this grid. In the Hilbert space L^2(\Omega), the inner product of two random variables is \langle X, Y\rangle = \mathbb{E}[XY], so each off-diagonal cell is literally the inner product of two of the summands X_i = H_{t_i}\Delta W_i. The cells being zero says the summands are mutually orthogonal — perpendicular vectors, one per time slice. And for orthogonal vectors, the squared length of a sum is the sum of squared lengths:

So the Itô isometry is the Pythagorean theorem, applied once per interval of the partition: each slice of the strategy contributes its own perpendicular leg \|X_i\|^2 = \mathbb{E}[H_{t_i}^2]\,\Delta t_i of "variance budget", and the legs add with no cross terms. Independence of Brownian increments is what makes the time slices perpendicular; the isometry is what perpendicularity buys you.

This is also why the word isometry is exact, not decorative: an isometry is a map that preserves distances. Here the map sends H \in L^2(dt\times d\mathbb{P}) to \int_0^T H\,dW \in L^2(\Omega), and the identity says every norm — hence, by linearity applied to differences, every distance \|H - G\| — comes through unchanged. The stochastic integral is a rigid embedding of one L^2 space inside another: no stretching, no crushing, anywhere.

Putting it to work

The isometry is also the everyday tool for computing second moments of stochastic integrals, since the right-hand side is an ordinary integral. Two quick examples for a deterministic integrand (where the outer expectation is trivial):

With H_s \equiv 1,

which is just \mathbb{E}[W_T^2] = T — a sanity check, since \int_0^T dW = W_T. With H_s = s,

No path-by-path heroics: a hard mean-square of a random integral became a one-line calculus exercise. In fact for deterministic f the integral \int_0^T f(s)\, dW_s (the Wiener integral) is a limit of sums of independent Gaussians, hence itself exactly Gaussian — so the isometry pins down its entire distribution: N\!\big(0,\ \int_0^T f(s)^2 ds\big).

The chart below shows the family H(s) = s^n: the dashed curve is the rate at which variance is bought, H(s)^2 = s^{2n}, and the solid curve the accumulated budget \operatorname{Var}\big(\int_0^t s^n dW_s\big) = t^{2n+1}/(2n+1). Slide n: a bigger exponent is quiet early and loud late, so the budget hugs zero then explodes — variance is bought through time, at the deterministic rate H^2.

Worked example: the variance of \int_0^T W_s\, dW_s

Deterministic integrands are the warm-up; the isometry earns its keep when the integrand is itself random. Take H_s = W_s — the strategy that holds a position equal to the current level of the Brownian path — and compute the variance of I_T = \int_0^T W_s\, dW_s, every step spelled out.

Step 1 — variance = second moment. W_s is adapted with \mathbb{E}\int_0^T W_s^2\, ds = \int_0^T s\, ds < \infty, so I_T is a genuine Itô integral and has mean zero. Hence \operatorname{Var}(I_T) = \mathbb{E}[I_T^2], which is exactly what the isometry computes.

Step 2 — apply the isometry.

Step 3 — swap \mathbb{E} and \int. The integrand W_s^2 is non-negative, so Tonelli's theorem lets the expectation pass inside the time-integral:

Step 4 — an ordinary integral. \mathbb{E}[W_s^2] = \operatorname{Var}(W_s) = s, so

A random integral of a random integrand, and the whole variance calculation was \int_0^T s\, ds — the conversion rate at work.

Cross-check (optional but satisfying). Itô's calculus gives the closed form \int_0^T W_s\, dW_s = \tfrac12(W_T^2 - T). Its variance can be computed the classical way: with W_T \sim N(0, T) the fourth moment is \mathbb{E}[W_T^4] = 3T^2, so \operatorname{Var}(W_T^2) = 3T^2 - T^2 = 2T^2, and

Same answer, twice the effort — and the classical route needed a closed form that most integrands simply don't have. The isometry never needs one.