Here is a game you can play with any function of time. Chop the interval [0, t] into tiny steps, take the increment over each step, square it, and add all the squares up. Then shrink the steps and ask: what does the total do?

For every smooth function you have ever met — polynomials, sines, exponentials — the total collapses to zero. Over a step of length \Delta t a differentiable f moves by \Delta f \approx f'\,\Delta t, so its squared increment is of order (\Delta t)^2: cut [0,t] into n pieces and you are summing n terms of size \sim 1/n^2 — a total of order 1/n, which dies. Squaring punishes small increments brutally, and smooth increments are very small.

Brownian motion breaks the game. Its increment over \Delta t has size \sim \sqrt{\Delta t} — the square-root scaling you met when you first built W — so the squared increment has size \sim \Delta t, and n terms of size t/n sum to something that refuses to die:

Not zero, not infinity: exactly t, for (almost) every single path. This limit is the quadratic variation [W]_t, and it is the sharpest roughness detector in analysis: it reads 0 on anything smooth and reads the elapsed time on Brownian noise. It is also the single reason stochastic calculus is not ordinary calculus — every second-order term that classical calculus throws away (\tfrac12 f''\,(dX)^2 in a Taylor expansion) survives when (dX)^2 accumulates at rate dt. The extra term in Itô's lemma is quadratic variation making itself felt.

Worked example: why [W]_t = t, honestly

The claim deserves a proof sketch, not just scaling folklore. Fix a partition 0 = t_0 < t_1 < \dots < t_n = t with mesh \|\Pi\| = \max_i \Delta t_i, and study the random sum

Step 1 — the mean is exactly t, for every partition. Each increment is Gaussian, \Delta W_i \sim N(0, \Delta t_i), so \mathbb{E}\big[(\Delta W_i)^2\big] = \Delta t_i. Summing:

No limit needed yet — on average the squared wiggles always add up to the elapsed time. The question is whether the randomness around that average survives the limit.

Step 2 — the variance dies with the mesh. For a Gaussian Z \sim N(0, \sigma^2) the fourth moment is \mathbb{E}[Z^4] = 3\sigma^4, so \operatorname{Var}(Z^2) = 3\sigma^4 - \sigma^4 = 2\sigma^4. Increments over disjoint intervals are independent, so variances add:

Step 3 — conclude. A random quantity whose mean is pinned at t and whose variance tends to zero converges to t in L^2. (Upgrading to almost-sure convergence along refining partitions takes a Borel–Cantelli argument — that is the honest fine print — but the L^2 statement already carries all the intuition.) So

Contrast the smooth case one more time, now quantitatively: if f is continuously differentiable with |f'| \le M, then \sum_i (\Delta f_i)^2 \le M^2 \sum_i (\Delta t_i)^2 \le M^2\,\|\Pi\|\,t \to 0. Every finite-variation path — every trend, every drift, everything an ordinary integral can build — has quadratic variation zero. Only genuine noise registers.

From one process to two

The quadratic variation [X]_t measures how much a single process wiggles. Its two-process cousin, the quadratic covariation [X, Y]_t, measures how much two processes wiggle together. For a partition 0 = t_0 < t_1 < \dots < t_n = t of [0, t] it is the limit of cross-products of increments as the mesh shrinks:

It is built exactly like a covariance, but pathwise and in time. Two immediate structural facts, straight from the definition: it is symmetric (\Delta X_i \Delta Y_i = \Delta Y_i \Delta X_i, so [X,Y] = [Y,X]) and bilinear (the sum is linear in each factor, so [aX + bZ,\, Y] = a[X,Y] + b[Z,Y]). Setting Y = X recovers the quadratic variation [X, X]_t = [X]_t.

The sign carries meaning, just as it does for covariance: if X and Y tend to jump the same way over the same tiny intervals, the products \Delta X_i\,\Delta Y_i are mostly positive and [X,Y]_t climbs; if they jump opposite ways it falls; and if their wiggles ignore each other the products cancel and the covariation flatlines at zero.

Worked example: [X]_t for a general Itô process

Take the general Itô process

Its quadratic variation accumulates as (dX_t)^2, so expand the square:

Why the first two terms vanish — this is worth seeing once with orders of magnitude, not just as a rule. Over n equal steps of size \Delta t = t/n:

• (dt)^2 terms: each is of size (\Delta t)^2 = t^2/n^2, and n of them total \sim t^2/n \to 0. Too small.
• dt\,dW terms: each is of size \Delta t \cdot \sqrt{\Delta t} = (\Delta t)^{3/2}, and n of them (with random signs, cancelling further) total \sim t^{3/2}/\sqrt{n} \to 0. Also too small.
• (dW)^2 terms: each is of size \Delta t — and n of them total \sim t. This one survives, and by the previous section it survives as exactly dt.

Only the diffusion term registers:

The drift \mu_t contributes nothing — you can double it, reverse it, make it enormous, and [X]_t does not move. Quadratic variation is completely blind to the trend and sees only the noise. A concrete payoff: for geometric Brownian motion, d(\log S_t) = \big(\mu - \tfrac{\sigma^2}{2}\big)dt + \sigma\,dW_t, so

The quadratic variation of the log-price is the squared volatility times time — which is precisely why summing squared log-returns from market data estimates \sigma^2. Hold that thought for the vignettes below.

The covariation of two Itô processes, line by line

Let two Itô processes be driven by the same Brownian motion W:

We want d[X, Y]_t, which by the definition is the leading part of the cross-product of increments dX_t\,dY_t. Evaluate that product with the box algebra (dW_t)^2 = dt, dt\,dW_t = 0, (dt)^2 = 0 — each rule now justified by the order-of-magnitude count above.

Step 1 — expand the product term by term:

Step 2 — kill the small terms. Three of the four products are negligible by the box algebra: (dt)^2 = 0 and both mixed dt\,dW_t = 0. Only the last survives, with (dW_t)^2 = dt:

Step 3 — read off the covariation (its increment is this leading term):

Only the diffusion coefficients enter — the drifts a^X, a^Y contribute nothing, because finite-variation (drift) parts are too smooth to accumulate covariation. Setting X = Y = W (so b^X = b^Y = 1) gives the headline special case [W, W]_t = t.

Polarisation: covariation from variations alone

There is a slick way to define [X, Y] using only single-process quadratic variations, the polarisation identity. It is the same algebra as pq = \tfrac14[(p+q)^2 - (p-q)^2], or equivalently:

Step 1 — expand the quadratic variation of the sum, using bilinearity:

Step 2 — solve for the cross term:

So covariation is fully determined by the variations of X, Y and their sum — no new machinery required. Run it on two correlated processes to see it earn its keep: take dX = \sigma_X\,dW^1 and dY = \sigma_Y\,dW^2 with dW^1 dW^2 = \rho\,dt. Then [X] = \sigma_X^2\,t, [Y] = \sigma_Y^2\,t, and d(X+Y) = \sigma_X dW^1 + \sigma_Y dW^2 gives [X+Y] = (\sigma_X^2 + 2\rho\,\sigma_X\sigma_Y + \sigma_Y^2)\,t. Polarisation hands back

— the exact stochastic analogue of \operatorname{Cov}(P,Q) = \rho\,\sigma_P\sigma_Q, delivered per unit time.

Independent Brownian motions: [W^1, W^2]_t = 0

Now let W^1, W^2 be independent Brownian motions. Their covariation is the limit of \sum_i \Delta W^1_i\,\Delta W^2_i.

Step 1 — each summand has mean zero. Over the step [t_i, t_{i+1}] the increments \Delta W^1_i, \Delta W^2_i are independent and each mean-zero, so

Step 2 — the sum has mean zero and vanishing variance. The expected sum is 0, and a short computation shows its variance is \sum_i (t_{i+1}-t_i)^2 \le \|\Pi\|\,t \to 0 as the mesh shrinks. A mean-zero quantity whose variance goes to zero converges (in L^2) to 0:

In box-algebra shorthand this is the rule dW^1_t\,dW^2_t = 0 for independent drivers — the multidimensional twin of (dW_t)^2 = dt. Notice the pattern of the whole argument — it is the same mean-plus-vanishing-variance sketch that proved [W]_t = t, just with a different mean.

The multiplication table: the working summary

Everything on this page compresses into one small table — the Itô multiplication table, the practical engine that working quants and every stochastic-calculus computation run on. To find d[X,Y] for any pair of Itô processes: write out dX\,dY, multiply through, and evaluate every product of differentials with:

In words: dt annihilates everything ((dt)^2 = 0, dt\,dW = 0); a driver squared gives time ((dW)^2 = dt); two different drivers give their correlation times time (dW^1 dW^2 = \rho\,dt, with \rho = 0 when independent). Behind each entry is an honest limit theorem — the mean-plus-vanishing-variance arguments above — but once proved, you never re-derive them: you look them up in the table.

Mini-example, table-driven. Two stocks: dS_1 = \mu_1 S_1\,dt + \sigma_1 S_1\,dW^1 and dS_2 = \mu_2 S_2\,dt + \sigma_2 S_2\,dW^2, with dW^1 dW^2 = \rho\,dt. Multiply, keep only the surviving cell:

Every drift term hit a zero cell; the single dW^1 dW^2 cross term carried the whole answer. That one-line computation is exactly what feeds the cross term of the two-asset Itô lemma, the pricing of spread and basket options, and the covariance matrix of a portfolio's instantaneous returns.

Coupled versus independent, seen

Below are two pairs of Brownian paths, and the whole page in one picture. The top pair is coupled: both processes are driven by the same W, so over every tiny interval their increments are identical and each cross-product is a square, \Delta X_i\,\Delta Y_i = (\Delta W_i)^2 > 0 — a sum of relentlessly positive terms that accumulates to exactly [X,Y]_t = t. The other pair is independent: increment signs agree and disagree at random, the products cancel in the long run, and the covariation flatlines ([W^1,W^2]_t = 0).

Refresh for a fresh \omega and watch the geometry: the coupled pair moves in perfect lockstep, every jag mirrored, while the independent paths wander with no memory of each other — sometimes drifting the same way for a stretch (exactly why a finite-sample covariation estimate is noisy), but never systematically. Correlated drivers with 0 < \rho < 1 would sit between these extremes.