Quadratic Covariation

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:

[X, Y]_t = \lim_{\|\Pi\|\to 0} \sum_{i} \big(X_{t_{i+1}} - X_{t_i}\big)\big(Y_{t_{i+1}} - Y_{t_i}\big) = \lim_{\|\Pi\|\to 0} \sum_i \Delta X_i\,\Delta Y_i.

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 covariation of two Itô processes, line by line

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

dX_t = a^X\,dt + b^X\,dW_t, \qquad dY_t = a^Y\,dt + b^Y\,dW_t.

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.

Step 1 — expand the product term by term:

dX_t\,dY_t = \big(a^X\,dt + b^X\,dW_t\big)\big(a^Y\,dt + b^Y\,dW_t\big), = a^X a^Y\,(dt)^2 + a^X b^Y\,dt\,dW_t + b^X a^Y\,dW_t\,dt + b^X b^Y\,(dW_t)^2.

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:

dX_t\,dY_t = b^X b^Y\,(dW_t)^2 = b^X b^Y\,dt.

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

\boxed{\,d[X, Y]_t = b^X b^Y\,dt, \qquad [X, Y]_t = \int_0^t b^X_s\,b^Y_s\,ds.\,}

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:

[X + Y,\, X + Y] = [X, X] + 2[X, Y] + [Y, Y].

Step 2 — solve for the cross term:

[X, Y] = \tfrac{1}{2}\Big([X + Y] - [X] - [Y]\Big).

So covariation is fully determined by the variations of X, Y and their sum — no new machinery required.

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

\mathbb{E}\big[\Delta W^1_i\,\Delta W^2_i\big] = \mathbb{E}[\Delta W^1_i]\,\mathbb{E}[\Delta W^2_i] = 0\cdot 0 = 0.

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:

[W^1, W^2]_t = 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.

For Itô processes X, Y the quadratic covariation [X, Y]_t = \lim \sum_i \Delta X_i\,\Delta Y_i satisfies:

The polarisation identity above is worth seeing once with every bracket expanded. Starting from [X+Y] and using bilinearity and symmetry:

[X+Y] = [X+Y,\,X+Y] = [X,X] + [X,Y] + [Y,X] + [Y,Y] = [X] + 2[X,Y] + [Y],

and isolating the middle term gives [X,Y] = \tfrac12([X+Y] - [X] - [Y]), exactly as claimed. This is the stochastic echo of the variance identity \operatorname{Cov}(P, Q) = \tfrac12(\operatorname{Var}(P+Q) - \operatorname{Var}(P) - \operatorname{Var}(Q)).

Between the two extremes — perfectly coupled (same W) and independent ([W^1,W^2]=0) — sit correlated Brownian motions with correlation \rho \in [-1, 1]. Their box rule interpolates:

dW^1_t\,dW^2_t = \rho\,dt, \qquad\text{so}\qquad [W^1, W^2]_t = \rho\,t.

This is the seed of the multidimensional Itô lemma: when a function depends on several correlated drivers, every pair contributes a cross-correction f_{x_i x_j}\,d[X^i, X^j], and the matrix of these \rho_{ij} values is what couples a basket of assets together in multi-asset pricing.

Coupled versus independent, seen

Below are two pairs of Brownian paths. On a coupled pair driven by the same W the increments move in lockstep, so their cross-products accumulate ([X,Y]_t = t). On an independent pair the cross-products cancel out and accumulate to nothing ([W^1,W^2]_t = 0). Refresh to draw a fresh \omega.