Covariation Sums and the Polarization Identity

From Increment Products to Covariation

Quadratic variation records the accumulated squares of increments of one process. To compare two processes, the corresponding quantity is the accumulated product of their increments. The key algebraic fact is that a product can be recovered from squares: $4ab=(a+b)^2-(a-b)^2$. Applied to increments, this gives an exact identity on every partition, before any limit is taken. The identity is useful because quadratic-variation sums may be easier to identify than product sums.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Let $X_t$ and $Y_t$ be real-valued continuous adapted processes. The sums below are pathwise finite sums; they are not stochastic integrals, so no integrand class is required. For $T>0$, take a deterministic partition $\Pi=\{0=t_0<t_1<\cdots<t_n=T\}$, and write $\Delta_iX=X_{t_i}-X_{t_{i-1}}$ and $\Delta_iY=Y_{t_i}-Y_{t_{i-1}}$.

Definition: The quadratic-variation sums of $X$ and $Y$, and their covariation sum, along $\Pi$ are $$ Q_\Pi(X)=\sum_{i=1}^n(\Delta_iX)^2,\qquad Q_\Pi(Y)=\sum_{i=1}^n(\Delta_iY)^2,\qquad C_\Pi(X,Y)=\sum_{i=1}^n\Delta_iX\,\Delta_iY. $$ If, along a specified sequence of partitions whose mesh tends to zero, $C_\Pi(X,Y)$ converges in probability to a random variable, that limit is the covariation at time $T$ along that sequence, denoted $[X,Y]_T$. The definition is sequence-specific unless independence of the partition sequence is established.

Convergence in probability means that for every $\varepsilon>0$, the probability that the absolute error exceeds $\varepsilon$ tends to zero. Almost-sure convergence can also be used in the identities below. We do not assume that every pair of continuous processes has a covariation limit along every sequence of partitions. The point here is to show exactly how existence of appropriate quadratic-variation limits gives existence and a formula for the covariation limit.

The Exact Polarization Identity

Proposition: For every finite partition $\Pi$ and every pair of real-valued paths $X,Y$, $$ C_\Pi(X,Y)=\frac14\left(Q_\Pi(X+Y)-Q_\Pi(X-Y)\right). $$

Proof. For each interval of the partition, the increment of $X+Y$ is $\Delta_iX+\Delta_iY$, and the increment of $X-Y$ is $\Delta_iX-\Delta_iY$. Therefore $(\Delta_iX+\Delta_iY)^2-(\Delta_iX-\Delta_iY)^2=4\Delta_iX\,\Delta_iY$. Summing this equality over $i=1,\ldots,n$ gives $Q_\Pi(X+Y)-Q_\Pi(X-Y)=4C_\Pi(X,Y)$. Dividing by $4$ proves the identity. It holds without probabilistic assumptions and includes cases such as $X=Y$, $X=-Y$, or zero increments. $\square$

The same partition must be used for both quadratic-variation sums in the difference. The identity is an equality of finite sums, and each summand compares increments over the same time interval. Mixing partitions would not give the displayed algebra interval by interval.

Worked Example: A Finite-Partition Calculation

Suppose a three-interval partition gives the increments $(\Delta_1X,\Delta_2X,\Delta_3X)=(1,2,-1)$ and $(\Delta_1Y,\Delta_2Y,\Delta_3Y)=(2,-1,3)$. Compute the covariation sum directly and by polarization.

Directly, the three increment products are $1\cdot2=2$, $2\cdot(-1)=-2$, and $(-1)\cdot3=-3$. Thus $C_\Pi(X,Y)=2-2-3=-3$.

For the sum path, the increments are $(3,1,2)$, so $Q_\Pi(X+Y)=3^2+1^2+2^2=9+1+4=14$. For the difference path, the increments are $(-1,3,-4)$, so $Q_\Pi(X-Y)=(-1)^2+3^2+(-4)^2=1+9+16=26$. Polarization gives $(14-26)/4=-12/4=-3$.

Check: the polarized answer agrees with the direct sum of products. The negative result is possible because products of increments can be negative; covariation sums are not sums of squares.

Passing the Identity to a Limit

Theorem: Fix a sequence $(\Pi_n)$ of deterministic partitions of $[0,T]$ whose mesh tends to zero. If $Q_{\Pi_n}(X+Y)$ and $Q_{\Pi_n}(X-Y)$ converge in probability to $A$ and $D$, respectively, then $C_{\Pi_n}(X,Y)$ converges in probability and $$ [X,Y]_T=\frac14(A-D). $$

Proof. The finite-partition proposition gives $C_{\Pi_n}(X,Y)=\tfrac14(Q_{\Pi_n}(X+Y)-Q_{\Pi_n}(X-Y))$ for every $n$. For any $\varepsilon>0$, if $\left|C_{\Pi_n}(X,Y)-\tfrac14(A-D)\right|>\varepsilon$, then the triangle inequality implies that at least one of $\left|Q_{\Pi_n}(X+Y)-A\right|>2\varepsilon$ or $\left|Q_{\Pi_n}(X-Y)-D\right|>2\varepsilon$ holds. Hence the probability of the first event is at most the sum of the probabilities of those two events. Both probabilities tend to zero by the assumed convergence in probability. This proves the claimed convergence and identifies the limit. The argument also covers limits that are random variables. $\square$

For example, if $Q_{\Pi_n}(X)$, $Q_{\Pi_n}(Y)$, and $C_{\Pi_n}(X,Y)$ all have limits denoted $[X]_T$, $[Y]_T$, and $[X,Y]_T$, then expanding the squares on each interval and taking limits gives $[X+Y]_T=[X]_T+2[X,Y]_T+[Y]_T$ and $[X-Y]_T=[X]_T-2[X,Y]_T+[Y]_T$. These relations require the stated limits along the same partition sequence. They are not a claim that the limits exist for arbitrary continuous paths.

The probability-limit step in this argument uses only finite linear combinations. In particular, no independence assumption is needed. The same proof applies to almost-sure limits: on the event where both quadratic-variation sums converge, the exact finite-sum identity passes to the limit. If the sums are viewed as processes in time and converge uniformly in probability on $[0,T]$, the identity passes to that process limit as well, by the same linearity and triangle-inequality argument applied to the supremum over time.

Brownian Examples

Worked Example: Covariation of Two Scaled Brownian Paths

Let $X_t=2B_t$ and $Y_t=3B_t$. Find the limit of their covariation sums on $[0,T]$, assuming $T>0$.

On each partition interval, $\Delta_iX=2\Delta_iB$ and $\Delta_iY=3\Delta_iB$, so direct multiplication gives $C_\Pi(X,Y)=6\sum_i(\Delta_iB)^2=6Q_\Pi(B)$. Equivalently, $X+Y=5B$ and $X-Y=-B$. Therefore $Q_\Pi(X+Y)=25Q_\Pi(B)$ and $Q_\Pi(X-Y)=Q_\Pi(B)$, and polarization gives $\tfrac14(25Q_\Pi(B)-Q_\Pi(B))=6Q_\Pi(B)$.

Brownian quadratic-variation sums converge in probability to $T$ along deterministic partitions whose mesh tends to zero. Consequently, $[X,Y]_T=6T$.

Check: at $T=2$, the limit is $6\cdot2=12$. The direct product calculation and polarization both give $6Q_\Pi(B)$, whose limit is $6\cdot2=12$.

Worked Example: Opposite Brownian Paths

Let $X_t=B_t$ and $Y_t=-B_t$. Find their covariation limit on $[0,T]$.

Here $X+Y$ is the zero path, so $Q_\Pi(X+Y)=0$. Also $X-Y=2B$, so $Q_\Pi(X-Y)=4Q_\Pi(B)$. Polarization yields $C_\Pi(X,Y)=\tfrac14(0-4Q_\Pi(B))=-Q_\Pi(B)$. Taking the Brownian quadratic-variation limit gives $[X,Y]_T=-T$.

Check: directly, each increment product is $\Delta_iB\,(-\Delta_iB)=-(\Delta_iB)^2$, so the sum is $-Q_\Pi(B)$ and has limit $-T$. The sign agrees with the fact that the two paths move in exactly opposite directions on every interval.

These examples also show why covariation should not be confused with statistical covariance. Statistical covariance compares centered random variables at specified times. A covariation sum instead multiplies increments over matching intervals and then takes a partition limit. The definition is about the joint pathwise increment structure; its limit can be negative, as in the opposite-path example.

A Common Normalization Error

In the finite-partition example, $Q_\Pi(X+Y)=14$ and $Q_\Pi(X-Y)=26$. A common error is to divide their difference by $2$, giving $(14-26)/2=-6$. This is wrong: expanding a single pair of squares gives $(a+b)^2-(a-b)^2=4ab$, not $2ab$. The correct divisor is $4$, so the covariation sum is $(14-26)/4=-3$, matching the direct products $2-2-3=-3$.

A second practical safeguard is to keep the partition fixed while comparing the two quadratic-variation sums. Their difference cancels the squared increments of each individual path and leaves exactly four times the product sum. Once limits are available along the same sequence, the same cancellation explains the limiting identity. The method matters because it can establish a covariation limit using quadratic-variation information about $X+Y$ and $X-Y$, without separately estimating every product increment.

Check Your Understanding

Use the finite-sum identity first, then consider what hypotheses allow the identity to pass to a limit.

For one interval with increments $\Delta X=4$ and $\Delta Y=-2$, compute $Q(X+Y)$, $Q(X-Y)$, and the covariation sum. Verify polarization.
Suppose $X=B$ and $Y=4B$. Express both quadratic-variation sums of $X+Y$ and $X-Y$ in terms of $Q_\Pi(B)$, then find the covariation limit on $[0,T]$.
Explain why the two quadratic-variation sums in the polarization formula must use the same partition.
If $Q_{\Pi_n}(X+Y)$ and $Q_{\Pi_n}(X-Y)$ converge in probability, state the resulting covariation limit.
Give a direct increment calculation showing why the covariation of $B$ and $-B$ is negative.