Covariation of Correlated Brownian Motions

From Correlation to Covariation

For independent Brownian motions, the sum of products of corresponding increments converges to zero. When the motions are correlated, those products have a nonzero mean, and their accumulated limit records the strength and sign of that correlation. The calculation also shows why the coefficient in a representation by independent Brownian motions must be chosen carefully.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. A standard Brownian motion relative to this filtration starts at zero, has continuous paths, and has increments independent of the past with $B_t-B_s\sim N(0,t-s)$ for $s<t$. A pair $(B^1,B^2)$ is a pair of correlated Brownian motions with parameter $\rho$ if each component is Brownian relative to the filtration and, on every interval $[s,t]$, the two increments are jointly normal with variances $t-s$, covariance $\rho(t-s)$, and are jointly independent of $\mathcal{F}_s$. Increments of the pair on disjoint time intervals are independent. Necessarily $|\rho|\le1$, since the covariance of two variables with variance $t-s$ cannot have absolute value greater than $t-s$.

Definition: For a deterministic partition $\Pi=\{0=t_0<t_1<\cdots<t_n=T\}$, write $\Delta_i B^j=B^j_{t_i}-B^j_{t_{i-1}}$. The covariation sum is $C_\Pi(T)=\sum_{i=1}^n\Delta_i B^1\Delta_i B^2$. When these sums converge in probability as the partition mesh tends to zero, their limit is the covariation $[B^1,B^2]_T$.

This tutorial establishes convergence in the stronger mean-square sense. For a random variable $X$, mean-square convergence to $c$ means $\mathbb{E}[(X-c)^2]\to0$. The ordinary covariance of the endpoints, $\operatorname{Cov}(B^1_T,B^2_T)$, is also $\rho T$, but that endpoint calculation alone does not establish the limit of the partition sums.

Representation by Independent Brownian Motions

Theorem: If $U$ and $V$ are independent standard Brownian motions on the same filtered probability space and $|\rho|\le1$, then $$ B^1_t=U_t,\qquad B^2_t=\rho U_t+\sqrt{1-\rho^2}\,V_t $$ defines a pair of correlated Brownian motions with parameter $\rho$. Conversely, every pair of correlated Brownian motions with parameter $\rho$ has this representation using independent Brownian motions when $|\rho|<1$. At $\rho=1$, $B^2=B^1$ almost surely for all times; at $\rho=-1$, $B^2=-B^1$ almost surely for all times.

Proof. First suppose $U,V$ are independent standard Brownian motions and put $a=\sqrt{1-\rho^2}$. Each increment of $B^2=\rho U+aV$ is normal, has mean zero, and has variance $\rho^2(t-s)+a^2(t-s)=t-s$. Its increments on disjoint intervals are independent, and they are independent of the past because the corresponding increments of $U,V$ have those properties. Thus $B^2$ is Brownian. The increment covariance is $\operatorname{Cov}(\Delta U,\rho\Delta U+a\Delta V)=\rho(t-s)$, so the pair has the stated correlation, including when $a=0$.

For the converse when $|\rho|<1$, set $U=B^1$, $a=\sqrt{1-\rho^2}$, and $V=(B^2-\rho B^1)/a$. On an interval of length $h=t-s$, $\Delta V$ is normal with mean zero and variance $$ \operatorname{Var}(\Delta V)=\frac{h+\rho^2h-2\rho(\rho h)}{1-\rho^2}=h. $$ Also $\operatorname{Cov}(\Delta U,\Delta V)=(\rho h-\rho h)/a=0$. The increments $(\Delta U,\Delta V)$ are jointly normal, so zero covariance makes them independent. The pair's increments on disjoint intervals are independent and independent of the past, by the corresponding properties of $(B^1,B^2)$. Hence $U,V$ are independent Brownian motions relative to the filtration, and rearranging the definition of $V$ gives the representation. At $\rho=1$, each increment of $B^2-B^1$ has variance $h+h-2h=0$, so it is zero almost surely. At $\rho=-1$, each increment of $B^2+B^1$ has variance $h+h-2h=0$, so it too is zero almost surely. These equalities hold simultaneously at rational times on an event of probability one; continuity extends them to all times. $\square$

Worked Example: Constructing a Pair with Correlation Three-Fifths

Let $U,V$ be independent standard Brownian motions and choose $\rho=\tfrac35$. The representation gives $$ B^1_t=U_t,\qquad B^2_t=\tfrac35U_t+\tfrac45V_t, $$ because $1-\rho^2=1-\tfrac9{25}=\tfrac{16}{25}$, whose nonnegative square root is $\tfrac45$.

For an interval of length $h$, the variance of the second increment is $$ \left(\tfrac35\right)^2h+\left(\tfrac45\right)^2h =\left(\tfrac9{25}+\tfrac{16}{25}\right)h=h. $$ The covariance with the first increment is $\tfrac35h$: the $U$-increment contributes $\tfrac35h$, while the independent $V$-increment contributes zero. Thus the increments have the required marginal variances and correlation parameter.

Check: the correlation is $(\tfrac35h)/\sqrt{h\cdot h}=\tfrac35$ for $h>0$, as specified. The coefficient squares add to one, so $B^2$ has unit Brownian variance per unit time.

Covariation from Increment Sums

Theorem: Let $B^1,B^2$ be correlated Brownian motions with parameter $|\rho|\le1$. For every $T\ge0$ and every sequence of deterministic partitions of $[0,T]$ whose mesh tends to zero, $$ C_\Pi(T)\longrightarrow \rho T\quad\text{in }L^2(\mathbb{P}). $$ Consequently, $[B^1,B^2]_T=\rho T$.

Proof. The assertion is immediate for $T=0$, since the sum is empty and equals zero. Fix $T>0$. Let $h_i=t_i-t_{i-1}$, and write $X_i=\Delta_i B^1$, $Y_i=\Delta_i B^2$. By the representation theorem, the pair of increments has the same distribution as $X_i=\Delta_i U$, $Y_i=\rho\Delta_i U+a\Delta_i V$, where $a=\sqrt{1-\rho^2}$ and the two increments on the right are independent $N(0,h_i)$ variables. Therefore $$ \mathbb{E}[X_iY_i]=\rho h_i. $$ Using the centered normal moments $\mathbb{E}[X_i^2]=h_i$, $\mathbb{E}[X_i^4]=3h_i^2$, and independence, $$ \mathbb{E}[(X_iY_i)^2] =\rho^2(3h_i^2)+2\rho a\,\mathbb{E}[(\Delta_i U)^3\Delta_i V] +a^2\mathbb{E}[(\Delta_i U)^2]\mathbb{E}[(\Delta_i V)^2] =(3\rho^2+a^2)h_i^2. $$ The middle term is zero because the centered normal variable $\Delta_i V$ has mean zero and is independent of $\Delta_i U$. Since $a^2=1-\rho^2$, subtracting $(\mathbb{E}[X_iY_i])^2=\rho^2h_i^2$ gives $$ \operatorname{Var}(X_iY_i)=(1+\rho^2)h_i^2. $$ The products $X_iY_i$ for distinct intervals are independent. Hence $\mathbb{E}[C_\Pi(T)]=\rho\sum_i h_i=\rho T$, and $$ \mathbb{E}[(C_\Pi(T)-\rho T)^2] =\operatorname{Var}(C_\Pi(T)) =(1+\rho^2)\sum_i h_i^2 \le (1+\rho^2)T|\Pi|. $$ Here $h_i^2\le h_i|\Pi|$, and $\sum_i h_i=T$. The upper bound tends to zero with the mesh, proving mean-square convergence and therefore convergence in probability. By the definition of covariation, the limit is $[B^1,B^2]_T=\rho T$. $\square$

The result includes the familiar independent case: when $\rho=0$, the covariation is zero. For positive $\rho$, the limiting covariation increases at rate $\rho$; for negative $\rho$, it decreases. In differential notation, the conclusion is recorded as $dB^1_t\,dB^2_t=\rho\,dt$. This notation summarizes the limit of increment products; it is not a claim that the product of two individual increments equals $\rho$ times the interval length.

Worked Example: A Two-Interval Sum with Negative Correlation

Suppose $\rho=-\tfrac12$ and partition $[0,1]$ into two intervals of length $h_1=h_2=\tfrac12$. Compute the mean and variance of $C_\Pi(1)$.

The mean is the sum of the increment-product means: $$ \mathbb{E}[C_\Pi(1)]=-\tfrac12\cdot\tfrac12-\tfrac12\cdot\tfrac12=-\tfrac12. $$ For each interval, the variance of its increment product is $(1+\rho^2)h_i^2=(1+\tfrac14)(\tfrac12)^2=\tfrac54\cdot\tfrac14=\tfrac5{16}$. The two products are independent, so $$ \operatorname{Var}(C_\Pi(1))=\tfrac5{16}+\tfrac5{16}=\tfrac58. $$ Check: the theorem's mesh bound is $(1+\rho^2)T|\Pi|=\tfrac54\cdot1\cdot\tfrac12=\tfrac58$, equal to the computed variance here. As the mesh decreases, this variance bound tends to zero while the mean remains $-\tfrac12$.

Worked Example: The Perfectly Anticorrelated Endpoint Case

Take $\rho=-1$, so the representation theorem gives $B^2=-B^1$. For any partition of $[0,1]$, $$ C_\Pi(1)=\sum_i\Delta_i B^1(-\Delta_i B^1)=-\sum_i(\Delta_i B^1)^2. $$ Thus the covariation sum is exactly the negative of the quadratic variation sum of $B^1$.

For two equal intervals, each of length $\tfrac12$, the mean of $C_\Pi(1)$ is $-(\tfrac12+\tfrac12)=-1$. Its variance is $2\sum_i h_i^2=2(\tfrac14+\tfrac14)=1$, using the normal-square variance $2h_i^2$ and independence of the increments. The general theorem gives the same mean $\rho T=-1$ and the same variance formula $(1+\rho^2)\sum_i h_i^2=2(\tfrac14+\tfrac14)=1$.

Check: along partitions with mesh tending to zero, the quadratic variation sum of $B^1$ converges in mean square to $1$. Its negative therefore converges to $-1=\rho T$, in agreement with the covariation theorem.

A Common Representation Error

A tempting but incorrect formula is $B^2_t=\rho U_t+\sqrt{1-\rho}\,V_t$. The coefficient of $V$ must compensate for the variance contributed by $\rho U$, so its square must be $1-\rho^2$, not $1-\rho$. For example, at $\rho=\tfrac35$, the incorrect formula gives variance per unit time $$ \left(\tfrac35\right)^2+\left(1-\tfrac35\right)=\tfrac9{25}+\tfrac25=\tfrac{19}{25}, $$ not $1$. The corrected coefficient is $\sqrt{1-\tfrac9{25}}=\tfrac45$, and $\tfrac9{25}+\tfrac{16}{25}=1$.

Do not confuse endpoint covariance with covariation, even though both equal $\rho T$ in this setting. Endpoint covariance is a property of the joint distribution of $(B^1_T,B^2_T)$; covariation is defined by limits of sums along partitions. The increment calculation proves that the latter limit exists and identifies it. The degenerate parameters $\rho=\pm1$ are included: they describe identical or sign-reversed paths, not a representation with two genuinely independent contributions.

Check Your Understanding

Use the representation and increment-sum calculation to answer the following.

For $\rho=\tfrac45$, find the coefficient of $V_t$ in the independent-Brownian representation and verify that the squared coefficients sum to one.
For a partition of $[0,2]$ into four equal intervals and $\rho=\tfrac12$, compute the mean and variance of the covariation sum.
Explain why zero covariance of jointly normal increments implies that the transformed Brownian motions in the converse representation are independent.
State what the representation becomes when $\rho=1$ and when $\rho=-1$.
Why does computing $\operatorname{Cov}(B^1_T,B^2_T)=\rho T$ not, by itself, prove the covariation-sum limit?

Covariation of Correlated Brownian Motions

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