Zero Covariation of Independent Brownian Motions

Why Independence Makes Product-Increment Sums Vanish

For one Brownian motion, the sum of squared increments over a fine partition converges to elapsed time. For two independent Brownian motions, the corresponding sum of products of increments instead converges to zero. The difference comes from the second moment: a squared increment has expectation proportional to the time step, while the product of independent increments has mean zero and second moment proportional to the square of the time step.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying two independent standard Brownian motions $B_t$ and $C_t$. Assume both are Brownian motions with respect to the filtration, and that they are independent as processes. In particular, each has independent increments over disjoint time intervals, and increments of $B$ are independent of the entire process $C$. This tutorial concerns partition sums; no stochastic integral or integrand is used.

Definition: For $T>0$, a deterministic partition of $[0,T]$ is $\Pi=\{0=t_0<t_1<\cdots<t_n=T\}$, with mesh $|\Pi|=\max_{1\le i\le n}(t_i-t_{i-1})$. Write $\Delta t_i=t_i-t_{i-1}$, $\Delta B_i=B_{t_i}-B_{t_{i-1}}$, and $\Delta C_i=C_{t_i}-C_{t_{i-1}}$. The covariation sum is $S_\Pi=\sum_{i=1}^n\Delta B_i\Delta C_i$.

The result below applies to every deterministic partition, without requiring equal interval lengths. Convergence in probability to zero means that for every $\varepsilon>0$, the probability of the strict tail event $\{|S_\Pi|>\varepsilon\}$ tends to zero as the mesh tends to zero.

The Second-Moment Calculation

Theorem: For independent standard Brownian motions $B$ and $C$, and every deterministic partition $\Pi$ of $[0,T]$, $$ \mathbb{E}[S_\Pi]=0,\qquad \mathbb{E}[S_\Pi^2]=\sum_{i=1}^n(\Delta t_i)^2 \le T|\Pi|. $$ Thus $S_\Pi\to0$ in $L^2(\mathbb{P})$, and hence in probability, as $|\Pi|\to0$. If $T=0$, the sum is empty and is defined to be zero.

Proof. Set $X_i=\Delta B_i\Delta C_i$, so $S_\Pi=\sum_i X_i$. Each increment has mean zero, and the two Brownian motions are independent. Therefore $\mathbb{E}[X_i]=\mathbb{E}[\Delta B_i]\mathbb{E}[\Delta C_i]=0$, which gives $\mathbb{E}[S_\Pi]=0$. For the diagonal terms, process independence and the Brownian second-moment formula give $$ \mathbb{E}[X_i^2] =\mathbb{E}[(\Delta B_i)^2]\mathbb{E}[(\Delta C_i)^2] =(\Delta t_i)(\Delta t_i) =(\Delta t_i)^2. $$ For distinct indices $i\ne j$, independence of increments on disjoint intervals gives $\mathbb{E}[\Delta B_i\Delta B_j]=0$ and $\mathbb{E}[\Delta C_i\Delta C_j]=0$. Independence of the two processes then yields $$ \mathbb{E}[X_iX_j] =\mathbb{E}[\Delta B_i\Delta B_j]\mathbb{E}[\Delta C_i\Delta C_j]=0. $$ Expanding the square of the finite sum now gives $\mathbb{E}[S_\Pi^2]=\sum_i\mathbb{E}[X_i^2]=\sum_i(\Delta t_i)^2$. Finally, $\Delta t_i\le|\Pi|$, so $$ \sum_i(\Delta t_i)^2\le|\Pi|\sum_i\Delta t_i=T|\Pi|. $$ The right-hand side tends to zero with the mesh; since the mean is zero, this is exactly $L^2$ convergence to zero. $\square$

Corollary: Under the theorem's assumptions, for every $\varepsilon>0$, $$ \mathbb{P}(|S_\Pi|>\varepsilon)\le\frac{T|\Pi|}{\varepsilon^2}. $$

Proof. Apply Markov's inequality to the nonnegative random variable $S_\Pi^2$, using the strict event $\{S_\Pi^2>\varepsilon^2\}$. This gives $$ \mathbb{P}(|S_\Pi|>\varepsilon) =\mathbb{P}(S_\Pi^2>\varepsilon^2) \le\frac{\mathbb{E}[S_\Pi^2]}{\varepsilon^2} \le\frac{T|\Pi|}{\varepsilon^2}. $$ For fixed $T$ and $\varepsilon>0$, the final bound tends to zero as $|\Pi|\to0$, proving convergence in probability. $\square$

Worked Applications

Worked Example: A Nonuniform Partition

For $T=1$, take the partition $0<\tfrac14<\tfrac12<1$. Find the mean and second moment of $S_\Pi$, and obtain a probability bound for $\mathbb{P}(|S_\Pi|>1)$.

The three time increments are $\Delta t_1=\tfrac14$, $\Delta t_2=\tfrac14$, and $\Delta t_3=\tfrac12$. The mean is zero. Substituting these interval lengths into the theorem gives $$ \mathbb{E}[S_\Pi^2] =\left(\frac14\right)^2+\left(\frac14\right)^2+\left(\frac12\right)^2 =\frac1{16}+\frac1{16}+\frac14 =\frac38. $$ The mesh is $\tfrac12$, so the general bound is $T|\Pi|=1\cdot\tfrac12=\tfrac12$. Thus the exact second moment, $\tfrac38$, is no larger than the mesh bound, $\tfrac12$.

With $\varepsilon=1$, the probability estimate is $$ \mathbb{P}(|S_\Pi|>1)\le\frac{\mathbb{E}[S_\Pi^2]}{1^2}=\frac38. $$ Check: the exact second moment is the sum $\tfrac1{16}+\tfrac1{16}+\tfrac14=\tfrac38$, while the bound from the mesh is $\tfrac12$; indeed, $\tfrac38\le\tfrac12$.

Worked Example: Four Equal Intervals on a Longer Time Horizon

Partition $[0,2]$ into four equal intervals. Compute the second moment of the covariation sum, then bound the probability that its absolute value exceeds $2$.

Each interval has length $\Delta t_i=\tfrac24=\tfrac12$. There are four terms, so $$ \mathbb{E}[S_\Pi^2] =4\left(\frac12\right)^2 =4\cdot\frac14 =1. $$ The mean is zero, and the mesh is $\tfrac12$. For the strict tail event with $\varepsilon=2$, the second-moment inequality gives $$ \mathbb{P}(|S_\Pi|>2) \le\frac{1}{2^2} =\frac14. $$ Check: the mesh estimate for the second moment is $T|\Pi|=2\cdot\tfrac12=1$, matching the exact value. The probability bound uses that value divided by $2^2=4$, giving $\tfrac14$.

Worked Example: Constantly Rescaled Brownian Motions

Let $X_t=2B_t$ and $Y_t=-\tfrac12 C_t$, with $B$ and $C$ independent standard Brownian motions. On the partition $0<\tfrac12<1$, compute the mean and second moment of $\sum_i\Delta X_i\Delta Y_i$.

Since $\Delta X_i=2\Delta B_i$ and $\Delta Y_i=-\tfrac12\Delta C_i$, their product is $\Delta X_i\Delta Y_i=-\Delta B_i\Delta C_i$. Its mean is zero. Each interval has length $\tfrac12$, so the second moment of the sum is $$ \mathbb{E}\left[\left(\sum_{i=1}^2\Delta X_i\Delta Y_i\right)^2\right] =(-1)^2\left[\left(\frac12\right)^2+\left(\frac12\right)^2\right] =\frac14+\frac14 =\frac12. $$ The negative scaling changes the sign of the sum but not its second moment.

Check: on one interval, $\mathbb{E}[(\Delta X_i)^2]=2^2\cdot\tfrac12=2$, and $\mathbb{E}[(\Delta Y_i)^2]=(-\tfrac12)^2\cdot\tfrac12=\tfrac18$. Independence gives $\mathbb{E}[(\Delta X_i\Delta Y_i)^2]=2\cdot\tfrac18=\tfrac14$. The two cross terms have expectation zero, so the sum has second moment $\tfrac14+\tfrac14=\tfrac12$, as calculated above.

What the Zero-Covariation Statement Means

In the quadratic-variation framework, the calculation shows that the covariation sums converge in probability to zero at each fixed endpoint $T$, along every deterministic sequence of partitions whose mesh tends to zero. The notation $[B,C]_T=0$ records this fixed-time covariation. In stochastic differential notation, the corresponding shorthand is $dB_t\,dC_t=0$. It describes the limiting covariation, not a claim that the product of two individual increments is identically zero.

This statement is different from saying that the terminal values have zero covariance. Independence does imply $\mathbb{E}[B_TC_T]=0$, but that single-time fact alone does not give the limit of partition sums. The second-moment calculation supplies the missing information: it controls every partition through $\sum_i(\Delta t_i)^2\le T|\Pi|$.

Independence is essential. If the two processes are the same Brownian motion, the product-increment sum becomes $\sum_i(\Delta B_i)^2$, whose limit is $T$, not zero. Thus one must not infer zero covariation merely because two processes are Brownian motions; the independence assumption is doing real work in the calculation.

Common Mistake: Using the Wrong Increment Scale

A tempting but incorrect calculation is to treat $\Delta B_i\Delta C_i$ as though its variance were of order $\Delta t_i$. For $n$ equal intervals on $[0,1]$, that error would give $\mathbb{E}[S_\Pi^2]=n\cdot\tfrac1n=1$, suggesting no convergence to zero. The correct calculation uses independence: $$ \mathbb{E}[(\Delta B_i\Delta C_i)^2] =\mathbb{E}[(\Delta B_i)^2]\mathbb{E}[(\Delta C_i)^2] =\frac1n\cdot\frac1n =\frac1{n^2}. $$ The cross terms between distinct intervals have expectation zero, so the true second moment is $n\cdot\tfrac1{n^2}=\tfrac1n$, which tends to zero. The error came from assigning the product of two increments the scale of a single squared increment rather than multiplying their second moments.

The same distinction explains why $dB_t\,dC_t=0$ for independent Brownian motions while $dB_t\,dB_t=dt$: the former is a statement about cross-covariation, and the latter records the nonvanishing quadratic variation of one Brownian path. Both are shorthand for limits of sums along partitions, not ordinary differential multiplication rules.

Check Your Understanding

Use the second-moment calculation and its hypotheses to answer the following.

For a partition of $[0,3]$ with interval lengths $\tfrac12,1,\tfrac32$, calculate the exact second moment and the mesh bound for the covariation sum of independent Brownian motions.
Explain why $\mathbb{E}[\Delta B_i\Delta C_i]=0$ does not by itself prove that the full sum converges to zero in probability.
For $n$ equal intervals on $[0,T]$, simplify the exact second moment and state its limit as $n\to\infty$.
Identify the independence fact used to show that the cross terms $\mathbb{E}[X_iX_j]$ vanish when $i\ne j$.
What does the product-increment sum become if $C=B$, and why does this show the need for the independence hypothesis?