Cross-Variation of Independent Brownian Motions

Products of Increments Along a Partition

For one Brownian motion, squared increments over finer and finer partitions accumulate to elapsed time. With two independent Brownian motions, the corresponding products of increments behave differently: their signed sum tends to zero. This distinction is the basic contrast between quadratic variation and cross-variation. It is also important to specify the partitions and the mode of convergence; convergence in mean square along arbitrary deterministic partitions is not, by itself, a claim of almost-sure convergence along every such sequence.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((B_t)_{t\ge0}\) and \((W_t)_{t\ge0}\) be independent standard Brownian motions, each with continuous time index \(t\) and state space \(\mathbb{R}\). Fix \(T>0\). A deterministic partition of \([0,T]\) is a finite strictly increasing list \(0=t_0<t_1<\cdots<t_m=T\). Write \(h_k=t_k-t_{k-1}\) for its interval lengths and \(|\Pi|=\max_{1\le k\le m}h_k\) for its mesh. Define the cross-increment sum

We use the standard Brownian increment facts: an increment over an interval of length \(h\) is normal with mean \(0\), variance \(h\), and fourth moment \(3h^2\); increments on disjoint intervals of the same Brownian motion are independent. Independence of the two Brownian motions means that their increments, considered as processes, are independent as well.

Proof. Set \(\Delta_k B=B_{t_k}-B_{t_{k-1}}\), \(\Delta_k W=W_{t_k}-W_{t_{k-1}}\), and \(U_k=\Delta_k B\,\Delta_k W\). The two increments in \(U_k\) are independent, each has mean zero, and each has variance \(h_k\). Thus \(\mathbb{E}[U_k]=\mathbb{E}[\Delta_k B]\mathbb{E}[\Delta_k W]=0\), and \(\mathbb{E}[U_k^2]=\mathbb{E}[(\Delta_k B)^2]\mathbb{E}[(\Delta_k W)^2]=h_k^2\). For distinct \(k,\ell\), the pairs \((\Delta_k B,\Delta_k W)\) and \((\Delta_\ell B,\Delta_\ell W)\) are independent, by independent increments and independence of the two motions. Therefore \(\mathbb{E}[U_kU_\ell]=\mathbb{E}[U_k]\mathbb{E}[U_\ell]=0\). Expanding the square of the sum gives \(\mathbb{E}[C(\Pi)]=0\) and \(\mathbb{E}[C(\Pi)^2]=\sum_k\mathbb{E}[U_k^2]=\sum_k h_k^2\). Since \(h_k\le|\Pi|\), \(\sum_k h_k^2\le|\Pi|\sum_k h_k=T|\Pi|\). This bound tends to zero along any sequence whose mesh tends to zero, proving mean-square convergence. Applying Markov’s inequality to \(C(\Pi_r)^2\) gives \(\mathbb{P}(|C(\Pi_r)|>\varepsilon)\le\mathbb{E}[C(\Pi_r)^2]/\varepsilon^2\le T|\Pi_r|/\varepsilon^2\), which proves convergence in probability. \(\square\)

The theorem does not require equal-length intervals. Its key quantity is the sum of squared interval lengths, bounded by the mesh times the total length. The exact variance also gives a useful calculation for a particular partition, while the bound gives a uniform estimate over all partitions with mesh no larger than a specified threshold.

Almost-Sure Convergence on Dyadic Grids

Mean-square convergence implies convergence in probability, but it does not alone establish almost-sure convergence along a chosen sequence. A summable probability bound can yield the stronger conclusion for dyadic partitions. For \(n\ge0\), divide \([0,T]\) into \(2^n\) equal intervals and denote the resulting sum by \(C_n\). Each interval has length \(T/2^n\), so the exact variance is \(2^n(T/2^n)^2=T^2/2^n\).

Proof. Fix \(\varepsilon>0\). The variance formula and Chebyshev’s inequality give \(\mathbb{P}(|C_n|>\varepsilon)\le T^2/(2^n\varepsilon^2)\). The sum over \(n\ge0\) of these upper bounds is \(2T^2/\varepsilon^2\), which is finite because \(\sum_{n\ge0}2^{-n}=2\). The first Borel–Cantelli lemma therefore implies that, with probability one, \(|C_n|>\varepsilon\) occurs only finitely often. Apply this statement to the countable thresholds \(\varepsilon=1,1/2,1/3,\ldots\). The intersection of the resulting probability-one events still has probability one. On that intersection, for every positive integer \(q\), eventually \(|C_n|\le1/q\); this is precisely \(C_n\to0\). \(\square\)

The proof uses summability specific to this sequence of grids. For an arbitrary sequence of deterministic partitions with mesh tending to zero, the theorem above guarantees convergence in probability; the variance bound alone need not be summable along that sequence, so this argument does not establish almost-sure convergence there.

Contrast with Each Motion’s Quadratic Variation

For the same partition, the sum of squared increments of \(B\) is \(Q_B(\Pi)=\sum_k(\Delta_kB)^2\). Each term has expectation \(h_k\). Since an increment of length \(h_k\) is normal with variance \(h_k\), its squared value has variance \(3h_k^2-h_k^2=2h_k^2\). The squared increments over distinct intervals are independent. Consequently, \(Q_B(\Pi)\) has mean \(T\) and variance \(2\sum_kh_k^2\). The same statements hold for \(W\). In particular, both quadratic sums converge to \(T\) in mean square as the mesh tends to zero, while the cross-increment sum converges to zero.

Proof of the quadratic-sum comparison. For each interval, \(\mathbb{E}[(\Delta_kB)^2]=h_k\), and the fourth-moment formula gives \(\operatorname{Var}((\Delta_kB)^2)=3h_k^2-(h_k)^2=2h_k^2\). Adding expectations gives \(\mathbb{E}[Q_B(\Pi)]=\sum_kh_k=T\). Independence of increments makes the variance of the sum \(2\sum_kh_k^2\). Hence \(\mathbb{E}[(Q_B(\Pi)-T)^2]=2\sum_kh_k^2\le2T|\Pi|\), which tends to zero as the mesh tends to zero. Replacing \(B\) by \(W\) gives the same calculation. \(\square\)

On dyadic partitions, these quadratic sums also converge almost surely to \(T\). Indeed, for each \(\varepsilon>0\), Chebyshev’s inequality gives \(\mathbb{P}(|Q_B(\Pi_n)-T|>\varepsilon)\le2T^2/(2^n\varepsilon^2)\), a summable sequence in \(n\); the first Borel–Cantelli lemma and the countable-threshold argument used above yield almost-sure convergence. The same applies to \(W\). Thus on these grids, the two individual square sums approach \(T\), whereas the sum of products of independent motions’ increments approaches zero.

Dependence Changes the Cross-Increment Limit

The zero limit depends on independence. For a concrete comparison, let \(B\) and \(W\) be independent standard Brownian motions and set \(D_t=(3/5)B_t+(4/5)W_t\). This is a standard Brownian motion, but it is not independent of \(B\). On any partition, the cross sum between \(B\) and \(D\) is exactly \((3/5)\sum_k(\Delta_kB)^2+(4/5)C(\Pi)\). The first term tends to \((3/5)T\) in mean square by the quadratic-sum calculation, and the second tends to zero in mean square by the cross-increment theorem. Therefore the cross sum between \(B\) and \(D\) tends to \((3/5)T\), not zero.

A Common Mistake

A tempting but incorrect calculation is to replace \(\Delta_kB\,\Delta_kW\) by \((\Delta_kB)^2\), then conclude that the cross sum approaches \(T\). That replacement treats increments from two different motions as if they were the same random variable. For one interval of length \(h>0\), independence gives \(\mathbb{E}[\Delta B\,\Delta W]=\mathbb{E}[\Delta B]\mathbb{E}[\Delta W]=0\), whereas \(\mathbb{E}[(\Delta B)^2]=h\). The expectations already disagree, so the substitution is invalid. The correct calculation keeps the product of the two increments: its mean is zero and its variance is \(h^2\). Across a partition these variances add, yielding \(\sum_kh_k^2\), which vanishes as the mesh shrinks.