Cross-Variation of Independent and Correlated Brownian Motions
Summary
The document discusses the cross-variation of two Brownian motions and the role of their dependence. For independent Brownian motions, their increments have zero covariance, leading to zero cross-variation. For correlated motions, the answer represents one process as a component proportional to the other plus an independent orthogonal Brownian component, with the coefficient set by their correlation.
This decomposition gives the standard relationship in which the cross-variation accumulates at a rate proportional to the correlation. The excerpt offers a sketch rather than a fully rigorous proof: it first reasons from covariance at a fixed time, then asserts the link to quadratic covariation, and does not detail the limiting sums over increments. Its dependence argument also presumes a constant correlation and a suitable Brownian representation. These concepts are useful in stochastic models where correlated sources of randomness drive asset prices or interest rates.
Key ideas
- Independent Brownian motions have zero cross-variation.
- A correlated Brownian motion can be decomposed into a scaled component of another motion and an orthogonal component.
- The correlation coefficient determines the rate of cross-variation under the stated constant-correlation setup.
- Zero covariance at a fixed time motivates, but alone does not provide, a complete proof using quadratic covariation.
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Full text
# Brownian Motions theorems
# Brownian Motions theorems
I know that if $W$ and $W′$ are two independent brownian motions, then $dWt \ dWt′$ = 0. How can I prove/demonstrate this theorem?
Additionaly, how can we prove that if $W$ and $W′$ are dependent, then $dWt \ dWt′ = \rho \ dt$?
## Answer by FunnyBuzer (score 4, accepted)
https://quant.stackexchange.com/a/45916
For the first part looks quite obvious, since independence implies that the covariance is zero and since the correlation is just the covariance divided by the product of the standard deviations, it will be zero, too. $$\text{Cov}(W_t,W_t^\prime)=\mathbb E [W_t,W_t^\prime]-\mathbb E [W_t]\mathbb E[W_t^\prime]$$ By law of iterated expactation $$\mathbb E [W_t,W_t^\prime]=\mathbb E[\mathbb E[W_tW_t^\prime|W_t^\prime]]=\mathbb E[W_t^\prime\underbrace{\mathbb E[W_t|W_t^\prime]}_{W_t\perp W_t^\prime}]=\mathbb E[W_t^\prime\mathbb E[W_t]]=\mathbb E[W_t]\mathbb E[W_t^\prime]$$ $$\Rightarrow \text{Cov}(W_t,W_t^\prime)=\mathbb E [W_t,W_t^\prime]-\mathbb E [W_t]\mathbb E[W_t^\prime]=\mathbb E [W_t]\mathbb E[W_t^\prime]-\mathbb E [W_t]\mathbb E[W_t^\prime]=0$$ $$\text{Corr}(W_t,W_t^\prime)=\frac{\text{Cov}(W_t,W_t^\prime)}{\sqrt{\text{Var}[W_t]}\sqrt{\text{Var}[W_t^\prime]}}=\frac{0}{t}=0$$ $$\Rightarrow d\langle W_t,W_t^\prime\rangle\underbrace{=}_{\text{by indep.}}d\langle W_t\rangle d\langle W_t^\prime\rangle=0$$
When $W$ and $W^\prime$ are dependent, one can re-write the Brownian motion as a linear combination of the other Brownian motion and another Brownian motion under the same filtration, that is orthogonal to the other, i.e. $$dW_t^\prime=\rho dW_t+\sqrt{1-\rho^2}dW_t^\perp, \quad W_t\perp W_t^\perp$$ then compute $d\langle W_t,W_t^\prime\rangle$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.