1
$\begingroup$

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 $dWt \ dWt′ = \rho \ dt$?

$\endgroup$
4
$\begingroup$

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 $d\langle W_t,W_t^\prime\rangle$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.