Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.


$${\rm Cov}[dW_t,dB_t]=\rho dt$$

then what is

$$\mathbb{E} \left[\int_0^t\sigma_{1s}dW_s \int_0^t\sigma_{2s}dB_s\right]$$

where $\sigma_{1s}$ and $\sigma_{2s}$ are two deterministic functions of $t$?

share|cite|improve this question

1 Answer 1

I assume that you mean that $(W_t,B_t)$ is 2D correlated Wiener process. By the property of Ito integral $\mathbb{E}\left( \int_0^t \sigma_{1s} \mathrm{d} W_s \right) = 0$ and $\mathbb{E}\left( \int_0^t \sigma_{2s} \mathrm{d} S_s \right) = 0$. Hence: $$ \begin{eqnarray} \mathbb{E}\left( \int_0^t \sigma_{1}(s) \mathrm{d} W_s, \int_0^t \sigma_{2}(u) \mathrm{d} B_u \right) &=& \mathbb{Cov} \left( \int_0^t \sigma_{1}(s) \mathrm{d} W_s, \int_0^t \sigma_{2}(u) \mathrm{d} B_u \right) \\ &\stackrel{\text{Ito isometry}}{=}& \rho \int_0^t \sigma_1(s) \sigma_2(s) \mathrm{d} s \end{eqnarray} $$

share|cite|improve this answer
Times $\rho$, in the last line? – Did Jul 1 '12 at 21:19
@did Thanks! Corrected. – Sasha Jul 1 '12 at 22:26
Upvoted. $ $ $ $ – Did Jul 1 '12 at 22:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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