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If

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

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 $t$?

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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: $$ \begin{eqnarray} \mathbb{E}\left( \int_0^t \sigma_{1}(s) \mathrm{d} W_s, \int_0^t \sigma_{2}(u) \mathrm{d} B_u \right) &=& \mathbb{Cov} \left( \int_0^t \sigma_{1}(s) \mathrm{d} W_s, \int_0^t \sigma_{2}(u) \mathrm{d} B_u \right) \\ &\stackrel{\text{Ito isometry}}{=}& \rho \int_0^t \sigma_1(s) \sigma_2(s) \mathrm{d} s \end{eqnarray} $$

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1  
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

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