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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 $t$?

1 Answers 1

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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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    Upvoted. $ $ $ $2012-07-01