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Right now I dont know how to find the expectation and the variance for the following stochastic integral: $$\int_{0}^t B(s) dW(s)$$ where $B(t)$ and $W(t)$ are correlated standard Brownian Motions with $d\langle B, W \rangle_t = \gamma dt$.

Usually, when $B(t)$ and $W(t)$ are independent, we know $$\mathbb{E}\left(\int_{0}^t B(s) dW(s)\right) = 0$$ and by Ito's isometry, $$Var\left(\int_{0}^t B(s) dW(s)\right) = \mathbb{E}\left(\int_{0}^t B(s) dW(s)\right)^2 = \int_{0}^t \mathbb{E}B(s)^2 ds.$$

Does these results still apply?

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Decompose your Brownian motion $W$ as $W_t = \gamma B_t + \bar \gamma B_t^\bot$, where $\bar \gamma = \sqrt{1-\gamma^2}$ and $B^\bot$ is independent from $B$. Then:

$$\int_0^t B_s \, \mathrm dW_s = \gamma \int_0^t B_s \, \mathrm dB_s + \bar\gamma \int_0^t B_s \, \mathrm dB_s^\bot$$

The first term can be computed using Itô's formula, the second one involves two independent Brownian motions.

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One other way to understand the integral $\int_0^tB(s)\,dW(s)$ is to understand the $(B(t),W(t))^{\top}$ as a 2-d Wiener process with covariance function $\begin{pmatrix}t&\gamma t\\ \gamma t& t\end{pmatrix}$. Using these notations, the Ito's isometry is effective too and following equalities are held: $$ \mathsf{E}\Bigl[\int_0^tB(s)\,dW(s)\Bigr]=0,\qquad \mathsf{E}\Bigl(\int_0^tB(s)\,dW(s)\Bigr)=\int_0^t\mathsf{E}[B(s)]^2\,ds=\frac{t^2}2.$$

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  • $\begingroup$ pretty old question, but still, isn't there a missing square in the last line? The stochastic integral should be squared inside the expectation sign. $\endgroup$
    – Chaos
    Commented Jan 30, 2020 at 8:16

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