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I need to find the expectation of this stochastic integral.

$$E\left[W(t) \int_0^t e^{3W(s)} dW(s)\right]$$

Obviously I cannot put the Expectation inside the integral because it is stochastic. Also I cannot separate $W(t)$ with the integral. Additionally, I know that the expectation of the $dW(t)$ term is $0$ but with the term outside the integral this does not hold. My intuition is to use the Ito formula, however I do not know which expression to use since part of it has an integral and is already Ito and part isn't.

jdods
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Shu Tiang wei
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1 Answers1

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Apply the Ito isometry:

$$ \begin{aligned} E\left(W(t) \cdot \int_0^t e^{3W(s)}dW(s)\right) &=E\left(\int_0^tdW(s) \cdot \int_0^t e^{3W(s)}dW(s)\right) \\ &=E\left(\int_0^te^{3W(s)}ds\right) \end{aligned}$$

The we use the fact that $W(s)\sim N(0,s)$ and property of stochastic integral that $E(\int_0^t X_s ds \mid \mathcal F_0)=\int_0^t E(X_s\mid \mathcal F_0) ds $.

$$ \begin{aligned} E\left(\int_0^te^{3W(s)}ds\right) &=\int_0^tE\left(e^{3W(s)}\right)ds \\ &=\int_0^t \frac{1}{\sqrt{2\pi s}}\int_{-\infty}^\infty e^{3x}e^{-\frac{x^2}{2s}}dxds\\ &=\int_0^t \frac{1}{\sqrt{2\pi s}}\int_{-\infty}^\infty \exp\left(-\frac{(x-3s)^2} {2s}\right) e^{\frac32s} dxds\\ &=\int_0^t e^{\frac32s} ds\\ &=\frac23(e^{\frac32t}-1) ds\\ \end{aligned}$$

jdods
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