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this problem has come up in my research and is confusing me immensely, any light you can shed would be deeply appreciated.

Let $B(t)$ denote a standard Brownian motion (Wiener process), such that the difference $B(t)-B(s)$ has a normal distribution with zero mean and variance $t-s$.

I am seeking an expression for

$E\left[ \cos(B(t))\int\limits_0^t \sin(B(s))\,\textrm{d}B(s) \right],$

where the integral is a stochastic It$\hat{\textrm{o}}$ integral. My first thought was that the expectation of the integral alone is zero, and that the two terms are statistically independent, hence the whole thing gives zero. However, I can't prove this.

To give you a little background: this expression arises as one of several terms in a calculation of the second moment of the integral

$\int\limits_{0}^{t}\cos(B(s))\,\textrm{d}s,$

after applying It$\hat{\textrm{o}}$'s lemma and squaring. I can simulate this numerically, so I should know when I get the right final expression!

Thanks.

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    @Michael Hardy: This is one of the most cryptic explanations I've ever received for a correction, but I eventually worked it out! Thanks for your edit, and the edifying link to Karl-Heinz Normal's brief and false bio.2012-06-20

1 Answers 1

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This addresses the question cited as a motivation.

For every $t\geqslant0$, introduce $X_t=\int\limits_{0}^{t}\cos(B_s)\,\textrm{d}s$ and $m(t)=\mathrm E(\cos(B_t))=\mathrm E(\cos(\sqrt{t}Z))$, where $Z$ is standard normal. Then $\mathrm E(X_t)=\int\limits_{0}^{t}m(s)\,\textrm{d}s$ and $\mathrm E(X_t^2)=\int\limits_{0}^{t}\int\limits_{u}^{t}2\mathrm E(\cos(B_s)\cos(B_u))\,\textrm{d}s\textrm{d}u$.

For every $s\geqslant u\geqslant0$, one has $2\cos(B_s)\cos(B_u)=\cos(B_s+B_u)+\cos(B_s-B_u)$. Furthermore, $B_s+B_u=2B_u+(B_s-B_u)$ is normal with variance $4u+(s-u)=s+3u$ and $B_s-B_u$ is normal with variance $s-u$. Hence, $2\mathrm E(\cos(B_s)\cos(B_u))=m(s+3u)+m(s-u)$, which implies $ \mathrm E(X_t^2)=\int\limits_{0}^{t}\int\limits_{u}^{t}(m(s+3u)+m(s-u))\,\textrm{d}s\textrm{d}u. $ Since $m(t)=\mathrm e^{-t/2}$, this yields after some standard computations, $\mathrm E(X_t)=2(1-\mathrm e^{-t/2})$ and $ \mathrm E(X_t^2)=2t-\frac13(1-\mathrm e^{-2t})-\frac83(1-\mathrm e^{-t/2}). $ Sanity check: When $t\to0^+$, $\mathrm E(X_t^2)=t^2+o(t^2)$.


To compute the integral $J_t=\mathrm E\left[ \cos(B_t)\int\limits_{0}^{t} \sin(B_s)\,\textrm{d}B_s \right]$, one can start with Itô's formula $ \cos(B_t)=1-\int\limits_{0}^{t} \sin(B_s)\,\textrm{d}B_s-\frac12\int\limits_{0}^{t} \cos(B_s)\,\textrm{d}s, $ hence $ J_t=\mathrm E(\cos(B_t))-\mathrm E(\cos^2(B_t))-\frac12\int\limits_{0}^{t} \mathrm E(\cos(B_t)\cos(B_s))\,\textrm{d}s, $ and it seems each term can be computed easily.

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    Every integral (with respect to the Lebesgue measure) of a function null everywhere except possibly on the diagonal $0\leqslant u=s\leqslant t$ is zero because the measure of the diagonal is zero.2012-06-20