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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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1  
At springerlink.com/content/f2m7114uw22851q2 you can read about Karl-Heinz Normal, eponym of the Normal subgroup, the Normal bundle, Normal crossings, Normal sheaves, Normal varieties, Normal forms, and the Normal distribution. However, it is fiction. I've set the word "normal" in lower case in the question above. –  Michael Hardy Jun 20 '12 at 17:29
    
@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. –  Gabriel Jun 20 '12 at 18:09

1 Answer 1

up vote 5 down vote accepted

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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@ did : as a remark $J_t=E[-\int_0^t sin^2(B_s)ds]$ is another expression for $J_t$ (simpler IMO) coming from Itô's isometry and your expression of $cos(B_t)$ Best regards. –  TheBridge Jun 20 '12 at 12:51
    
@did: Thanks so much! I'll focus on the first half of your answer, since it turns out to be a much more elegant solution than my attempt to rewrite the integral. I'm a little confused about two aspects: 1) why are the limits $0\rightarrow t$ and $u\rightarrow t$, as opposed to both being $0\rightarrow t$?; 2) why do we choose to rewrite $B_s + B_u = 2B_u - (B_s - B_u)$? Is this to avoid issues of correlation? Thanks, no needs for detailed answers but a reference would be appreciated. –  Gabriel Jun 20 '12 at 13:40
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1) The integral of any symmetric function $f(u,s)$ over $(u,s)\in[0,t]^2$ is twice the integral of $f(u,s)$ over $0\leqslant u\leqslant s\leqslant t$. 2) Yes, this rewriting is the most direct way (that I know) to compute the variance of $B_s+B_u$. –  Did Jun 20 '12 at 13:48
    
@did: Thanks again. To clarify my confusion, the domain of integration $(u,s)\in [0,t]^2$ contains an infinite number of points (of combined measure zero?) where $s=u$. I had previously attempted a solution in a similar way to your suggestion, but had neglected to make the two changes mentioned above, hence my answer was incorrect. It still isn't fully clear to me why we can't change the domain of integration to $(u,s)\in [0,t]\times [u,t]$ but then maintain $B_s + B_u$ as is, since we no longer have to worry about the case $s=u$ in the new domain. –  Gabriel Jun 20 '12 at 14:03
    
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. –  Did Jun 20 '12 at 20:14

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