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How much is

$$\int_0^T tB_t \, dt$$

where $B_t$ is Brownian motion and $T$ an universal constant?

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yes, its independent of $t$ and $B_t$ –  Troy McClure Jan 12 '13 at 16:04
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1 Answer

up vote 3 down vote accepted

It is a Gaussian variable with expectation zero and variance $$ \mathbb{E}\Bigl[\Bigl(\int_0^T tB_t\,dt\Bigr)^2\Bigr] =\int_0^T\int_0^T\mathbb{E}[stB_sB_t]\,ds\,dt =\int_0^T\int_0^Tst\min(s,t)\,ds\,dt. $$ I expect you can compute the final integral yourself, by dividing the square into the two triangles given by $s<t$ and $s>t$.

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I was going to write a more detailed, multiline calculation, but MathJax wouldn't cooperate – it refused to deal with the aligned environment for some odd reason. So I ran out of energy, sorry about that. –  Harald Hanche-Olsen Jan 12 '13 at 16:10
    
thanks! i can calculate the last integral myself. I understand you used Ito isometry –  Troy McClure Jan 12 '13 at 16:29
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Not quite as fancy as that, since these are not Itō integrals. It's basically just Fubini's theorem, relying on the fact that the expectation is itself an integral. And then I use $\mathbb{E}[B_sB_t]=\min(s,t)$, which is a fundamental property of Brownian motion. In the first equality, I just write out the square as a product of the inner integral with itself, using different integration variables $s$ and $t$ in the two factors. –  Harald Hanche-Olsen Jan 12 '13 at 16:35
    
are you sure that the last step is valid? factoring the expectation where $t$ and $B_t$ have something in common? –  Troy McClure Jan 12 '13 at 17:14
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As far as the expectation is concerned, $s$ and $t$ are just constants. And $\mathbb{E}[aX]=a\mathbb{E}[X]$ for deterministic $a$ and r.v. $X$. –  Eckhard Jan 12 '13 at 18:02
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