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I'm trying to solve a problem that's now doing my head in a bit. I'll share with you the question and let's see if somebody can shed some light into the matter:

Let B be a standard Brownian Motion started at zero, and let M be a stochastic process defined by: $$ M_t = \int_0^{\log(\sqrt{1 + 2t})} e^{s}\mathrm dB_s\,. $$

  1. Show that M is a Standard Brownian Motion.
  2. Calculate $$E\left(\int_0^tM^6_s\mathrm dM_s\right).$$
  3. Calculate $$E\left(\left(\int_0^tM_s\mathrm dM_s\right)^3\right).$$

Hope you can give me some hints, I have the feeling it's actually not that tricky. I would solve question number one using Lévy's characterization theorem for Brownian Motion, not so sure about questions 2 and 3.

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    $\begingroup$ So does Levy's characterisation work for part 1? $\endgroup$
    – Lost1
    Dec 31, 2013 at 12:30
  • $\begingroup$ It does actually, you just need to calculate the quadratic variation of M and the result is t, which is any Brownian Motion's quadratic variation. My problem is still 2 and 3. $\endgroup$
    – Adam
    Dec 31, 2013 at 12:37
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    $\begingroup$ so now, treat $M$ as a brownian motion. I am trying to have a go myself. $\endgroup$
    – Lost1
    Dec 31, 2013 at 12:40

1 Answer 1

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For part 2:

Is a square-integrable continuous local martingale a true martingale?

I think 'lemma 3' in the first answer tells you how to solve question 2. It shows the stochastic integral inside your expectation is a true martingale, which means the expectation is 0.

For part 3:

The integral is $(M_t^2-t)/2$, where $M$ is a Brownian motion. Then you just cube this and use the fact $E[M_t^n] = t^{n/2}(n-1)!$ for $n$ is even, $E[M_t^n]=0$ for $n$ odd. In fact we could have used it for part 2 but this would be too much hassle...

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  • $\begingroup$ I'm happy with the answer for question 2. $\endgroup$
    – Adam
    Jan 2, 2014 at 12:00
  • $\begingroup$ For question 3: couldn't we just say that $\int_0^tM_s\mathrm dM_s$ has as a result a Brownian Motion (which follows a normal distribution), and the expectation of a normally distributed random variable to the power of 3 is always $0$? $\endgroup$
    – Adam
    Jan 2, 2014 at 12:05
  • $\begingroup$ No... How is it a normal distribution? Only when you integrate something normal against dt, it is normal. I already said it is $1/2(M^2-t)$, this is clearly not normal. You can check this by Ito. Also, please please please do NOT accept this answer in 10 days. I need it not to be accepted to get the unsung hero badge. $\endgroup$
    – Lost1
    Jan 2, 2014 at 12:42
  • $\begingroup$ @Adam see above comment. Also please do NOT press the 'check mark' on my question :P $\endgroup$
    – Lost1
    Jan 2, 2014 at 12:49
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    $\begingroup$ Same as part2, show it is a proper martingale, so has mean 0 $\endgroup$
    – Lost1
    Jan 2, 2014 at 13:39

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