4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ So does Levy's characterisation work for part 1? $\endgroup$ – Lost1 Dec 31 '13 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 '13 at 12:37
  • 1
    $\begingroup$ so now, treat $M$ as a brownian motion. I am trying to have a go myself. $\endgroup$ – Lost1 Dec 31 '13 at 12:40
2
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ I'm happy with the answer for question 2. $\endgroup$ – Adam Jan 2 '14 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 '14 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 '14 at 12:42
  • $\begingroup$ @Adam see above comment. Also please do NOT press the 'check mark' on my question :P $\endgroup$ – Lost1 Jan 2 '14 at 12:49
  • 1
    $\begingroup$ Same as part2, show it is a proper martingale, so has mean 0 $\endgroup$ – Lost1 Jan 2 '14 at 13:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.