Martingales

Laurent

MHF Hall of Honor
Aug 2008
1,174
769
Paris, France
Please read attachment
I guess you know Itô's formula; you should try it (the two-variable version : with \(\displaystyle f(t,x)=x^3-3tx\) for the first one)
 

Laurent

MHF Hall of Honor
Aug 2008
1,174
769
Paris, France
Thanks for the quick answer. Below I have tried to do the calculations. Please see attachment

Are they correct?

Can I get hint on how to show, that \(\displaystyle M_3\) and \(\displaystyle M_4\) are true martingales?
Once you get \(\displaystyle dM_3(t)=H(t) dB(t)\), you are done proving that \(\displaystyle M_3\) is a local martingale, whatever the right-continuous adapted process \(\displaystyle H(t)\) is. You don't have to specify that \(\displaystyle H(t)\) is a martingale or some kind of product of martingales.

As for proving that \(\displaystyle M_3,M_4\) are true martingales, you have the following result: if \(\displaystyle M\) is a local martingale with \(\displaystyle M_0=0\), the following is equivalent :
(i) \(\displaystyle M\) is a true square-integrable martingale (\(\displaystyle E[M_t^2]<\infty\))
(ii) for all \(\displaystyle t>0\), \(\displaystyle E[\langle M\rangle_t]<\infty\).

Remember that \(\displaystyle E[\langle \int_0^\cdot H(s)dB(s)\rangle_t]=\int_0^t E[H(s)^2] ds\) so that you can easily check this criterion in your case.
 
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