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Let $M_t$ be a continuous local martingale with $M_0=0$ and define $I_t^0=1$ and $I_t^n= \int_0^t I_s^{n-1}\; dM_s$. Prove that we have $$n I_t^n= I_t^{n-1}- \int_0^t I_s^{n-2} \;d((M)_t) $$ where $(M)_t$ is the quadratic variation of $M$.

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integration by parts should work. – bgins Nov 29 '11 at 12:58
There is a problem with your statement. If you compute the case $n=2$ (with integration by parts), you obtain $2I^2 = (I^1)^2- \int I^0 d(M)$, which is different from your formula. – pgassiat Nov 29 '11 at 16:34
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@phil Could you tell us where you found this formula? – Byron Schmuland Dec 2 '11 at 12:58

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