# Funny problem about stochastic integrals and Ito' s lemma

Consider a probability filtred space $(\Omega, \mathcal F, \mathcal F_ t, \mathbb P)$ and a continuous $\mathcal F _t$-martingal starting from $0$, $M = (M_t)_{t \geq 0}$, such that $\left \langle M \right \rangle_\infty \leq 1$ $\mathbb P$-ps. Now, we define by recurence $\forall n \in \mathbb{N}$ $$I^{(o)}_t \equiv 1, \ I^{(n+1)}_t = \int _0 ^t I^{(n)}_s d M_s \ , \ t \geq 0$$

The question: How to show the following relation ?

$$\forall n \geq 2 : \ \ n I ^{(n)}_t = I ^{(n-1)}_t M_t - I ^{(n-2)}_t \left \langle M \right \rangle_t$$

Let's suppose by induction hypothesis that $(n -1) I ^{(n-1)}_t = I ^{(n-2)}_t M_t - I ^{(n-3)}_t \left \langle M \right \rangle_t$

By Ito's lemma, we have that

\begin{align} I ^{(n-1)}_t M_t &= \int _0 ^t I ^{(n-1)}_s dM_s+ \int _0 ^t M_s \ d I ^{(n-1)}_s + \left \langle I ^{(n-1)},M \right \rangle_t \\& =I ^{(n)}_t +\int _0 ^t M_s \ I ^{(n-2)}_s d M_s+ \int _0 ^t \ I ^{(n-2)}_s \ I ^{(0)}_s d \left \langle M \right \rangle_s \\&= I ^{(n)}_t +\int _0 ^t \left[ (n -1) I ^{(n-1)}_t+ I ^{(n-3)}_t \left \langle M \right \rangle_t\right] d M_s+ \int _0 ^t \ I ^{(n-2)}_s \ I ^{(0)}_s d \left \langle M \right \rangle_s \\& = nI ^{(n)}_t + \int _0 ^t I ^{(n-3)}_t \left \langle M \right \rangle_t d M_s+ \int _0 ^t \ I ^{(n-2)}_s \ I ^{(0)}_s d \left \langle M \right \rangle_s \\ & \overset{\text{Ito's lemma}}{=} nI ^{(n)}_t +I ^{(n-2)}_t \left \langle M \right \rangle_t -\left \langle I ^{(n-2)},\left \langle M \right \rangle\right\rangle_t\end{align}

which is almost the proof except the fact that I still don't know how to show that $$\left \langle I ^{(n-2)},\left \langle M \right \rangle\right\rangle_t=0$$

Someone can help me on it, please?

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what makes it funny?!! –  Koushik Dec 31 '12 at 3:42
I don't know the definition of Ito's integral, but this feels like integration by parts to me. –  Patrick Da Silva Dec 31 '12 at 3:54
@PatrickDaSilva, just keep in mind the chain rule in stochastic calculus is more involved, with a quadratic component. –  alancalvitti Dec 31 '12 at 5:32
It's probably a standard notation because no one here seems to ask, but what do the $\langle \rangle$ stand for here? I wish to understand what just happened. I know about stochastic processes and filtrations but I've never used Ito's integral. –  Patrick Da Silva Jan 2 '13 at 13:35

$\left \langle I ^{(n-2)},\left \langle M \right \rangle\right\rangle_t=0$
You only need to observe that $< M >_t$ is a continuous finite variation process, so its quadratic covariation with any process is 0.