Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

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

Elements of answer:

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?

share|improve this question
6  
what makes it funny?!! –  Koushik Dec 31 '12 at 3:42
1  
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
3  
@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

1 Answer 1

up vote 2 down vote accepted

It takes one line of argument to prove that :

$\left \langle I ^{(n-2)},\left \langle M \right \rangle\right\rangle_t=0$

You only need to observe that $< M >_t $ is a continuous finite variation process, so its quadratic covariation with any process is 0.

Best regards

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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