4
$\begingroup$

If $(M_t)_{t \geq 0}$ is a continuous local martingale, one can define the iterated integrals $I_0=1$, $I_1(t)=M_t$ and for $n \geq 2$ $$I_{n}(t) = \int_0^t I_{n-1} (s) \mathrm{d} M_s.$$ By noting that $n! I_n(t) = H_n(M_t,\langle M,M \rangle_t)$, where $H_n(x,t)=t^{n/2} h_n(x/\sqrt{t})$ and $h_n$ is the $n$th Hermite polynomial, one can use Ito's formula to derive the Kailath-Segall identity $$n I_n = I_{n-1} M - I_{n-2} \langle M,M \rangle,$$ valid for $n \geq 2$ and also for $n=1$ if one defines $I_{-1}=0$.

I'm reading the paper http://projecteuclid.org/euclid.aop/1176990549 where on p.3 the authors say that this identity can also be "derived inductively by making two integrations by parts". For $n=2$, the identity is just a statement of the Ito formula for the function $f(x)=x^2$ but I'm stuck at making the induction step. Does somebody see how to do this without using the polynomial representation and can help me out?

$\endgroup$
1

1 Answer 1

3
$\begingroup$

The case $n=1$ is straighfoward. Now, applying Ito's formula and the definition of $\{I_n\}$ to integrate by parts we have: \begin{align*} I_n = \int_0^t I_{n-1} (s) \ d M_s = I_{n-1}M - \int_0^t I_{n-2}M\ dM - \int_0^t I_{n-2}\ d\langle M,M \rangle\ = \ = I_{n-1}M - \int_0^t I_{n-2}M\ dM - (\ I_{n-2}\langle M,M \rangle\ - \int_0^t I_{n-3}\langle M,M\rangle \ dM \ ) \end{align*} because by prorties of the stochastic integral, \begin{align*} d\langle I_{n-1},M \rangle = I_{n-2}d\langle M,M\rangle \ \end{align*} \begin{align*} d\langle I_{n-2}, \langle M,M \rangle \rangle = I_{n-3} d\langle M,\langle M,M \rangle \rangle \ \end{align*} but $\langle M,M \rangle $ is increasing, hence of finite variation, $M$ is continuous, so $ \langle M,\langle M,M \rangle \rangle =0 $.

Thus, $I_n = I_{n-1}M -\ I_{n-2}\langle M,M \rangle\ - \int_0^t (I_{n-2}M -I_{n-3}\langle M,M\rangle) \ dM . $
By induction, the parenthesis under the last integral equals $(n-1)I_{n-1}$, hence: $I_n = I_{n-1}M -\ I_{n-2}\langle M,M \rangle\ - \int_0^t (n-1)I_{n-1}\ dM = $
$ = I_{n-1}M -\ I_{n-2}\langle M,M \rangle\ - (n-1)I_{n} $ , which after rearranging terms gives the desired identity.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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