# Stochastic integration by parts to obtain Kailath Segall identity for iterated stochastic integrals?

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?

• – Did Aug 20 '14 at 13:54

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.