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?



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


