# Stochastic integration by parts formula to prove identity between iterated 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 and can help me out?

For simplicity, I'll assume that $(M_t)_{t \geq 0}$ is a continuous martingale. The argumentation is similar for a local martingale, but more technical.

Set $$I_n(t) := \int_0^t I_{n-1}(s) \, dM_s, \qquad n \in \mathbb{N}, t \geq 0, \tag{1}$$

and suppose that $$(n-1) I_{n-2}(t) = I_{n-2}(t) M_t - I_{n-3}(t) \langle M \rangle_t, \qquad t \geq 0. \tag{2}$$

If $(X_t)_{t \geq 0},(Y_t)_{t \geq 0}$ are two continuous martingales, then we know from Itô's formula that

$$X_t \cdot Y_t - X_0 Y_0 = \int_0^t X_s \, dY_s + \int_0^t Y_s \, dX_s+ \langle X,Y \rangle_t.$$

If we apply this for $X=I_{n-1}$ and $Y=M$, then we see

\begin{align*} I_n(t) &\stackrel{(1)}{=} \int_0^t I_{n-1}(s) \, dM_s \\ &= I_{n-1}(t) M(t) - \int_0^t M(s) \, dI_{n-1}(s) - \langle I_{n-1},M \rangle_t \\ &\stackrel{(1)}{=} I_{n-1}(t) M(t) - \int_0^t I_{n-2} M(s) \, dM_s - \langle I_{n-1},M \rangle_t. \\ &\stackrel{(2)}{=} I_{n-1}(t) M(t) - (n-1) \underbrace{\int_0^t I_{n-1}(s) dM_s}_{I_{n}(t)} - \int_0^t I_{n-3} \langle M \rangle_t dM_s - \langle I_{n-1},M\rangle_t. \tag{3} \end{align*}

Recall that $\langle M \rangle_t$ is a non-decreasing (hence in particular of bounded variation) and continuous (since $M$ is a continuous martingale; see e.g. Ikeda/Watanabe for a proof). Therefore, it follows from Itô's formula that

\begin{align*} \int_0^t I_{n-3}(s) \langle M \rangle_s \, dM_s &= \int_0^t \langle M \rangle_s dI_{n-2}(s) \\ &\stackrel{\text{Itô}}{=} I_{n-2}(t) \langle M \rangle_t - \int_0^t I_{n-2}(s) d\langle M \rangle_s \\ &= I_{n-2}(t) \langle M \rangle_t - \langle I_{n-1},M \rangle_t. \tag{4} \end{align*}

For the last equality, we have used that $$\langle \int_0^{\cdot} f(s) \, dX_s, Y \rangle_t = \int_0^t f(s) \, d\langle X,Y \rangle_s$$ for any two martingales $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$. Combining $(3)$ and $(4)$ finishes the proof.

• Thank you VERY much , I was stuck at (3) and now that I see how to do it I don't understand why :-) – herrsimon Aug 20 '14 at 13:15