# Can this integrability assumption be weakened in a way that the stochastic integration by parts still holds?

$$\newcommand{\Ex}{\mathbb E} \newcommand{\diff}{~\mathrm d}$$Recently, I have read the integration by parts formula for a continuous semi-martingale in these notes.

Theorem Let $$X$$ and $$Y$$ be continuous semi-martingales such that $$\Ex \bigg [ \int_0^t (X_s + Y_s)^2 \diff \langle X + Y \rangle_s \bigg ] <\ \infty \quad \forall t \ge 0. \quad \quad (\star)$$ Then $$X_t Y_t - X_0 Y_0 = \int_0^t X_s \diff Y_s + \int_0^t Y_s \diff X_s + \langle X, Y \rangle_t \quad \text{a.s.} \quad \forall t \ge 0.$$ Proof Let $$f(x):=x^2$$. Then $$f'(x) = 2x$$ and $$f''(x)=2$$ for all $$x \in \mathbb R$$. Notice that $$Z _t :=X_t+Y_t$$ is a continuous semi-martingale. By Itô's lemma, \begin{align} (X_t+Y_t)^2 - (X_0+Y_0)^2 &= 2 \int_0^t (X_s+Y_s) \diff (X_s+Y_s)+\langle X+Y\rangle_t, \\ (X_t-Y_t)^2 - (X_0-Y_0)^2 &= 2 \int_0^t (X_s-Y_s) \diff (X_s-Y_s)+\langle X-Y\rangle_t. \end{align} Subtracting these two formulas gives: $$4 X_t Y_t-4 X_0 Y_0=4 \int_0^t X_s \diff Y_s+4 \int_0^t Y_s \diff X_s + (\underbrace{\langle X+Y\rangle_t - \langle X-Y\rangle_t}_{=4\langle X, Y\rangle_t} ),$$ which completes the proof.

The integrability condition $$(\star)$$ is imposed so that the required hypothesis of Itô's lemma is satisfied.

Can $$(\star)$$ be weakened in a way that the integration by parts still holds?

Related definition: Let $$(\Omega, \mathcal F, \mathbb P)$$ be a probability space and $$\mathfrak F = (\mathcal{F}_t, t \ge 0)$$ a filtration.

Let $$M$$ be a continuous square-integrable martingale w.r.t. $$\mathfrak F$$. Then $$M^2$$ is a continuous sub-martingale w.r.t. $$\mathfrak F$$. By Doob's decomposition theorem, there exists a unique continuous increasing process $$\langle M \rangle$$ adapted to $$\mathfrak F$$ such that $$\langle M \rangle_0 = 0$$ a.s. and that $$M^2 - \langle M \rangle$$ is a continuous martingale w.r.t. $$\mathfrak F$$. Then $$\langle M \rangle$$ is called the quadratic variation of $$M$$.

A process $$X$$ is called a continuous semi-martingale w.r.t. $$\mathfrak F$$ is that can be written as $$X_t = M_t+V_t$$ where

• $$M$$ is a continuous square-integrable martingale w.r.t. $$\mathfrak F$$.
• $$V$$ is a continuous process that has bounded variation and is adapted to $$\mathfrak F$$ such that $$V_0 = 0$$ a.s.

Then the quadratic variation of $$X$$ is defined as $$\langle X \rangle := \langle M \rangle.$$

Let $$H$$ be a continuous adapted process such that $$\mathbb E \bigg [ \int_0^t H_s^2 \mathrm d \langle X \rangle_s \bigg ] < \infty \quad \forall t\ge 0.$$

Then the stochastic integral of $$H$$ with respect to $$X$$ is defined as $$(H \cdot X)_t := \int_0^t H_s \mathrm d X_s := \underbrace{\int_0^t H_s \mathrm d M_s}_{\text{Itô's integral}} + \underbrace{\int_0^t H_s \mathrm d V_s}_{\text{Riemann-Stieltjes's integral}}.$$

Itô's lemma. Let $$X$$ be a continuous semi-martingale and $$f:\mathbb R \to \mathbb R$$ twice continuously differentiable such that $$\mathbb E \bigg [ \int_0^t (f'(X_s))^2 \mathrm d \langle X \rangle_s \bigg ] < \infty \quad \forall t\ge 0.$$ Then for all $$t \ge 0$$, $$f(X_t)-f(X_0) = \int_0^t f'(X_s) \mathrm d X_s + \frac{1}{2} \int_0^t f''(X_s) \mathrm d \langle X \rangle_s \quad \text{a.s.}$$

I suspect the lectures notes introduce these because they want to stay in the realm of $$L^2$$-martingales.