I'm reading a theorem (about integration by parts) from page 9 of these notes, i.e.,

Theorem Let $X$ and $Y$ be continuous semi-martingales such that $$ \mathbb E \bigg [ \int_0^t X_s^2 \mathrm d \langle Y \rangle_s + \int_0^t Y_s^2 \mathrm d \langle X \rangle_s\bigg ] <\ \infty \quad \forall t \ge 0. \quad (\star) $$ Then $$ X_t Y_t - X_0 Y_0 = \int_0^t X_s \mathrm d Y_s + \int_0^t Y_s \mathrm d X_s + \langle X, Y \rangle_s \quad \text{a.s.} \quad \forall t \ge 0. $$

and its proof, i.e.,

Proof. By using Itô's lemma, we have: $$ \begin{aligned} \left(X_t+Y_t\right)^2-\left(X_0+Y_0\right)^2 &= 2 \int_0^t\left(X_s+Y_s\right) \mathrm d\left(X_s+Y_s\right)+\langle X+Y\rangle_t \\ \left(X_t-Y_t\right)^2-\left(X_0-Y_0\right)^2 &= 2 \int_0^t\left(X_s-Y_s\right) \mathrm d\left(X_s-Y_s\right)+\langle X-Y\rangle_t \end{aligned} $$ Subtracting these two formulas gives: $$ 4 X_t Y_t-4 X_0 Y_0=4 \int_0^t X_s \mathrm d Y_s+4 \int_0^t Y_s \mathrm d X_s+\underbrace{\left(\langle X+Y\rangle_t-\langle X-Y\rangle_t\right)}_{=4\langle X, Y\rangle_t} $$ which completes the proof.

In the proof, the author applies Itô's lemma on the continuous semi-martingale $X+Y$ with function $f(x) = x^2$. Itô's lemma requires $$ \mathbb E \bigg [ \int_0^t (f'(X_s + Y_s))^2 \mathrm d \langle X+Y \rangle_s \bigg ] = 4\mathbb E \bigg [ \int_0^t (X_s + Y_s)^2 \mathrm d \langle X+Y \rangle_s \bigg ] < \infty \quad \forall t\ge 0. $$

Could you explain how this requirement is satisfied from the assumption $(\star)$?

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 $t \ge 0$, $$ f(X_t)-f(X_0) = \int_0^t f'(X_s) \mathrm d X_s + \frac{1}{2} \int_0^t f''(X_s) \mathrm d \langle X \rangle_s \quad \text{a.s.} $$


1 Answer 1


I sent an email to the author and have got the reply. The assumption should be $$ \mathbb E \bigg [ \int_0^t X_s^2 \mathrm d \langle Y \rangle_s + \int_0^t Y_s^2 \mathrm d \langle X \rangle_s + \int_0^t X_s^2 \mathrm d \langle X \rangle_s + \int_0^t Y_s^2 \mathrm d \langle Y \rangle_s \bigg ] <\ \infty \quad \forall t \ge 0. $$


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