# Differentiation under the integral sign for the Itō integral

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space
• $T>0$
• $I:=(0,T]$
• $(\mathcal F_t)_{t\in\overline I}$ be a complate and right-continuous filtration on $(\Omega,\mathcal A,\operatorname P)$
• $M$ be a real-valued continuous square-integrable $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$
• $\mu_M$ denote the Doléans measure corresponding to $M$
• $d\in\mathbb N$
• $\Lambda\subseteq\mathbb R^d$ be open

Now, let $F:\Omega\times\overline I\times\Lambda\to\mathbb R$ with $$F_t\in C^1(\Lambda)\;\;\;\text{almost surely for all }t\in\overline I\tag1$$ and $$F(x)\in\mathcal L^2(\mu_M)\;\;\;\text{for all }x\in\Lambda\tag2.$$ Moreover, let $i\in\left\{1,\ldots,d\right\}$ and assume that $$\int\sup_{x\in K}\left|\frac{\partial F}{\partial x_i}(x)\right|^2\:{\rm d}\mu_M<\infty\;\;\;\text{for all compact }K\subseteq\Lambda\tag3.$$

Let $$N(x):=F(x)\cdot M$$ denote the Itō integral process of $F(x)$ with respect to $M$ for $x\in M$. Are we able to conclude that $N$ is partially differentiable with respect to the $i$th-variable?

For the sake of simplicity, assume $\Lambda=\mathbb R^d$. Let $x\in\Lambda$ and $$G(h):=\frac{N(x+he_i)-N(x)}h\;\;\;\text{for }h\in\mathbb R\setminus\left\{0\right\}.$$ Clearly, $$G(h)=\frac{F(x+he_i)-F(x)}h\cdot M\;\;\;\text{almost surely for all }h\in\mathbb R\setminus\left\{0\right\}\tag4$$ and $$\frac{F_t(x+he_i)-F_t(x)}h\xrightarrow{h\to0}\frac{\partial F_t}{\partial x_i}(x)\;\;\;\text{almost surely for all }t\in\overline I\tag5$$ From $(3)$ and $(5)$, we should be able to conclude that the convergence in $(5)$ holds in $L^2(\mu_M)$ by applying Lebesgue's dominated convergence theorem. This should yield that the corresponding Itō integral processes converge in the space of square-integrable martingales.

Am I missing something?

• 1. Wouldn't it be more natural to assume that $\sup_{x \in K} |\partial_{x_i} F(x)|$ is square-integrable with respect to $\mu_M$...? If this is not true, the stochastic integral $(\partial_{x_i} F(x) \bullet M)$ might not even be well-defined? 2. If I understand you correctly, then the bound in (3) may depend on $\omega \in \Omega$, yes? – saz Aug 1 '18 at 12:54
• @saz 1. There was indeed a square missing. 3. No, the integral is with respect to the Doléans measure $\mu_M$ which is a measure on the predictable $\sigma$-algebra on $\Omega\times\overline I$. – 0xbadf00d Aug 3 '18 at 20:51
• Thanks for the clarification. – saz Aug 4 '18 at 5:07

By the definition of the Doléans measure, we have

\begin{align*} \left\| G(h)- (\partial_{x_i} F(x)) \bullet M \right\|_{L^2(\mathbb{P})}^2 &= \left\| \left( \frac{F(x+h e_i)-F(x)}{h}- \partial_{x_i} F(x) \right) \bullet M \right\|_{L^2(\mathbb{P})}^2 \\ &= \int \left| \frac{F(x+h e_i)-F(x)}{h}- \partial_{x_i} F(x) \right|^2 \, d\mu_M. \tag{6} \end{align*}

Since

$$\frac{F_t(x+h e_i)-F_t(x)}{h}- \partial_{x_i} F_t(x) \xrightarrow[]{h \to 0} 0 \quad \text{a.s.}$$

and, by the mean value theorem,

\begin{align*} \left| \frac{F_t(x+h e_i)-F_t(x)}{h}- \partial_{x_i} F_t(x) \right|^2 &\leq \left( \left|\frac{F_t(x+h e_i)-F_t(x)}{h} \right| + |\partial_{x_i} F_t(x)| \right)^2 \\ &\leq 4 \sup_{y \in B(x,1)}|\partial_{x_i} F_t(y)|^2 \in L^1(\mu_M) \end{align*}

for any $h \in (0,1)$, it follows from the dominated convergence theorem that the right-hand side of $(6)$ converges to $0$ as $h \to 0$, and therefore $N(x) = F(x) \bullet M$ is partially differentiable (in the $L^2(\mathbb{P})$-sense) and

$$\partial_{x_i} N(x) = (\partial_{x_i} F(x)) \bullet M$$