I'm currently self-studying Oksendal's Intro to Stochastic Differential Equations, and I'm confused as to how Oksendal proves one of the limits. To give context, let $\mathcal{H} = \{\mathcal{H}_t\}_{t\geq 0}$ be a filtration and $\mathcal{V}_{\mathcal{H}}$ be the class of functions $f:[0,\infty)\times \Omega\rightarrow\mathbb{R}$ such that

(i) $f$ is progressively measurable, i.e. it is $\mathcal{B}\times\mathcal{F}$-measurable on the product measure space $([0,\infty)\times\Omega, \mathcal{B}\times\mathcal{F},\lambda\times P)$.

(ii) $f$ is adapted to $\mathcal{H}$, and $B = \{B_t\}_{t\geq 0}$ be a standard Brownian motion that is a martingale with respect to $\mathcal{H}$.

(iii) $\mathbb{E}\left[\int_0^t \, f^2(s,\omega)\, ds\right]<\infty$ for all $t$.

Oksendal then defines an Ito process as the stochastic process $X = \{X_t\}_{t\geq 0}$ where

$X_t = X_0 + \int_0^t u(s,\omega)\, ds + \int_0^t v(s,\omega)\, dB_s$,

where $v(s,\omega)\in \mathcal{V}_{\mathcal{H}}$, and $u(t,\omega)$ is adapted to $\mathcal{H}$ and satisfies

$P\left[\int_0^t|u(s,\omega)|\, ds \text{ for all }t\geq 0\right] = 1$.

Oksendal proves that Ito's lemma applies for these kinds of Ito processes (in fact, he weakens condition (iii) in the definition of $\mathcal{V}_{\mathcal{H}}$, but then Ito's integral will exist only as a limit in probability rather than an object in $L^2(\Omega,\mathcal{F},P)$.) The strategy is to use Taylor's Theorem on simple functions $u$ and $v$ for some function $g$ that is once continuously differentiable in $t$ and twice continuously differentiable in $t$. I'm getting stuck on this limit

$\sum_j \dfrac{\partial^2 g}{\partial x^2}(t_j,\omega) v_j^2(\Delta B_j)^2 \rightarrow \int_0^t \dfrac{\partial^2 g}{\partial x^2}(s,\omega)v^2(s,\omega)\, ds$,

where the $j$ ranges over the partition $t_0 = 0< t_1 <\cdots<t_k = t$, $v_j = v(t_j,\omega)$ (recall that this is a simple function) and $\Delta B_j = B_{t_{j+1}} - B_{t_j}$ is an increment of a Brownian motion. In general, $\Delta$ will denote a forward difference. To simplify notation, let

$a_j = \dfrac{\partial^2 g}{\partial x^2} v_j^2$.

In particular, observe that $a_j$ is $\mathcal{H}_{t_j}$ measurable but not necessarily measurable w.r.t. $\mathcal{F}_{t_j}$, where $\mathcal{F} =\{\mathcal{F}_t\}_{t\geq 0}$ is the natural filtration generated by $B_t$. This follows from the fact that $v$ is only required to be $\mathcal{H}_{t_j}$-measurable, and $\mathcal{F}_{t_j}\subset \mathcal{H}_{t_j}$ since $B_t$ is a martingale w.r.t. $\mathcal{H}$.

We want to show this convergence in $L^2(P)$, thus

$\mathbb{E}\left[\left(\sum_j a_j(\Delta B_j)^2 -\sum_j a_j\Delta t_j\right)^2\right] = \sum_{i,j} \mathbb{E}[a_ia_j((\Delta B_i)^2-\Delta t_i)((\Delta B_j)^2 - \Delta t_j)]$.

Suppose $i<j$. Oksendal claims that $(\Delta B_j)^2 - \Delta t_j$ is independent of $a_ia_j((\Delta B_i)^2 - \Delta t_i)$. Now, if we required $u$ and $v$ of the Ito process to be adapted to $\mathcal{F}$, then this makes complete sense to me since $\Delta B_j$ is independent of all $B_s$ for $s\leq t_j$, and $\{B_s\}_{s\leq t_j}$ is precisely the collection of generators of $\mathcal{F}_{t_j}$.

However, $a_j$ is not necessarily measurable $\mathcal{F}_{t_j}$, only measurable $\mathcal{H}_{t_j}$, although we do know that $B_t$ is a martingale w.r.t $\mathcal{H}$. Therefore, how do I know independence still holds? Or do I need to use some other property of martingales?

For example, I know that

$\mathbb{E}[a_j \cdot \Delta B_j ] = \mathbb{E}[ \mathbb{E}[a_j \cdot \Delta B_j| \mathcal{H}_{t_j}]] = \mathbb{E}[ a_j \mathbb{E}[\Delta B_j| \mathcal{H}_{t_j}]] = \mathbb{E}[a_j \cdot 0] = 0$

by the tower property, $\mathcal{H}_{t_j}$-measurability of $a_j$, and the fact that $B_t$ is a martingale w.r.t $\mathcal{H}$.

  • 1
    $\begingroup$ Typically one assumes that $(\mathcal{H}_t)_t$ is an admissible filtration for the Brownian motion, i.e. that $B_t-B_s$ is independent from $\mathcal{H}_s$ for all $s \leq t$. (In my experience, Oksendals book is, unfortunately, not too careful about these kind of "details"...) $\endgroup$ – saz Mar 14 '18 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.