I quote Øksendal (2003).

Itô integral. Let $\mathcal{V}=\mathcal{V}(S,T)$ be the class of functions $f(t,\omega):[0,\infty)\times\Omega\to\mathbb{R}$ such that $(t,\omega)\to f(t,\omega)$ is $\mathcal{B}\times\mathcal{F}$-measurable (where $\mathcal{B}$ denotes the Borel $\sigma$-algebra on $[0,\infty)$), $f(t,\omega)$ is $\mathcal{F}_t$-adapted and $\mathbb{E}\bigg[\int_{S}^T f(t,\omega)^2 dt\bigg]<\infty$.
[...] For functions $f\in\mathcal{V}$ we will now show how to define the Itô integral $$\mathcal{I}[f](\omega)=\int_{S}^{T}f(t,\omega)dB_t(\omega)$$ where $B_t$ is $1-$dimensional Brownian motion.
[...] The idea is natual: First we define $\mathcal{I}[\phi]$ for a simple class of functions $\phi$. Then, we show that each $f\in\mathcal{V}$ can be approximated by such $\phi$'s and we use this to define $\int fdB$ as the limit of $\int\phi dB$ as $\phi\to f$.
Recall that a function $\phi\in\mathcal{V}$ is called elementary if it has the form $$\phi(t,\omega)=\sum_j e_j(\omega)\cdot\chi_{[t_j, t_{j+1}]}(t)\tag{1}$$ Lemma (Itô isometry). If $\phi(t,\omega)$ is bounded and elementary then $$\mathbb{E}\bigg[\bigg(\int_{S}^{T}\phi(t,\omega)dB_t(\omega)\bigg)^2\bigg]=\mathbb{E}\bigg[\int_{S}^T\phi(t,\omega)^2dt\bigg]\tag{2}$$

Proof Set $\Delta B_j=B_{t_{j+1}}-B_{t_j}$. Then $$\mathbb{E}\left[e_ie_j\Delta B_i\Delta B_j\right]=\begin{cases}0\hspace{3.74cm}\text{if }i\ne j\\ \mathbb{E}\left[e_j^2\right]\cdot (t_{j+1}-t_j)\hspace{0.5cm}\text{if }i=j\end{cases}$$ using that $e_ie_j\Delta B_i$ and $\Delta B_j$ are independent if $i<j$. Thus: $$\mathbb{E}\left[\left(\int_S^T \phi dB\right)^2\right]=\sum_{i,j}\mathbb{E}\left[e_ie_j\Delta B_i\Delta B_j\right]=\color{blue}{\sum_j\mathbb{E}\left[e_j^2\right]\cdot(t_{j+1}-t_j)}\\\color{red}{=}\mathbb{E}\left[\int_S^T\phi^2 dt\right]$$

My question refers to the $\color{red}{\text{red}}$ equality above:
starting from $(1)$, I would say that $$\mathbb{E}\left[\int_S^T \phi^2 dt\right]=\mathbb{E}\left[\int_S^T\left(\sum_j e_j(\omega)\cdot\chi_{[t_j,t_{j+1})}(t)\right)^2dt\right]=\color{orange}{\mathbb{E}\left[\left(\sum_j e_j(\omega)\right)^2(t_{j+1}-t_j)\right]}$$ So, why does it hold true that: $$\color{blue}{\sum_j\mathbb{E}\left[e_j^2\right]\cdot(t_{j+1}-t_j)}=\color{orange}{\mathbb{E}\left[\left(\sum_j e_j(\omega)\right)^2(t_{j+1}-t_j)\right]}$$?

  • $\begingroup$ $\sum_j\mathbb{E}\left[e_j^2\right]\cdot(t_{j+1}-t_j)=\mathbb{E}\sum_j\left[e_j^2\right]\cdot(t_{j+1}-t_j)= \mathbb{E}\left[\int_S^T\phi^2 dt\right].$ The first equality is because of the linearity of the expectation and the second because of the definition of the integral. $\endgroup$ – UBM Oct 17 '20 at 15:00
  • $\begingroup$ As to your second equality: in my opinion, according to $(1)$, $\phi^2=\left[\sum_{j}e_j\right]^2$, and not $\phi^2=\sum_{j}\left[e_j^2\right]$. Isn't it? If so, your second equality would not hold true I think @UBM $\endgroup$ – Strictly_increasing Oct 17 '20 at 15:03
  • $\begingroup$ The intervals $[t_j,t_{j+1})$ are disjoint. $\endgroup$ – UBM Oct 17 '20 at 15:51

Note that with $I_j = [t_{j},t_{j+1})$,

$$\phi^2 =\left(\sum_{j}e_j \chi_{I_j} \right)^2 = \sum_{j}e_j \chi_{I_j}\sum_{k}e_k\chi_{I_k}= \sum_je_j^2 \chi^2_{I_j} + \underset{k\neq j}{\sum\sum}e_je_k \chi_{I_j} \chi_{I_k} \\= \sum_je_j^2 \chi_{I_j},$$

since $\chi^2_{I_j} = \chi_{I_j}$ and $\chi_{I_j} \chi_{I_k} = 0$ for disjoint intervals.

  • 1
    $\begingroup$ Easy to check that $\chi^2_{I_j}(\omega) = 1^2 = 1 = \chi_{I_j}(\omega)$ for $\omega \in I_j$, etc. $\endgroup$ – RRL Oct 17 '20 at 16:02

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.