# A doubt along the proof of Itô isometry

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 $$i. Thus: $$\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: $$\color{blue}{\sum_j\mathbb{E}\left[e_j^2\right]\cdot(t_{j+1}-t_j)}=\color{orange}{\mathbb{E}\left[\left(\sum_j e_j(\omega)\right)^2(t_{j+1}-t_j)\right]}$$?

• $\sum_j\mathbb{E}\left[e_j^2\right]\cdot(t_{j+1}-t_j)=\mathbb{E}\sum_j\left[e_j^2\right]\cdot(t_{j+1}-t_j)= \mathbb{E}\left[\int_S^T\phi^2 dt\right].$ The first equality is because of the linearity of the expectation and the second because of the definition of the integral. – UBM Oct 17 '20 at 15:00
• 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 – Strictly_increasing Oct 17 '20 at 15:03
• The intervals $[t_j,t_{j+1})$ are disjoint. – 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.
• Easy to check that $\chi^2_{I_j}(\omega) = 1^2 = 1 = \chi_{I_j}(\omega)$ for $\omega \in I_j$, etc. – RRL Oct 17 '20 at 16:02