I am attempting the following proof but two aspects of the solution confuse me:
Given \begin{align} I^{n}_{t} = \int^t_0 \Delta_u^ndW_u = \sum_{j=0}^{k-1}\Delta_{t_{j}}(W_{t_{j+1}}-W_{t_{j}}) + \Delta_{t_{k}}(W_{t} - W_{t_{k}}) \end{align} Prove \begin{align} E[I^n_t|\mathcal{F}_s] = I^{n}_{s} \end{align} where \begin{align} 0\le s <t \end{align}
$\textbf{Answer}$
Let $i<k$
\begin{align} E[I^{n}_t|\mathcal{F}_{t_{i}}] &= E\left[\sum^{k-1}_{j=0}\Delta_{t_{j}}(W_{t_{j+1}} - W_{t_{j}})+\Delta_{t_{k}}(W_t - W_{t_{k}})\big|\mathcal{F}_{t_{i}} \right]\\ &= \sum^{k-1}_{j=0}E\left[\Delta_{t_{j}}(W_{t_{j+1}} - W_{t_{j}})|\mathcal{F}_{t_{i}}\right]+E\left[\Delta_{t_{k}}(W_t - W_{t_{k}})\big|\mathcal{F}_{t_{i}}\right]\qquad \text{(1)} \end{align} Let $j\ge i$, take the first expectation and use Tower property of conditional expectations: \begin{align} E\left[\Delta_{t_{j}}(W_{t_{j+1}} - W_{t_{j}})|\mathcal{F}_{t_{i}}\right] &= E\left[\Delta_{t_{j}}E\left[(W_{t_{j+1}} - W_{t_{j}})\big|\mathcal{F}_{t_{j}}\right]|\mathcal{F}_{t_{i}}\right]\\ &= E\left[\Delta_{t_{j}}(E\left[W_{t_{j+1}}|\mathcal{F}_{t_{j}}\right] - W_{t_{j}})|\mathcal{F}_{t_{i}}\right] = 0 \qquad \text{(2)} \end{align}
$\textbf{Question 1: }$ Does this equal zero because $E[W_{t_{j+1}}|\mathcal{F}_{t_{j}}] = W_{t_{j}}$ and hence the $W_{t}$'s cancel?
Let $j\ge i$ and take second expectation: \begin{align} E\left[\Delta_{t_{k}}(W_t - W_{t_{k}})\big|\mathcal{F}_{t_{i}}\right] &= E\left[\Delta_{t_{k}}(E\left[W_t\big|\mathcal{F}_{t_{k}}\right] - W_{t_{k}})\big|\mathcal{F}_{t_{i}}\right] = 0 \qquad \text{(3)} \end{align}
Equations (1) and (2) imply that: \begin{align} E[I_{t}^{n}|\mathcal{F}_{t_{i}}] = E\left[\sum_{j=0}^{i-1} \Delta_{t_{j}}(W_{t_{j+1}} - W_{t_{j}})\big|\mathcal{F}_{t_{i}}\right] = I^{n}_{t_{i}} \end{align}
$\textbf{Question 2: }$ I'm confused by this last equation. If the expectation of equation (2) was zero, and the expectation of equation (3) was zero, then how is this equal to $I_{t_{i}}^{n}$ ?
I feel like this is something obvious. Please could someone explain what I'm missing?
Many thanks,
John
Question 1: Yes, that's it - because the Brownian motion $(W_t)_{t \geq 0}$ is a martingale.
Question 2: Fix $i<k$. By the definition of $I_t^n$ we have $$\begin{align*} I_t^n &= \sum_{j=0}^{i-1} \Delta_{t_j}(W_{t_{j+1}}-W_{t_j}) + \sum_{j=i}^{k-1} \Delta_{t_j}(W_{t_{j+1}}-W_{t_j}) + \Delta_{t_k}(W_t-W_{t_k}) \\ &=: J_1+J_2+J_3 \end{align*}$$ Now $(2)$ shows that
$$\mathbb{E}(J_2 \mid \mathcal{F}_{t_i}) = \sum_{j=i}^{k-1} \mathbb{E} \bigg( \Delta_{t_j}(W_{t_{j+1}}-W_{t_j}) \mid \mathcal{F}_{t_i} \bigg) = 0$$
and $(3)$ shows
$$\mathbb{E}(J_3 \mid \mathcal{F}_{t_i}) = 0.$$
Since $J_1=I_{t_i}^n$ is $\mathcal{F}_{t_i}$-measurable, we get
$$\mathbb{E}(I_t^n \mid \mathcal{F}_{t_i}) = \underbrace{\mathbb{E}(J_1 \mid \mathcal{F}_{t_i})}_{J_1}+\underbrace{\mathbb{E}(J_2 \mid \mathcal{F}_{t_i})}_{0} + \underbrace{\mathbb{E}(J_3 \mid \mathcal{F}_{t_i})}_{0} = J_1 = I_{t_i}^n.$$