Step in proof of Ito isometry

Question

Question

I am following the course advanced stochastic processes online but I'm struggling with an equality in the proof of Ito iseometry of theorem 1. On the third line of page 3 it is used that: $$ \mathbb{E}[X_{t_j}^2 \mathbb{E}[(B_{t_{j+1}} - B_{t_j})^2 \mid \mathcal{F}_{t_j}]] = \mathbb{E}[X_{t_j}^2 (t_{j+1} -t_j)]. $$ where $X_t$ is a simple process and $(B_t)_t$ a Brownian motion, both adopted to the filtration $(\mathcal{F}_t)_t$. I'm pretty sure this step isn't very hard to show but I can't figure it out.

Side Question

I have a mathematical background and I'm looking at this theory to tackle inventory theory problems in a mathematical way, if you have some more interesting further references this could also be interesting. I'm thinking about Applied Probability and Queues

Answer 1

Due to the settings of a brownian motion $(B_{t_{j+1}} - B_{t_j})$ is independent of $F_{t_j}$ so for every measurable function is holds $$E[f(B_{t_{j+1}} - B_{t_j}) \mid \mathcal{F}_{t_j}] = E[f(B_{t_{j+1}} - B_{t_j})]$$

Taking $f(x) = x^2$ and consider that $(B_{t_{j+1}} - B_{t_j}) \sim \mathcal{N}(0,t_{j+1} - t_j)$ we get

$$E[(B_{t_{j+1}} - B_{t_j})^2 \mid \mathcal{F}_{t_j}] = E[(B_{t_{j+1}} - B_{t_j})^2] = t_{j+1} - t_j$$

Answer 2

This follows from the definition of the Brownian motion:

1. Independent increments
2. Increments over interval $\Delta t$ are distributed according to $N(0,\Delta t)$

Thus, $$E[(B_{t_j+1}−B_{t_j})^2∣F_{t_j}] = E[(B_{t_j+1}−B_{t_j})^2]=t_{j+1}-t_{j}$$