$E[M_t|H_t]$ is a martingale with respect to $H=(H_t)_{t\geq 0}$, $H_t \subset \mathcal{F}_t \forall t$

Question

Being $(M_t)_{t \geq 0}$ an $\mathcal{F}$-martingale, I have to show that $E[M_t|H_t]$ is a martingale with respect to $H$ ($H=(H_t)_{t\geq 0}$, $H_t \subset \mathcal{F}_t \forall t$).

I proceded this way:

1) since $E[M_t]<+\infty$, then $E[M_t|H_t] <+\infty \forall$ subsets of $\mathcal{F}_t$

2)$E[M_t|H_t]$ is $H_t$-measurable by definition of conditional expectation

3)$E[E[M_t|H_t]|H_s]=E[E[M_t|H_s]|H_t]$ (by tower rule) = $E[E[M_t-M_s+M_s|H_s]|H_t]=E[E[M_t-M_s]+M_s|H_t]=E[M_s|H_t]=E[M_s|H_s]$

In particular I'm not sure on the last equality.

Answer 1

Let $W_t=E[M_t|H_t]$. For $s\leq t$, iterated conditioning gives $$ E[W_t|H_s]=E[E[M_t|H_t]|H_s]=E[M_t|H_s]. $$ Using iterated conditioning again, the RHS above equals $$ E[M_t|H_s]=E[E[M_t|F_s]|H_s]=E[M_s|H_s]=W_s $$ where the next to last equality above uses the Martingale property of $M$. Thus we indeed have $$ E[W_t|H_s]=W_s\text{ for }s\leq t. $$

Answer 2

For $s \leq t$, $E[E[M_t|F_s]|H_s] = E[M_t|H_s]$ due to iterated conditioning.

Furthermore, $E[E[M_t|F_s]|H_s] = E[M_s|H_s] = M_s$ since $M_t$ is an $F$-martingale and $M_s$ is $H_s$ measurable.

Therefore, $E[M_t|H_s] = M_s$.