Doob's stopping time proof (discrete)

Question

I have difficulty in understanding the transition in following statement:

$\mathbb{E}\left( X_{\tau }\right) = \sum _{k=0}^{K}\mathbb{E}[ \mathbb{E}(X_{k}|\mathscr{F}_{k})1_{\tau=k}] = \sum _{k=0}^{K}\mathbb{E}[X_K1_{\tau=k}] $

Could you please explain what property is used to replace $\mathbb{E}(X_{k}|\mathscr{F}_{k})$ by $X_K$?

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Further details on are from (Bingham, Kiesel 2004). I have highlighted the part, which I cannot grasp :

Answer 1

Since $\{X_t\}$ is martingale we know that $E[X_K \mid \mathcal{F}_k]=X_k$.

Then, reemplacing $X_k$ we obtain

$E[X_k 1_{(\tau=k)}]=E[E[X_K \mid \mathcal{F}_k] 1_{(\tau=k)}]$. (*)

On the other hand, since $\tau$ is stopping time we have that $1_{(\tau=k)}$ is $\mathcal{F}_k$-measurable. Then , due to an elemental property of conditional expectation, we have $E[X_K \mid \mathcal{F}_k]1_{(\tau=k)}=E[X_K 1_{(\tau=k)}\mid \mathcal{F}_k]$.

In view of (*), we conclude that $E[X_k 1_{(\tau=k)}]=E[E[X_K 1_{(\tau=k)} \mid \mathcal{F}_k] ]$

Finally, due to the definition of conditional expectation $E[X_K 1_{(\tau=k)}]=E[E[X_K 1_{(\tau=k)}\mid\mathcal{F}_k]]$.

We obtain, $E[X_k 1_{(\tau=k)}]=E[X_K 1_{(\tau=k)}]$.

Answer 2

If $\mathcal{F}_k$ is the $\sigma$-algebra generated by $X_1, \dots, X_k$, the Martingale properties gives you that:

$\mathbb{E}[X_{n+1}|\mathcal{F}_n] = X_n$ and $X_n \in \mathcal{F}_n$.