The Martingale-Part of the Doob decomposition for a stochastic process $(X_n)_n$ and filtration $(\mathcal F_n)_n$ is
$$M_n=X_0+\sum_{k=1}^n\bigl(X_k-\mathbb{E}[X_k\,|\,\mathcal{F}_{k-1}]\bigr)$$ (see e.g. here).
I want to prove that this is indeed a Martingale and suceeded by showing $$\mathbb E[M_n-M_{n-1}| \mathcal F_{n-1}] =0$$.
However, I failed to show $\mathbb E[M_n| \mathcal F_{n-1}] =M_{n-1}$ and wanted to know where my mistake is.
My attempt is:
$$\mathbb E[M_n| \mathcal F_{n-1}] = X_0 + \sum_{k=1}^n(\mathbb E[X_k | \mathcal F_{n-1}] – \mathbb E\big[ \mathbb E[X_k | \mathcal F_{k-1}] | \mathcal F_{n-1}\big]) = (*).$$
Now, note that for $k<n$:
$$ \mathbb E\big[ \mathbb E[X_k | \mathcal F_{k-1}] | \mathcal F_{n-1}\big]) = \mathbb E[X_k | \mathcal F_{n-1}] = X_k,$$
and for $k=n$:
$$\mathbb E\big[ \mathbb E[X_k | \mathcal F_{k-1}] | \mathcal F_{n-1}\big]) = \mathbb E[X_n | \mathcal F_{n-1}],$$
so that all terms of the sum in $(*)$ are 0. Therefore, we get
$$(*) = X_0.$$
However, we should get $(*) = M_{n-1}$.
Where am I making my mistake?
Best Answer
When $k=n$, both terms become $X_n$, thus it is $X_n-X_n=0$. When $k<n$, it remains same. Therefore, you will have $M_{n-1}$.