Martingale: why $\mathbb E[S_{n+m}\mid X_1,...,X_n]= S_n$.

Question

Let $(S_n)$ a martingale by ratio to $(X_n)$ (I'm not sure if the terme "by ratio" is correct, I hope you'll understand). A lemma of my lecture say:

$$\mathbb E[S_{n+m}\mid X_1,...,X_n]= S_n,\quad n,m\geq 1.$$

The proof goes like:

$$\mathbb E[S_{n+m}\mid X_1,...,X_n]\underset{(1)}{=}\mathbb E\big[\mathbb E[S_{n+m}\mid X_1,...,X_{n+m-1}]\ \big|\ X_1,...,X_n\big]=\mathbb E[S_{n+m-1}\mid X_1,...,X_n]$$

To me, for $(1)$, the fact that $(S_{n})$ is a martingale should imply that $$S_{n+m}=\mathbb E[S_{n+m+1}\mid X_1,...,X_{n+m}],$$ so why do we have $$S_{m+n}=\mathbb E[S_{n+m}\mid X_1,...,X_{n+m-1}]\ \ ?$$

Answer 1

We want to prove that $\mathbb E[S_{n+m}\mid X_1,...,X_n]= S_n,\quad n,m\geq 1$

Use induction on $m$.

By the definition of the martingale, $\mathbb E[S_{n+1}\mid X_1,...,X_n]= S_n$ so the base case is true.

Suppose $\mathbb E[S_{n+m}\mid X_1,...,X_n]= S_n$ for some $m \geq 1$. Then:

$$S_n = \mathbb E[S_{n+m}\mid X_1,...,X_n]$$ (by the induction hypothesis)

$$=\mathbb E [\mathbb E[S_{n+m+1}\mid X_1,...,X_{n+m}]| X_1,...,X_n]$$ (since $S_n$ is a martingale, we can replace $S_{n+m}$ with $\mathbb E[S_{n+m+1}\mid X_1,...,X_{n+m}]$ )

$$=\mathbb E[S_{n+m+1}| X_1,...,X_n]$$

(by the tower property - http://en.wikipedia.org/wiki/Law_of_total_expectation#Proof_in_the_general_case).

So the inductive step is done.

Answer 2

If $\mathcal{B}\subset\mathcal{A}$ are measurable, then $\mathbb{E}\left[\mathbb{E}\left[Y\mid\mathcal{A}\right]\mid\mathcal{B}\right]=\mathbb{E}\left[Y\mid\mathcal{B}\right]$ for any random variable $Y$. Now use it for $\mathcal{A}=\sigma\left(X_1,\ldots,X_{n+m-1}\right)$ and $\mathcal{B}=\sigma\left(X_1,\ldots,X_{n}\right)$.