# Show that S_n is a martingale.

Let $\{X_n:n \geq 1\}$ be a sequence of random variables. Let $S_n = \sum_{i=1}^{n} (X_i-E(X_i|X_1,...,X_{i-1})).$ Show that {S_n} is a martingale.

This is what I have so far:

To show it's martingale then : $E(S_{n+1}|F_n)$

=$E(S_{n+1}|S_1,S_2,..,S_n) \rightarrow E(S_{n+1}|X_1,X_2,..,X_n) =$

$E(X_i -E(X_i|X_1,....,X_{i-1}))|X_1,X_2,...,X_n)$ =

$X_i-E(X_i|X_1,...,X_{i-1})|X_1,X_2,...,X_n)$.

Am I going in the right path? So with the feedback of angryavian, something like this?

$X_i -X_{n+1}-E[X_{n+1}|X_1,...,X_n]$.?

• Did you note the confusion in indices in your question? If you correct these, the solution should pop up. – Did May 7 '17 at 4:52
• I'm new to proof and martingale questions so I wouldn't be surprised If i confused some parts. – Killer May 7 '17 at 5:14
• Even being new, you might want to correct the confusions of $X_i$ for $X_n$. – Did May 7 '17 at 7:38
• Since $X_n: n\geq 1$ then $X_n=X_i$? – Killer May 7 '17 at 14:47
• What? Sorry but you are not making any sense. – Did May 7 '17 at 14:50

## 1 Answer

You want to show $E[S_{n+1} \mid X_1,\ldots,X_n] = S_n$, or equivalently $E[S_{n+1} - S_n \mid X_1,\ldots,X_n]=0$.

Note that $S_{n+1} - S_n = X_{n+1} - E[X_{n+1} \mid X_1,\ldots,X_n]$ and plug this into the above expectation.