2
$\begingroup$

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?

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

$\endgroup$
  • $\begingroup$ Did you note the confusion in indices in your question? If you correct these, the solution should pop up. $\endgroup$ – Did May 7 '17 at 4:52
  • $\begingroup$ I'm new to proof and martingale questions so I wouldn't be surprised If i confused some parts. $\endgroup$ – Killer May 7 '17 at 5:14
  • $\begingroup$ Even being new, you might want to correct the confusions of $X_i$ for $X_n$. $\endgroup$ – Did May 7 '17 at 7:38
  • $\begingroup$ Since $X_n: n\geq 1$ then $X_n=X_i$? $\endgroup$ – Killer May 7 '17 at 14:47
  • $\begingroup$ What? Sorry but you are not making any sense. $\endgroup$ – Did May 7 '17 at 14:50
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.