Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X_k$ be iid with $P(X_k=1)=p$, $P(X_k=-1)=q$, where $p\ne q$. Now define $S_n=\sum\limits^n_{k=1}X_k$ with $S_0=0$ and $Y'_n=S^2_n$, and $Y_n=Y'_n-\alpha_n =S^2_n-\alpha_n$, we want to find $\alpha_n$ such that $Y_n$ is a martingale.

I found $\alpha_n=1+2S_n(p-q)$ but according to Doob's decomposition the compensator should be $F_{n-1}$ measurable, but we don't have that here.


share|cite|improve this question
up vote 2 down vote accepted

The process $(S_n^2-A_n)_{n\geqslant0}$ is a martingale if $A_0=0$ and $A_{n+1}=A_n+2(p-q)S_n+1$, for example $$ A_n=\sum\limits_{k=0}^{n-1}(2(p-q)S_k+1)=n+2(p-q)\sum\limits_{k=1}^{n-1}S_k. $$ For every $n\geqslant1$, $A_n$ is $F_{n-1}$ measurable hence $(A_n)_{n\geqslant1}$ is a compensator of $(S_n^2)_{n\geqslant0}$.

share|cite|improve this answer
So $\alpha_n n$ is actually $F_{n-1}$ measurable? – John Klein Nov 12 '11 at 22:19
It seems your $\alpha_nn$ should actually be $\alpha_n$. Anyway, as I said, $A_n$ (which is $\alpha_n$ if the typo in your post is actually as I described) is $F_{n-1}$ measurable. – Did Nov 12 '11 at 22:22
@John, did you read my answer? Please do (carefully), and then we'll talk. – Did Nov 12 '11 at 22:25
Sorry, Didier, I am not sure how you calculated the $A_{n+1}$, my answer is similar to yours at the top, but I Don't have it in a recursive definition. thanks – John Klein Nov 12 '11 at 22:32
Thanks so much, I figured it out now! – John Klein Nov 12 '11 at 22:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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