Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am a bit confused in my reading of martingales. I am using Breiman's book and there is an example that doesn't quite make sense to me. Let us define a sequence of random variables in this way:

Let $Y_0=0$ and $Y_1,Y_2,\ldots$ i.i.d. with distribution $P[Y_i=1]=p$ and $P[Y_i=-1]=1-p$ for some $p\in(0,1)$ . Define $X_n:=\sum_{i=0}^n Y_i$. Then he says: If $p=\frac{1}{2}$ , then $(X_n)_{n\in\mathbb{N}}$ is a martingale with respect to its natural filtration $\mathcal{F}_n^X:=\sigma(X_0,\ldots,X_n)$. I dont really understand this statement, because I have to show that $E[X_{n}\mid\mathcal{F}_{s}]=X_{s}$ whenever $n\geq s$ , and I have no idea why this has to be true. By definition of conditional expectation, for any $A\in\mathcal{F}_{s}$ , we have $\int_{A}E[X_{n}\mid\mathcal{F}_{s}]=\int_{A}X_{n}\, dP=\int_{A}\, X_{s}\, dP$ . For some reason, $p=\frac{1}{2}$ should play a role here. I don't really see it.

share|improve this question
    
To be more concrete $s=1, n=2$ ... Let $A = \{X_1=1\}$ and show that $\int_A X_2\,dP = \int_A X_1\,dP$. –  GEdgar Dec 14 '11 at 16:29
    
[By definition of conditional expectation, for any $A\in\mathcal{F}_{s}$ , we have $\int_{A}E[X_{n}\mid\mathcal{F}_{s}]=\int_{A}X_{n}\, dP=\int_{A}\, X_{s}\, dP$]: in fact, the first equal sign is the definition of conditional expectation $E[X_{n}\mid\mathcal{F}_{s}]$, the second is the relation you want to show. –  Did Dec 14 '11 at 18:03
    
@Didier Does my answer below make sense? –  r.g. Dec 14 '11 at 18:39
    
@r.g. Since you ask... a crucial argument is missing when you assert that $\mathrm E(Y_{n+1}:A)=0$ if $p=\frac12$, namely, the fact that $Y_{n+1}$ and $A$ are independent (do you see why?), hence $\mathrm E(Y_{n+1}\mathbf 1_A)=\mathrm E(Y_{n+1})\mathrm P(A)=(2p-1)\mathrm P(A)$. This shows that $\mathrm E(X_{n+1}\mid\mathcal F_n)=X_n+2p-1$ and you are done. –  Did Dec 14 '11 at 20:00
    
yes. that's an important point that I forgot to state in my proof. Of course, $Y_n+1$ and $A$ are independent, because $A$ is in the sigma algebra that is not generated by $Y_n+1$. –  r.g. Dec 15 '11 at 2:31

1 Answer 1

The trick here is to write recursively. You want to show that $E[X_{n+1}\mid\mathcal{F}_{n}]=X_{n}$ . To see this, write $X_{n+1}=X_{n}+Y_{n+1}$ . Then, by the linearity of conditional expectation (which follows from the linearity of the Lebesgue integral), one has $E[X_{n+1}\mid\mathcal{F}_{n}]=E[X_{n}+Y_{n+1}\mid\mathcal{F}_{n}]=E[X_{n}\mid\mathcal{F}_{n}]+E[Y_{n+1}\mid\mathcal{F}_{n}]$ . By defintion of a martingale, $X_{n}$ is $\mathcal{F}_{n}$ -measurable, and so by another property of conditional expectations, $E[X_{n}\mid\mathcal{F}_{n}]=X_{n}$ . It remains to show that $E[Y_{n+1}\mid\mathcal{F}_{n}]=0$ almost surely (with respect to the probability measure $P$ ) , and this is where $p=\frac{1}{2}$ comes into play. Note by the definition of conditional expectation that for any $A\in\mathcal{F}_{n}$ , $\int_{A}E[Y_{n+1}\mid\mathcal{F}_{n}]\, dP=\int_{A}Y_{n+1}\, dP$ . Note that $\int_{A}Y_{n+1}\, dP=0$ , since we are given that $P[Y_{n+1}=1]=\frac{1}{2}$ and $P[Y_{n+1}=-1]=\frac{1}{2}$ . Now note that if it were that case for $p<\frac{1}{2}$ , then it is easy to see that $\int_{A}\, Y_{n+1}\, dP<0$ , and in fact, $E[X_{n+1}\mid\mathcal{F}_{n}]<X_{n}$ , which makes the stochastic process $(X_{n})_{n\in\mathbb{N}}$ a submartingale. Analogously, when $p>\frac{1}{2}$ , then we have a supermartingale.

share|improve this answer

Your Answer

 
discard

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.