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 studying martingales and I have a few conceptual questions regarding why we need stopping times. My book (Probability and Computing by Mitzenmacher and Upfal) defines a martingale as follows:

A sequence of random variables $Z_0,Z_1,\ldots$ is a martingale with respect to the sequence $X_0,X_1,\ldots$ if $\forall n\geq 0$, the following condition holds:

  1. $Z_n$ is a function of $X_0,X_1,\ldots, X_n$
  2. $\mathbb{E}(|Z_n|) < \infty$
  3. $\mathbb{E}(Z_{n+1}\mid X_0,\ldots, X_n) = Z_n$

Here is what I don't get: It seems to me, you could just pick any random variable and symbolically assert the following:

$\forall n \geq 0, \mathbb{E}(Z_n)=\mathbb{E}(Z_0) $ using the tower of expectations property recursively, so how do I symbolically verify the need to worry about the stopping time and develop the martingale stopping theorem?

PS: Is the following guess as to why we need the stopping theorem correct? We need it because the original theorem might be defined for a countably infinite number of random variables and stopping it at a random time might break the conditions under which it holds?

share|improve this question
    
A crucial point is that condition (1) asks that $Z_n$ be a function of $X_0$, $X_1$, ..., $X_n$ only (using no $X_k$ for $k\ge n+1$). –  Did Apr 21 '11 at 23:30
    
Good point. I have edited the post. –  EVK Apr 22 '11 at 4:38
    
I guess a more crisper variant of the question could be (in case someone was not clear on what I was mumbling about): Why do we need to care about Stopping Times, when given any time $T$, I can easily find $\mathbb{E}(Z_T)$ recursively ? –  EVK Apr 22 '11 at 4:40
1  
Given any deterministic time $T$ you can find $E(Z_T)$ easily. But a stopping time is a random time. –  GWu Apr 22 '11 at 4:59
    
Well, I guess I don't understand clearly why the fact that the stopping time is random prevents you from finding $\mathbb{E}(Z_T)$ easily? I mean even though the time is random, doesn't the fact that this behavior is true for all times ensure that $\mathbb{E}(Z_T)$ is available to us ? For example, suppose I toss a coin and irrespective of whether I get a head or tail, I roll a die, doesn't it mean that the die roll is invariant of the coin toss? –  EVK Apr 22 '11 at 5:15

1 Answer 1

When $T$ is not a stopping time, $\mathrm E(Z_T)$ can be well defined but very different from $\mathrm E(Z_0)$. Consider for instance the symmetric random walk $(Z_n)_{n\geqslant0}$ on the integer line which starts from $Z_0=0$ and whose steps are $\pm1$. Let $R$ denote the time of the first return to $0$, and $T$ the last time before $R$ such that $Z$ is maximal on the time span $[0,R]$, that is, such that $Z_T\geqslant Z_n$ for every $0\leqslant n\leqslant R$.

Then $Z_R=0$ almost surely hence $\mathrm E(Z_R)=0$. On the other hand, $Z_T\geqslant0$ almost surely and $Z_T\geqslant1$ with positive probability hence $\mathrm E(Z_T)\gt0$. One sees that $\mathrm E(Z_R)=\mathrm E(Z_0)$ and $\mathrm E(Z_T)\ne\mathrm E(Z_0)$. Of course, $R$ is a stopping time while $T$ is not.

Edit In the case described above, one can even show that $Z_T$ is not integrable. To see this, note that $Z_T=0$ on $[Z_{-1}=-1]$, which happens with probability $\frac12$, and that, for each integer $z\geqslant1$, $Z_T\geqslant z$ if and only if $Z_1=1$ and, starting from $1$, one hits the level $z$ before the level $0$. This last event has probability $\frac1z$ hence, for every $z\geqslant1$, $\mathrm P(Z_T\geqslant z)=\frac1{2z}$. In particular $\mathrm E(Z_T^-)=0$ and $\mathrm E(Z_T^+)=+\infty$ hence $\mathrm E(Z_T)=+\infty$.

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.