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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?

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    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 $k\ge n+1$).2011-04-21
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    Good point. I have edited the post.2011-04-22
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    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 ?2011-04-22
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    Given any deterministic time $T$ you can find $E(Z_T)$ easily. But a stopping time is a random time.2011-04-22
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    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?2011-04-22
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    I think I get it now. So basically if $T$ is a random time such that $T=n$ does not depend just on $X_1,\ldots, X_n$, then $\mathbb{E}(Z_{T}\mid X_0,\ldots, X_n-1)$ will have hidden dependencies due to the random time $T$ and my condition 3 does not hold. Does this make sense?2011-04-24

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