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All the problems I've seen treat martingales as a sum of independent rvs, $X_t=\sum_{k=1}^{n}B_k$, where $\mathbf{E}[B_{k}|\mathcal{F}_{k-1}]=\mathbf{E}B_{k}$. What is a good approach to study (e.g. find bounds on the first passage time etc) for martingales that are sums (or other functions) of rvs that are not independent? E.g. in the previous example $\mathbf{E}[B_{k}|\mathcal{F}_{k-1}] \neq B_{k}$.

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up vote 3 down vote accepted

There seems to be two questions in your post. A martingale which is not a sum of independent increments? Try $X_n=a^{S_n}$, where $(S_n)$ is a $\pm1$ random walk on the integer line with probability $p\ne\frac12$ for the step $+1$ and $1-p$ for the step $-1$, for a suitable value of $a$ which I will let you discover. A good place to learn the basics of martingale theory? Try the so-called blue book written by David Williams and titled Probability with Martingales.

Edit Another classical example: let $X_n=S_n^2-n$ where $S_n=Y_1+\cdots+Y_n$, $(Y_n)$ is a $\pm1$ valued i.i.d. sequence with probability $\frac12$ for $+1$ and $\frac12$ for $-1$, and $\mathcal{F}_n=\sigma(Y_k,k\le n)$. Since $X_{n+1}=X_n+2Y_{n+1}S_n$, you see that the increment $X_{n+1}-X_n$ is not independent on $\mathcal{F}_n$, nevertheless $E(X_{n+1}|\mathcal{F}_n)=X_n$ almost surely.

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Does it cover the case with dependent RVS? –  sigma.z.1980 Mar 31 '11 at 6:20
    
Did you read my post? –  Did Mar 31 '11 at 7:35
    
I mean, are the cases similar to the one you provided above covered in this book in addition to the basics of theory –  sigma.z.1980 Mar 31 '11 at 10:04
    
Yes. $ $ $ $ $ $ –  Did Mar 31 '11 at 10:12
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There is a bit of confusion in your question. If you think that your book is treating a martingale as a sum of independent random variables, then you are almost certainly misunderstanding your book. Non-independent increments is not just another case to be considered, it is the general case, and I will bet this is what your book does; its theorems should apply just fine. You just may not have seen examples without independent increments.

But I bet that you have. The simplest and most natural example is perhaps a stopped random walk. Take $X_n = \sum_{i=1}^n \xi_i$, where $\xi_i$ are iid fair coin flips (so $\xi_i = \pm 1$), let $T = \min\{n : X_n \ge 1\}$, and let $M_n = X_{n \wedge T}$. $M_n$ is certainly a martingale, but I'll leave it to you to check that its increments are not independent.

It does tend to be the case that most martingales you encounter in practice are somehow cooked up from a process with independent increments. Indeed, there is a "martingale representation theorem" that asserts that this is actually unavoidable, in some sense.

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Good example. What happens if the probability to toss Head changes and depends on past coin flips? In some sense we have another conditional distribution $E[\xi_{k}|\xi_{k-1}]$ –  sigma.z.1980 Mar 31 '11 at 21:47
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