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The question is as follows:

Exercise 5.2.4. Give an example of a martingale $X_n$ with $X_n \to -\infty$ a.s. Hint: Let $X_n = \xi_1 + · · · + \xi_n$, where the $\xi_i$ are independent (but not identically distributed) with $E\xi_i = 0$.

My guess is that we should not make $\xi_n$ having a distribution such that it has zero expectation but are more and more likely to be negative. For example, we can have $\mathbb{P}(\xi_n = 1) = 1/n$ and $\mathbb{P}(\xi_n = \frac {-1} {n-1}) = \frac {n-1} n$. But I am not sure how to prove $\sum \xi_n \to -\infty$.

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Could you please state which book this exercise of from? Thanks – Aubrey Apr 6 '15 at 16:32
up vote 1 down vote accepted

Hint: you can let each $\xi_n$ be either $-1$ or positive (and like you said, each with zero expectation and more and more likely to be negative) and if you choose the distributions right then you can use the Borel–Cantelli lemma to make sure that the probability that infinitely many $\xi_n$'s are positive is zero.

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I see. How about let $\xi_{n}$ be $-1$ with probability $1-\frac{1}{2^{n}}$, and be $2^{n}-1$with probability $1/2^{n}$.Since $\sum_{n=1}^{\infty}1/2^{n}<\infty,$we have $\mathbb{P}\{\xi_{n}>0 \, \text{i.o.}\}=0$. In other words, $\mathbb{P}\{\xi_{n}<0 \, \text{i.o.}\}=1$. So we have $\sum_{m=1}^n \xi_n \to -\infty$. – ablmf Feb 20 '13 at 15:25
@ablmf Yep, that would work. – Trevor Wilson Feb 20 '13 at 18:10

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