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Let $x_i$ be a uniformly random real in $(-1,1)$. And let $f(x)$ be a strictly increasing positive and unbounded function.
Let $S_j(f)=x_0/f(0)+x_1/f(1)+x_2/f(2)+\cdots+x_j/f(j)$

Does $S_j(f)$ converge for every $f$?

Is there an $f$ such that with probability 1, the partial sums $S_j$ change sign infinitely many times and $S_j$ converges?

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You may find section 5 of this paper useful: calpoly.edu/~kmorriso/Research/RandomWalks.pdf –  Byron Schmuland Nov 24 '11 at 19:02

1 Answer 1

up vote 4 down vote accepted

The answer to both questions is no.

By Kolmogorov's three series theorem, $S_j(f)$ will converge almost surely if and only if $\sum_j {1/f(j)^2}<\infty$. This fails, for instance, if $f(j)=1+\sqrt{j}$.

If $S_j(f)$ converges, it can only change sign infinitely many times provided the limit is zero. If that were true, the equation $x_0=-f(0)\ [x_1/f(1)+x_2/f(2)+\cdots]$ would show that the random variable $x_0$ is independent of itself. But this is impossible for a non-constant random variable.

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Can any sum of random variables converge to 0 with probability 1 ? –  user1708 Nov 24 '11 at 20:50
    
For a sum of independent random variables, it can only happen if the summands are constants (variance equal to zero). –  Byron Schmuland Nov 24 '11 at 21:11

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