# Martingale convergence almost surely

Let $(X_i)$ be a sequence of r.v. and define $Y_i^n:=X_i\mathbf1\{|X_i|\le n\}$ and $Z_i:=T^i(Y_i^i)$, where $T^i$ is defined as

$$T^i(X) :=\sum_{l=-\infty}^\infty l2^{-i}\mathbf1\{X\in(l2^{-i},(l+1)2^{-i}]\}$$

So $T^i$ is like an approximation from below and therefore $T^i(X) \le X$ for every r.v. $X$.

assume that

$$W_n:=\sum_{k=1}^n \frac{1}{k}(Z_k-E(Z_k|Z_1,\dots,Z_{k-1}))$$

and we have proved the following:

1. $(W_n)$ is a square integrable martingale.
2. $|E(Z_k|Z_1,\dots,Z_{k-1}))|\le 2^{-k+1}$
3. $W_n$ converges almost surely as $n \to \infty$, denote this limit by $W$.

Now why does the series

$$\sum_{k=1}^\infty \frac{1}{k}(Y_k^k)$$

converges a.s.?

The question reduces to this: I can prove that $Y_k^k-2^{-(k+1)}\le Z_k-E(Z_k|Z_1,\dots,Z_{k-1}) \le Y_k^k + 2^{-(k+1)}$. Hence I know

$$W-\sum_{k=1}^\infty 2^{-(k+1)} \le \sum_{k=1}^\infty \frac{1}{k}(Y_k^k) \le W+\sum_{k=1}^\infty 2^{-(k+1)}$$

Here's the point where I'm unsure. Both sums on the left and right are converging. However this does not imply the convergence of the series in the middle, since there are bounded sequences which do not converge (for example if they oscillate). Perhaps this estimate is not helpful for my proof.

Can I argue like this: Since $W_n$ is defined through this approximation r.v. and $W_n$ converges, it can not oscillate? However there's still a problem, since I don't how to handle the conditional expectation part in the definition of $W_n$. Hopefully someone could help me! Thanks a lot!

cheers

math

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@ math : Hi, in the definition of $W_n$ is the index $i$ of $Z_i$s fixed for all $n$ ? If this is the case $W_n$ is constant for any $n>i$, right ? – TheBridge Mar 9 '12 at 12:46
@ TheBridge: That was a typo! I'm very sorry for that! I edited my post – math Mar 9 '12 at 13:29
@ math : this is what I expected but I wanted to be sure. Thx and regards. – TheBridge Mar 9 '12 at 15:00
@ Byron Schmuland: I'm so sorry! There should be a $k$ instead of $i$. So I have a bound for every term in the sum. – math Mar 9 '12 at 16:36