Let $X_n$, $n \in \mathbb{N}$, be a sequence of i.i.d random variables with $\mathbb{E}|X_1| < \infty$. I've been thinking about proving the strong law of large numbers using the following decomposition: \begin{equation} Y_N = \sum_{n=1}^N \frac{X_n}{N} = \sum_{n=1}^N \frac{X_n}{N} \mathbb{I}(X_n \leq N) + \sum_{n=1}^N \frac{X_n}{N} \mathbb{I}(X_n > N). \end{equation} I've tried to justify the almost sure convergence of the latter part to zero using Fatou's lemma: \begin{align} \mathbb{P} ( \lim_{N \rightarrow \infty} \sum_{n=1}^N \frac{X_n}{N} \mathbb{I}(X_n > N) \neq 0) &= \mathbb{P} ( \bigcup_{k \in \mathbb{N}} \{ \lim_{N \rightarrow \infty} \sum_{n=1}^N \frac{X_n}{N} \mathbb{I}(X_n > N) > \frac{1}{k} \}) \\ &\leq \sum_{k \in \mathbb{N}} \mathbb{P}( \lim_{N \rightarrow \infty} \sum_{n=1}^N \frac{X_n}{N} \mathbb{I}(X_n > N) > \frac{1}{k}) \\ &\leq \sum_{k \in \mathbb{N}} k \mathbb{E}( \lim_{N \rightarrow \infty} \sum_{n=1}^N \frac{X_n}{N} \mathbb{I}(X_n > N)) \\ &\leq \sum_{k \in \mathbb{N}} k \liminf_{N \rightarrow \infty} \mathbb{E}( \sum_{n=1}^N \frac{X_n}{N} \mathbb{I}(X_n > N)) \quad \mbox{(Fatou)} \\ &= \sum_{k \in \mathbb{N}} k \liminf_{N \rightarrow \infty} \mathbb{E}( X_1 \mathbb{I}(X_1 > N)), \end{align} but $X_1 \mathbb{I}(X_1 > N)$ is dominated by $X_1$ and converges pointwise to zero, so we should be able to use dominated convergence to see that $\mathbb{E} (X_1 \mathbb{I}(X_1 > N)) \rightarrow 0$, which implies that the sum over $k$ is also zero.

Does this make sense to you? This is not the usual approach taken in books, which makes me a little suspicious of my reasoning (and my earlier attempt at a similar problem here was pointed out to be wrong).

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    $\begingroup$ You are implicitly assuming that $\lim_N \sum_{n=1}^N \dots$ exists (almost everywhere). Without this assumption, your first step does not really make sense. Furthermore, Fatou's Lemma usually needs $f_n \geq 0$. Finally, you probably mean to decompose as $|X_n| >N $, not $X_n > N$. Or are you assuming $X_n \geq 0$ for all $n$? $\endgroup$ – PhoemueX Jul 20 '15 at 13:06
  • $\begingroup$ My two perplexities on the passages you made: 1) If I recall correctly, Fatou's lemma works either for sequences of positive functions or for dominated sequences of functions. 2) 2 lines before usin Fatou, you use Markov's inequality, which works for positive random varibles. $\endgroup$ – user228113 Jul 20 '15 at 13:10
  • $\begingroup$ Decomposing as $|X_n| > N$ might be more useful, but since $N$ is the number of terms in the sum, the terms in the sums are nonnegative. Good point about the existence of the limit, I was wondering about that myself. $\endgroup$ – jhk Jul 20 '15 at 13:16

You can assume that $X_n \geq 0$. Because we can always split the original X into positive and negative parts (FYI you cannot center the variables though).

The probability that the partial sums $\lim_N \sum_{n=1}^N$ is not equal to zero does NOT imply that there exists a number $1/k$ such that the limit is greater than $1/k$. It does not even imply that the liminf is bigger than $1/k$. So you cannot apply the Fatou's Lemma here.

If you use a fixed $N$ instead of letting $N$ vary as $n$, then the last term does not disappear fast enough for convergence a.s. to occur.


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