2
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.