# Can one construct a monotone law of large numbers?

let $X_1, X_2, \dots$ be a sequence of IID random variables defined on a probability space $( \Omega, F, P)$ with mean $E[X_1] = \mu$, define

$$\bar{X}_n = \frac{1}{n}(X_1+ \dots+ X_n)$$

then $\bar{X}_n \xrightarrow{a.s.} \mu$.

This is the standard strong law of large number, the almost sure convergence implies that given an $\omega \in \Omega \setminus A$ (where $A$ is the set of probability zero for which the law of large numbers does not work) there always exists a monotone subsequence of $\bar{X}_n( \omega)$, that we call $\hat{X}_n(w)$, s.t. $\hat{X}_n(w) \rightarrow \mu$, correct?

I was wondering if we can always mimic this sequence by eliminating the values that would break the monotonicity of $\bar{X}_n( \omega)$ and if the direction of the monotonicity implies that we are overestimating or underestimating the mean , e.g.

assume we are tossing fair coins and we obtain a sample $\{ 1,1,0,1,0,0 \}$, where $1$ represents a tails outcome and $0$ represents a heads, here we would have that $\{ \bar{X}_1 = 1, \bar{X}_2 = 1, \bar{X}_3 = 2/3, \bar{X}_4 = 3/4, \bar{X}_5 = 3/5, \bar{X}_6= 1/2 \}$ so we can construct the decreasing subsequence $\{ \hat{X}_1 = 1, \hat{X}_3 = 2/3, \hat{X}_5 = 3/5, \hat{X}_6= 1/2 \}$ can we be sure then that the sequence $\hat{X}_n$ will eventually overestimate the mean with probability one?

(we choose to construct $\hat{X}_n$ with the same monotonicity of the tail of the $\bar{X}_n$).

My attempted solution to my own doubt:

I think the answer is yes because the set of sequences of counterexamples shrinks in size as $n$ increases. To prove this I would proceed by cases:

• If we fix an $\omega$ s.t. the sequence $X_1(\omega) , X_2(\omega), \dots$ oscillates around $\mu$ forever then there exists $n_1 \in N$ s.t. the sequence has oscillated around $\mu$ twice. Thus for every $n > n_1$ the monotone sequence $\hat{X}_n$ always overestimates or underestimates the mean (notice that we could construct $\hat{X}_n$ as decreasing or increasing in this case but for the construction to not be ambiguous in the upcoming second case we need to construct $\hat{X}_n$ with the same monotonicity as the tail of the sample).
• If we fix an $\omega$ s.t. the sequence $X_1(\omega) , X_2(\omega), \dots$ oscillates around $\mu$ for a fixed $n_2 \in N$ number of times then there exists a $n_3 \ge n_2$ s.t. for every $n > n_3$ the sequence $\bar{X}_n$ is monotone so the construction of $\hat{X}_n$ will be correctly overestimating or underestimating after $n_3.$

I think with that I have exhausted all the cases, so if this construction of $\hat{X}_n$ is always possible and correct(?), after a certain sample size, why is it not more used? I would assume knowing if one is likely to be underestimating or overestimating the mean is valuable information.

EDIT: added that $\omega \in \Omega \setminus A$ and not only $\omega \in \Omega$.

• I just noticed that my construction of the sequence $\hat{X}_n$ is ambiguous, could it be fixed? – Monolite Mar 14 '18 at 19:55
• "the almost sure convergence implies that given an $\omega \in \Omega$ there exists a monotone subsequence of $\bar{X}_n$, that we call $\hat{X}_n(w)$, s.t. $\hat{X}_n(w) \rightarrow \mu$, correct?" Well, no. Never seen this version, and it seems obviously wrong. What is your source for this? – Did Mar 14 '18 at 20:10
• Yeah, I think this is almost true; every sequence that converges has a monotonic subsequence that converges mathonline.wikidot.com/the-monotone-subsequence-theorem. Now, if I am not mistaken, a.s. convergence tells you that for almost all $w$, you can do this. Uh, I cannot really see your argument for the over and under-estimating the mean. After all, law of large numbers is a statement in the limit, so any finite samples shouldn't really affect the convergence. – E-A Mar 14 '18 at 20:16
• @Did I should have said that fixed $\omega$ then one can extract a monotone subsequence from $\bar{X}_n(\omega)$. – Monolite Mar 14 '18 at 20:25
• Every real valued sequence shares this property, yes, but there is no guarantee that any subsequence, monotone or not, converges to $\mu$. – Did Mar 14 '18 at 20:43