# Dependent Bernoulli sequence for which the strong law fails to hold

Background:

The strong law of large numbers (SLLN) is a powerful result in probability, and there has been extensive literature on when the SLLN holds. However, constructing nontrivial examples for which the SLLN fails to hold seems to be (very) hard.

K.L. Chung's famous paper "The strong law of large numbers" talks about the necessary and sufficient conditions for the SLLN to hold. However, it is for independent random variabes. On the other hand, R. Lyons' paper "Strong laws of large numbers for weakly correlated random variables" is probably the best on when the SLLN holds under dependence. One of the theorems of Lyon's is quoted below:

Let $\left\{ X_{n}\right\} _{n=1}^{\infty}$ be a sequence of real-valued, zero mean random variables such that $\mathbb{E}\left[ \left\vert X_{n}\right\vert ^{2}\right] \leq1$. If $\left\vert X_{n}\right\vert \leq1$ a.s. and $$\sum\nolimits_{N=1}^{\infty}N^{-1}\mathbb{E}\left[ \left\vert N^{-1} \sum\nolimits_{n=1}^{N}X_{n}\right\vert ^{2}\right] <\infty\text{,}$$ then $\lim_{m\rightarrow\infty}N^{-1}\sum_{n=1}^{N}X_{n}=0$ a.s.

Question:

However, could anyone please give an example for which the SLLN fails to hold in the following setting? Let $\left\{ X_{n}\right\} _{n\geq1}$ be a sequence of Bernoulli random variables such that

1. $\mathbb{P}\left(X_{n}=1\right) =p_{n}$ for which $p_{i}\neq p_{j}$ whenever $i\neq j$, and $\mathbb{P}\left(X_{n}=0\right) =1-p_{n}$
2. $\lim\inf_{n\rightarrow\infty}p_{n}>0$ and $\lim\sup_{n\rightarrow\infty }p_{n}<1$
3. $X_{n}$'s are not wekly dependent in Lyon's sense (see the displayed equation above)
4. no significant proportion (compared to $m^2$ when $m$ is large) of the sequence $X_n$ for $1 \leq n \leq m$ should contain elements that are a scalar multiple of one another (this is to avoid types of examples given by AnthonyQuas)

Follow-up

Anthony and Fedor each provided an example of dependent Bernoulli's for which the SLLN fails. Their examples have special structures, i.e., $X_n = f(Y_1, \cdots, Y_{n_0}, Y_n)$ for a fixed $n_0$ and independent Bernoulli's $Y_n$, where $f$ is a linear mapping whose functional norm is bounded from above and away from $0$. It seems that the key is to introduce infinite-range dependence (which is in sharp constrast to finite-range dependence or exchangeability; see a related post Is there a McDiarmid-type inequality for sequences with a finite range of dependence? )

Is this a general technique to construct examples for the failure of SLLN for dependent Bernoulli's?

• Here's an example with $p_n=0.5$ for each $n$. It is possible, but annoying to modify it to make all the $p_n$'s distinct. Let $X_1$ be a random variable taking the values $\pm 1$ with equal probability. Then set $X_n=X_1$ for each $n$. – Anthony Quas Mar 5 '16 at 1:05
• @AnthonyQuas: Thanks! Your example has a special feature, i.e., the sequence of random variables are totally linearly dependent. (It can be classified as a "trivial" example). I have edited the question. – Chee Mar 5 '16 at 1:21
• What is $m^2$ in your 4th condition? I'm wondering if the example can be weakened to have "fairly large" blocks of totally dependent random variables that still satisfies your condition. – Anthony Quas Mar 5 '16 at 7:04
• Example may be modified to satisfy condition 4. For example, choose independent $Y_i$'s and set $X_n=[(Y_1+Y_2+Y_n)/3]$. – Fedor Petrov Mar 5 '16 at 13:02
• @Anthony: I have clarified $m^2$ in condition 4; Anthony and Fedor: I have summarized your examples into an abstract formulation. Maybe the techniques you guys have presented is a general way to construct dependent Bernoulli's for which SLLN fails to hold. Thank you! – Chee Mar 5 '16 at 14:27

Let $(S_n, n\geq 0)$ be the one-dimensional simple random walk started at the origin. Set $X_n=\mathbf{1}\{S_n\geq 0\}$. Then $n^{-1}(X_1+\cdots+ X_n)$ is the proportion of time that the SRW is non-negative; it doesn't converge a.s. anywhere (although converges in distribution to a non-trivial r.v., cf. the arcsine law).
• thank you for providing this elegant example! Can you please provide another example of dependent Bernoulli's which fails the SLLN on the whole sample space $\Omega$ but for which a SLLN holds on a nonempty subset of $\Omega$ of positive probability measure? Thanks! – Chee Mar 6 '16 at 22:21