# Strong law of large numbers without independence

Let $$(X_n)_n$$ be a sequence of independent random variables and identically distributed, $$d \in \mathbb{N},$$ $$f: \mathbb{R^{d+1}} \rightarrow \mathbb{R}$$ a measurable function, $$Y_n=f(X_n,...,X_{n+d}),W_n=\frac{1}{n}\sum_{k=1}^nY_k.$$

1. a) Prove that $$Y_1 \in L^1$$ if and only if $$(W_n)_n$$ converges a.s.

In this case, Show that $$(W_n)_n$$ converges also in $$L^1.$$

b) If $$k_1,...,k_{d+1} \in \mathbb{N},U_n=f(X_{n+k_1},...,X_{n+k_{d+1}}),$$ deduce that a) remains true with $$R_n=\frac{1}{n}\sum_{l=1}^n U_l.$$

2. We suppose that there exists a sequence $$(x_n)_n$$ such that $$W_n-x_n$$ converges a.s. Is it true that $$Y_1 \in L^1?$$

Attempt : In this problem, $$(Y_n)_n$$ are not independent, so we have to work with subsequences, and grouping terms.

For the first part, $$W_n$$ converges a.s this implies that $$\frac{Y_n}{n}$$ converges a.s to $$0,$$ and that $$\frac{Y_{(d+1)n}}{(d+1)n}$$ converges a.s to $$0$$, so $$\frac{Y_{n(d+1)}}{n}$$ a.s to $$0$$, and since $$(Y_{(d+1)n})_n$$ is a sequence of i.i.d random variables, which means that $$Y_1 \in L^1.$$

If $$Y_1 \in L^1,$$ then we should write $$W_n=\frac{1}{n}\sum_{k=0}^d\sum_{l=0}^{ \left \lfloor{\frac{n-k}{d+1}}\right \rfloor }Y_{l(d+1)+k}$$ and we conclude using the strong law of large numbers.

b) is simple, taking the projection, and considering $$k=\max(k_{1},..,k_{d+1})$$ and we apply a)

Having problems with 2), if only, we can remove $$x_n.$$ Any ideas?

• $(Y_n)_n$ is strong stationary and ergodic. Jun 28, 2020 at 2:44
• I was thinking in another way, I don't know if it works, supposing that $W_n-x_n$ converges a.s to $X,$ then $X$ is constant a.s, $X=c$ a.s, since $\limsup_n(W_n-x_n)$ is a tail function and $(X_n)_n$ is independent, also $(X_1,...,X_{n+p+d})$ and $(X_2,...,X_{n+p+d+1})$ have the same distribution, so $P(\max_{1 \leq k \leq p}|\frac{1}{k+n}\sum_{l=1}^{k+n}Y_l-x_{k+n}-c|>\epsilon)=P(\max_{1 \leq k \leq p}|\frac{1}{k+n}\sum_{l=1}^{k+n}Y_{l+1}-x_{k+n}-c|>\epsilon)$ taking the increasing limit $p \to +\infty$ then $n \to +\infty$ so $\lim_n \frac{1}{n}\sum_{k=1}^nY_{k+1}-x_n=c$ a.s Jun 28, 2020 at 14:00
• $\frac{1}{n}\sum_{k=1}^n(Y_{k+1}-Y_k)=\frac{Y_{n+1}}{n}-Y_1/n$ converges a.s to $0,$ which means that $(Y_{n(d+1)})_n,$ which is a sequence of independent and identically distributed random variables, converges a.s to $0$, so $Y_1 \in L^1$ Jun 28, 2020 at 14:06

1. a) If $$Y_1$$ is integrable then $$(W_n)$$ converges a.s and in $$L^1$$ by the ergodic theorem. The limit is a.s. equal to $$\Bbb E[Y_1]$$ by the Kolmogorov zero-one law.
If $$(W_n)$$ converges a.s, then $$Y_n/n\to 0$$ a.s. because $$W_n = {n-1\over n}W_{n-1}+{1\over n}Y_n$$. Therefore $$\lim_kY_{2dk}/(2dk)=0$$ a.s, so by the second Borel-Cantelli lemma, $$\sum_{k=1}^\infty \Bbb P[|Y_{2dk}|>2dk]<\infty.$$
Consequently, $$(2d)^{-1}\Bbb E[|Y_1|] =\Bbb E\left[{|Y_1|\over 2d}\right]\le 1+\sum_{k=1}^\infty \Bbb P[|Y_1|>2d\cdot k]=1+\sum_{k=1}^\infty \Bbb P[|Y_{2dj}|>2d k]<\infty.$$ That is, $$Y_1\in L^1$$.
1. If $$W_n-x_n$$ converges a.s., then by the earlier reasoning you must have $${Y_n\over n}-x_n+{n-1\over n}x_{n-1}\to 0,\qquad\hbox{a.s.}$$ From this it follows (because all the $$Y_n$$ have the same distribution) that for each $$\epsilon>0$$, $$\lim_n\Bbb P[Y_1/n-x_n+{n-1\over n}x_{n-1}>\epsilon]=0,$$ forcing $$\limsup_n [(n-1)x_{n-1}/n-x_n]\le 0$$. Similarly, $$\liminf_n [(n-1)x_{n-1}/n-x_n]\ge 0$$. Thus $$\lim_n[(n-1)x_{n-1}/n-x_n]=0$$, so $$Y_n/n\to 0$$ a.s., and then $$Y_1\in L^1$$ by Borel-Cantelli as before.
• Ergodic theorem is not in our program, but it's better to know all the ways. In part 2), you mentioned that $Y_n$ and $Y_1$ have the same distribution, did you use that the convergence in distribution of $Y_1/n-x_n+\frac{n-1}{n}x_{n-1}$ to $0$ implies convergence in probability to $0$? (We can also use that, for a sequence $(u_n)_n$ of real numbers, if $(\delta_{u_n})_n$ converges weakly to $\delta_u,$ then $\lim_nu_n=u$). Jul 3, 2020 at 20:29
• Also, in the above comment, I suggested a way to remove $x_n,$ without proving that $(n-1)x_{n-1}/n-x_n$ converges to $0$, so is this suggestion true? Jul 3, 2020 at 20:32
• And in question 1) b), we should deduce it taking the suggested $k=\max(k_{1},...,k_{d+1}),$ right? Jul 3, 2020 at 20:34
• @Kurt.W.X: 1. Rather, the convergence (a.s) of $Y_1/n-x_n+{n-1}\over n}x_{n-1}$ to 0 implies its convergence in distribution; 2. What is your choice of $u_n$?; 3. Yes, to the choice of $k$. Jul 4, 2020 at 16:05
• $u_n=(n-1)x_{n-1}/n-x_n$ Jul 4, 2020 at 16:53