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, $Y_n=f(X_n,...,X_{n+d}),W_n=\frac{1}{n}\sum_{k=1}^nY_k.$

*

*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 $R_n=\frac{1}{n}\sum_{l=1}^n U_l.$


*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?
 A: *

*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, $Y_1\in L^1$.


*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.

