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?

  • $\begingroup$ $(Y_n)_n$ is strong stationary and ergodic. $\endgroup$
    – JGWang
    Jun 28, 2020 at 2:44
  • $\begingroup$ 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 $\endgroup$
    – Kurt.W.X
    Jun 28, 2020 at 14:00
  • $\begingroup$ $\frac{1}{n}\sum_{k=1}^n(Y_{k+1}-Y_k)=\frac{Y_{n+1}}{n}-Y_1/n$ 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 $Y_1 \in L^1$ $\endgroup$
    – Kurt.W.X
    Jun 28, 2020 at 14:06

1 Answer 1

  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.
  • $\begingroup$ 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 $\lim_nu_n=u$). $\endgroup$
    – Kurt.W.X
    Jul 3, 2020 at 20:29
  • $\begingroup$ 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? $\endgroup$
    – Kurt.W.X
    Jul 3, 2020 at 20:32
  • $\begingroup$ And in question 1) b), we should deduce it taking the suggested $k=\max(k_{1},...,k_{d+1}),$ right? $\endgroup$
    – Kurt.W.X
    Jul 3, 2020 at 20:34
  • $\begingroup$ @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 $k$. $\endgroup$ Jul 4, 2020 at 16:05
  • $\begingroup$ $u_n=(n-1)x_{n-1}/n-x_n$ $\endgroup$
    – Kurt.W.X
    Jul 4, 2020 at 16:53

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