1
$\begingroup$

Let $(X_i)_{i\in\mathbb{N}}$ be a sequence of $L^2$ random variables in a probability space $(\Omega, \mathcal{A}, P)$. Define $\overline{X_n}:=\frac{1}{n}\sum_{i=1}^n X_i$. Show that (2) follows from (1):

(1) $\mathrm{Cov}(X_i,X_j)\leq c_{|i-j|}$ for a sequence $c_n$ with limit $0$.

(2) $\overline{X_n} $ converge in probability to the expected value of it (weak law of large numbers).

Without loss of generality I take $E(X_j)=0 \ \forall j$.

Be $ \epsilon>0$. It exists an $N: \forall n \ge N: c_n<\epsilon$. A convergent sequence is bounded: $\exists M: c_n< M < \infty \forall n \in \mathbb{N}$.

\begin{align} \mathrm{Var}(\overline{X_n}) &= \frac{1}{n^2} *( \sum_{i=1}^n \sum_{j=1}^n Cov(X_j,X_k)) \\ &= \frac{1}{n^2} *( \sum_{j,k: |j-k|\le N} Cov(X_j,X_k) + \sum_{j,k:|j-k|>N} Cov(X_j,X_k)) \\ &\le \frac{1}{n^2} * ( \sum_{j,k: |j-k|\le N}M + \sum_{j,k:|j-k|>N} *\epsilon) \\ &\le \frac{1}{n^2} (M*n*(1+2N)+ \epsilon* n^2) \\ &= \frac{M(1+2N)}{n}+ \epsilon \\ &\rightarrow \epsilon \quad (n\rightarrow \infty) \end{align}

Because I count for the entries with $|j-k|\le N $ the number $$n+2*(n-1)+2*(n-2)+..+2*(n-N)\\=n+2nN-2*(1+2+..+N)= n+2nN-2(\frac{(N+1)N}{2}) \le n*(1+2N)$$

I get $Var(\overline{X_n}) \rightarrow 0 \quad ( n \rightarrow \infty)$.

Be $\epsilon_1, \delta_1 >0$, so $ \exists N: Var(\overline{X_n}) < \epsilon_1^2 \delta_1 $ for $n \ge N $.

$$P(|\overline{X_n}|>\epsilon_1) \le \frac{Var(\overline{X_n})}{\epsilon_1^2} < \frac{\delta_1}{\epsilon_1^2} *\epsilon_1^2= \delta_1$$

I think i do some mistakes because the covariance don't must be positive so also the sequence? I think only if all is positive the proof is right.

$\endgroup$
  • $\begingroup$ With no hypothesis on the expectations, it is impossible to conclude that $\bar X_n$ converges (and note that $\bar X_n\to E(\bar X_n)$ is absurd unless $E(\bar X_n)$ is constant). Hence (1) does not imply (2). What is your source? $\endgroup$ – Did Sep 28 '16 at 15:49
  • $\begingroup$ ((No answer, this was only to be expected, I guess.)) $\endgroup$ – Did Sep 28 '16 at 21:09
  • $\begingroup$ thanks for your answer. Sry my source is not in english.. $\endgroup$ – Lauren Veganer Sep 29 '16 at 6:15
  • $\begingroup$ And why does this prevent you from giving it? $\endgroup$ – Did Sep 29 '16 at 6:16
  • $\begingroup$ Oh sry: mat.univie.ac.at/~gerald/teaching/wts_ue.pdf exercise 72 $\endgroup$ – Lauren Veganer Sep 29 '16 at 6:24
1
$\begingroup$

Your approach works. The covariances may not be positive but it is not a problem when we write $$ \operatorname{Var}\left(\overline{X_n}\right)=\frac 1{n^2}\sum_{i,j=1}^n \operatorname{Cov}\left(X_i,X_j\right).$$ The sum is non-negative but the terms are not necessarily all non-negative .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.