2
$\begingroup$

Let $X_1,...,X_n,...$ be independent variable satisfying $P(X_i=0)=P(X_i=1)=\frac{1}{2}$ for all i then denote $Z_i=X_iX_{i+1}$ for all i .I want to show that $\lim_{n\to \infty}\frac{Z_1+Z_2+...+Z_n}{n}=\frac{1}{4}$ a.s by using law of large number. I'm new to this area and struck immediately since $\{Z_i\}$is not independent. Perhaps Borel-Cantelli lemma can be used here?

really thanks for your help

$\endgroup$
  • $\begingroup$ Odd $Z_i$'s are independent and even $Z_i$'s are independent so two applications of SSLN yield $\lim=\frac 1 2(\frac 1 4+\frac 1 4)=\frac 1 4$. $\endgroup$ – A.S. Dec 3 '15 at 22:34
  • $\begingroup$ @A.S. From where did you get the 1/2? (re my answer) $\endgroup$ – BCLC Dec 5 '15 at 19:35
-1
$\begingroup$

Hint: $Z_n$'s are not independent but

what about $Z_{2n}$'s or $Z_{2n+1}'s$?

I am not quite sure the answer is 1/4 considering $Z_n$'s are not independent. I think it is 1/2

Got the idea from here, w/c involves Borel-Cantelli Lemmas, but I don't think you need those.

$\endgroup$
  • $\begingroup$ Dear anonymous downvoter: Is it wrong to say that $Z_{2n}$'s are independent and $Z_{2n+1}$'s are independent? $\endgroup$ – BCLC Dec 3 '15 at 13:27
  • $\begingroup$ It seems to me that the weak law of large numbers applies to the sequence of (dependent) random variables $Z_k$: $\frac{1}{n}\sum_{k=1}^{n} Z_k$ converges in probability to $\frac14$, as can be shown by applying Chebyshew"s inequality and using the independence of the $X_i$. $\endgroup$ – user164118 Dec 3 '15 at 15:53
  • $\begingroup$ @user164118 How to compute variance when using Chebyshev's inequality? $\endgroup$ – BCLC Dec 3 '15 at 16:08
  • $\begingroup$ It suffices to bound the variance. Using the independence of the $X_i$ and the assumption that $E(X_i)$ and $\sigma^2(X_i)$ do not depend on $i$, it is not difficult to verify that for some constant $c>0$ the variance of $\sum_{k=1}^{n}Z_k$ is bounded by $cn$ for all $n$ and so $\hbox{var}(\frac{1}{n}\sum_{k=1}^{n}Z_k)=\frac{1}{n^2}\hbox{var}(\sum_{k=1}^{n}Z_k)$ tends to 0 as $n$ goes to infinity. $\endgroup$ – user164118 Dec 3 '15 at 17:19
  • $\begingroup$ @user164118 Any idea how my idea is wrong? If the odd z's are indp, limit of average is 1/4 I think.same for even z's. Add em up to get 1/2 $\endgroup$ – BCLC Dec 3 '15 at 17:49

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.