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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

BCLC
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user63416
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  • Odd $Z_i$'s are independent and even $Z_i$'s are independent so two applications of SSLN yield $\lim=\frac 1 2(\frac 1 4+\frac 1 4)=\frac 1 4$. – A.S. Dec 03 '15 at 22:34
  • @A.S. From where did you get the 1/2? (re my answer) – BCLC Dec 05 '15 at 19:35

1 Answers1

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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.

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BCLC
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  • Dear anonymous downvoter: Is it wrong to say that $Z_{2n}$'s are independent and $Z_{2n+1}$'s are independent? – BCLC Dec 03 '15 at 13:27
  • 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 $X_i$. – user164118 Dec 03 '15 at 15:53
  • @user164118 How to compute variance when using Chebyshev's inequality? – BCLC Dec 03 '15 at 16:08
  • 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. – user164118 Dec 03 '15 at 17:19
  • @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 – BCLC Dec 03 '15 at 17:49
  • @ BCLC You should have used pen and pencil. For ease of presentation, let's take $n=1000$. Then $\frac{1}{1000}(Z_1+Z_2+\cdots+Z_{1000})=\frac{1}{2}( \frac{1}{500}(Z_2+Z_4+\cdots+Z_{1000})+\frac{1}{2}( \frac{1}{500}(Z_1+Z_3+\cdots+Z_{999}). – user164118 Dec 03 '15 at 18:54
  • @user164118 I'm going to type it out. FYI i wasn't notified $\frac{1}{1000}(Z_1+Z_2+\cdots+Z_{1000})=\frac{1}{2}( \frac{1}{500}(Z_2+Z_4+\cdots+Z_{1000})+\frac{1}{2}( \frac{1}{500}(Z_1+Z_3+\cdots+Z_{999})$ – BCLC Dec 05 '15 at 19:33
  • @user164118 It doesn't hold true in the finite case obviously. Let both sides approach infinity. – BCLC Dec 05 '15 at 19:34