Let ${X_n}$ be a sequence of independent and identically distributed, square integrable random variables. Write $ u = E(X_n)$. Study the almost sure convergence, as $n \rightarrow \infty$, $$S_n = (X_1X_2 + X_2X_3 + ... + X_{n-1}X_{n})/n$$

Since $X_iX_{i+1}$ are not independent, it seems we cannot directly use law of large number for that, so anyone can give me some idea?

  • $\begingroup$ Before $u=E(X_n)$ you had a symbol that I could recognize. Is the editing I did correct? Please let me know! $\endgroup$ – Jimmy R. Dec 15 '14 at 12:53
  • $\begingroup$ @Stef yes it should be correct $\endgroup$ – annimal Dec 15 '14 at 12:54
  • $\begingroup$ Slightly related: math.stackexchange.com/q/430930/66096 $\endgroup$ – Gabriel Romon Apr 26 at 9:06

Hint: define $Y_j:=X_{2j-1}X_{2j}$ and $Z_j:=X_{2j}X_{2j+1}$. Notice that the sequence $(Y_j)_{j\geqslant 1}$ is independent, as well as $(Z_j)_{j\geqslant 1}$. After having expressed $S_{2n}$ and $S_{2n+1}$ in terms of $Y_j$ and $Z_j$, apply the strong law of large numbers.


You need to consider separately two terms: $$S_n^1 = \frac{X_1 X_2 + X_3 X_4 + \ldots + X_{2n - 1} X_{2 n}}{n}$$ and $$S_n^2 = \frac{X_2 X_3 + X_4 X_5 + \ldots + X_{2n} X_{2 n + 1}}{n}.$$ Then $$S_{2 n + 1} = S_n^1 + S_n^2$$ and for $S_n^1$ and $S_n^2$ you can apply Law of large numbers. Then you can proceed sum of two random variables by definition of almost sure convergence.


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