Assume $X_1,X_2,\ldots,X_N$ are identically distributed, non-negative, not independent. If $E\left[\sum\limits_{k=1}^N \frac{X_k}{N}\right]=\mu$, is there a way that I can conclude $$\lim\limits_{N \to \infty}P\left(\sum\limits_{k=1}^N \frac{X_k}{N}>\frac{\mu}{2}\right)=1$$

If $X_i$s are independent, we can use the weak law of large numbers. But, how about the non-independent version?

I know one counter example is to set $X_1=X_2=\cdots=X_N$. Then, $$\lim\limits_{N \to \infty}P\left(\sum\limits_{k=1}^N \frac{X_k}{N}>\frac{\mu}{2}\right)=\lim\limits_{N \to \infty}P\left( X_k>\frac{\mu}{2}\right)$$ which is not necessaruly true.

Is there any condition to add which makes it possible to prove what I need? i.e, $\lim\limits_{N \to \infty}P\left(\sum\limits_{k=1}^N \frac{X_k}{N}>\frac{\mu}{2}\right)=1$?

  • $\begingroup$ Lots of ways, for example: (i) They could be pairwise uncorrelated with finite variance, and with $\mu > 0$ (just use Chebyshev inequality). (ii) They could be values associated with a finite-state ergodic Markov chain that starts with an initial distribution given by the steady state distribution. (iii) They could be pairwise duplicated, i.e., $\{Y_1, Y_1, Y_2, Y_2, Y_3, Y_3, ...\}$ with $\{Y_i\}_{i=1}^{\infty}$ independent and identically distributed. $\endgroup$ – Michael Apr 25 '17 at 4:40
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – BGM Apr 25 '17 at 5:14

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