Hey guys I need some help. Consider $X_3,...,X_N$ real valued, independent random variables and let $X_1,X_2$ be real valued dependent random variables. But only $X_1,X_2$ are dependent. The series $X_1,X_3,...,X_N$ and $X_2,X_3,...,X_N$ are still independent. I now want to show a weak law of large numbers. Because of convergence rates I have to use markovs inequality for higher moments, i.e. of order 2m (All even numbers). By applying this we get.

$$ \mathbb{P} (\frac{1}{N} \sum_{j=1}^N (X_j - \mu_j )> N^{-\epsilon}) \leq N^{2m\epsilon} \mathbb{E} \left[ (\sum_{j=1}^N (X_j - \mu_j))^{2m} \right] $$

The problem now is as follows. If we take $2m=6$ as an example. Then we have to deal with terms like

$$ \mathbb{E} \left[(X_1-\mu_1)*(X_2-\mu_2)*(X_3-\mu_3)*(X_4-\mu_4)*(X_5-\mu_5)*(X_6 - \mu_6) \right]$$

$$ \mathbb{E} \left[(X_1-\mu_1)^2*(X_2-\mu_2)^2*(X_3-\mu_3)^2 \right] $$

If the random variables would all be independent, I of course could use $\mathbb{E}(X*Y) = \mathbb{E}(X) *\mathbb{E}(Y)$ and the terms with exponent 1 would vanish. If I just had order 2 I could use the covariance to manage the dependency. But how can I manage these terms in this case?

  • $\begingroup$ Atm this is above me. But i was wondering. Could you not use a joint distribution of $X_1$ and $X_2$ and then use the fact that this would be indepedent to the rest? If this was unhelpful. My apologises. $\endgroup$ – Vaas Aug 29 '17 at 8:46
  • $\begingroup$ I see your point. But I think it is not true that if $X$ and $Z$ are independent and $Y$ and $Z$ are independent then also $X*Y$ is independent of $Z$. I found a counterexample here math.stackexchange.com/questions/1956664/… $\endgroup$ – Jack_Stiller10 Aug 29 '17 at 9:26

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