1
$\begingroup$

Consider a sequence of random variables $X_1,X_2,...,X_n$. No assumptions abou independence is made. Only joint probability density function is known, i.e. $f(x_1,...,x_n)$. Then Markov's theorem states that for any constant $\epsilon>0$: $$\underset{n\rightarrow \infty }{lim} \mathbb{P}\{ \left | \frac{1}{n}\sum_{k=1}^{n}X_k-\frac{1}{n}\sum_{k=1}^{n}\textbf{E}X_k \right |<\epsilon \}=1$$ if $$\frac{1}{n^2}Var\left ( \sum_{k=1}^n X_k \right )\rightarrow 0.$$ This is a very general formulation of law of the large numbers. This formulation with arbitrary dependence structure can by found in other books as well. But I am interested in the classes of distributions or a model structure when the condition $\frac{1}{n^2}Var\left ( \sum_{k=1}^n X_k \right )\rightarrow 0$ does not hold. For example does this condition fail for heavy tailed distributions? Does the distribution have to decay at certain rate at infinity? Or maybe it cannot have strong coorelations between too many variables? etc.

I could not find any source on this matter. Any information on this issue would be highly appreciated.

UPDATE

Variance of the sum i.e. $Var\left ( \sum_{k=1}^n X_k \right )$ is in fact simply the sum of all entries of the covariance matrix of random variables $X_1,...,X_n$. Denote this covariance matrix by $C$. In addition, denote by $e$ a vector with all entries being equal to 1. Then it is true that $$Var\left ( \sum_{k=1}^n X_k \right )=e^T C e$$ We have also the following bound for the quadratic forms: $$\lambda_{min} x^T x\leq x^T A x\leq \lambda_{max}x^Tx$$ Hence, with our notation we have that $$n\lambda_{min}\leq Var\left ( \sum_{k=1}^n X_k \right )\leq n\lambda_{max},$$ and then deviding by $n^2$ and taking the limit: $$0\leq lim \frac{1}{n^2}Var\left ( \sum_{k=1}^n X_k \right )\leq lim \frac{\lambda_{max}}{n}$$

And the question boils down to the question whether maximal eigenvalue stays bounded for the covariance matrix when the dimension grows. Any suggestions how to attack this problem is welcome.

$\endgroup$
  • 1
    $\begingroup$ The only way this condition can fail is if the variance of each $X_k$ is growing in $k$ sufficiently fast, if it's not already infinite. It will fall for any distributions, fat-tail or heavy so long as the variance is infinite, like the Cauchy distribution. $\endgroup$ – Alex R. Aug 11 '15 at 16:14
  • 2
    $\begingroup$ Of course it fails for heavy tailed distributions with infinite variance... $\endgroup$ – Nate Eldredge Aug 11 '15 at 16:14
  • $\begingroup$ But is the bounded variance is sufficient for the condition to hold? $\endgroup$ – Tomas Aug 11 '15 at 16:34
  • 1
    $\begingroup$ If $X_n=X$ for all $n$, where $Var(X)=1$, then it fails (you said that you did not necessarily have independence). $\endgroup$ – Michael Aug 11 '15 at 17:03
  • 1
    $\begingroup$ @Tomas: No. In your sum, you will have $n$ varianece terms and around $n(n-1)/2$ covariance terms, which grows like $n^2$ and hence when divided by $n^2$ won't go to 0. You need "most" of the covariance terms to contribute much less. $\endgroup$ – Alex R. Aug 11 '15 at 17:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.