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The different laws of large numbers that I've seen state that

  1. If conditions $K$ holds

  2. Then the sample average of a process X_n converges in probability/almost-surely, to $\mu$.

Is there an inverse of this?

  1. If conditions $L$ do not hold

  2. Then the sample averges does not converge in probability nor almost surely, to anything.

Or even better:

  1. If and only if conditions $M$ hold

  2. ....

Basically, what I'd like to know is, whether it is known under what conditions the law of large numbers does not apply.

user56834
  • 12,925
  • So you want a statement which is equivalent to "The sample average converges [in probability / almost surely] to $\mu$" but not trivially so? – Arthur Feb 12 '18 at 07:56
  • @Arthur, basically, what I want to know is, whether it is known under what conditions the law of large numbers does not apply. – user56834 Feb 12 '18 at 08:18

2 Answers2

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An interesting converse is the following: if $\{X_n\}$ is an i.i.d sequence of non-negative random variables such that $\frac {S_n} n \to a$ almost surelyfor some $a>0$ then $E|X_n| < \infty$ and $a=EX_1$. Thus finiteness of the mean is a necessary condition for SLLN's. To prove that just apply ordinary SLLN to the i.i.d. sequence $\min \{{X_n, k}\}$ and note that $\min \{{X_n, k}\} \leq X_n$ to conclude that $E\min \{{X_n, k}\}$ is bounded as $k \to \infty$. By fatou's Lemma we get $EX_1 <\infty$.

Kavi Rama Murthy
  • 311,013
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A trivial counterexample is obtained if you drop the independence condition: take all samples being equal to the first one, and convergence will be to that value instead of to the mean. (More generally, dependence can imply a bias on the mean or even non-convergence.)