# Do the converses of [weak law of large numbers / central limit theorem] hold?

Let $\; X_0,X_1,X_2,X_3,...\;$ be independent and identically distributed (real-valued) random variables.

1. Suppose $\frac1n \cdot\sum\limits_{m=0}^n X_m$ converges in probability. Does it follow that $\operatorname{E}(X_0)$ exists?

2. Suppose $\operatorname{E}(X_0) = 0$ and that $\frac1{\sqrt n} \cdot\sum\limits_{m=0}^n X_m$ converges in distribution to a normal random variable. Does it follow that $\operatorname{E}((X_0)^2)$ is finite?

(I already found that the converse of the strong law of large numbers holds.)

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a remark. The weak law fails for the Cauchy distribution. –  Gerald Edgar Aug 25 '11 at 12:01
A classical example for 1'. is a symmetric integer-valued X with P(X=n)=P(X=-n)=c/(n^2log(n)). Then phi is C^1 but X is not integrable. On the other hand, if phi is C^2 then X^2 is integrable. –  Did Aug 25 '11 at 12:44
Didier, this gives a counterexample to 1, right? I think the last line in the question means that if one replaces, in 1, convergence in probability by a.s. convergence, then the answer is yes (by say the converse to Borel-Cantelli). –  Mikael de la Salle Aug 25 '11 at 12:59
Necessary and sufficient conditions (in terms close to those you want) for the WLLN and the CLT can be found, e.g., in "Foundations of modern probability" by Kallenberg (Theorems 4.16 and 4.17). –  Yvan Velenik Aug 25 '11 at 14:09
@Ricky in the 2nd Edition it's Theorems 5.16 and 5.17 –  pgassiat Aug 25 '11 at 23:48