show/hide this revision's text 2 deleted 191 characters in body

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

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

2. Suppose $\hspace{.02 in}\operatorname{E}(X_0) \operatorname{E}(X_0) = 0\hspace{.02 in}$ 0$ and that $\; \frac1{\sqrt \frac1{\sqrt n} \cdot \left(\displaystyle\sum_{0\leq m\lt n} X_m\right)$ cdot\sum\limits_{m=0}^n X_m$ converges in distribution to a normal random variable. Does it follow that $\; \operatorname{E}((X_0)^2) < +\infty \;$ \operatorname{E}((X_0)^2)$ is finite?

(I already found that the converse of the strong law of large numbers holds.)
show/hide this revision's text 1

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 $\; \frac1n \cdot \left(\displaystyle\sum_{0\leq m\lt n} X_m\right)$ converges in probability. $\;\;$ Does it follow that $\operatorname{E}(X_0)$ exists?



2.


Suppose $\hspace{.02 in}\operatorname{E}(X_0) = 0\hspace{.02 in}$ and that $\; \frac1{\sqrt n} \cdot \left(\displaystyle\sum_{0\leq m\lt n} X_m\right)$ converges in distribution to a normal random variable.

Does it follow that $\; \operatorname{E}((X_0)^2) < +\infty \;$ ?




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