5

Assuming that the distribution has finite variance (a condition not required for the LLN), then doesn't the LLN follow from the CLT?

gung - Reinstate Monica
  • 132,789
  • 81
  • 357
  • 650

1 Answers1

6

WLLN, yes.

Here is a general claim: Suppose $\{ f_n \}$, $f$, and $g$ are random variables, and

$$ \sqrt{n} (f_n - f) \stackrel{d}{\mapsto} g. $$

Let's say the CDF of $g$ is continuous everywhere. Then $f_n \rightarrow f$ in probability. This is because $\sqrt{n} (f_n - f)$ is bounded in probability/uniformly tight.

Michael
  • 2,853
  • 10
  • 15