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


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.