Assuming that the distribution has finite variance (a condition not required for the LLN), then doesn't the LLN follow from the CLT?
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gung - Reinstate Monica
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tetradeca7tope
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2Possibly of interest: [central-limit-theorem-versus-law-of-large-numbers](http://stats.stackexchange.com/questions/22557/). – gung - Reinstate Monica Dec 06 '13 at 23:59
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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
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