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In a university statistics course, we were presented with a "proof" of the Weak Law of Large Numbers (as it applies to population samples) based on Chebyshev's inequality. The steps the professor took are nearly identical to those linked in the following Wikipedia article: https://en.wikipedia.org/wiki/Law_of_large_numbers#Proof_using_Chebyshev.27s_inequality

The main difference in the proofs appeared when my professor began with Chebyshev's inequality:

$P(|\bar{X_n}-\mu|< k\sigma)\geq1-\frac{1}{k^2}$

Next, she replaced $\sigma$ with $\frac{\sigma}{\sqrt{n}}$ and $\frac{k\sigma}{\sqrt{n}}$ with $c$, so

$P(\mu-c<\bar{X_n}< \mu+c)\geq1-\frac{\sigma^2}{nc^2}$

This aligns identically with the proof on Wikipedia. We examine this as the sample number (n) approaches infinity and claim the probability approaches 1 that the sample average ($\bar{X_n}$) approaches the expected value ($\mu$).

Here are my problems:

1: As n approaches infinity, c approaches 0, so $P(\mu-c<\bar{X_n}< \mu+c)=P(\mu<\bar{X_n}< \mu)$ This doesn't make sense, since $\mu$ is not less than $\mu$.

Given this question, does this proof have any merit? Or am I perhaps misunderstanding something algebraically or conceptually? Thanks for your time!

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    $\begingroup$ A first quick comment: even outside any probabilistic setting, strict inequalities are not preserved by taking the limit. $0< \frac{1}{n}$ for all $n$, yet $\frac{1}{n} \xrightarrow[n\to\infty]{} 0$. (Strict inequalities "become inequalities" when taking the limit.) $\endgroup$ – Clement C. Mar 4 '16 at 20:37
  • $\begingroup$ Why does $P(\mu-c<\bar{X}_n < \mu+c)$ equal the other stuff when $c$ is approaching $0$ but is not quite there yet? $\endgroup$ – Dilip Sarwate Mar 4 '16 at 20:39
  • $\begingroup$ Maybe it helps to write the probability at issue with $P\{|\bar X_n - \mu| \leq c\},$ where the $c$ in question is positive. I think it might have been $\leq$, not $<$, throughout, which does no harm to the logic. And it's always a good idea to go through class notes with a skeptical eye. $\endgroup$ – BruceET Mar 4 '16 at 20:54
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In the proof, $c$ is held constant as $n\to\infty$; it doesn't tend to zero. The argument shows that: $$ \text{for every $c>0$, }\lim_{n\to\infty}P(\mu-c<\bar X<\mu+c)=1\tag1 $$ By taking the complement of the event $\{\mu-c<\bar X<\mu+c\}$ in (1), we get $$ \text{for every $c>0$, }\lim_{n\to\infty}P(|\bar X-\mu|>c)=0\tag2 $$ which is precisely what the weak law of large numbers states. (So the proof has merit!)

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