How does this interpretation of Chebyshev's theorem fail for the sample mean?

Solution 1:

Why do you think $P[|\bar{x}-\mu|>k\frac{\sigma}{\sqrt{n}}]$ should tend to zero? From Central Limit Theorem we know $\sqrt{n}(\bar{x}-\mu)$ will converge to Gaussian distribution with mean zero and variance $\sigma^2$, which means $P[|\bar{x}-\mu|>k\frac{\sigma}{\sqrt{n}}]=P[\sqrt{n}|\bar{x}-\mu|>k\sigma]$ should be of constant scale(actually tend to $P[|Z|>k]$ for $Z\sim N(0,1)$). So your work with Chebyshev gives you a bound on this non-zero quantity--and actually, it is the Chebyshev-type bound for $Z$.

For the law of large numbers, note that here $\bar{x}=\frac{S_n}{n}$ where the latter is what is shown in the note you provide.

$$P[|\bar{x}-\mu|>k\sigma]\leq\frac{Var(\bar{x})}{k^2\sigma^2}=\frac{\sigma^2/n}{k^2\sigma^2}=\frac{1}{kn},$$

which goes to zero. This proves that $\bar{x}$ converges to $\mu$ in probability, which is the (weak)law of large numbers.