Chebyshev's Theorem

or Tchebysheff's Theorem

Let $X$ be a random variable with finite mean $μ$ and variance $σ^{2} \neq 0$ , then $\forall k > 0$

P (| X - μ | < k σ) \geq 1 - \frac{1}{k^{2}}, or P (| X - μ | \geq k σ) \leq \frac{1}{k^{2}}

Probability that the sample points sit within $k$ standard deviations of the mean get squeezed to 1 as $k$ increases

At $k = 2$ , probability for any random variable under any distribution is at least 75% (for the normal, we have 95%)

Applies to a new random variable $Y = g (X)$ , as long as $μ_{y}$ and $σ_{y}^{2}$ exists

If the exact value is not needed when finding $P (X > c)$ , we can use the theorem to quickly calculate a bound instead of doing a laborious integration

Just plugging in $k = \frac{k}{σ} :$

P (| X - μ | \geq k) \leq \frac{σ^{2}}{k^{2}}

Law of large numbers

For normal random variable $X$ and Sample Proportion $Y = \frac{X}{n} = \hat{p}$

$μ_{Y} = p, σ_{y} = \sqrt{p (1 - p) / n},$ let $c = k σ \to k = c / σ$

\begin{aligned} P (| Y - p | \leq c) = P (| Y - p | \leq k σ) & \geq 1 - \frac{1}{k^{2}} \\ \geq 1 - \frac{1}{(c / σ)^{2}} \\ as n \to \infty & \geq 1 - \frac{p (1 - p)}{c^{2} n} \\ squeezed at 1 by axioms & \geq 1 - 0 \\ P (| Y - p | \leq c) & = 1 \end{aligned}

for any arbitrary positive integer $c$

With the equivalent symmetrical inequality and $V a r (\bar{x}) = \frac{σ^{2}}{n} :$

\begin{aligned} P (| \bar{x} - μ | \geq k) & \leq \frac{V a r (\bar{x})}{k^{2}} = \frac{σ^{2}}{n k^{2}} \\ P (| \bar{x} - μ | \geq k) & \leq 0 as n \to \infty \end{aligned}

sample mean converges to population mean with large sample sizes