Chebyshev's inequality

Chebyshev's inequality is an inequality that sets bounds to the probability of deviation of a random variable. Given a RV $X$, Chebyshev's inequality states

where $\lambda$ is a positive real number, $\sigma$ is the standard deviation of $X$ and $\mu$ is its mean. Only the cases $\lambda>1$ is significant. When $\lambda\leq1$, then $1/\lambda ^{2}\geq 1$, which is a trivial bound since $P\leq 1$ always.

If we call $k=\lambda ^{2}\sigma ^{2}$ and $h(x)=(x-\mu)^{2}$, we get the inequality.

This inequality says is that the probability that $X$ deviates from its mean by $\lambda \sigma$ is at most $1/\lambda ^{2}$. For instance, it states that there is at most a $1/2^{2}=0.25=25\%$ chance that $X$ assumes a value that is $2\sigma$ or higher from the mean. Seeing it the other way around, it says that at least $75\%$ of values sampled from $X$ must be within $2\sigma$ of the mean.

This inequality is almost universal: it applies to all RVs with a finite mean and variance, regardless of their probability distribution. This provides universal bounds across almost all distributions, but because it's so general, it is rather imprecise compared to distribution-specific arguments. For example, this inequality states that there must be at least $0\%$, $75\%$ and $88.89\%$ of values within $1\sigma$, $2\sigma$ and $3\sigma$. The Gaussian distribution on the other hand is known to instead require a minimum of $68\%$, $95\%$ and $99.7\%$ of values within the same ranges.