Chebyshev's Inequality

Chebyshev's inequality tightens Markov by incorporating variance. For any random variable $X$ with mean $\mu$ and variance $\sigma^2$: $$P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}$$ Equivalently: $P(|X - \mu| \geq t) \leq \frac{\sigma^2}{t^2}$. Intuition: Markov says "the mean controls the tail." Chebyshev says "the spread controls the tail even better." If a distribution has small variance, most of its mass must be near the mean — Chebyshev quantifies exactly how much. Proof: Apply Markov's inequality to $Y = (X - \mu)^2$, which is non-negative. Then $P(|X-\mu| \geq t) = P(Y \geq t^2) \leq E[Y]/t^2 = \sigma^2/t^2$. Concrete example: An exam has mean score 70 and standard deviation 10. What fraction of students score below 40 or above 100? That's $|X - 70| \geq 30 = 3\sigma$. By Chebyshev: $P \leq$$\frac{1}{9}$ \approx 11.1\%. When to use: Bounding tail probabilities when you know the mean and variance but not the full distribution. Proving concentration results. In interviews, Chebyshev often appears as: "Without assuming normality, bound the probability that..." Alternative approach: If you know the moment generating function, Chernoff bounds are exponentially tighter. Chebyshev is the go-to when you only have two moments.