TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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Chebyshev's inequality on a finite probability space

Statement

For every finite real random variable $X$ and every $t > 0$ , $P (∣ X - E [X] ∣ \geq t) \leq \frac{Var (X)}{t^{2}} .$ Equivalently, if $σ_{X} > 0$ and $c > 0$ , then $P (∣ X - E [X] ∣ \geq c σ_{X}) \leq 1 / c^{2}$ . If $σ_{X} = 0$ , the first form remains valid for every $t > 0$ .

Facts & Assumptions

Given: A finite real random variable $X$ and a real threshold $t > 0$ .

[L1]

Variance is $E [(X - E [X])^{2}]$ and $σ_{X}^{2} = Var (X)$ (Variance, standard deviation, and covariance on a finite probability space).

[L2]

Markov's inequality states $P (Y \geq a) \leq E [Y] / a$ for nonnegative $Y$ and $a > 0$ (Markov's inequality on a finite probability space).

Proof

technique · direct

1.1

The variable $Y = (X - E [X])^{2}$ is nonnegative, and for $t > 0$ the events ${Y \geq t^{2}}$ and ${∣ X - E [X] ∣ \geq t}$ are equal.

givenalgebra

2.1

Apply [L2] to $Y$ at the positive threshold $t^{2}$ and use [L1] to obtain the first inequality.

step 1.1L1L2

3.1

If $σ_{X} > 0$ , substitute $t = c σ_{X}$ in step 2.1 and cancel $σ_{X}^{2}$ to obtain $1 / c^{2}$ . Conversely, given $t > 0$ , choose $c = t / σ_{X} > 0$ in the standard-deviation form to recover step 2.1. If $σ_{X} = 0$ , step 2.1 gives probability at most zero for every $t > 0$ .

step 2.1L1choosealgebra∎

Depends on

Used by

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Sources

C. M. Grinstead and J. L. Snell, Introduction to Probability, 2nd ed., Theorem 8.1 (standard reference, not scraped)
H. Pishro-Nik, Introduction to Probability, Statistics, and Random Processes, Section 6.2.2 (standard reference, not scraped)