ProbabilityConcept reference

Chebyshev's inequality

Chebyshev bounds P(|X−μ|≥d) by Var(X)/d² for positive d and finite variance.

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Overview
Chebyshev bounds how often a variable lies at least a distance d>0 from its mean μ
For a symmetric open band, measure from the center to either endpoint
Chebyshev provides a ceiling on the tail probability
Key takeaway
Sources & further reading
Concept connections

01Chebyshev bounds how often a variable lies at least a distance d>0 from its mean μ#

Chebyshev bounds how often a variable lies at least a distance $d>0$ from its mean $\mu$ . It needs finite variance, without requiring a normal distribution.

\begin{gathered}P(|X-\mu|\ge d)\\\le\frac{\operatorname{Var}(X)}{d^2}\end{gathered}

With mean 10 and variance 4, distance 4 gives a bound of 0.25. This includes both tails: values at most 6 or at least 14.

Figure 1Mean is 10 and variance is 4, so SD is 2. Deviation at least 4 selects X≤6 or X≥14, including both boundary points. Chebyshev bounds this tail probability by .25; no distribution shape or exact tail mass is specified.

Check your reasoning

Mean $3$ , variance $2$ . Bound $P(|X-3|\ge 4)$ by Chebyshev, capped at $1$ .

Show answer and explanation

0.125

Distance is $4$ ; variance divided by its square is $2/16=0.125$ . Use upper bound $0.125$ .

02For a symmetric open band, measure from the center to either endpoint#

For a symmetric open band, measure from the center to either endpoint. The whole band width would double the distance incorrectly.

The same bound follows by applying Markov to the nonnegative squared deviation. If the ratio exceeds 1, retain the trivial bound 1.

Check your reasoning

Mean $12$ , variance $4$ . Bound the probability outside the open band $(8,16)$ by Chebyshev, capped at $1$ .

Show answer and explanation

0.25

Distance is $4$ ; variance divided by its square is $4/16=0.25$ . Use upper bound $0.25$ .

03Chebyshev provides a ceiling on the tail probability#

Chebyshev provides a ceiling on the tail probability. It cannot turn a mean and variance into an exact tail probability for every distribution.

The endpoints count in the deviation event because it uses $\ge d$ . The inside event is the open band $|X-\mu|<d$ .

Check your reasoning

Mean $5$ , variance $9$ . Report: $P(|X-5|\ge 6)=0.25$ . Capped Chebyshev correction?

AExactly 0.25
BAt most 1
CAt most 0.25

Show answer and explanation

At most 0.25

$9/6^2=0.25$ .

Key takeaway

Divide variance by squared distance; cap at 1 and keep the upper-bound interpretation.

Bound a deviation probability using finite variance.

Sources & further reading

[1]
Pishro-Nik: Markov and Chebyshev Inequalities ↗Pishro-Nik · Article

01Chebyshev bounds how often a variable lies at least a distance d>0 from its mean μ#

02For a symmetric open band, measure from the center to either endpoint#

03Chebyshev provides a ceiling on the tail probability#

Divide variance by squared distance; cap at 1 and keep the upper-bound interpretation.

Sources & further reading

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