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ProbabilityConcept reference

Markov's inequality

For nonnegative X and positive a, Markov bounds P(X≥a) by E[X]/a.

2 min read1 figureSources & further reading
On this page 7 sections
  1. Overview
  2. How much probability can sit far above the mean?
  3. The variable can be a nonnegative count, duration or size
  4. A bound limits possible probabilities; it does not usually give the exact probability
  5. Key takeaway
  6. Sources & further reading
  7. Concept connections

01How much probability can sit far above the mean?#

How much probability can sit far above the mean? If a variable is nonnegative, the mean limits how often it can reach a high threshold.

P(X≥a)≤E[X]aP(X\ge a)\le\frac{E[X]}{a}

Markov requires X≥0X\ge0 and a>0a>0.

If the mean is 66, outcomes at least 2424 each contribute at least 2424 times their probability to that mean. Their probability cannot exceed 6/24=0.256/24=0.25.

For nonnegative X with mean 6, Markov gives P(X≥24)≤6/24=.25. The highlighted ray starts at 24 and continues right. Its length is a domain indicator, not probability area; no distribution is specified.For nonnegative X with mean 6, Markov gives P(X≥24)≤6/24=.25. The highlighted ray starts at 24 and continues right. Its length is a domain indicator, not probability area; no distribution is specified.
Figure 1For nonnegative X with mean 6, Markov gives P(X≥24)≤6/24=.25. The highlighted ray starts at 24 and continues right. Its length is a domain indicator, not probability area; no distribution is specified.
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Check your reasoning

X≥0X\ge0, E[X]=4E[X]=4. Bound P(X≥10)P(X\ge 10) using Markov and the cap 11.

Show answer and explanation
0.4

Mean divided by threshold is 4/10=0.44/10=0.4. The upper bound is 0.40.4.

02The variable can be a nonnegative count, duration or size#

The variable can be a nonnegative count, duration or size. Translate the requested event into a positive threshold, then divide its mean by that threshold.

If the ratio exceeds 1, use 1 as the tighter probability bound. This says nothing more than every probability being at most 1.

Check your reasoning

A nonnegative duration TT has mean 77 seconds. Bound P(T≥14)P(T\ge 14) using Markov and the cap 11.

Show answer and explanation
0.5

Mean divided by threshold is 7/14=0.57/14=0.5. The upper bound is 0.50.5.

03A bound limits possible probabilities; it does not usually give the exact probability#

A bound limits possible probabilities; it does not usually give the exact probability. Two distributions can share a mean and have different tails.

The sign assumption matters. Negative values could offset a large positive tail, invalidating this direct calculation.

Check your reasoning

Report: P(X≥20)=0.25P(X\ge20)=0.25 for X≥0X\ge0, E[X]=5E[X]=5. Repair it.

  1. AExactly 0.25
  2. BAt least 0.25
  3. CAt most 0.25
Show answer and explanation
At most 0.25

5/20=0.255/20=0.25 is an upper bound.

Key takeaway

Check nonnegativity, divide mean by positive threshold, and report an upper bound capped at 1.

  • Bound a nonnegative variable's upper tail from its mean.

Sources & further reading

  1. [1]

Reference this concept

Link to this page, a section, or an individual figure.

Glacius. “Markov's inequality.” Math behind ML. /learn/p-markov-bound