The Chebyshev inequality goes like this. Suppose that \(x\) is an \(n\)-vector, and that \(k\) of its entries satisfy \(|x_i| \geq a\), where \(a > 0\). Then \(k\) of its entries satisfy \(x_i^2 \geq a^2\). It follows that

\[||x||^2 = x_1^2 + \cdots + x_n^2 \geq ka^2\]

since \(k\) of the numbers in the sum are at least \(a^2\), and the other \(n - k\) numbers are nonnegative. We conclude that \(k \leq ||x||^2 / a^2\), which is the Chebyshev inequality.

When \(||x||^2 / a^2 \geq n\), the inequality tells us nothing, since we always have \(k \leq n\). In other cases, it limits the number of entries in a vector that can be large. In general, the inequality states that no entry of a vector can be larger in magnitude than the norm of the vector. Another interpretation, using the RMS value, looks like this

\[\frac{k}{n} \leq \bigg( \frac{\mathbf{rms}(x)}{a} \bigg)^2\]

where \(k\) is the number of entries of \(x\) with absolute value at least \(a\). The left-hand side is the fraction of entries of the vector that are at least \(a\) in absolute value. The right hand side is the inverse square of the ratio of \(a\) to \(\mathbf{rms}(x)\). It says, for example, that no more than 1/25 = 4% of the entries of a vector can exceed its RMS value by more than a factor of 5. Some statements that follow from the inequality are that:

  • not too many of the entries of a vector can be much bigger (in absolute value) than its RMS value
  • at least one entry of a vector has absolute value as large as the RMS value of the vector
import numpy as np

a = 6

x = np.array([3, 4, 1, 2, 4, 8, 5])
n = x.size
k = np.count_nonzero(x >= a)
rms = np.sqrt(np.mean(x ** 2))


k / n <= (rms / a) ** 2
## True

Relation To Standard Deviation

The Chebyshev inequality can be transcribed to an inequality expressed in terms of the mean and standard deviation: If \(k\) is the number of entries of \(x\) that satisfy \(|x_i - \mathbf{avg}(x)| \geq a\), then \(k/n \leq (\mathbf{std}(x)/a)^2\).

In probability theory the Chebyshev inequality is used to state something along the lines of “at most, approximately 11.11% of a distribution will lie at least three standard deviations away from the mean.”.

Another way to state this is: The fraction of entries of \(x\) within \(\alpha\) standard deviations of \(\mathbf{avg}(x)\) is at least \(1 - 1/\alpha^2\) (for \(a > 1\)).

As an example, if we consider a time series of return on an investment, with a mean return of 8%, and a risk (standard deviation) of 3%. By the Chebyshev inequality, the fraction of periods with a loss (\(x_i \leq 0\)) is no more than \((3/8)^2\), or 14.1%.