How likely is a random variable to be far from its center?

There are many answers to the question in the title: How likely is a random variable to be far from its center?

The answers depend on how much you’re willing to assume about your random variable. The more you can assume, the stronger your conclusion. The answers also depend on what you mean by “center,” such as whether you have in mind the mean or the mode.

Chebyshev’s inequality says that the probability of a random variable X taking on a value more than k standard deviations from its mean is less than 1/k². This of course assumes that X has a mean and a standard deviation.

If we assume further that X is unimodal, and k ≥ √(8/3), then the conclusion of Chebyshev’s inequality can be strengthened to saying that the probability of X being more than k standard deviations from its mean is less than 4/9k². This is the Vysochanskiĭ-Petunin inequality. More on this inequality here.

If k ≤ √(8/3) the Vysochanskiĭ-Petunin inequality says the probability of X being more than k standard deviations from its mean is less than

4/3k² − 1/3.

Gauss’ inequality is similar to the Vysochanskiĭ-Petunin inequality. It also assumes X is unimodal, and for convenience we will assume the mode is at zero (otherwise look at Y = X – m where m is the mode of X). Gauss’ inequality bounds the probability of X being more than k standard deviations away from its mode rather than its mean.

Let τ² be the expected value of X². If the mean value of X is zero then τ² = σ² and the equations below are similar to the Vysochanskiĭ-Petunin inequality. But Gauss does not require that X has mean zero.

Gauss’ inequality says that

P(|X| > kτ) ≤ 4/9k²

if if k ≥ √(4/3) and

P(|X| > kτ) ≤ 1 − k/(√3 τ)

otherwise.

Gauss’ inequality is stronger than the Vysochanskiĭ-Petunin inequality when X has zero mean, but it is also assuming more, namely that the mean and mode are equal.

Related post: Strengthening Markov’s inequality with conditional probability.

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