chebyshev inequality??

Question

how do we know when to use 1/a^2>P(|X-u|>a*standard deviation or
1/a^2>1-P(|X-u|

Answer 1

|X-μ|>a iff (X-μ)²>a², so P(|X-μ|>aσ) = P((X-μ)²>a²σ²), and by Markov's inequality P((X-μ)²>a²σ²) ≤ E((X-μ)²)/(a²σ²) = σ²/(a²σ²) = 1/a²

Of course, that raises the question, how do we prove Markov's inequality? Suppose X is a nonnegative random variable, and let I be the random variable given by I={1 if X≥a, 0 if X

Answer 2

Chebyshev's inequality (sometimes spelled Tchebysheff) is used to give a conservative estimate of the fraction of a population that lies between +- k (or "a" as you use) standard deviations when the form of the distribution is not known, but does have only one mode and is mound-shaped. The rule is-- at least (1-1/k^2)of a set of n measurements will lie within k standard deviations of their mean.
For example-- 3/4 of the measurements will lie within mu+-2s (when k=2) and 8/9 of the measurements will lie within mu+-3s (when k = 3), etc.