Bienaymé-Chebyshev inequality

Apply Markov's inequality with a congruent k^2 to obtain P[(x - μ)^2>=k^2]<=(〈(x - μ)^2 〉)/k^2 = σ^2/k^2. Therefore, if a random variable x has a finite mean μ and finite variance σ^2, then for all k>0, P( left bracketing bar x - μ right bracketing bar >=k) | <= | σ^2/k^2 P( left bracketing bar x - μ right bracketing bar >=k σ) | <= | 1/k^2.