[The Bizarre Bernoulli] The Bernoulli distribution px (u) is one of the only int

Question

[The Bizarre Bernoulli] The Bernoulli distribution px (u) is one of the only interesting probability distributions that is strictly bounded, on at least one side, by a one-standard-deviation bound, in the following sense. Recall that the Bernoulli distribution is defined by: Prove the following statement: for every value of p, either E[X] - sigma x le 0, or E[X] + sigma x ge 1, or both, depending on the value of p. (Note: this statement is interesting and worth proving because it implies that either P{X ge E[X] - sigma x} = 1 or P{X le E[X] + sigma x) = 1 or both, depending on the value of p. There are not many types of random variables that can be so tightly bounded.)

Explanation / Answer

E(X)= 1*px(1)+0*px(0) = 1*px(1)= p For a bernoulli random variable, X^2= X E(X^2)= 1* px(1)+ 0* px(0) = 1* px(1)= p Hence Var(X)= E(X^2)-(E(X))^2= p-p^2= p(1-p) Thus the standard deviation sx = sqrt(p(1-p)) Let E(X)-sx = p- sqrt(p(1-p))= f(p) and let E(X)+ sx = p+ sqrt(p(1-p))= g(p) to show either f(p) =1 or both depending on p. for p=1/2, f(1/2)=0 and g(1/2)= 1.. hence both the inequalities are satisfied. for p =1. for p = p => p(1-p) >= p^2 => sqrt(p(1-p)) >= p => p- sqrt(p(1-p)) f(p) =1/2, 1-p p(1-p) >= (1-p)^2 => sqrt(p(1-p)) >=(1- p ) => p+ sqrt(p(1-p)) >= p+1-p=1 => g(p)

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