The Markov inequality states thatwhen P(Y ≥ 0) = 1, then P(Y ≥ t) ≤ E(Y )~t.

(a) When Y is discreteoverthenonnegativeintegers,provethisbyexplainingwhy E(Y ) ≥

Σy≥t yf(y) ≥ Σy≥t tf(y) = tP (Y ≥ t).

(b) If X is anyrandomvariablewithmean μ and variance σ2, applytheMarkovinequality with Y = (X − μ)2 to prove Chebyshev’s inequality that P(SX − μS ≥ kσ) ≤ 1~k2.

(c) Consider k = 1 in Chebyshev’sinequality.Specifyaprobabilitydistributionforwhich P(SX − μS ≥ σ) = 1.