Prove Theorem 2.6 (Markov) for the case when \(X\) is a discrete random variable on \(\mathbb{N}\) with probability distribution function \(p(x)\).

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Theorem 2.6 (Markov). Consider a random variable X where E(X]") < for some r > 0 and let 8 > 0. Then P(|X|>8) 8"E(|X|"). Proof. Assume for simplicity that X is a continuous random variable with distribution function F. If 8> 0 then E(|X|") = = 2dF(x) { |x|dF(x) + |x|dF(x) {x:|x|>8} since {x:|x|8} 52:270} |x|dF(x), |x| dF(x) 0. {x:|x|6} Note that x 8" within the set {x: |x| > 8}, so that |x|"dF(x) > SdF(x)=8" P(|X|> 8). {x:|x|>8} {x:|x|>6} Therefore E(X") 8" P(|X|> 6) and the result follows.