reverse Markov inequality

Let $X$ be a random variable that satisfies $\\Pr(X\\leq a)=1$ for some constant $a$ .Then, for $d<E[X]$ ,

\\Pr(X>d)\\geq\\frac{E[X]-d}{a-d}

Proof:Apply the Markov’s inequality to the random variable $\\tilde{X}=a-X$ ,

\\Pr(X\\leq d)=\\Pr(\\tilde{X}\\geq a-d)\\leq\\frac{E[\\tilde{X}]}{a-d}=\\frac{a-E[X]}{%a-d}.

Hence

\\Pr(X>d)\\geq 1-\\frac{a-E[X]}{a-d}=\\frac{E[X]-d}{a-d}.

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