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I am trying to show that if $\epsilon > 0$ and $X$ is a random variable and with $0 \leq X \leq 1$ and $E[X] \geq 1-\epsilon$, then I can estimate that $P[X < 2/3] < 3 \epsilon$

I tried using markov's inequality and also proving the converse, but I cannot get it done.

I would appreciate any help!

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1 Answer 1

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Consider $Y \equiv 1 - X$ such that $0 \leq Y \leq 1$ and $E[Y] \leq \epsilon$. By Markov inequality $$P\left[ X < \frac23 \right] = P\left[ Y > \frac13 \right] \leq \frac{ E[Y]}{1/3} \leq 3\epsilon$$

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