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A one-tailed version of Chebyshev's inequality is that for t>0 $$P[X-E(X) \geq t] \leq \frac{1}{1 + t^2/Var(X)}$$ i.e. $$P[X-E(X)\geq t] \leq \frac{Var(X)}{Var(X) + t^2}$$

I heard someone claimed that the inequality is also true for the other side $$P[X-E(X)\leq t] \leq \frac{Var(X)}{Var(X) + t^2}$$

I briefly looked through the proof of the one-tailed version of Chebyshev's inequality, and it doesn't seem to work for the other side's version. So I was wondering if the other side isn't true?

Has the other side version been studied before?

Thanks!

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up vote 2 down vote accepted

The other side cannot be true in general, as then we would have for $X \in L^2(P)$ \begin{align*} 1 &\le P(X - E(X)\le t) + P(X - E(X)\ge t)\\ &\le \frac{2V(X)}{V(X) + t^2}\\ &\to 0, \quad t \to \infty \end{align*} which is absurd.

But of course for $t < 0$ we will have (applying the first version to $-X$) \begin{align*} P(X - E(X) \le t) &= P\bigl(-X - E(-X) \ge -t\bigr)\\ &\le \frac{V(-X)}{V(-X) + (-t)^2}\\ &= \frac{V(X)}{V(X) + t^2} \end{align*}

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