Let $X_1,X_2,\ldots,X_n$ be i.i.d real-valued random variables with finite variance $\sigma^2>0$. Can we non-trivially upper bound the probability $$ \mathbb{P}\bigl(\max_{1\leq i\leq n} X_i < \mathbb{E}[x]\bigr)\text{, or equivalently }\mathbb{P}\bigl(X_i < \mathbb{E}[x]\bigr) $$ It seems that a one-sided Chebyshev bound doesn't give any information without some buffer between the mean. We know by the law of large numbers is that $$\mathbb{P}(\max X_i < \mathbb{E}[x]) = \mathbb{P}(X_1 < \mathbb{E}[x])^n \to 0$$ as $n\to\infty$ which would imply that $\mathbb{P}(X_1 < \mathbb{E}[x])<1$ but I'd like a quantitative result.


1 Answer 1


You can't, because that would require you to upper-bound the probability $\mathbb{P}(X_1 < \mathbb{E}[X])$, which you can't – at least not given only the mean and the variance. Consider $P(X=(m-1)a)=\frac1m$ and $P(X=-a)=\frac{m-1}m$, with $E[X]=0$ and $\operatorname{Var}[X]=(m-1)a^2$. For $a=(m-1)^{-\frac12}$, you get a series of distributions with zero mean and unit variance for which the probability to obtain a value below the expected value gets arbitrarily close to $1$ as $m\to\infty$.

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    $\begingroup$ Ah effectively for the same reason Chebyshev is tight. Thank you. I guess I’ll have to look into those pesky anti-concentration results. :) $\endgroup$
    – cdipaolo
    Aug 11, 2018 at 8:48

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