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I'm looking for an upper bound on $E_X\left[\left\vert X^2-(E[X])^2\right\vert\right]$, where $X$ is a nonnegative random variable, with known $E[X]$ and $E[X^2]$, but unknown higher moments.

A lower bound is $E[X^2]-(E[X])^2$, and I wonder if a there is a constant multiple of this as an upper bound?

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    $\begingroup$ Upper bound in terms of what, exactly? The absolute value sign is superfluous, because $E[X^2] \ge E[X]^2$ in any event, so you are asking for an upper bound for the variance, given that you know the variance. $\endgroup$
    – whuber
    Commented Oct 4, 2023 at 2:41
  • $\begingroup$ The issue is that the absolute value is a nonlinear function. $\endgroup$
    – axk
    Commented Oct 4, 2023 at 2:52
  • $\begingroup$ @Whuber - the absolute value sign is not superfluous: for non-constant random variables, there is a positive probability that $X<E[X]$ in which case $X^2 -(E[X])^2<0$ $\endgroup$
    – Henry
    Commented Oct 4, 2023 at 8:48
  • $\begingroup$ @Henry Thank you -- I had misread that by distributing expectation over the absolute value! $\endgroup$
    – whuber
    Commented Oct 4, 2023 at 13:14

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I would have thought you could say:

$E\big|X^2-(E[X])^2\big| \le E\big|X^2+(E[X])^2\big| = E\big[X^2\big]+(E[X])^2$

As an example to show it can come close:

  • suppose $X=1000$ with probability $\frac1{100}$ and $X=0$ otherwise,

  • so $E[X]=10$ and $E[X^2]=10000$,

  • we get $E[X^2]-(E[X])^2=9900$ and $E[X^2]+(E[X])^2=10100$

  • with $E\big|X^2-(E[X])^2\big|=10098$ closer to the upper bound.


As to your question of a constant multiple of the variance, the answer seems to be no. Adjusting the previous example, suppose $X=1000$ with probability $\frac1{100}$ and $X=999$ otherwise, so $E[X]=999.01$ and $E[X^2]=998020.99$, you have $E[X^2]-(E[X])^2=0.0099$ and $E\big|X^2-(E[X])^2\big|=39.570498$, almost four thousand times as big. This is not the worst case.

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  • $\begingroup$ You would have thought but...? Or do you think it still? $\endgroup$ Commented Oct 4, 2023 at 10:30
  • $\begingroup$ @RichardHardy It was a quick response giving a bound which was not of the form requested. I often make mistakes when looking at something new (here I initially persuaded myself $E[X^2]$ was a bound and then found a counterexample), so I used a weaker tense and mood to reflect this. I still think it correct, but I am happy to leave my original words. $\endgroup$
    – Henry
    Commented Oct 4, 2023 at 10:37
  • $\begingroup$ Thanks, I understand. $\endgroup$ Commented Oct 4, 2023 at 10:38

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