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Let $C > c > 0$ and $K > 1$ be constants. Does there exist, for all small enough $\varepsilon > 0$ depending on $c, C, K$, some bound of the following form?

For all random variables $X$ with $c \leq \mathbb E[X^2] \leq C$ and $E[|X|] \leq \varepsilon c^{\frac{1}{2}}$, we have $$\mathbb P\left (|X| \geq K \mathbb E[|X|] \right ) \geq \delta \mathbb E[|X|]^2$$ for some constant $\delta > 0$ depending only on $c, C, K, \varepsilon$.

Upon some googling, it would seem if we allow $K < 1$, then this is the Paley Zygmund inequality. I wonder if it can be improved to arbitrarily large $K$ in this scenario.

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  • $\begingroup$ Replace $X$ by $X / \sqrt{c}$ you get rid of one constant that you need to quantify over. $\endgroup$ Commented May 13 at 13:54
  • $\begingroup$ @an_ordinary_mathematician ah, you are right. This needs to be refined more if there is really anything here... will delete the post shortly once I confirm you have recieved this. $\endgroup$
    – Nate River
    Commented May 13 at 14:04
  • $\begingroup$ @an_ordinary_mathematician Uh, actually I think I found the fix... would you mind if I edited the post? $\endgroup$
    – Nate River
    Commented May 13 at 14:07
  • $\begingroup$ sure no problem , I will delete the comment $\endgroup$ Commented May 13 at 15:29
  • $\begingroup$ @an_ordinary_mathematician I am confused, $\delta$ was allowed to depend on $C$, $c$ and $\varepsilon$, so your example is not a counterexample (in it $\varepsilon \to 0$). Anyways, the answer to the original question is no for any $K \ge 1$, I will now type an answer. $\endgroup$ Commented May 13 at 15:55

2 Answers 2

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As you noticed, for $K < 1$ (some form of) it is true. I will show that it is false for $K \ge 1$. It is enough to construct functions $f_n$ on $[0, 1]$ with $\int |f_n| = \varepsilon$ and $\int |f_n|^2=1$ for some $0 < \varepsilon < 1$, such that $|\{ x: |f_n(x)| \ge 1\}| \le \gamma_n$, where $\gamma_n \to 0$. Indeed, the function $f_n$ would only take two values: on $[0, 1-1/n]$ it will be equal to $a$ and on $(1-1/n, 1]$ it will be equal to $b$. We have $$\begin{cases}a(1-1/n) + b/n = \varepsilon,\\ a^2(1-1/n)+b^2/n = 1.\end{cases}$$

From the first equation $b = n\varepsilon - a(n-1)$, plugging this into the second one we get a quadratic equation, which I was taught how to solve in middle school. We have (I chose the smaller root) $$a = \varepsilon - \frac{\sqrt{\varepsilon^2(n-1)^2-(n-1)(\varepsilon^2 n - 1)}}{(n-1)},$$ from which we get $0 < a < \varepsilon$ as long as $\varepsilon^2 n > 1$ and also in this case $b > \varepsilon$. Now it is clear that our function is bigger than $\varepsilon=\int |f_n|$ only on the set of measure $\frac{1}{n}$.

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    $\begingroup$ $+1$, though you should not say "Indeed", since that makes it seem like you are justifying the previous sentence (about why "it is enough"). $\endgroup$ Commented May 13 at 16:22
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    $\begingroup$ Huh, I only learnt it in high school. Nice solution! $\endgroup$
    – Nate River
    Commented May 13 at 16:49
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$\newcommand\ep\varepsilon$Here is a version of Alexei's answer which is arguably more transparent (and does not require solving quadratic equations :-)).

Take any $\ep\in(0,1)$ and then any $a\in(0,\ep)$. Let $b:=\ep+\dfrac1{\ep-a}>\ep$. Let $P(X=b)=\dfrac{\ep-a}{b-a}$ and $P(X=a)=1-P(X=b)=\dfrac{b-\ep}{b-a}$. Then $EX=\ep$, $EX^2=1+\ep^2\in[1,2]$, and for any $K\ge1$ $$P(X\ge K\,EX)\le P(X\ge EX)=\dfrac{\ep-a}{b-a}\to0$$ if $a\uparrow\ep$.

So, for any $K\ge1$, the best lower bound on $P(X\ge K\,EX)$ is $0$.

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