# Reverse Markov inequality

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.

• Replace $X$ by $X / \sqrt{c}$ you get rid of one constant that you need to quantify over. Commented May 13 at 13:54
• @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. Commented May 13 at 14:04
• @an_ordinary_mathematician Uh, actually I think I found the fix... would you mind if I edited the post? Commented May 13 at 14:07
• sure no problem , I will delete the comment Commented May 13 at 15:29
• @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. Commented May 13 at 15:55

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 $$\frac{1}{n}$$.
• $+1$, though you should not say "Indeed", since that makes it seem like you are justifying the previous sentence (about why "it is enough"). Commented May 13 at 16:22
$$\newcommand\ep\varepsilon$$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 $$0$$.