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Assume that $X_n$ is a sequence of a zero-mean and unit variance random variables (and maybe having density w.r.t. to Lebesgue). Can we conclude that $ P(X_n \in [0,R_n]) $ is bounded away from zero eventually, say $$\liminf_{n \to \infty} P(X_n \in [0,R_n]) > 0$$ assuming that $R_n \to \infty$. Intuitively, $X_n$ should put some nonvanishing amount of mass on the positive real line because of the zero-mean assumption and the condition of unit variance should prevent that mass from escaping to infinity (?)

Here is a partial argument: We have \begin{align*} P(X_n \in [0,R_n]) &= 1 - P(X_n > R_n) - P(X_n < 0) \\ &\ge 1 - P(|X_n| > R_n) - P(X_n < 0) \\ &\ge 1 - \frac{E X_n^2}{R_n^2} - P(X_n < 0) \\ &= 1 - o(1) - P(X_n < 0) \end{align*} Thus, the problem reduces to arguing that $P(X_n < 0)$ stays away from 1.

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    $\begingroup$ No. It's possible to put $X_n=\pm 2^{n+1}$ with probability $2^{-2n}$ and 0 with the remaining probability. $\endgroup$ Commented Mar 7, 2016 at 18:54
  • $\begingroup$ @AnthonyQuas, thanks. What if we assume that all the moments are uniformly bounded? (Another option is to assume $P(X_n = 0) = 0$ for all $n$, but something tells me that a modified version of your counterexample, works in that case.) $\endgroup$
    – passerby51
    Commented Mar 7, 2016 at 19:46
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    $\begingroup$ Probably having uniformly bounded 3rd moment as well mean 0 and variance 1 would suffice. I'll have a think about the details. $\endgroup$ Commented Mar 7, 2016 at 19:51
  • $\begingroup$ @AnthonyQuas: Thanks, 3rd moment would be great. I was thinking about writing $X_n = X_n^+ - X_n^{-}$ and trying to force something using the moments, but it wasn't that successful. $\endgroup$
    – passerby51
    Commented Mar 7, 2016 at 20:16

3 Answers 3

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Here's a proof if a third moment condition is satisfied.

Suppose that $\mathbb EX=0$, $\mathbb EX^2=1$ and $\mathbb E|X|^3\le K$. Then let $Z=|X|$. We use Cauchy-Schwarz: $Z^2=Z^{1/2}Z^{3/2}$, so that $(\mathbb EZ^2)^2\le \mathbb EZ\cdot \mathbb EZ^3$. This gives $\mathbb EZ\ge \frac 1K$. Hence $\mathbb EX\mathbf 1_{X>0}\ge \frac{1}{2K}$.

Also $\mathbb EX^2\mathbf 1_{X>0}\le 1$. Now $\mathbb EX\mathbf 1_{X>4K}\le (1/4K)\mathbb EX^2\mathbf 1_{X>4K}\le 1/(4K)$, so that $\mathbb EX\mathbf 1_{X\in [0,4K]}\ge 1/(4K)$ and $\mathbb P(X\in [0,4K])\ge 1/(16K^2)$.

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  • $\begingroup$ Thanks. I have tried to fill in the details of the second moment argument below. Not sure if this was intended argument, but all seem to be good. $\endgroup$
    – passerby51
    Commented Mar 7, 2016 at 21:38
  • $\begingroup$ I have to take that back. The last line in my argument doesn't seem to be correct. The inequality goes the wrong way... $\endgroup$
    – passerby51
    Commented Mar 7, 2016 at 21:53
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A standard technique to lower-bound $\mathbb P(0 < X < r)$ as a function of $\mathbb E[|X|^3]$ subject to $\mathbb E[X] = 0$ and $\mathbb E[X^2] = 1$ is to consider a suitable linear combination of $|X|^3$, $X$, $X^2$ and $1$ that is negative on $(0,r)$ and nonnegative outside that interval and has a negative expected value. In this case try $$ f(x) = |x|^3 - b r x^2 - (1-b) r^2 x$$ For $x \ge 0$ we have $f(x) = x (x-r) (x-(b-1)r)$, so this satisfies the sign requirements there if $b < 1$. $f(x) > 0$ for $x < 0$ if $b^2 + 4 b - 4 < 0$, which is true if $0 \le b \le 2\sqrt{2}-2$. We have $\mathbb E[f(X)] = \mathbb E[|X|^3] - b r$, which we want to be negative. The minimum value of $f(x)$ on $[0,r]$ turns out to be $$ v = - \dfrac{r^3}{27} \left( 2 (b^2 - 3 b + 3)^{3/2} + 2 b^3 - 9 b^2 + 9 b\right)$$ The conclusion is then that if $\mathbb E[|X|^3] < b r$ where $0 < b \le 2\sqrt{2}-2$, $$ \mathbb P(0 < X < r) \ge \frac{ br - \mathbb E[|X|^3]}{-v} $$ Taking $b = 2\sqrt{2}-2$, we get $$\mathbb P(0 < X < t E[|X|^3]) \ge c \dfrac{t-(1+\sqrt{2})/2}{t^3 \mathbb E[|X|^3]^2}$$ for $t > (1+\sqrt{2})/2$, where $c \approx 4.43035992$.

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  • $\begingroup$ Thanks. This is interesting. I am intrigued by the word "standard". In what context is this standard? (I thought anything involving a cubic polynomial is fairly nonstandard!) $\endgroup$
    – passerby51
    Commented Mar 8, 2016 at 15:48
  • $\begingroup$ Here is another observation which is intriguing: The lower bound is decreasing in $t$ while the probably (LHS) is increasing, so it seems the bound is good for small values of $t$. With $a = (1+\sqrt{2})/2$, the lower bound seems to be maximized at $t = 3a/2$. So for all $t > 3a/2$, we should just use the bound for $t = 3a/2$. $\endgroup$
    – passerby51
    Commented Mar 8, 2016 at 16:29
  • $\begingroup$ Very cool, so basically we use the Markov like inequality $E[f(X)] \ge \Pr[0 \le x\le r] \min_{0\le x \le r} f(x)$ for a cleverly chosen $f$. $\endgroup$ Commented Apr 10, 2020 at 18:56
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Filling in the details for Anthony's argument:

Assume that $\mathbb E |X|^3 \le c$ for numerical constant $c > 0$.

Let $X^+ = X \mathbf 1_{X > 0}$ and $X^- = (-X) \mathbf 1_{X < 0}$. Then, $X = X^+ - X^-$. Let $\mu = \mathbb E X^+ = \mathbb E X^-$. Then, by Anthony's argument $2 \mu = \mathbb E |X| \ge 1/c$.

For $a \le \mu$, we have $\mathbb E X^+ \le a + \mathbb{E} X^+ \mathbb1_{X^+ > a} \le a + [\mathbb E (X^+)^2]^{1/2} [\mathbb P(X^+ > a)]^{1/2} $ or \begin{align} \mathbb P(X^+ > a) \ge \frac{(\mu-a)^2}{\mathbb E (X^+)^2} \ge (\mu-a)^2 \quad (*) \end{align} using $\mathbb E (X^+)^2 \le \mathbb E X^2 \le 1$. Take $a = \frac1{4c}$ so that $a \le \frac1{2c} \le \mu$. Then, $\mathbb P(X^+ > \frac1{4c}) \ge \frac1{16c^2}$. Same bound holds for $X^-$ by symmetry.

We can conclude that $X$ does not belong to $[-\frac{1}{4c},\frac{1}{4c}]$ with probability at least $1 -\frac{1}{8c^2}$.

EDIT: Apparently $(*)$ is called Paley-Zygmund inequality if one takes $a = \theta \mu$ for $\theta \in [0,1]$.

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    $\begingroup$ I've added some details to the end of the proof. $\endgroup$ Commented Mar 7, 2016 at 22:21

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