# Lower bounding the probability that a zero-mean sequence of random variables stays positive

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.

• No. It's possible to put $X_n=\pm 2^{n+1}$ with probability $2^{-2n}$ and 0 with the remaining probability. Commented Mar 7, 2016 at 18:54
• @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.) Commented Mar 7, 2016 at 19:46
• Probably having uniformly bounded 3rd moment as well mean 0 and variance 1 would suffice. I'll have a think about the details. Commented Mar 7, 2016 at 19:51
• @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. Commented Mar 7, 2016 at 20:16

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 $\mathbb P(X\in [0,4K])\ge 1/(16K^2)$.

• 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. Commented Mar 7, 2016 at 21:38
• I have to take that back. The last line in my argument doesn't seem to be correct. The inequality goes the wrong way... Commented Mar 7, 2016 at 21:53

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 $c \approx 4.43035992$.

• 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!) Commented Mar 8, 2016 at 15:48
• 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 $t = 3a/2$. Commented Mar 8, 2016 at 16:29
• 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 $f$. Commented Apr 10, 2020 at 18:56

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 $$\theta \in [0,1]$$.

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