# Is $\inf \{P(|X| > \varepsilon) : E[X] = 0, \operatorname{Var}(X) = 1\} = 0$ for all $\varepsilon > 0$?

If $$\varepsilon > 0$$, is it true that the infimum of $$\{P(|X| > \varepsilon) : E[X] = 0, \operatorname{Var} (X) = 1\}$$ is zero? That is, for every $$a > 0$$, can we always find a random variable $$X$$ with zero mean and unit variance that satisfies $$P(|X| > \varepsilon) < a$$?

My first attempt to prove this statement was by using the Chebyshev's inequality: if $$E[X] = 0$$ and $$\operatorname{Var} (X) = 1$$, then $$E[X^2] = \operatorname{Var} (X) + E[X]^2 = 1$$, so we have $$P(|X| > \varepsilon) \leq \frac{1}{\varepsilon^2}.$$ However, this inequality seems to be unhelpful, especially if $$\varepsilon \leq 1$$. Moreover, the right hand side does not even depend on $$X$$ or any other variable, so I cannot make the upper bound of $$P(|X| > \varepsilon)$$ to approach zero in this inequality.

Next, I tried to construct a sequence $$\{X_n\}_{n\in\mathbb{N}}$$ of random variables based on the uniform and U quadratic distributions. The idea is to reduce the support of $$X_n$$ to be less than $$[-\varepsilon, \varepsilon]$$ as $$n \to \infty$$, so that $$P(|X_n| > \varepsilon) = 0$$ eventually. More precisely, each $$X_n$$ has a probability density function defined by $$f_{X_n} = \frac{2n+1}{2a_n^{2n+1}}x^{2n} \quad (x \in [-a_n,a_n]),$$ where $$a_n = \sqrt{1+\frac{2}{2n+1}}$$, and zero otherwise (this formulation ensures that $$E[X_n] = 0$$ and $$\operatorname{Var}(X_n) = 1$$ for all $$n$$). Unfortunately, $$a_n$$ converges to $$1$$ in this case, and $$P(|X_n| > \varepsilon) \neq 0$$ for $$\varepsilon < 1$$.

After these attempts, I began to doubt the validity of this proposition. After all, wouldn't a random variable with variance of $$1$$ surely "spread over" to the value of $$1$$ if it is centered at $$0$$? This is only my intuition and I have yet to prove this. But if that's the case, then $$P(|X| > \varepsilon)$$ surely can't approach $$0$$ for $$0 < \varepsilon < 1$$.

Any help on proving or disproving the statement in my question is appreciated.

• The flaw in your intuition is that you can force the mean to be zero and the variance to be $1$ by having a very small probability assigned to a very large value of one sign and then assigning the rest of the probability to a small value of the other sign. The variance "sees" the largeness of the value in a way that $P(|X|>\varepsilon)$ doesn't. (This is related to the fact that convergence in probability doesn't imply convergence in $L^p$.)
– Ian
Commented Feb 13, 2021 at 17:15
• @Ian Alright, thank you for the clarification. Indeed, I overlooked that possibility. Commented Feb 14, 2021 at 5:34

You can use a random variable $$X$$ that only takes two possible values (one with large and one with small absolute value):
Let $$a>0$$ and set $$p=\frac{1}{1+a^2} \in (0,1)$$. Now define a random variable $$X$$ such that \begin{align*}P(X=a) &= p \\ P(X=-\frac{1}{a}) &= 1-p \end{align*}
Then $$E[X] = ap+(-\frac{1}{a})(1-p) = \frac{a}{1+a^2} - \frac{1}{a}\cdot\frac{a^2}{1+a^2} = 0$$ and $$\operatorname{Var}(X) = a^2 p + (-\frac{1}{a})^2 (1-p) = \frac{a^2}{1+a^2} + \frac{1}{a^2}\cdot\frac{a^2}{1+a^2} = 1$$ so $$X$$ has zero mean and unit variance.
Now let $$a \to \infty$$ so that $$p \to 0$$ and $$P(|X|>\varepsilon) \to 0$$ for any fixed $$\varepsilon > 0$$.
Note that you cannot have $$P(|X|>\varepsilon) = 0$$ exactly, when $$\varepsilon < 1$$, because if $$X$$ only takes values in $$[-\varepsilon, \varepsilon]$$ and $$E[X] = 0$$, then $$\operatorname{Var}(X) = E[X^2] \leq \varepsilon^2 < 1$$ and you cannot have unit variance.