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.

  • 1
    $\begingroup$ 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 $L^p$.) $\endgroup$
    – Ian
    Commented Feb 13, 2021 at 17:15
  • $\begingroup$ @Ian Alright, thank you for the clarification. Indeed, I overlooked that possibility. $\endgroup$ Commented Feb 14, 2021 at 5:34

1 Answer 1


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.

  • $\begingroup$ Ah, I see. Thank you for your answer, this is exactly what I was looking for. $\endgroup$ Commented Feb 14, 2021 at 5:32

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