Let $X$ and $Y$ be two independent non-negative random variables such that $E[X] = E[Y] = \mu$ but $\operatorname{Var}[X] < \operatorname{Var}[Y]$. Can we use only these two moments to show that for $z>\mu$,

$$ \operatorname{Pr}(X > z) < \operatorname{Pr}(Y > z). $$

If not, can we show a counterexample or weaker claim?


1 Answer 1


It's false. Just let $X$ have a tiny mass at a big number and have moderate value otherwise. Then you can construct $Y$ with the same mean and bigger variance but which is always smaller than the "big" number.

Here is a more explicit example:

Suppose the r.v. $X$ takes value $5$ with probability $p$ and value $10$ otherwise.

The mean $\mu:=\mathbb{E}[X] = 5p+10(1-p)$ can be made arbitrarily close to $5$ by taking $p$ close to $1$. Also, the variance is $Var[X] = (5(1-p))^2 p + (5p)^2(1-p)$ can be made arbitrarily close to $0$. Choose $p<1$ large enough that the mean $\mu < 6$ and the variance less than $3$. Note that $\mathbb{P}(X>9) = 1-p > 0$.

Now suppose $Y$ is Uniform$[\mu-3,\mu+3]$. Then, $\mathbb{E}[Y] = \mu$, $Var[Y] = 3 > Var[X]$ and $\mathbb{P}(Y>9)=0 < \mathbb{P}(X>9)$.


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