This is a problem I stumbled upon in my research. Consider $n$ Gaussian random variables $x_i \sim \mathcal{N} (\mu_i, \sigma_i^2)$, each with its own mean $\mu_i$ and variance $\sigma_i^2$. Can we say that \begin{equation} \text{Median} \left( \sum_{i=1}^n x_i^2 \right) \ge \sum_{i=1}^n \text{Median} \left( x_i^2 \right) \; ? \end{equation} If the $x_i$ were standard normal variables ($\mu_i = 0$ and $\sigma_i^2=1$) we could compute $\text{Median} (x_i^2) \simeq 0.4549$, and the sum $\sum_{i=1}^n x_i^2 $ would follow the $\chi^2$-distribution with $n$ degrees of freedom, which is reported to have median $\simeq n \left( 1 - \frac{2}{9n}\right)^3$. The above inequality would be \begin{equation} \left( 1 - \frac{2}{9n}\right)^3 \ge 0.4549 \end{equation} which is satisfied $\forall n \in \mathbb{N}$. The question is whether the inequality is true even for non standard normal variables. The fact that it holds for standard normal variables is related to the heavy-tailedness of the distribution of $x_i^2$, so I think it might hold also for general Gaussian variables.

  • $\begingroup$ One possible approach: the means of both sides are equal, but the skewness of the left side must be less than the skewnesses of any of the components of the right side. By relating smaller skewness (a standardized third moment) to a smaller relative difference between the mean and the median, the inequality would result. $\endgroup$
    – whuber
    Commented Jan 21, 2022 at 15:02
  • 1
    $\begingroup$ I think a first step would be to prove the statement for normal variables centered in zero, i.e. $\mu_i = 0$. $\endgroup$
    – MrRobot
    Commented Jan 21, 2022 at 16:14

1 Answer 1


Here is a simple proof, by comparison with appropriate symmetric variables.

For a normal variable $X$, let $m$ be the median of $X^2$. The graphs show $X\sim N(4,1)$.

Now we take a variable $Y$ which is a version of $X^2$ but symmetrized from right to left about $m$:

enter image description here

enter image description here

In formulas: $$ f_Y(y) = \begin{cases} f_{X^2}(y)\phantom{2m-\, }\ \text{ if }\, y>m\\ f_{X^2}(2m-y)\ \text{ if }\, y<m\\ \end{cases}$$ $$F_Y(y) = \begin{cases} \phantom{1-\, }F_{X^2}(y)\phantom{2m-\, }\ \text{ if }\, y>m\\ 1-F_{X^2}(2m-y)\ \text{ if }\, y<m\\ \end{cases} $$

Since the $Y_i$'s are symmetric with median $m_i$, the values near $(y_1, \ldots y_n)$ and $(2m_1-y_1, \ldots 2m_n-y_n)$ are equally probable and on opposite sides of $m_1+\cdots+m_n$. So $m_1+\cdots+m_n$ must be the median of their sum.

Also $X^2$ dominates $Y$, i.e. $F_{X^2}(y)\le F_Y(y)$, as can be seen in graphs like the above. Thus:

$$ \begin{align} \text{median}\left(\sum X_i^2\right) &\ge \text{median}\left(\sum Y_i\right)\ &\text{ (by the dominance of the }X\text{'s)}\phantom{\ \square}\\ &=\sum\text{median}(Y_i)\ &\text{ (by the symmetry of the }Y\text{'s)}\phantom{\ \square}\\ &=\sum\text{median}(X_i^2)\ &\text{ (by the construction of the }Y\text{'s)}\ \square \end{align} $$


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