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Let $X$ and $Y$ be two independent, continuous random variables with cumulative distribution functions $F_X$ and $F_Y$. Suppose that $X$ and $Y$ are not identically distributed, i.e. $F_X \ne F_Y$, and that

$$\mathbb{P}[X<Y] = \mathbb{P}[X>Y] = 1/2. \qquad (*)$$

It seems that a necessary condition for $(*)$ is that $X$ and $Y$ have the same medians, i.e. there exists some $m \in \mathbb{R}$ such that $F_X(m) = F_Y(m) = 1/2$. Must $X$ and $Y$ be symmetric around $m$?

jII
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  • Are you interested only in distributions which density is nonzero over the entire $\mathbb{R}$? Or will you take those with zero density outside a bounded interval? – Lee David Chung Lin Jan 03 '19 at 17:36
  • The distributions of $X$ and $Y$ can be arbitrarily defined i.e. they need not have full support. If the median is not unique then the question intends to ask whether there exists some median $m$ around which $X$ and $Y$ are symmetric. – jII Jan 03 '19 at 18:12

1 Answers1

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Here's a counter-example where both $X$ and $Y$ are supported only in $[0,1]$. The medians are different, and obviously neither $X$ nor $Y$ is symmetric with respect to their own median.

\begin{align} F_X(t) &= t^2 \\ F_Y(t) &= 1 - (1 - t)^b \quad \text{where} \quad b = \frac{-3 + \sqrt{17}}2 \approx 0.561553 \end{align} That is, $X$ has a linear density $f_X(t) = 2t$ and $Y$ has a power law density flipped $f_Y(t) = (1-t)^b$ where the power $b$ shall by design render $$\mathbb{P}[X > Y] = \int_{x = 0}^1 f_X(x) \cdot \Pr\{Y < x\} \,\mathrm{d}x = \frac12 $$ Or, as $\Pr\{Y < x\} = F_Y(x)$, more explicitly \begin{align} \mathbb{P}[X > Y] &= \int_{x = 0}^1 2x\left( 1 - (1-x)^b\right) \,\mathrm{d}x \\ &= 1-\frac2{b^2 + 3b + 2} \\ \end{align} with $b$ being a solution to $b^2 + 3b + 2 = 4$ (the root that is larger than $-1$; for $L_1$ integrability) such that $\mathbb{P}[X > Y] = 1-\frac24 = \frac12$ as desired.

The medians are numerically close but emphatically not identical. \begin{align} m_X &= \frac1{ \sqrt{2} } \approx 0.707107 \\ m_Y &= 1 - \frac1{ 2^{1/b} } \approx 0.708973 \end{align}

One can systematically generate a family of counter-examples of this combination: $f_X \propto t^a$ versus $f_Y \propto (1-t)^b$, with the two densities being power-laws "opposing each other".

Lee David Chung Lin
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