Conditions for uniqueness of the median A median of a random variable is defined as any $m \in \mathbb{R}$ such that 
$P(X \le m) \ge 1/2$ and $P(X \ge m) \ge 1/2$. 
Alternatively, in terms of the CDF $F$ of $X$ defined by $F(x) := P(X \le x)$, we need 
$F(m) \ge 1/2$ and $F(m^-) \le 1/2$, where the latter is the left limit of $F$ at $m$.
Under what conditions is the median unique? It seems that either 


*

*there should be a point $m$ with $F(m) = 1/2$ such that the  CDF is continuous and inceasing in a neighborhood of $m$, 

*or there should be a point $m$ where $F(m-) < 1/2$ and $F(m) = 1/2$ and $F$ is increasing on $[m,m+\epsilon)$ for some small $\epsilon > 0$,

*or there should be a point $m$ where $F(m-) < 1/2$ and $F(m) > 1/2$.


Are these all the possibilities?
 A: $\lambda(F_X^{-1}(0.5))=0 \iff X\text { is unique }$, where $\lambda()$ is the lebesgue measure and $F^{-1}()$ is the inverse image of the CDF. This condition implies that $\lambda[ \{x:P(X\leq x)\geq 1/2\}\cap\{x: P(X\geq x)\geq 1/2\}]=0$
A: For absolutely continuous radom variables you always have a unique median. Indeed, let $F$ denote the distribution function of a continuous random variable $X$, and $f$ its density function. We have
$$F(x) = \int_{-\infty}^x f(y)dy.$$
The median is given by the value $x^\ast$ such that
$$F(x^\ast)=1/2.$$
Definte the function $G(x) := F(x) - 1/2$. We have (by the Properties of distribution functions)
$$\lim_{x\to -\infty} G(x) = -1/2, \, \mbox{and} \, \lim_{x\to \infty}G(x) = 1/2$$
hence by Bolzano's theorem (aka Intermediate value theorem) there is a value $x^\ast \in (-\infty,\infty)$ such that $G(x^\ast)=0$, hence $x^\ast$ is a median. It is unique since
$$G'(x) = F'(x) = f(x)\geq 0.$$
A: The following conditions guarantee the exsitence of a unique median
$1.$ CDF of the random variable $X$ is continuous.
$2.$ CDF of the random variable $X$ is strictly increasing.
Let $$a(m)=\int_{-\infty}^{m}f_X(x)\mathrm{d}x,\quad \mathcal{A}=\{m:a(m)\geq 1/2\}$$ and $$b(m)=\int_{m}^{\infty}f_X(x)\mathrm{d}x,\quad \mathcal{B}=\{m:b(m)\geq 1/2\}$$
If $a(m^*)=1/2$ for some $m^*$ then, $a(m)\geq 1/2$ for $m\geq m^*$ because $a$ is an increasing function.
Simlarly, if $b(m^{**})=1/2$ for some $m^{**}$ then, $b(m^{**})\geq 1/2$ for $m\leq m^{**}$ because $b$ is a decreasing function.
Hence, $$m^{**}\geq m\geq m^*\quad\quad (1)$$
We also know what $b(m)=1-a(m)\forall m$. Assume $m=m^{**}$. Then $$1/2=b(m^{**})=1-a(m^{**})=1-a(m^{*})$$ which implies $$a(m^{*})=b(m^{**})=1/2$$ Since both $a$ and $b$ are continuos. They intersect only at a single point at the co-domain of the CDF (even if the functions are non-increasing, non-decreasing type). If $a$ and $b$ are strictly increasing (decreasing resp.) functions, then the domain of intersection is also a single point, which is $m^*=m^{**}$. With this result Equation $1$ is also a single point and we are done.
A: I would have thought you need a point $m$ such that $$\left(\tfrac12-F(m-\delta)\right)\left(F(m+\delta)-\tfrac12\right)\gt 0 \text{ for all } \delta \gt 0$$
You could split this into either of  


*

*$F(m)=\tfrac12$ and $F(x)$ is strictly increasing at $x=m$

*$F(m) \gt \tfrac12$ and $F(x)\lt \tfrac12$ for all $x \lt m$


perhaps splitting (2) into  


*$F(m) \gt \tfrac12$ and $\displaystyle \lim_{x\nearrow \,m-} F(x)\lt \tfrac12$ 

*$F(m) \gt \tfrac12$ and $\displaystyle \lim_{x\nearrow \,m-} F(x)= \tfrac12$ and $F(x)\lt \tfrac12$ for all $x \lt m$

A: From the definitions
$$P(X \le m) \ge \frac12 \land P(X \ge m) \ge \frac12,$$
$$F(x) := P(X \le x),$$
we infer
$$F(m)\ge \frac12\land 1-F(m)+P(X=m)\ge\frac12,$$
hence
$$\frac12\le F(m)\le\frac12+P(X=m).$$
The solution set is $m\in F^{-1}([\frac12,\frac12+P(X=m)])$.
For continuous distributions, $F^{-1}(\frac12)$. $F$ must be injective for the image $\frac12$.
