Does a median always exist for a random variable?

Does a median always exist for a random variable?

Note that a median of a random variable $X$ is defined as a number $m \in \mathbb{R}$ such that $P(X \leq m) \geq \frac{1}{2}$ and $P(X \geq m) \geq \frac{1}{2}$.

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Solution 1:

I will assume that $X$ is finite w.p. $1$. Then we claim that $m := \sup \ \{ a \mid P(X \geqslant a) \geqslant \frac12 \}$ is always a median. We just need to verify the two defining properties:

• For $a > m$, from the definition of $m$, the tail probability $P(X \geqslant a)$ is strictly less than $\frac12$. Now since $\lim \limits_{a \to m+} P(X \geqslant a)$ exists and equals $P(X \gt m)$, it follows that $P(X \gt m) \leqslant \frac12$. That is, $P(X \leqslant m) \geqslant \frac12$.

• For the other direction, we reverse the above argument. For any $a \lt m$, we have $P(X \geqslant a) \geqslant \frac12$. Now since $\lim \limits_{a \to m-} P(X \geqslant a) = P(X \geqslant m)$, it follows that $P(X \geqslant m) \geqslant \frac12$.

Remarks.

• The median that we defined above is the largest median of $X$. Correspondingly, the smallest median is given by $\inf \ \{ a \mid P(X \leqslant a) \geqslant \frac12 \}$.

• The set of medians of $X$ is a convex set, i.e., interval in $\mathbb R$. Moreover it is also compact (i.e., closed and bounded), because its infimum and supremum are finite and are contained in the set.