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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 $P(X \geq m) \geq \frac{1}{2}$.

Thanks and regards!

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What would need to hold for a median not to exist? –  Henning Makholm Nov 26 '11 at 3:14
    
It always exists but sometimes it might have more than one value. –  geraldgreen Nov 26 '11 at 3:23
    
@HenningMakholm: by its definition, that is the best I can say now. –  Tim Nov 26 '11 at 3:25
    
@John.Mathew: Why does it always exist? –  Tim Nov 26 '11 at 3:26

1 Answer 1

up vote 3 down vote accepted

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.

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Thanks! Yes, I think I need more information. Also is the set of medians closed? The one you gave looks like the max median, does it? –  Tim Nov 26 '11 at 3:50
    
Thanks! So in $\mathbb{R}$, is interval defined as a convex subset? –  Tim Nov 26 '11 at 4:07
    
No, it's not defined that way, but in $\mathbb R$, these two notions coincide. An interval is defined on a universe with total or partial order; a set $I$ is an interval if whenever $x, z \in I$, $\{ y \mid x < y < z \} \subseteq I$. OTOH a convex set is defined in a (say) vector space. A set $C$ is convex if whenever $x,y \in C$, we have $\{ t x + (1-t)y \mid 0 \leqslant t \leqslant 1 \} \subseteq C$. –  Srivatsan Nov 26 '11 at 4:12

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