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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5What would need to hold for a median not to exist? – hmakholm left over Monica Nov 26 '11 at 03:14
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1It always exists but sometimes it might have more than one value. – geraldgreen Nov 26 '11 at 03:23
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@HenningMakholm: by its definition, that is the best I can say now. – Tim Nov 26 '11 at 03:25
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@John.Mathew: Why does it always exist? – Tim Nov 26 '11 at 03:26
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If both conditions were not true you would have a contradiction since you know that either $X(\omega)\le m$ or $X(\omega) \ge m$ for any $m$. – copper.hat Sep 18 '18 at 20:55
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It might not exist if $X$ is a random variable with complex values or if $X$ takes values in ${\text{red}, \text{green}, \text{blue}}$. A median needs an order for your definition to be meaningful – Henry Feb 24 '20 at 08:37
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@copper.hat what does this argument bring a contradiction? – JacobsonRadical Oct 22 '23 at 21:38
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@JacobsonRadical The sum of probabilities of $X(\omega) \le m$ and $X(\omega) \ge m$ must add up to at least one. – copper.hat Oct 22 '23 at 21:52
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@copper.hat braindead for a second. Forgive me. – JacobsonRadical Oct 22 '23 at 22:36
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@JacobsonRadical I did not establish the existence of a median, in fact, now that I look at I am not sure what my comment was about, it may have been a response to a subsequently deleted comment or I may have had too much wine. The answer by Srivatsan establishes the existence. – copper.hat Oct 22 '23 at 22:37
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1@copper.hat agree. I deleted my previous reply which got denied by myself but now turns out actually correct. Thx for the discussion! – JacobsonRadical Oct 22 '23 at 22:40
1 Answers
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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.
Srivatsan
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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 03:50
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1No, 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 04:12
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