Median of symmetric distribution, where median is defined using inverse of CDF

Question

Let $X$ be a symmetric random variable, that is $X$ and $-X$ have the same distribution function $F$. Suppose that $F$ is continuous and strictly increasing in a neighborhood of $0$. Then prove that the median $m$ of $F$ is equal to $0$, where we define $m:=\inf\{x\in \mathbb{R}|F(x)\ge \frac{1}{2}\}$.

This definition of the median kind of annoys me. I could easily show that $\mathbb{P}(X\le 0)\ge \frac{1}{2}$ and $\mathbb{P}(X\ge 0)\ge \frac{1}{2}$ and by the usual definition of the median I would be done, but I don't know how to deal with that $\inf$. I could only observe that my first equality implies that $m<0$. I think that the point of the qustion is to use that $F$ is invertible on that neighborhood, but I can't make any progress.

Answer 1

If you know how to prove that $P(X\le 0)=1/2$, then you can prove that $0$ satisfies the "inf" definition of median by contradiction. If you assume that $\textrm{inf}\, \{x, F(x)\ge 1/2\}>0$ then, by definition of infimum, there will be some $x_0>0$ such that $F(x_0)=P(X\le x_0)<1/2$. But then $P(X\le 0)\le P(X\le x_0)<1/2$, contradicting the fact that $P(X\le 0)=1/2$. A similar contradiction arises if you assume that the infimum is negative. So it has to be zero.

Answer 2

Define $M:=\{x\in {\mathbb R}\mid F(x)\ge\frac12\}$. You've shown that $0\in M$. It remains to show that no number less than zero is a member of $M$. To do this:

0. You've already shown $P(X\le0)\ge\frac12$.

1. Since $F$ is continuous at $x=0$, you know $P(X=0)=0$; otherwise there would be a jump discontinuity in $F$ at $x=0$.

2. Since $F$ is strictly increasing in a neighborhood of $0$, there exists $\delta>0$ such that $F(t_1)<F(t_2)$ whenever $-\delta<t_1<t_2<\delta$.

3. If $t\in(-\delta, 0)$ then $-\delta<t<0<\delta$ so $$F(t)\stackrel{(2)}<F(0)=P(X\le 0)\stackrel{(1)}=P(X<0)=1-P(X\ge0)\stackrel{(0)}\le \frac12.$$ Hence $F(t)<\frac12$ and $t$ cannot be a member of $M$.

4. If $t\in (-\infty, -\delta]$, then $t\le -\delta<-\delta/2<0$ so $F(t)\le F(-\delta/2)\stackrel{(3)}<\frac12$ and again $t\not\in M$.

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Note that you need to use the condition that $F$ is strictly increasing in a neighborhood of zero. If $F$ is flat in a neighborhood of zero, then $\inf M$ will be negative.