Median of a truncated distribution

Question

I found $$F^{-1}\left(\frac{F(a)+F(b)}{2}\right)$$ as median of a truncated distribution on Wikipedia. How do you derive this statement for any truncated distribution? The references don't show and it is somewhat counterintuitive.

Answer 1

Suppose that $X$ has absolutely continuous distribution with pdf $g(x)$ and $a<b$ are those that $F(b)-F(a)>0$ where $F$ is the cdf of $X$. Then $$ \mathbb P(X\leq x\mid a<X\leq b) = \int\limits_a^x \frac{g(x)}{F(b)-F(a)}\,dx =\frac{F(x)-F(a)}{F(b)-F(a)}=\frac12 $$ implies $$ F(x) = F(a)+\frac{F(b)-F(a)}{2}=\frac{F(b)+F(a)}2. $$ A solution of this equation is a median.