I am given the problem,
For a distribution function $F$, we define $F^{-1}: [0,1] \to \mathbb{R}$, where $F^{-1}(x)=\inf \{y \in \mathbb{R}: F(y) \geq x\}$. Show $F^{-1}$ is right-continuous.
I know $F: \mathbb{R} \to [0,1]$ is a right-continuous, non-decreasing function s.t. $\lim_{x \to \infty}F(x)=1$ and $\lim_{x \to -\infty}F(x)=0$.
To show right-continuous, I want that $\lim_{y \downarrow x}F^{-1}(y)=F^{-1}(x).$
Best Answer
As defined $F^{-1}$ actually must be left-continuous.
Fix $a \in (0,1]$ and consider a sequence $x_n \uparrow a$. Since $F^{-1}$ is nondecreasing it follows that
$$y_n =F^{-1}(x_n) \uparrow y \leqslant F^{-1}(a)$$
For any $\epsilon > 0$, we have $F(y_n - \epsilon) < x_n \leqslant F(y_n + \epsilon) $.
Now suppose $y < F^{-1}(a)$. Taking $\epsilon = (F^{-1}(a)-y)/2$ we get
$$x_n \leqslant F(y_n+ \epsilon) \leqslant F(F^{-1}(a) + \epsilon) < a,$$ since $y_n\leq F^{-1}(a)$
which implies the contradiction $x_n \uparrow a < a$.