Point between left and right limits of a CDF

Question

This is from the Chapter 15 text of Gourieroux and Monfort's Statistics and Econometric Models II:

Set Up: Suppose that there are 2 possible parameter values $\theta_0$ and $\theta_1$ from which there are 2 density functions $l_{\theta_0}(y)$ and $l_{\theta_1}(y)$ on the data. Define $$ F(k)=P_{\theta_0}\left(\frac{l_{\theta_1}(Y)}{l_{\theta_0}(Y)}\leq k\right). $$ Then, $F$ is right continuous, increasing, and $\lim_{k\to+\infty}F(k)=1$.

Question: If $F(k-)$ means $\lim_{x\uparrow k}F(x)$, why is it that: given any $\alpha\in(0,1)$, there exists a real number $k_0$ satisfying $F(k_0-)\leq\alpha\leq F(k_0)$?

This actually seems be a general property of Càdlàg functions. Very appreciative if someone could show me or point me to some proof. Thank you very much!

Answer 1

It doesn't hold in general, you need to use the facts that $\lim_{k \to \infty}F(k) = 1$ and $F(0) = 0$. It does hold in general for distribution functions however. The following is adapted from Billingsley's Probability and Measure:

Define the quantile function to be $$ \phi(u) = \inf\{x \mid u \leq F(x) \}, \ \ \ u \in (0,1). $$ Since $F$ is non-decreasing $\{x \mid u \leq F(x) \}$ is an interval stretching to $\infty$, and since $F$ is right-continuous the interval is closed on the left. So for $u \in (0,1)$ $$ \{x \mid u \leq F(x) \} = [\phi(u),\infty). $$ This means that $\phi(u) \leq x$ if and only if $u \leq F(x)$.

So for $\alpha \in (0,1)$, let $k_0 = \phi(\alpha)$. Then $\alpha \leq F(k_0)$. Let $x_n$ be a sequence satisfying $x_n \uparrow k_0$. By the fact that $F$ has limits on the left and is non-decreasing, $\lim_{x \to k_0}F(x) = F(k_0^-) \leq \alpha$

P.S. be cool to those North Koreans.

Answer 2

Because $F$ is increasing and tends to $1$, given any $\alpha \in (0,1)$ there exists an $a \in \mathbb{R}$ such that $F(a) \geq \alpha$. This shows that the set $A:= \{ x \in \mathbb{R} : F(x) \geq \alpha \} $ is nonempty. It is obviously bounded below because $\lim_{x \rightarrow -\infty}F(x)= 0$. So we can set $k_0:= \inf A.$

But then $$ F(x) \leq \alpha \quad \forall x < k_0. $$

Taking limits as $x \uparrow k_0$ in the above inequality implies that $F(k_0-) \leq \alpha$.

NB We have used all the properties of the distribution function $F$ in the above. So we can conclude that this is a general property of distribution functions.

The same proof as above does however supply the following result:

Let $F$ be any increasing cadlag function. Then we can define $\alpha_{-}:= \lim{x \rightarrow -\infty}F(x)$ and $\alpha_{+}:= \lim{x \rightarrow \infty}F(x)$, where we may have $\alpha_{\pm} = \pm \infty$ respectively. Then

$$ \forall \alpha \in (\alpha_{-}, \alpha_{+}) \quad \exists k_0 \in \mathbb{R} \quad \textrm{such that} \quad F(k_0-) \leq \alpha \leq F(k_0). $$