We define the set of quantiles as:
Let $F$ be a distribution function for a probability $P$ on $(\mathbb{R},\mathcal{B}(\mathbb{B}))$. Let $0<p<1$. Then
$$Q(p)=\{x\in \mathbb{R} | F(x-)\leq p \leq F(x)\}$$
to be a $p$-quantile of $P$ (or $F$) and $x\in Q$ is said to be a $p$-quantile.
Let $F$ be a distribution for a probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{B}))$. Show that $Q(p)\neq\emptyset$
I am quite new to measure theory and its applications on probability theory. I know that intuitively this result makes sense since $0<p<1$ and from my (very basic) understanding the quantile is basically quantiles are cut points dividing the range of a probability distribution into continuous intervals with equal probabilities.
However, showing this is far beyond my capabilities. How would this be proven?
Best Answer
Sketch Look at $T=\{t | F(t) \geq p\}$. Observe that this is a closed interval of the form $[q,\infty)$ where $q$ is the inf over the aforementioned set.
You might question the inclusion of $q$ in the interval. Look at $F(q +{1/n})$ and recall that $F$ is right cts to show that $q$ is indeed included.
Now show that $q$ is the element you seek. Hint: take $(x_n)_{n \in \mathbb{N}}$ with $x_n\uparrow q$ (like for instance $x_n=q-\frac{1}{n}$) and look at $F(x_n)$ as $n \rightarrow \infty$.
Proof: Let $T=\{t | F(t) \geq p\}$. If $t \in T$ then $t'>t \implies t' \in T$ since $F$ is non-decreasing. Since $F(t)\rightarrow 0$ as $t\rightarrow -\infty$ there is some $r \in \mathbb{R}$ such that $F(r) \leq p\in (0,1)$. So $\inf{T}>-\infty$. Set $q=\inf{T}$. Then by the greatest lower bound property: $q+1/n \in T$ for every $n \in \mathbb{N}$. Thus we have $p \leq F(q+1/n)$ for every $n \in \mathbb{N}$. If it is true for every $n$, then it remains true in the limit. That is (here we use right-cts);
$$p\leq F(q)$$
We want to show $F(q^-)\leq p$. Because $q$ is a lower bound for $T$, $q-1/m \notin T$ for any $m$. So $F(q-1/m)\leq p$ for every $m$. Now $F(q-1/m)\rightarrow F(q^-)$ as $m \rightarrow \infty$. So (as before) $$F(q^-)\leq p$$ Putting the two inequalities togther yields; $$F(q^-)\leq p \leq F(q)$$ so that $q \in Q(p)$ and we are done.