Suppose $\theta \in [\underline\theta, \bar\theta]$ is distributed with CDF F(.). What does it mean when I say that this F is bounded away from 1? Does it mean that F can never take the value 1 in this interval or that F never takes the value 1? What kind of probability distributions will act in this way?
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3$\begingroup$ The expression 'bounded away from' has a specific meaning in mathematics. For a definition you can look at these questions on Math SE math.stackexchange.com/questions/1340637/… math.stackexchange.com/questions/1641798/… $\endgroup$– BakerStreetJan 3 at 10:41
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1$\begingroup$ Where is it from? How is it possible for a CDF to be bounded away from 1? Isn't $\lim_{x \rightarrow \bar \theta} F(x) = 1$ by definition? Or am I missing something? $\endgroup$– Ishan Kashyap HazarikaJan 3 at 11:50
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2 Answers
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Does it mean that F can never take the value 1 in this interval or that F never takes the value 1?
In a sense yes, but it must be specified. Roughly speaking, it means that it is 'enough away' from $1$, but this statement must be made rigorous.
Observe that a function bounded away from 'something' is not the same as a bounded function.
A function is bounded, for instance bounded above, if there exists a $M\in \mathbb {R}$ such that $f(x)<M$, for all $x$ in its domain. But a function can be bounded away from 'something' and be unlimited elsewhere.
I quote the clearest answer to MathES questions I reported in the comments (an accepted answer and with $27$ upvotes), which can explain better than I can do the concept (1):
If a set $S \subset \mathbb R$ is bounded away from zero, it means that there exists $m > 0$ such that $|x| > m$ for all $x \in S$.
If a function $f$ is bounded away from zero, it means that its range is bounded away from zero: there exists $m > 0$ such that $|f(x)| > m$ for all $x$.
Edited to clarify: When we say a set is bounded away from zero, we are not saying that away from zero, it is bounded. What would that even mean? We are saying that its distance from zero is bounded below by a strictly positive number. I see now that this is not self-evident, but that's what it means.
Of course, what @Varulle says (and Ishan Kashyap Hazarika in the comments) is true, if $F$ is a CDF, it should be $F(\bar {\theta})=1$ (and a CDF is always bounded).
So, it is necessary to read the paper you quoted to understand what the author means in that context. Maybe they uses a restriction of $F$? Why? Or a mistake? It is impossible to say anything without reading the original paper.
(1) https://math.stackexchange.com/questions/1340637/what-does-bounded-away-from-zero-actually-mean
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1$\begingroup$ Thanks, I really appreciate the help! In the paper that I am reading, $\theta$ is a random variable that gets realized at t=2. Depending on a variable x, a particular value of $\theta$ is calculated at t=1, $\hat\theta(x)$ which is increasing in x. $F(\hat\theta(x))$ that is the probability that the realized $\theta \leq \hat\theta(x)$ is bounded away from 1. $\endgroup$– Pc1Jan 4 at 8:27
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1$\begingroup$ I think it means that $\hat\theta(x)$ never takes the value $\bar\theta$. So instead of writing that the CDF is bounded away from 1, it should have been $\hat\theta(x)$ is bounded away from $\bar\theta$. $\endgroup$– Pc1Jan 4 at 8:49
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$\begingroup$ Your answers on this site are very helpful @BakerStreet! $\endgroup$– CormJackFeb 10 at 13:35
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$\begingroup$ Thank you very much @CormJack, you are very kind. $\endgroup$ Feb 10 at 13:38
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This would mean that $\sup\{F(\theta):\,\underline\theta\le\theta\le\bar\theta\}<1$, which makes no sense as $F(\bar\theta)=1$.