<p>This would mean that <span class="math-container">$\sup\{F(\theta):\,\underline\theta\le\theta\le\bar\theta\}&lt;1$</span>, which makes no sense as <span class="math-container">$F(\bar\theta)=1$</span>.</p>

CC BY-SA 4.0
2023-01-03T13:51:04.803Z
2023-01-03T13:51:04.803Z
VARulle
<blockquote>
<p>Does it mean that F can never take the value 1 in this interval or
that F never takes the value 1?</p>
</blockquote>
<p>In a sense yes, but it must be specified. Roughly speaking, it means that it is 'enough away' from <span class="math-container">$1$</span>, but this statement must be made rigorous.</p>
<p>Observe that a function <em>bounded away from 'something'</em> is not the same as a <em>bounded</em> function.</p>
<p>A function is <em>bounded</em>, for instance <em>bounded above</em>,  if there exists a <span class="math-container">$M\in \mathbb {R}$</span> such that <span class="math-container">$f(x)&lt;M$</span>, for all <span class="math-container">$x$</span> in its domain. But a function can be <em>bounded away from 'something'</em> and be unlimited elsewhere.</p>
<p>I quote the clearest answer to MathES questions I reported in the comments (an accepted answer and with <span class="math-container">$27$</span> upvotes), which can explain better than I can do the concept (1):</p>
<blockquote>
<p>If a set <span class="math-container">$S \subset \mathbb R$</span> is bounded away from zero, it means that there
exists <span class="math-container">$m &gt; 0$</span> such that <span class="math-container">$|x| &gt; m$</span> for all <span class="math-container">$x \in S$</span>.</p>
<p>If a function <span class="math-container">$f$</span> is bounded away from zero, it means that its range
is bounded away from zero: there exists <span class="math-container">$m &gt; 0$</span> such that <span class="math-container">$|f(x)| &gt; m$</span>
for all <span class="math-container">$x$</span>.</p>
<p><strong>Edited to clarify:</strong> 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 <em>its distance from zero is bounded below
by a strictly positive number</em>. I see now that this is not
self-evident, but that's what it means.</p>
</blockquote>
<p>Of course, what @Varulle says (and Ishan Kashyap Hazarika
in the comments) is true, if <span class="math-container">$F$</span> is a CDF, it should be  <span class="math-container">$F(\bar {\theta})=1$</span> (and a CDF is always bounded).</p>
<p>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 <span class="math-container">$F$</span>? Why? Or a mistake? It is impossible to say anything without reading the original paper.</p>
<hr />
<p>(1) <a href="https://math.stackexchange.com/questions/1340637/what-does-bounded-away-from-zero-actually-mean">https://math.stackexchange.com/questions/1340637/what-does-bounded-away-from-zero-actually-mean</a></p>

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2023-02-10T13:35:49.320Z
Your answers on this site are very helpful @BakerStreet!
CormJack
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2023-02-10T13:38:25.593Z
Thank you very much @CormJack, you are very kind.
BakerStreet
CC BY-SA 4.0
2023-01-04T08:27:44.607Z
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.
Pc1
CC BY-SA 4.0
2023-01-04T08:49:59.300Z
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 $\bar\theta$.
Pc1
CC BY-SA 4.0
2023-01-03T14:53:25.980Z
2023-01-03T15:00:30.550Z
2023-01-03T15:00:30.550Z
BakerStreet
<p>Suppose <span class="math-container">$\theta \in [\underline\theta, \bar\theta]$</span> 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?</p>

CC BY-SA 4.0
2023-01-03T10:41:06.503Z
The expression 'bounded away from' has a specific meaning in mathematics. For a definition you can look at these questions on Math SE https://math.stackexchange.com/questions/1340637/what-does-bounded-away-from-zero-actually-mean

https://math.stackexchange.com/questions/1641798/sequence-bounded-away-from-0-and-2
BakerStreet
CC BY-SA 4.0
2023-01-03T11:50:29.607Z
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?
Ishan Kashyap Hazarika
<mathematical-economics><probability>
What does it mean when I say that CDF is bounded away from 1?
Pc1
# What does it mean when I say that CDF is bounded away from 1?

(Asked by: username_0 on 2023-01-03)

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?

**Tags:** mathematical-economics, probability

> **Comments:**
> **username_1**: The expression 'bounded away from' has a specific meaning in mathematics. For a definition you can look at these questions on Math SE https://math.stackexchange.com/questions/1340637/what-does-bounded-away-from-zero-actually-mean  https://math.stackexchange.com/questions/1641798/sequence-bounded-away-from-0-and-2  
> **username_2**: 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?  

---

## 2 answers

### Answer by username_1 on 2023-01-03. Score: 5 | Accepted answer

> 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), 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 username_2
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>

> **Comments:**
> **username_0**: 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.  
> **username_0**: 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 $\bar\theta$.  
> **username_4**: Your answers on this site are very helpful @username_1!  
> **username_1**: Thank you very much @username_4, you are very kind.  

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### Answer by username_3 on 2023-01-03. Score: 6

This would mean that $\sup\{F(\theta):\,\underline\theta\le\theta\le\bar\theta\}<1$, which makes no sense as $F(\bar\theta)=1$.


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