<p>It is not exactly clear what you mean by &quot;distribution functions that belong to the same class.&quot;
Consider a random variable <span class="math-container">$\widehat X$</span> with an arbitrary distribution on some support <span class="math-container">$[\underline x,\overline x]$</span>. We can then normalize the support to <span class="math-container">$[0,1]$</span> by looking at <span class="math-container">$X=\frac{\widehat X-\underline x}{\overline x-\underline x}$</span> and let <span class="math-container">$F$</span> be the corresponding cdf. Do you want to see if the conditional random variable <span class="math-container">$X\geq b$</span> FOSD <span class="math-container">$X$</span>? The answer is yes.</p>
<p>We can express the cdf of <span class="math-container">$X\geq b$</span> as <span class="math-container">$\frac{F(x)-F(b)}{1-F(b)}$</span> for all <span class="math-container">$x \in [b,1]$</span>.</p>
<p>Then we have,
<span class="math-container">$$F(x) \geq \frac{F(x)-F(b)}{1-F(b)} \quad \forall x \in [b,1] \mbox{ and } \forall b \in (0,1),$$</span>
and, moreover, we have
<span class="math-container">$$\frac{F(x)-F(b')}{1-F(b')} \geq \frac{F(x)-F(b)}{1-F(b)} \quad \forall x \in [b,1] \mbox{ if } b&gt;b' .$$</span></p>
<p>Hence, the distribution (or random variable) with some <span class="math-container">$b$</span> FOSD the distribution with a lower <span class="math-container">$b'&lt; b$</span>.</p>

CC BY-SA 4.0
2021-07-14T17:54:28.143Z
By "same class", I actually meant the same distribution but different parameter values. Sorry for the confusion.
Adam
CC BY-SA 4.0
2021-07-14T19:39:03.130Z
But what does that mean? In my answer it means comparing X and X>b. But it could also mean comparing  X and X+b for a constant b.
Bayesian
CC BY-SA 4.0
2021-07-14T20:38:58.967Z
Consider a uniform distribution on [b, 2], but each different value of b gives us a different (admittedly trivial) distribution.  Can we say their FOSD increases with b? If not, what concept captures that property? That was the original question.
Adam
CC BY-SA 4.0
2021-07-14T22:16:15.010Z
Yes, both answers here confirm this for the uniform distribution. I was wondering about different distributions. For instance, take CDF $x^2$ on [0,1]. What would be the "same" distribution on [b,1]?
Bayesian
CC BY-SA 4.0
2021-07-14T23:59:08.067Z
perhaps x^2/b^2 on [0,b]?
Adam
CC BY-SA 4.0
2021-07-15T08:34:41.550Z
In that case, $(x^2-b^2)/(1-b^2)$ FOSD $x^2$ FOSD $x^2/b^2$. That is $X>b$ FOSD $X$ FOSD $X<b$. So you are indeed interested in conditinal distributions?
Bayesian
CC BY-SA 4.0
2021-07-15T11:31:17.593Z
No,  I am really looking for a way to concisely describe how a distribution varies with a parameter value. In the two examples mentioned above (uniform and x^2/b^2 on [0,b]), how would you describe how the distribution varies with b?
Adam
CC BY-SA 4.0
2021-07-15T12:50:16.157Z
If X is distributed with $x^2$ on $[0,1]$, then the distribution of $X<b$ is $x^2/b^2$. That is the connection. Obviously, $X$ FOSD $X<b$, and $X<b$ FOSD $X<b'$ for $b>b'$.
Bayesian
CC BY-SA 4.0
2021-07-14T09:08:08.370Z
2021-07-14T09:08:08.370Z
Bayesian
<p>Let <span class="math-container">$X_b$</span> a random variable uniformly distributed on <span class="math-container">$[b,2]$</span> with <span class="math-container">$b&lt;2$</span>. For every <span class="math-container">$x\in\mathbb{R}$</span>,
<span class="math-container">$$\mathbb{P}[X_b\geq x]=\begin{cases}
0\text{ if }x&gt;2\\
\frac{2-x}{2-b}\text{ if } b\leq x\leq 2\\
1\text{ if }x&lt;b.
\end{cases}
$$</span>
Clearly, this probability is increasing in <span class="math-container">$b$</span>. So <span class="math-container">$b&gt;b'$</span> implies that <span class="math-container">$X_b$</span> first order stochastically dominates <span class="math-container">$X_{b'}$</span> for <span class="math-container">$b$</span> and <span class="math-container">$b'$</span> both smaller than <span class="math-container">$2$</span>.</p>

CC BY-SA 4.0
2021-07-14T00:05:26.943Z
Thanks, but can we say their FOSD increases with b? If not, what concept captures that property?
Adam
CC BY-SA 4.0
2021-07-14T03:39:03.380Z
@Adam So basically you are looking for one word that says "$b>b'$ implies that $X_b$ first order stochastically dominates $X_{b'}$ for $b$ and $b'$ both smaller than $2$"? Is there something wrong with saying with a sentence?
Giskard
CC BY-SA 4.0
2021-07-14T06:46:55.693Z
@Adam It is saying exactly that.
Michael Greinecker
CC BY-SA 4.0
2021-07-14T17:49:40.133Z
@Giskard There is nothing wrong with saying such a sentence, but it is not concise/intuitive. To use an analogy, if I could say "f(x) increases with x", I would not say "... implies that if for all x and x' such that x<x' one has f(x)<f(x')" even though there is nothing wrong with the second statement.
Adam
CC BY-SA 4.0
2021-07-14T17:51:48.113Z
@MichaelGreinecker Thanks, but I am really looking for a way to concisely describe how a distribution varies with a parameter value.
Adam
CC BY-SA 4.0
2021-07-13T22:59:27.103Z
2021-07-13T22:59:27.103Z
Michael Greinecker
<p>The actual distributions I am dealing with are not uniform, but to
keep it simple, consider two uniform distributions, one on [1, 2] and
the other on [0,2]. Can we say that the first FOSDs the second?
Further, if we have a multitude of distributions (suppose uniform
still) on [b, 2], can we say their FOSD increases with b? If not, what concept captures that property?</p>

<definition>
Can we rank stochastic dominance of distribution functions that belong to the same class?
Adam
# Can we rank stochastic dominance of distribution functions that belong to the same class?

(Asked by: username_0 on 2021-07-13)

The actual distributions I am dealing with are not uniform, but to
keep it simple, consider two uniform distributions, one on [1, 2] and
the other on [0,2]. Can we say that the first FOSDs the second?
Further, if we have a multitude of distributions (suppose uniform
still) on [b, 2], can we say their FOSD increases with b? If not, what concept captures that property?

**Tags:** definition

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## 2 answers

### Answer by username_2 on 2021-07-13. Score: 5

Let $X\_b$ a random variable uniformly distributed on $[b,2]$ with $b<2$. For every $x\in\mathbb{R}$,
$$\mathbb{P}[X\_b\geq x]=\begin{cases}
0\text{ if }x>2\\
\frac{2-x}{2-b}\text{ if } b\leq x\leq 2\\
1\text{ if }x
Clearly, this probability is increasing in $b$. So $b>b'$ implies that $X\_b$ first order stochastically dominates $X\_{b'}$ for $b$ and $b'$ both smaller than $2$.

> **Comments:**
> **username_0**: Thanks, but can we say their FOSD increases with b? If not, what concept captures that property?  
> **username_3**: @Adam So basically you are looking for one word that says "$b>b'$ implies that $X_b$ first order stochastically dominates $X_{b'}$ for $b$ and $b'$ both smaller than $2$"? Is there something wrong with saying with a sentence?  
> **username_2**: @Adam It is saying exactly that.  
> **username_0**: @Giskard There is nothing wrong with saying such a sentence, but it is not concise/intuitive. To use an analogy, if I could say "f(x) increases with x", I would not say "... implies that if for all x and x' such that x<x' one has f(x)<f(x')" even though there is nothing wrong with the second statement.  
> **username_0**: @MichaelGreinecker Thanks, but I am really looking for a way to concisely describe how a distribution varies with a parameter value.  

---

### Answer by username_1 on 2021-07-14. Score: 4

It is not exactly clear what you mean by "distribution functions that belong to the same class."
Consider a random variable $\widehat X$ with an arbitrary distribution on some support $[\underline x,\overline x]$. We can then normalize the support to $[0,1]$ by looking at $X=\frac{\widehat X-\underline x}{\overline x-\underline x}$ and let $F$ be the corresponding cdf. Do you want to see if the conditional random variable $X\geq b$ FOSD $X$? The answer is yes.

We can express the cdf of $X\geq b$ as $\frac{F(x)-F(b)}{1-F(b)}$ for all $x \in [b,1]$.

Then we have,
$$F(x) \geq \frac{F(x)-F(b)}{1-F(b)} \quad \forall x \in [b,1] \mbox{ and } \forall b \in (0,1),$$
and, moreover, we have
$$\frac{F(x)-F(b')}{1-F(b')} \geq \frac{F(x)-F(b)}{1-F(b)} \quad \forall x \in [b,1] \mbox{ if } b>b' .$$

Hence, the distribution (or random variable) with some $b$ FOSD the distribution with a lower $b'< b$.

> **Comments:**
> **username_0**: By "same class", I actually meant the same distribution but different parameter values. Sorry for the confusion.  
> **username_1**: But what does that mean? In my answer it means comparing X and X>b. But it could also mean comparing  X and X+b for a constant b.  
> **username_0**: Consider a uniform distribution on [b, 2], but each different value of b gives us a different (admittedly trivial) distribution.  Can we say their FOSD increases with b? If not, what concept captures that property? That was the original question.  
> **username_1**: Yes, both answers here confirm this for the uniform distribution. I was wondering about different distributions. For instance, take CDF $x^2$ on [0,1]. What would be the "same" distribution on [b,1]?  
> **username_0**: perhaps x^2/b^2 on [0,b]?  
> **username_1**: In that case, $(x^2-b^2)/(1-b^2)$ FOSD $x^2$ FOSD $x^2/b^2$. That is $X>b$ FOSD $X$ FOSD $X<b$. So you are indeed interested in conditinal distributions?  
> **username_0**: No,  I am really looking for a way to concisely describe how a distribution varies with a parameter value. In the two examples mentioned above (uniform and x^2/b^2 on [0,b]), how would you describe how the distribution varies with b?  
> **username_1**: If X is distributed with $x^2$ on $[0,1]$, then the distribution of $X<b$ is $x^2/b^2$. That is the connection. Obviously, $X$ FOSD $X<b$, and $X<b$ FOSD $X<b'$ for $b>b'$.  

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