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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?

Giskard
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Adam
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2 Answers2

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Let $X_b$ a random variable uniformly distributed on $[b,2]$ with $b<2$. For every $x\in\mathbb{R}$, $$\mathbb{P}[X_b\geq x]=\begin{cases} 0\text{ if }x>2\\ \frac{2-x}{2-b}\text{ if } b\leq x\leq 2\\ 1\text{ if }x<b. \end{cases} $$ 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 $2$.

Michael Greinecker
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  • Thanks, but can we say their FOSD increases with b? If not, what concept captures that property? – Adam Jul 14 '21 at 00:05
  • @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 Jul 14 '21 at 03:39
  • @Adam It is saying exactly that. – Michael Greinecker Jul 14 '21 at 06:46
  • @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 Jul 14 '21 at 17:49
  • @MichaelGreinecker Thanks, but I am really looking for a way to concisely describe how a distribution varies with a parameter value. – Adam Jul 14 '21 at 17:51
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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 $b'< b$.

Bayesian
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  • By "same class", I actually meant the same distribution but different parameter values. Sorry for the confusion. – Adam Jul 14 '21 at 17:54
  • 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 Jul 14 '21 at 19:39
  • 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 Jul 14 '21 at 20:38
  • 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 Jul 14 '21 at 22:16
  • perhaps x^2/b^2 on [0,b]? – Adam Jul 14 '21 at 23:59
  • 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 Jul 15 '21 at 08:34
  • 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 Jul 15 '21 at 11:31
  • 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 Jul 15 '21 at 12:50