6
$\begingroup$

Let $\mathcal{X}$ be some set of independent random variables. Define the ordering on $\mathcal{X}$ by $X_i \prec X_j$ if and only if $\mathcal{P}\left\{X_i \le X_j\right\} \ge \frac{1}{2}$. Are there known weak conditions on $\mathcal{X}$ such that this ordering is a total ordering? A partial ordering? (I am guessing there are multiple answers to this question.)

The counterexample of non-transitive dice illustrates that not all $\mathcal{X}$ have partial orderings, much less total orderings. However, if, for example, $\mathcal{X}$ consisted of Gaussian random variables, a total ordering is established by comparing the means.

${}\qquad{}$

$\endgroup$
2
  • $\begingroup$ "partially ordered" should be replaced by quasi ordered or pre-ordered. $\endgroup$ Commented Mar 25, 2015 at 8:01
  • $\begingroup$ @thomas right, I thought of that too late; the ordering does not form a poset if it is not transitive! editing. $\endgroup$
    – Steven Pav
    Commented Mar 25, 2015 at 16:32

1 Answer 1

2
$\begingroup$

The random variable $X$ stochastically dominates the random variable $Y$, written $X\succeq Y$, if $\Pr(X\ge y)\ge\Pr(Y\ge y)$ for all $y$. This relation is transitive.

Let's write $X\gtrsim Y$ if $\Pr(X\ge Y)\ge 1/2$. This relation is not transitive as the example of intransitive dice shows. However it is "total" in the sense that $X\lesssim Y$ or $Y\lesssim X$ holds.


Theorem. $X\succeq Y$ implies $X\gtrsim Y$.

Proof: In the notation of continuous random variables with pdfs $f_X$, $f_Y$, \begin{align} \Pr(X\ge Y)&=\int_{-\infty}^\infty \int_y^\infty f_X(x)f_Y(y)\,dx\, dy =\int_{-\infty}^\infty \Pr(X\ge y) \,f_Y(y)\, dy\\ &\ge\int_{-\infty}^\infty \Pr(Y\ge y) \,f_Y(y)\, dy=\int_{-\infty}^\infty \int_y^\infty f_Y(x)f_Y(y)\,dx\, dy= \frac12. \end{align}


By the Theorem, the intransitive dice phenomenon cannot occur when stochastic domination totally orders the elements of $\mathcal X$.

$\endgroup$
3
  • $\begingroup$ I think you need a $\le$ at the split of your equations. $\endgroup$
    – Steven Pav
    Commented Mar 25, 2015 at 16:50
  • $\begingroup$ This kind of kicks the can down the road: the question becomes, "are there sufficient conditions for stochastic dominance on $\mathcal{X}$ to be a total ordering?" $\endgroup$
    – Steven Pav
    Commented Mar 25, 2015 at 19:01
  • $\begingroup$ Yes, although there may be other sufficient conditions @StevenPav $\endgroup$ Commented Mar 25, 2015 at 20:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .