1
$\begingroup$

Let $X_1, \cdots, X_n$ be $n$ random variables (not necessarily independent) such that $E[X_i] > E[X_j]$ whenever $i < j$. I am interested in obtaining lower bounds on the following probability: $P[ (X_1 > X_2) \wedge (X_1 > X_3) \wedge \cdots \wedge (X_1 > X_n) ]$.

$\endgroup$
3
  • $\begingroup$ Identically distributed? $\endgroup$
    – Emre
    Commented Jun 23, 2011 at 19:12
  • $\begingroup$ @Emre, no they are not identically distributed $\endgroup$ Commented Jun 23, 2011 at 19:13
  • 3
    $\begingroup$ Note that "lowerbounding" is not a word. There is a single word for this -- minorizing -- but it is now somewhat obscure and old-fashioned. Good mathematical prose does not call attention to itself, so I have suggested an unobtrusive alternative. $\endgroup$ Commented Jun 24, 2011 at 1:19

1 Answer 1

2
$\begingroup$

The lower bound is $0$.

For example let $X_i = -(i^{k-1})$ with probability $\frac{i^2-1}{i^k-1}$ for some $k>2$, and $X_i=-\frac{1}{i}$ otherwise. Then $E[X_i]=-i$, satisfying the condition.

If $i <j$ then both possible values of $X_i$ are between the two possible values of $X_j$. In particular, $X_1 = -1$ with probability 1. Meanwhile $\frac{i^2-1}{i^k-1}$ is a decreasing function of $k$, so as $k$ increases the more negative value of $X_i$ becomes less likely.

So $\Pr(X_1 > X_2)$ and the longer expression can be made arbitrarily small by increasing $k$.

$\endgroup$

You must log in to answer this question.

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