17
$\begingroup$

Suppose that $a_1,\dots,a_n,b_1,\dots,b_n$ are iid random variables each with a symmetric non-atomic distribution. Let $p$ denote the probability that there is some real $t$ such that $t a_i \ge b_i$ for all $i$. It was shown that $$p=\frac{n+1}{2^n}.$$

Can this be proved by a combinatorial/symmetry argument?

Improve this question
$\endgroup$
3
  • $\begingroup$ What is your intuitive understanding as to why the answer is not simply $1/2^n$? $\endgroup$
    – thedude
    Commented Jan 31, 2022 at 20:35
  • $\begingroup$ @thedude : If you look at the existing proof of this result, you can see that $1/2^n$ is just the probability (say) that all $a_i$'s are $<0$, and for such $a_i$'s one can easily find a good $t$. But, of course, there are other possibilities for a good $t$ to exist. $\endgroup$
    – Iosif Pinelis
    Commented Jan 31, 2022 at 21:05
  • $\begingroup$ @thedude : Now there is a complete answer -- below -- to your question (which is also an answer to my question) $\endgroup$
    – Iosif Pinelis
    Commented Jan 31, 2022 at 21:31

1 Answer 1

Reset to default
25
$\begingroup$

If I understand correctly, $c_i := b_i/a_i$ should also be symmetric and non-atomic. Then the result holds if there exists $t$ so that for all $i$

  • $t \geq c_i$ if $a_i > 0$;
  • $c_i \geq t$ if $a_i < 0$.

Reorder the indices so that $|c_1| < |c_2| < \dots < |c_n|$. There are $2^n$ ways to assign signs to $a_i$ for each $i \in [n]$. The only assignments for which an appropriate $t$ exists are those where the signs of $a_1,\dots,a_k$ are positive and the signs of $a_{k+1},\dots,a_n$ are negative with $k \in \{0,1,2,\dots,n\}$. There are $n+1$ such assignments, so the desired probability is $(n+1)/2^n$.

Improve this answer
$\endgroup$
2

You must log in to answer this question.

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