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Article #168Inequality question
From: B. Strohmeier
Date: Thu, 17 Dec 1992 10:03
Date: Thu, 17 Dec 1992 10:03
21 lines
1468 bytes
1468 bytes
Can the following theorem be easily proofed?
Theorem:
Let
1. p_1\geq p_2\geq\ldots\geq p_n and q_1\geq q_2\geq\ldots\geq q_n
with 0\leq p_i\leq 1 and 0\leq q_i\leq 1 \forall 1\leq i\leq n
2. \sum_{i=1}^np_i=\sum_{i=1}^nq_i=1
(imagine p_i and q_i as probabilities in decreasing order)
3. \sum_{i=1}^kp_i\geq\sum_{i=1}^kq_i \forall 1\leq k\leq n
(the cumulated sum of p_i is always greater or equal than the
cumulated sum of q_i)
Then
\sum_{i=1}^np_i(1-q_i)^\alpha\leq\sum_{i=1}^nq_i(1-p_i)^\alpha
\forall real \alpha\geq 1
If this is true, it follows that a weaker expert can better predict a
stronger expert than vice versa.
Thanks in advance.
Bernhard Strohmeier
bstrohm@erasmus.hrz.uni-bielefeld.de
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Newsgroups: sci.math.research Path: arc1-ca.pugleaf.net!sparky!uunet!zaphod.mps.ohio-state.edu!sdd.hp.com!ux1.cso.uiuc.edu!news.cso.uiuc.edu!dan From: Berni@nw42.wiwi.uni-bielefeld.de (B. Strohmeier) Subject: Inequality question Nntp-Posting-Host: wiwipc07.uni-bielefeld.de Message-ID: <Berni.6.724586581@nw42.wiwi.uni-bielefeld.de> Originator: dan@symcom.math.uiuc.edu Sender: Daniel Grayson <dan@math.uiuc.edu> X-Submissions-To: sci-math-research@uiuc.edu Organization: Fakultaet fuer Wirtschaftswissenschaften, Uni Bielefeld, Germany X-Administrivia-To: sci-math-research-request@uiuc.edu Approved: Daniel Grayson <dan@math.uiuc.edu> Date: Thu, 17 Dec 1992 10:03:01 GMT Lines: 23