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- 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
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- Approved: Daniel Grayson <dan@math.uiuc.edu>
- Date: Thu, 17 Dec 1992 10:03:01 GMT
- Lines: 23
-
- 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
-
-
-