Comments on: A measure-theoretic definition of synergy? http://stochastix.wordpress.com/2012/09/14/a-measure-theoretic-definition-of-synergy/ Sun, 07 Apr 2013 23:13:55 +0000 hourly 1 http://wordpress.com/ By: Rod Carvalho http://stochastix.wordpress.com/2012/09/14/a-measure-theoretic-definition-of-synergy/#comment-73208 Sat, 15 Sep 2012 14:23:48 +0000 http://stochastix.wordpress.com/?p=8643#comment-73208 I never claimed novelty! Thanks for the reference.

]]> By: s. sinha http://stochastix.wordpress.com/2012/09/14/a-measure-theoretic-definition-of-synergy/#comment-73207 Sat, 15 Sep 2012 14:20:25 +0000 http://stochastix.wordpress.com/?p=8643#comment-73207 The concept is introduced in the Theory of Games by John von Neumann and Oscar Morgenstern in 1944. How old is this abstract Measure Theoretic Approach of Synergy??

]]> By: vfp15 http://stochastix.wordpress.com/2012/09/14/a-measure-theoretic-definition-of-synergy/#comment-73206 Sat, 15 Sep 2012 06:27:28 +0000 http://stochastix.wordpress.com/?p=8643#comment-73206 Like I said, your initial exposition was fine. I just threw a wrench into the works :) and yes, I think your refinement resolves that problem.

]]> By: Rod Carvalho http://stochastix.wordpress.com/2012/09/14/a-measure-theoretic-definition-of-synergy/#comment-73204 Sat, 15 Sep 2012 01:29:01 +0000 http://stochastix.wordpress.com/?p=8643#comment-73204 Excellent point! I had not thought of that possibility.

To borrow some terminology from Ecology, I would say that the productivity function you chose is an example of parasitic synergy, whereas the productivity function I chose is an example of symbiotic synergy.

A measure \mu is superadditive if it satisfies the condition

\mu (X \cup Y) \geq \mu (X) + \mu (Y)

for all X, Y that are disjoints subsets of X. To ensure that a superadditive measure is also symbiotic, the following conditions would have to be satisfied

\frac{1}{|X \cup Y|} \mu (X \cup Y) \geq \frac{1}{|X|} \mu (X)

\frac{1}{|X \cup Y|} \mu (X \cup Y) \geq \frac{1}{|Y|} \mu (Y)

or, equivalently

\mu (X \cup Y) \geq \frac{|X \cup Y|}{|X|} \mu (X)

\mu (X \cup Y) \geq \frac{|X \cup Y|}{|Y|} \mu (Y)

for non-empty subsets X and Y. Since X and Y are disjoint sets, we conclude that |X \cup Y| = |X| + |Y| and, thus

\mu (X \cup Y) \geq \left( 1 + \frac{|Y|}{|X|} \right) \mu (X)

\mu (X \cup Y) \geq \left( 1 + \frac{|X|}{|Y|} \right) \mu (Y)

In words: the average measure of the union must be greater or equal than the average measure of the two disjoint non-empty subsets in the union. Do you agree with this refinement?

]]> By: vfp15 http://stochastix.wordpress.com/2012/09/14/a-measure-theoretic-definition-of-synergy/#comment-73199 Fri, 14 Sep 2012 13:52:45 +0000 http://stochastix.wordpress.com/?p=8643#comment-73199 OK, but what if the synergy results in Bob’s portion of the total being lower?

  1. \pi(\{\text{Alice}\}) = 3
  2. \pi(\{\text{Bob}\}) = 10
  3. \pi(\{\text{Alice},\text{Bob}\}) = 15

Working together improves Alice’s performance a lot, it decreases Bob’s performance, and it slightly improves the total performance.

This does not invalidate the earlier analysis, but it does add a dimension to the model. In such a setting, what would Bob´s incentive be?

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