## A measure-theoretic definition of synergy?

Alice and Bob are fruit-pickers at an orange orchard.

Alice can pick 6 baskets of oranges in one hour. In contrast, Bob can pick 7 baskets of oranges in the same period of time. However, if Alice and Bob work together, then they can pick a total of 15 baskets in one hour. As part of a team:

• Alice is now picking 7.5 baskets per hour, a most impressive increase in productivity of 25%.
• Bob is now picking 7.5 baskets per hour as well, a modest increase in productivity of approximately 7%.

Both Alice and Bob benefit if they work together (though Alice benefits more). We thus have an example of synergy. To use a cliché: “the whole is greater than the sum of its parts”. If Alice and Bob work alone, they can only pick 13 baskets per hour in total, but if they work together they can pick 15 baskets per hour. Everyone is happy.

Let $S := \{\text{Alice}, \text{Bob}\}$, let $2^S$ denote the power set of $S$, and let $\mathbb{R}_0^{+}$ denote the set of nonnegative real numbers. We introduce a productivity function $\pi : 2^S \to \mathbb{R}_0^{+}$, enumeratively defined as follows

• $\pi (\emptyset) = 0$
• $\pi (\{\text{Alice}\}) = 6$
• $\pi (\{\text{Bob}\}) = 7$
• $\pi (\{\text{Alice}, \text{Bob}\}) = 15$

where $\pi (\emptyset) = 0$ because the productivity of the “empty team” is zero. Since we have that

$\pi (\{\text{Alice}, \text{Bob}\}) > \pi (\{\text{Alice}\}) + \pi (\{\text{Bob}\})$

we conclude that we have synergy. Note that the existence of a synergistic or synergetic effect is a property of the productivity function $\pi$. We could attempt to study such property in a more general setting.

__________

We now introduce a definition

Definition: Given a finite set $S := \{s_1, s_2, \dots, s_n\}$ and a function $\mu : 2^S \to \mathbb{R}_0^{+}$, if the following conditions are satisfied

• $\mu (\emptyset) = 0$
• $\mu (X \cup Y) \geq \mu (X) + \mu (Y)$ for all sets $X, Y \in 2^S$ such that $X \cap Y = \emptyset$

we say that $\mu$ is a superadditive measure [1].

Using the superadditivity property recursively, one can conclude that

$\mu (X) \geq \displaystyle\sum_{x \in X} \mu (\{x\})$

for every $X \in 2^S$. For example, if $S := \{s_1, s_2, s_3, s_4\}$, then we have that the measure of $X :=\{s_1, s_2, s_3\}$ is

$\begin{array}{rl} \mu (\{s_1, s_2, s_3\}) &= \mu (\{s_1, s_2\} \cup \{s_3\})\\\\ &\geq \mu (\{s_1, s_2\}) + \mu (\{s_3\})\\\\ &\geq \left( \mu (\{s_1\}) + \mu (\{s_2\}) \right) + \mu (\{s_3\}\\\\ &\geq \mu (\{s_1\}) + \mu (\{s_2\}) + \mu (\{s_3\})\end{array}$

Frankly, I have (accidentally) opened a can of worms. I started writing this post thinking about synergy and productivity, and I am now drowning in Measure Theory! As it turns out, the union of all my knowledge of Measure Theory is a set of measure zero ;-) Hence, I will abruptly finish this post with a passage from Wang & Klir [1]:

Observe that superadditive measures are capable of expressing a cooperative action or synergy between sets in terms of the measured property, while subadditive measures are capable of expressing inhibitory effects or incompatibility between sets in terms of the measured property. Additive measures, on the other hands, are not able to express either of these interactive effects. They are applicable only to situations in which there is no interaction between sets as far as the measured property is concerned.

I may return to this topic if I happen to have any interesting ideas.

__________

References

[1] Zhenyuan Wang, George J. Klir, Generalized Measure Theory, Springer, 2009.

### 5 Responses to “A measure-theoretic definition of synergy?”

1. vfp15 Says:

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?

• Rod Carvalho Says:

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?

2. vfp15 Says:

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.

3. s. sinha Says:

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??