# Games with non-uniform interaction rates

Background: Many models in evolutionary game theory assume uniform interaction rates. For instance, consider the $2\times 2$ game: \begin{array}{l c c} & A & B \\ A & a & b \\ B & c & d \end{array} The fitness of the strategy $A$ is $xa + yb$, where $x$ and $y$ are the frequencies of the strategies $A$ and $B$, respectively.

Taylor and Nowak "Evolutionary game dynamics with non-uniform interaction rates" (link to the paper) rewrite the fitness function to allow for non-uniform interaction. In particular, the fitness of the strategy $A$ and $B$ become: $$f_{A} = \frac{ar_{1}x + br_{2}y}{r_{1}x + r_{2}y}; \quad f_{B} = \frac{cr_{2}x + dr_{3}y}{r_{2}x + r_{3}y}$$ According to this model, [a]nalogous to a chemical reaction, an $A$ player interacts with another $A$ player with reaction rate $r_{1}$, an $A$ player and a $B$ player interact with reaction rate $r_{2}$, and a $B$ player interacts with another $B$ player with reaction rate $r_{3}$." (p.244).

Question: What would be an intuitive interpretation of $f_A$ and $f_B$ (as defined by Taylor and Nowak)?

Further comments: When $r_1 = r_2$ and $r_2 = r_3$, we have the situation with uniform interaction. But I am not sure how to interpret the above fitness function for when $r_1 \neq r_2$ (and $r_2 \neq r_3$). I thought of interpreting $r_1$ and $r_2$ as the conditional probabilities $P(A\mid A)$ and $P(A\mid B)$. However, if I'm right about this, $f_A$ would have to be $r_1 a + r_2 b$ instead. For this reason, I am under the impression that the $r$s should not be conditional probabilities. But if the $r$s are not conditional probabilities, what are they exactly?

• To make sure I understand your interpretation, does it involves that 1) $r_1+r_2=1$, 2) if $r_1=r_2$, then individuals interact randomly without preference, 3) If $r_1>r_2$, then individuals interact preferentially when individuals playing the same strategy and 4) if $r_1<r_2$, then individuals interact preferentially with individuals playing a different strategy? If the above is true I would expect the mean fitness in the population to be $\frac{1}{2 (x+y)(r_1x+r_2y)}$ while the equation suggests that the mean fitness is $r_1x+r_2y$. – Remi.b Nov 22 '15 at 21:12
• @Remi.b 1). Yes, if my interpretation is correct. 2). Yes. 3 and 4) Here is a quote from the paper "If r1 > r2 and r3 > r2, then players prefer to interact with their own kind. If, however, r1 < r2 and r3 < r2, then mixed interactions (between A and B) are more likely." – falsum Nov 23 '15 at 3:37
• @Remi.b Note that I added a formula for $f_B$ to make my question more clear. – falsum Nov 23 '15 at 3:50