Fitness Beats Truth in the Evolution of Perception

Abstract

Does natural selection favor veridical percepts—those that accurately (if not exhaustively) depict objective reality? Perceptual and cognitive scientists standardly claim that it does. Here we formalize this claim using the tools of evolutionary game theory and Bayesian decision theory. We state and prove the “Fitness-Beats-Truth (FBT) Theorem” which shows that the claim is false: If one starts with the assumption that perception involves inference to states of the objective world, then the FBT Theorem shows that a strategy that simply seeks to maximize expected-fitness payoff, with no attempt to estimate the “true” world state, does consistently better. More precisely, the FBT Theorem provides a quantitative measure of the extent to which the fitness-only strategy dominates the truth strategy, and of how this dominance increases with the size of the perceptual space. The FBT Theorem supports the Interface Theory of Perception (e.g. Hoffman et al. in Psychon Bull Rev https://doi.org/10.3758/s13423-015-0890-8, 2015), which proposes that our perceptual systems have evolved to provide a species-specific interface to guide adaptive behavior, and not to provide a veridical representation of objective reality.

Appendix: Calculations for the Numerical Example in Table 1

In this appendix we perform the Bayesian and expected-fitness calculations using the data given in Table 1.

To compute the Truth estimates, we first need the probability of each stimulation ${\mathbb{P}}({x}_{1})$ and ${\mathbb{P}}({x}_{2})$. These can be computed by marginalizing over the priors in the world as follows:

$${\mathbb{P}}({x}_{1})=p({x}_{1}\left|{w}_{1}\right)\mu \left({w}_{1}\right)+p\left({x}_{1}|{w}_{2}\right)\mu \left({w}_{2}\right)+p\left({x}_{1}|{w}_{3}\right)\mu \left({w}_{3}\right)$$

$$=\frac{1}{4}.\frac{1}{7}+\frac{3}{4}.\frac{3}{7}+\frac{1}{4}.\frac{3}{7}=\frac{13}{28}$$

$${\mathbb{P}}({x}_{2})=p({x}_{2}\left|{w}_{1}\right)\mu \left({w}_{1}\right)+p\left({x}_{2}|{w}_{2}\right)\mu \left({w}_{2}\right)+p\left({x}_{2}|{w}_{3}\right)\mu \left({w}_{3}\right)$$

$$=\frac{3}{4}.\frac{1}{7}+\frac{1}{4}.\frac{3}{7}+\frac{3}{4}.\frac{3}{7}=\frac{15}{28}$$

By Bayes’ Theorem, the posterior probabilities of the world states, given ${x}_{1},$ are

$$p({w}_{1}\left|{x}_{1}\right)=p\left({x}_{1}\left|{w}_{1}\right)\right).\frac{\mu \left({w}_{1}\right)}{P\left({x}_{1}\right)}=\frac{1}{4}.\frac{1}{7}/\frac{13}{28}=\frac{1}{13}$$

$$p({w}_{2}\left|{x}_{1}\right)=p\left({x}_{1}\left|{w}_{2}\right)\right).\frac{\mu \left({w}_{2}\right)}{P\left({x}_{1}\right)}=\frac{3}{4}.\frac{3}{7}/\frac{13}{28}=\frac{9}{13}$$

$$p({w}_{3}\left|{x}_{1}\right)=p\left({x}_{1}\left|{w}_{3}\right)\right).\frac{\mu \left({w}_{3}\right)}{P\left({x}_{1}\right)}=\frac{1}{4}.\frac{3}{7}/\frac{13}{28}=\frac{3}{13}$$

Thus the maximum a posteriori, or Truth estimate for stimulus ${x}_{1}$ is ${w}_{2}$.

Posterior probabilities of the world states, given ${s}_{2},$ are:

$$p({w}_{1}\left|{x}_{2}\right)=p\left({x}_{2}\left|{w}_{1}\right)\right).\frac{\mu \left({w}_{1}\right)}{P\left({x}_{2}\right)}=\frac{3}{4}.\frac{1}{7}/\frac{15}{28}=\frac{1}{5}$$

$$p({w}_{2}\left|{x}_{2}\right)=p\left({x}_{2}\left|{w}_{2}\right)\right).\frac{\mu \left(2\right)}{P\left({x}_{2}\right)}=\frac{1}{4}.\frac{3}{7}/\frac{15}{28}=\frac{1}{5}$$

$$p({w}_{3}\left|{x}_{2}\right)=p\left({x}_{2}\left|{w}_{3}\right)\right).\frac{\mu \left({w}_{3}\right)}{P\left({x}_{2}\right)}=\frac{3}{4}.\frac{3}{7}/\frac{15}{28}=\frac{3}{5}$$

Thus the maximum a posteriori, or Truth estimate for stimulus ${x}_{2}$ is ${w}_{3}.$

Finally, the expected-fitness values of the different sensory stimulations ${x}_{1}$ and ${x}_{2}$ are, respectively:

$$F\left({x}_{1}\right)=p({w}_{1}\left|{x}_{1}\right)f\left({w}_{1}\right)+p({w}_{2}\left|{x}_{1}\right)f\left({w}_{2}\right)+p({w}_{3}\left|{x}_{1}\right)f\left({w}_{3}\right)=\frac{1}{13}.20+\frac{9}{13}.4+\frac{3}{13}.3=5;$$

$$F\left({x}_{2}\right)=p({w}_{1}\left|{x}_{2}\right)f\left({w}_{1}\right)+p({w}_{2}\left|{x}_{2}\right)f\left({w}_{2}\right)+p({w}_{3}\left|{x}_{2}\right)f\left({w}_{3}\right)$$

$$={\frac{1}{5}}.20+{\frac{1}{5}}.4+{\frac{3}{5}}.3=6.6.$$

Thus ${x}_{2}$ has a larger expected fitness than ${x}_{1}$.