Practical Basis of the Geometric Mean Fitness and its Application to Risk-Spreading Behavior

Abstract

Temporal variations in population size under unpredictable environments are of primary concern in evolutionary ecology, where time scale enters as an important factor while setting up an optimization problem. Thus, short-term optimization with traditional (arithmetic) mean fitness may give a different result from long-term optimization. In the long-term optimization, the concept of geometric mean fitness has been received well by researchers and applied to various problems in ecology and evolution. However, the limit of applicability of geometric mean has not been addressed so far. Here we investigate this problem by analyzing numerically the probability distribution of a random variable obeying stochastic multiplicative growth. According to the law of large number, the expected value (i.e., arithmetic mean) manifests itself as a proper measure of optimization as the number of random processes increases to infinity. We show that the finiteness of this number plays a crucial role in arguing for the relevance of geometric mean. The geometric mean provides a satisfactory picture of the random variation in a long term above a crossover time scale that is determined by this number and the standard deviation of the randomly varying growth rates. We thus derive the applicability condition under which the geometric mean fitness is valid. We explore this condition in some examples of risk-spreading behavior.

Appendix

In the long-time limit t → ∞, S_t obeys the log-normal distribution with its µ and σ² parameters as given in the main text. In the log-normal distribution, the probability that S_t is less than K is given by ${\text{Prob}}\left( {S_{t} < K} \right) = {\text{Prob}}\left( {\frac{1}{t}\sum\limits_{t = 0}^{t - 1} {\log r_{t} < \frac{1}{t}\log K} } \right) = \Phi \left( {\frac{\log K - \mu t}{{\sigma \sqrt t }}} \right)$, where ${\Phi }\left( y \right)$ is the cumulative distribution function (cdf) of the normal distribution with mean 0 and standard deviation 1. Owing to ${\text{Prob}}\left( {S_{t} > S_{{{\text{max}}}} } \right) = 1/N$, or ${\text{Prob}}\left( {S_{t} < S_{{{\text{max}}}} } \right) = 1 - \frac{1}{N}$, we obtain $S_{{{\text{max}}}} = e^{\mu t + C\left( N \right)\sigma \sqrt t }$, where $C\left( N \right) = {\Phi }^{ - 1} \left( {1 - \frac{1}{N}} \right)$. On the other side, the conditional expectation is also expressed as $E\left[ {S_{t} {|}S_{t} < K} \right] = e^{{\mu t + \frac{{\sigma^{2} }}{2}t}} {\Phi }\left( {\frac{{\log K - \mu t - \sigma^{2} t}}{\sigma \sqrt t }} \right)/{\Phi }\left( {\frac{\log K - \mu t}{{\sigma \sqrt t }}} \right)$ $= e^{{\mu t + \frac{{\sigma^{2} }}{2}t}} {\Phi }\left( {C\left( N \right) - \sigma \sqrt t } \right)/\left( {1 - \frac{1}{N}} \right)$, where we substituted K = S_max in the second equation. For a sufficiently large t, we may use an asymptotic formula ${\Phi }\left( x \right) \simeq e^{{ - \frac{{x^{2} }}{2}}} /\left( {\left| x \right|\sqrt {2\pi } } \right)$ valid for x →  ± ∞. Consequently, we obtain $E\left[ {S_{t} {|}S_{t} < S_{{{\text{max}}}} } \right] \simeq e^{\mu t } /\left( {\sqrt {2\pi t} \sigma } \right)$, for $\sigma \sqrt t \gg C\left( N \right)$, or for $t \gg t_{{{\text{cr}}}}$ with $t_{{{\text{cr}}}} = \left( {C\left( N \right)/\sigma } \right)^{2} .$ Noting that $e^{\mu } = m_{{{\text{geo}}}}$ by Eqs. (2) and (10), we obtain $S_{tN} \sim \left( {m_{{{\text{geo}}}} } \right)^{t}$ for $t \gg t_{{{\text{cr}}}}$.