Estimating hominid life history: the critical interbirth interval

Abstract

Unlike any great apes, humans have expanded into a wide variety of habitats during the course of evolution, beginning with the transition by australopithecines from forest to savanna habitation. Novel environments are likely to have imposed hominids a demographic challenge due to such factors as higher predation risk and scarcer food resources. In fact, recent studies have found a paucity of older relative to younger adults in hominid fossil remains, indicating considerably high adult mortality in australopithecines, early Homo, and Neanderthals. It is not clear to date why only human ancestors among all hominoid species could survive in these harsh environments. In this paper, we explore the possibility that hominids had shorter interbirth intervals to enhance fertility than the extant apes. To infer interbirth intervals in fossil hominids, we introduce the notion of the critical interbirth interval, or the threshold length of birth spacing above which a population is expected to go to extinction. We develop a new method to obtain the critical interbirth intervals of hominids based on the observed ratios of older adults to all adults in fossil samples. Our analysis suggests that the critical interbirth intervals of australopithecines, early Homo, and Neanderthals are significantly shorter than the observed interbirth intervals of extant great apes. We also discuss possible factors that may have caused the evolutionary divergence of hominid life history traits from those of great apes.

Appendices

Appendix 1

Let $k$ and $x$ represent the age interval and cut-off age, respectively, adopted in each life table. Then, we have data, ${l_{15}}$, ${l_{15+k}}$, ${l_{15+2k}}$, $\cdots$, ${l_x}$, and we fix ${l_{x+k}}=0$. Standardizing ${l_{15}}$ as 1, the uncorrected life expectancy at age 15 (${e_{15({\text{u}})}}$) corresponds to the area of the survival curve obtained from the data. Thus, using the trapezoidal rule, we have

$$\begin{aligned}{e_{15\left( {\text{u}} \right)}}&=\left( {1+\frac{{{l_{15+k}}}}{{{l_{15}}}}} \right)\frac{k}{2}+\left( {\frac{{{l_{15+k}}}}{{{l_{15}}}}+\frac{{{l_{15+2k}}}}{{{l_{15}}}}} \right)\frac{k}{2}\\ & \quad +\,\cdots +\left( {\frac{{{l_x}}}{{{l_{15}}}}+0} \right)\frac{k}{2}~~=\frac{k}{2}+\mathop \sum \limits_{{i=1}}^{{\left( {x - 15} \right)/k}} \frac{{k{l_{15+ik}}}}{{{l_{15}}}}.\end{aligned}$$

(8)

Here, the life expectancy at the cut-off age is $k/2$ years, which is generally too short. Thus, we have to add something to ${e_{15\left( {\text{u}} \right)}}$ to obtain a more realistic life expectancy. Since the adult life expectancy generally decreases in age, the life expectancy at the cut-off age may be smaller than that at age 15. Therefore, we add a half of the (uncorrected) life expectancy at age 15 to the life expectancy at the cut-off age, and have the corrected life expectancy at age 15:

$${e_{15\left( {\text{c}} \right)}}={e_{15\left( {\text{u}} \right)}}+\frac{{{e_{15\left( {\text{u}} \right)}}{l_x}}}{{2{l_{15}}}}.{\text{~}}$$

(9)

That is, proportion ${l_x}/{l_{15}}$ of adult individuals have an additional lifetime ${e_{15\left( {\text{u}} \right)}}/2$, i.e., the life expectancy at the cut-off age is $({e_{15\left( {\text{u}} \right)}}+k)/2$ years.

Appendix 2

When the probability with which an individual having survived to age $15$ will survive to age $15+x$ is represented as $~l\left( x \right)={\text{exp}}\left( { - \mu x} \right)$, we have

$${e_{15}}=\int \limits_{{x=0}}^{\infty } l\left( x \right)dx=\frac{1}{\mu }.~$$

(10)

Then, the older adult ratio, i.e., the probability with which an individual having survived to age $15$ will survive to age $15+15=30$, is given by $r=l\left( {15} \right)={\text{exp}}\left( { - 15\mu } \right)$, i.e., $\log r= - 15\mu$, so that we have

$${e_{15}}= - \frac{{15}}{{\log r}}.$$

(11)

Appendix 3

As in Appendix 2, when $l\left( x \right)={\left( {1 - x/M} \right)^s}$ for $x<M$ and $l\left( x \right)=0$ for $x \geqslant M$, we have

$${e_{15}}=\int \limits_{{x=0}}^{M} l\left( x \right)dx=\frac{M}{{1+s}}.$$

(12)

Then, $r=l\left( {15} \right)={\left( {1 - 15/M} \right)^s}$, i.e., $\log r=s~\log \left( {1 - 15/M} \right)$, so that we have

$${e_{15}}=\frac{M}{{1+\frac{{\log r}}{{\log \left( {1 - 15/M} \right)}}}}.{\text{~}}$$

(13)

Appendix 4

Since the probability that an individual having survived to age $15$ will survive to age $15+x$ is $l\left( x \right)$ and that he/she will die between age $x$ and $x+{\Delta}x$ is $- l^{\prime}\left( x \right){\Delta}x$, the instantaneous mortality rate at age $15+x$ is given by

$$h\left( x \right)= - \frac{{l^{\prime}\left( x \right)}}{{l\left( x \right)}},~$$

(14)

i.e.,

$${\left[ {\log l\left( x \right)} \right]^\prime }= - h\left( x \right).$$

(15)

Therefore, when $h\left( x \right)=a{\text{exp}}\left( {bx} \right)$, we have

$$\log l\left( x \right)= - \int \nolimits^{} a{\text{exp}}\left( {bx} \right)dx=\frac{a}{b}\left[ {1 - {\text{exp}}\left( {bx} \right)} \right],$$

(16)

because $l\left( 0 \right)=1$, i.e.,

$$l\left( x \right)={\text{exp}}\left\{ {\frac{a}{b}\left[ {1 - {\text{exp}}\left( {bx} \right)} \right]} \right\}.$$

(17)

Since $r=l\left( {15} \right)={\text{exp}}\left\{ {(a/b)\left[ {1 - {\text{exp}}\left( {15b} \right)} \right]} \right\}$, i.e.,

$$\frac{a}{b}=\frac{{\log r}}{{1 - {\text{exp}}\left( {15b} \right)}},$$

(18)

we have

$${e_{15}}=\int \limits_{{x=0}}^{\infty } l\left( x \right)dx= \int \limits_{0}^{\infty } {\text{exp}}\left( {\frac{{1 - {\text{exp}}\left( {bx} \right)}}{{1 - \exp \left( {15b} \right)}}\log r} \right)dx.$$

(19)

Appendix 5

The methods of calculating the 95% prediction intervals are as follows. For the polynomial regression model of the life expectancies at age 15 (${e_{15}}$), since linear functions are selected, we simply adopt the standard prediction interval function for linear regression analysis (Kleinbaum et al. 2008). The 95% prediction intervals of the probability of survival from birth to age 15 (${l_{15}}$) are similarly obtained because a linear function is also selected when Taï chimpanzee population is removed. For the Gompertz model of the life expectancies at age 15 (${e_{15}}$), using Mathematica 7.0, we obtain the 95% confidence intervals of parameter b (the range of b in which the residual is smaller than $1+{F_{1,34,0.05}}/34$ times the minimum residual, where F is the F distribution and 34 is the degree of freedom) to calculate the 95% prediction intervals of ${e_{15}}$. The 95% prediction intervals of ${B_C}$ are obtained from the 95% prediction intervals of ${e_{15}}$ and ${l_{15}}$ because $\log {B_C}=\log {l_{15}}+\log {e_{15}} - \log 2$.