Optimal population and exhaustible resource constraints

Abstract

A large literature considers the optimal size and growth rate of the human population, trading off the utility value of additional people with the costs of a larger population. In this literature, an important parameter is the social weight placed on population size; a standard result is that a planner with a larger weight on population chooses larger population levels and growth rates. We demonstrate that this result is conditionally overturned when an exhaustible resource constraint is introduced: if the discount rate is small enough, the optimal population today decreases with the welfare weight on population size. That is, a more total-utilitarian social planner could prefer a smaller population today than a more average-utilitarian social planner. We also present a numerical illustration applied to the case of climate change, where we show that under plausible real-world parameter values, our result matters for the direction and magnitude of optimal population policy.

Appendices

Appendix A: Algebra and proofs

This appendix presents the algebra and proofs from our main model in Section 3 of the paper.

1.1 A.1 Simple two-period exogenous-utility model

Consider first the simplest possible version of the model, with only two relevant periods: t ∈{1, 2} (the population and utility in period 0 is exogenously fixed). The planner’s problem is:

$$\mathcal{L} = n_{1}^{\alpha}u_{1} + \beta n_{2}^{\alpha}u_{2} - \lambda\left( \frac{n_{1}}{T_{1}} + \frac{n_{2}}{T_{2}} - X\right). $$

Differentiating with respect to n ₁ and n ₂, we find:

$$\alpha n_{1}^{\alpha-1}u_{1} = \frac{\lambda}{T_{1}} $$

$$\beta\alpha n_{2}^{\alpha-1}u_{2} = \frac{\lambda}{T_{2}}. $$

Combining these expressions, and assuming that u ₂ = u ₁, we find:

$$\frac{n_{2}}{n_{1}} = \left( \beta\gamma\right)^{\frac{1}{1-\alpha}}. $$

Defining the growth factor as $g \equiv \left (\beta \gamma \right )^{\frac {1}{1-\alpha }}$, and combining n ₂ = g n ₁ with the resource constraint:

$$ n_{1} = \frac{T_{2}X}{g + \gamma} $$

(4)

$$ n_{2} = \frac{gT_{2}X}{g + \gamma}. $$

(5)

These expressions depend on α only through g, leading to the following simple result:

Proposition 3

n ₂ is increasing and n ₁ is decreasing with α if and only if β γ > 1.

Proof 3

Differentiate (4) and (5) with respect to g:

$$\frac{dn_{1}}{dg} = \frac{-T_{2}X}{(g+\gamma)^{2}} < 0 $$

$$\frac{dn_{2}}{dg} = \frac{\gamma T_{2}X}{(g+\gamma)^{2}} > 0. $$

Since n ₁and n ₂depend on α only throughg, and $\frac {dg}{d\alpha } > 0$if andonly if β γ > 1,then $\frac {dn_{1}}{d\alpha } < 0$and$\frac {dn_{2}}{d\alpha } > 0$if andonly if β γ > 1. □

Thus, the optimal population today decreases with α, and the population tomorrow increases, whenever the optimal population growth is positive. When facing a resource constraint with technological improvement, total utilitarians prefer to have fewer people today so as to defer population to the future, when it will be less costly in terms of resource use and thus a larger population can be supported.²⁸ In a sense, having a higher α is like having a lower discount rate over the number of people: as long as population is increasing over time, the future is valued more highly when α is larger and the planner wants more people to be alive in that future.

1.2 A.2 Simple infinite-horizon exogenous-utility model

As noted in Section 3.1, in the exogenous-utility model, the lagrangian for the planner’s problem can be written as follows:

$$\mathcal{L} = \sum\limits_{t=0}^{\infty}{\beta^{t}n_{t}^{\alpha}u_{t}} - \lambda\left( \sum\limits_{t=1}^{\infty}{\frac{n_{t}}{T_{t}}} - X\right) $$

and differentiating with respect to n _t and n _t+1 gives:

$$\beta^{t}\alpha n_{t}^{\alpha-1}u_{t} = \frac{\lambda}{T_{t}} $$

$$\beta^{t+1}\alpha n_{t+1}^{\alpha-1}u_{t+1} = \frac{\lambda}{T_{t+1}}. $$

Combining these two expressions leads to an equation for optimal population growth:

$$\frac{n_{t+1}}{n_{t}} = \left( \frac{\beta\gamma u_{t+1}}{u_{t}}\right)^{\frac{1}{1-\alpha}}. $$

which, if we assume u _t to be constant across time, simplifies to:

$$\frac{n_{t+1}}{n_{t}} = \left( \beta\gamma\right)^{\frac{1}{1-\alpha}} \equiv g. $$

Differentiating g with respect to α, we find:

$$\frac{dg}{d\alpha} = \ln\left( \beta\gamma\right)\left( \beta\gamma\right)^{\frac{1}{1-\alpha}}\left( \frac{1}{(1-\alpha)^{2}}\right) $$

and the only term that has an uncertain sign is the first one: if β γ > 1, and thus g > 1, then $\frac {dg}{d\alpha } > 0$.

Using the equation for population in any period $n_{t} = \left (\beta \gamma \right )^{\frac {t-1}{1-\alpha }}n_{1}$, and the resource constraint ${\sum }_{t=1}^{\infty }{\frac {n_{t}}{\gamma ^{t-1}T_{1}}} = X$, we can derive the starting value n ₁:

$$n_{1}\sum\limits_{t=1}^{\infty}{\left( \beta\gamma^{\alpha}\right)^{\frac{t-1}{1-\alpha}}} = XT_{1}. $$

For this to give a positive solution for n ₁, it must be the case that β γ ^α < 1.²⁹ Assuming this to be true, the solution for n ₁ is:

$$n_{1} = \left[1 - \left( \beta\gamma^{\alpha}\right)^{\frac{1}{1-\alpha}}\right]XT_{1} $$

and this can be used to give the value for any n _t :

$$n_{t} = \left( \beta\gamma\right)^{\frac{t-1}{1-\alpha}}\left[1 - \left( \beta\gamma^{\alpha}\right)^{\frac{1}{1-\alpha}}\right]XT_{1}. $$

Finally, we can differentiate n _t with respect to α:

$$\frac{dn_{t}}{d\alpha} = \frac{XT_{1}\ln\left( \beta\gamma\right)\left( \beta\gamma\right)^{\frac{t-1}{1-\alpha}}}{(1-\alpha)^{2}}\left[(t-1)\left( 1 - \left( \beta\gamma^{\alpha}\right)^{\frac{1}{1-\alpha}}\right) - \left( \beta\gamma^{\alpha}\right)^{\frac{1}{1-\alpha}}\right]. $$

If we set this to zero, we find that the critical value of t is:

$$t^{*} = \frac{1}{1 - \left( \beta\gamma^{\alpha}\right)^{\frac{1}{1-\alpha}}}. $$

Notice that the first part of $\frac {dn_{t}}{d\alpha }$—the fraction prior to the square brackets—is always positive given the assumption that β γ > 1. The part within the square brackets is clearly increasing with t given that β γ ^α < 1, and thus we have that $\frac {dn_{t}}{d\alpha } < 0$ for t < t ^∗ and $\frac {dn_{t}}{d\alpha } > 0$ if t > t ^∗, exactly as stated in Proposition 1.

1.3 A.3 Intermediate model

We want to prove the single-crossing property in the intermediate model; that is, the existence of a critical time t ^∗ such that $\frac {dn_{t}}{d\alpha } < 0$ if t < t ^∗ and $\frac {dn_{t}}{d\alpha } > 0$ if t > t ^∗.

The derivative of the planner’s problem with respect to n _t is:

$$\frac{\partial\mathcal{L}}{\partial n_{t}} = \beta^{t}\alpha n_{t}^{\alpha-1}u(c(n_{t})) + \beta^{t}n_{t}^{\alpha}u^{\prime}(c(n_{t}))c^{\prime}(n_{t}) - \frac{\lambda}{T_{t}} = 0 $$

which leads immediately to:

$$ \left( \beta\gamma\right)^{t}n_{t}^{\alpha-1}\left[\alpha u(c(n_{t})) + n_{t}u^{\prime}(c(n_{t}))c^{\prime}(n_{t})\right] = \frac{\lambda}{T_{0}}. $$

(6)

We first prove that $\frac {dn_{t}}{dt} > 0$, for which purpose we take the derivative with respect to t; after a bit of simplification, we have:

$$\begin{array}{@{}rcl@{}} &&\ln\left( \beta\gamma\right)n_{t}\left[\alpha u(c(n_{t})) + n_{t}u^{\prime}(c(n_{t}))c^{\prime}(n_{t})\right]\\ &&+ \left[(\alpha-1)\alpha u(c(n_{t})) + n_{t}\left( 2\alpha u^{\prime}(c(n_{t}))c^{\prime}(n_{t}) + n_{t}u^{\prime\prime}(c(n_{t}))c^{\prime}(n_{t})c^{\prime}(n_{t})\right.\right.\\ &&\quad\left.\left.+ n_{t}u^{\prime}(c(n_{t}))c^{\prime\prime}(n_{t})\right)\right]\frac{dn_{t}}{dt} = 0. \end{array} $$

All of the terms multiplying $\frac {dn_{t}}{dt}$ are at least weakly negative (and the first two are strictly negative), while if β γ > 1, the first term is positive. Therefore, the derivative must be positive: $\frac {dn_{t}}{dt} > 0$ if and only if β γ > 1.

Next, we take the derivative of the first-order condition with regard to α; whereas the multiplier λ is a constant with regard to t, it is not with regard to α: a change in α will necessitate a different λ to ensure that the resource constraint is binding. Therefore, we have:

$$\begin{array}{@{}rcl@{}} &&\left( \beta\gamma\right)^{t}\left[(\alpha-1)n_{t}^{\alpha-2}\left( \alpha u(c(n_{t})) + n_{t}u^{\prime}(c(n_{t}))c^{\prime}(n_{t})\right)\right.\\ &&\quad\quad\quad\left.+ n_{t}^{\alpha-1}\left( \alpha u^{\prime}(c(n_{t}))c^{\prime}(n_{t}) + u^{\prime}(c(n_{t}))c^{\prime}(n_{t})\right.\right.\\ &&\quad\quad\quad\left.\left. + n_{t}u^{\prime\prime}(c(n_{t}))c^{\prime}(n_{t})c^{\prime}(n_{t}) + n_{t}u^{\prime}(c(n_{t}))c^{\prime\prime}(n_{t})\right)\right]\frac{dn_{t}}{d\alpha}\\ &&+ \left( \beta\gamma\right)^{t}\left[\ln(n_{t})n_{t}^{\alpha-1}\left( \alpha u(c(n_{t})) + n_{t}u^{\prime}(c(n_{t}))c^{\prime}(n_{t})\right) + n_{t}^{\alpha-1}u(c(n_{t}))\right] = \frac{1}{T_{0}}\frac{d\lambda}{d\alpha}. \end{array} $$

Simplifying, the total derivative is:

$$\begin{array}{@{}rcl@{}} &&T_{0}\left( \beta\gamma\right)^{t}n_{t}^{\alpha-2}\left[(\alpha-1)\alpha u(c_{t}) + n_{t}\left( 2\alpha u^{\prime}(c_{t})c^{\prime}(n_{t}) + n_{t}u^{\prime\prime}(c_{t})c^{\prime}(n_{t})^{2}\right.\right.\\ &&\qquad\qquad\qquad\left.\left.+ n_{t}u^{\prime}(c_{t})c^{\prime\prime}(n_{t})\right)\right]\frac{dn_{t}}{d\alpha}\\ &&+ \lambda\ln(n_{t}) + T_{0}\left( \beta\gamma\right)^{t}n_{t}^{\alpha-1}u(c_{t}) = \frac{d\lambda}{d\alpha}. \end{array} $$

To further simplify notation, define g(t) as the terms multiplying $\frac {dn_{t}}{d\alpha }$; as every term in g(t) is at least weakly negative, and the first two are strictly negative, we can conclude that g(t) < 0 for all values of t. Similarly, define $f(t) \equiv \frac {d\lambda }{d\alpha } - \lambda \ln (n_{t}) - T_{0}\left (\beta \gamma \right )^{t}n_{t}^{\alpha -1}u(c_{t})$, and then we have:

$$\frac{dn_{t}}{d\alpha} = \frac{f(t)}{g(t)}. $$

The effect of α on n _t thus comes down to the expression f(t), and how it changes over time. Because g(t) is always negative, proving single-crossing of f(t) would also prove single-crossing of $\frac {dn_{t}}{d\alpha }$. We start by defining f(t) as a function of t only through n _t , using (6) to substitute for $T_{0}\left (\beta \gamma \right )^{t}n_{t}^{\alpha -1}$:

$$\begin{array}{@{}rcl@{}} f(t) &=& \frac{d\lambda}{d\alpha} - \lambda\ln(n_{t}) - \frac{\lambda u(c(n_{t}))}{\alpha u(c(n_{t})) + n_{t}u^{\prime}(c(n_{t}))c^{\prime}(n_{t})}\\ &=& \frac{d\lambda}{d\alpha} - \lambda\left[\ln(n_{t}) + \left( \alpha +\frac{n_{t}u^{\prime}(c((n_{t}))c^{\prime}(n_{t}))}{u(c(n_{t}))}\right)^{-1}\right]. \end{array} $$

Defining the derivative with respect to t as f ^′(t), we now have $f^{\prime }(t) = \frac {\partial f(t)}{\partial n_{t}}\frac {dn_{t}}{dt}$, and we have already shown that $\frac {dn_{t}}{dt} > 0$ if β γ > 1. Therefore, consider $\frac {\partial f(t)}{\partial n_{t}}$:

$$\begin{array}{@{}rcl@{}} \frac{\partial f(t)}{\partial n_{t}} \!\!&=&\!\! \!- \lambda\left[\frac{1}{n_{t}} \,-\, \left( \alpha \,+\, \frac{n_{t}u^{\prime}(c(n_{t}))c^{\prime}(n_{t})}{u(c(n_{t}))}\right)^{-2}\right.\\ &&\qquad\times\left.\left( \frac{\left( u^{\prime}(c(n_{t}))c^{\prime}(n_{t}) \,+\, n_{t}u^{\prime\prime}(c(n_{t}))c^{\prime}(n_{t})^{2} \,+\, n_{t}u^{\prime}(c(n_{t}))c^{\prime\prime}(n_{t})\right)u(c(n_{t})) \,-\, n_{t}u^{\prime}(c(n_{t}))^{2}c^{\prime}(n_{t})^{2}}{u(c(n_{t}))^{2}}\right)\right]. \end{array} $$

Every term in the numerator of the second line is negative, which means that $\frac {\partial f(t)}{\partial n_{t}} < 0$. Therefore, f ^′(t) < 0 if and only if β γ > 1.

Next, notice that $\frac {dn_{t}}{d\alpha }$ cannot be either always positive or always negative, as long as the resource constraint is binding: if α increases, any increase in population at some point in time must be offset by a decrease at some other point in time. Therefore, f(t) cannot be always positive or always negative; given that f ^′(t) < 0, f(t) must initially be positive and then become negative beyond some critical time t ^∗. Finally, we know that g(t) is always negative, and therefore $\frac {dn_{t}}{d\alpha } = \frac {f(t)}{g(t)}$ must be negative for t < t ^∗ and positive for t > t ^∗, if and only if β γ > 1. This is the result presented in Proposition 2.

1.4 A.4 Recursive dynasty model

Solving the planner’s problem with respect to n _t , we find:

$$\beta^{t}\alpha n_{t}^{\alpha-1}u(c_{t}) + \beta^{t}n_{t}^{\alpha-2}u^{\prime}(c_{t})\kappa n_{t+1} - \beta^{t-1}n_{t-1}^{\alpha-1}u^{\prime}(c_{t-1})\kappa = \frac{\lambda}{T_{t}}. $$

Using the first-order condition for n _t+1 to cancel out the λ, we then find:

$$\begin{array}{@{}rcl@{}} &&\beta\alpha n_{t}^{\alpha-1}u(c_{t}) + \beta n_{t}^{\alpha-2}u^{\prime}(c_{t})\kappa n_{t+1} - n_{t-1}^{\alpha-1}u^{\prime}(c_{t-1})\kappa\\ &=& \beta\gamma\left( \beta\alpha n_{t+1}^{\alpha-1}u(c_{t+1}) + \beta n_{t+1}^{\alpha-2}u^{\prime}(c_{t+1})\kappa n_{t+2} - n_{t}^{\alpha-1}u^{\prime}(c_{t})\kappa\right). \end{array} $$

and it is easy to rearrange this expression to find that, if a balanced growth path is optimal, g must be equal to $(\beta \gamma )^{\frac {1}{1-\alpha }}$.

Our goal is to use this equation to derive a similar result to that in Sections 3.1 or 3.2. For this purpose, we begin by considering the possibility of a balanced growth path, along which n _t+1 = g n _t . Given a fixed starting value of n ₀ = 1, there is only one possible constant growth rate g ^∗ that would exactly satisfy the resource constraint:

$$\begin{array}{@{}rcl@{}} \sum\limits_{t=0}^{\infty}{\frac{n_{t}}{T_{t}}} &=& X \rightarrow \sum\limits_{t=0}^{\infty}{\left( \frac{g}{\gamma}\right)^{t}} = XT_{0}\\ g^{*} &=& \left( \frac{XT_{0} - 1}{XT_{0}}\right)\gamma. \end{array} $$

(7)

These two conditions can only be satisfied at one particular value of α, specifically $\alpha ^{*} = 1 - \frac {\ln (\beta \gamma )}{\ln (g^{*})}$. This is the only α for which a balanced-growth equilibrium exists; at any other value, implementing the feasible growth path g ^∗ from Eq. 7 would make λ too small or too large to satisfy the first-order conditions.

Given the existence of the child-bearing cost κ, it is possible that having children could be sufficiently costly that desired population growth would be low and that the resource constraint would not be binding. Equation 3 holds regardless, but to ensure that the constraint is binding, we need $\frac {\partial \mathcal {L}}{\partial n_{t}} > 0$ at λ = 0 on the balanced growth path, or:

$$\beta\alpha u(c) + \beta u^{\prime}(c)\kappa g - g^{1-\alpha}u^{\prime}(c)\kappa > 0. $$

Using the fact that g ^1−α = β γ, this can be simplified to:

$$ \kappa < \frac{\alpha u(c)}{(\gamma - g)u^{\prime}(c)} $$

(8)

where c ≡ 1 − κ g is constant per-capita consumption. Greater intuition into condition (8) can be obtained if we assume that utility is linear in consumption, so that Eq. 8 simplifies to:

$$ \kappa < \frac{\alpha}{\gamma - (1-\alpha)g}. $$

(9)

This condition simply requires that the cost of raising children κ not be too large, and further note that it is quite a weak condition. γ is necessarily larger than (1 − α)g due to the assumption that β γ ^α < 1, and in most cases g will be of the same order of magnitude as γ; in fact, from Eq. 7 above, g ^∗ approaches γ as X T ₀ becomes large relative to one. If g ≃ γ, then Eq. 9 simplifies to $\kappa < \frac {1}{g}$, which is exactly the condition required for consumption 1 − κ g to be positive. Equation 9 could only be significantly restrictive in a situation in which either β or X T ₀ are very small, so that γ could be quite large and yet g could be close to 1.

Additionally, allowing for risk-aversion using a CRRA utility function further weakens this condition by raising utility relative to marginal utility. Let $u(c) = \frac {c^{1-R}}{1-R}$ where R < 1 is the coefficient of relative risk-aversion; we restrict R to be less than one to ensure that utility remains positive. Then Eq. 8 becomes:

$$\kappa < \frac{\alpha c}{(\gamma - g)(1-R)} $$

and it is clear that the right-hand side is increasing in R, so that risk-aversion makes this a weaker condition by raising utility relative to marginal utility.

Now, suppose that α takes the value α ^∗ defined above, the one that ensures that the balanced growth path described in Eq. 7 is optimal, and further assume that the value of κ is small enough that the resource constraint is binding. Consider the effect of a marginal increase in α. We will examine a case in which only two population values, n _t and n _t+1 for some t ≥ 1, are allowed to change from their balanced-growth-path values. We find the following condition for the effect of α on n _t and n _t+1:

Proposition 4

If α is initially at the value that ensures a balanced growth path and κ satisfies(8), and only n _t and n _t+1 are allowed to vary in response to a marginal change in α , $\frac {dn_{t}}{d\alpha } < 0$ and $\frac {dn_{t+1}}{d\alpha } > 0$ if and only if β γ > 1.

Proof 4

Totally differentiate (3) with respect to α, n _t and n _t+1:

$$\begin{array}{@{}rcl@{}} &&\left[\beta\alpha(\alpha-1)n_{t}^{\alpha-2}u(c_{t}) + 2\beta(\alpha-1)n_{t}^{\alpha-3}u^{\prime}(c_{t})\kappa n_{t+1}\right.\\ &&\quad\left.+ \beta n_{t}^{\alpha-4}u^{\prime\prime}(c_{t})\kappa^{2} n_{t+1}^{2} + n_{t-1}^{\alpha-2}u^{\prime\prime}(c_{t-1})\kappa^{2}\right]dn_{t}\\ &&-\left[\beta\alpha n_{t}^{\alpha-2}u^{\prime}(c_{t})\kappa + \beta n_{t}^{\alpha-3}u^{\prime\prime}(c_{t})\kappa^{2} n_{t+1} - \beta n_{t}^{\alpha-2}u^{\prime}(c_{t})\kappa\right]dn_{t+1}\\ &&+\left[\beta n_{t}^{\alpha-1}u(c_{t})(1 + \alpha\ln(n_{t})) + \beta n_{t}^{\alpha-2}u^{\prime}(c_{t})\kappa n_{t+1}\ln(n_{t})\right.\\ &&\quad\quad\left.- n_{t-1}^{\alpha-1}u^{\prime}(c_{t-1})\kappa\ln(n_{t-1})\right]d\alpha\\ &=& \beta\gamma\left[\beta\alpha(\alpha-1)n_{t+1}^{\alpha-2}u(c_{t+1}) + 2\beta(\alpha-1)n_{t+1}^{\alpha-3}u^{\prime}(c_{t+1})\kappa n_{t+2}\right.\\ &&\quad\quad\left.+ \beta n_{t+1}^{\alpha-4}u^{\prime\prime}(c_{t+1})\kappa^{2}n_{t+2}^{2} + n_{t}^{\alpha-2}u^{\prime\prime}(c_{t})\kappa^{2}\right]dn_{t+1}\\ &&-\beta\gamma\left[(\alpha-1)n_{t}^{\alpha-2}u^{\prime}(c_{t})\kappa + n_{t}^{\alpha-3}u^{\prime\prime}(c_{t})\kappa^{2}n_{t+1}\right]dn_{t}\\ &&+\beta\gamma\left[\beta n_{t+1}^{\alpha-1}u(c_{t+1})(1 + \alpha\ln(n_{t+1})) + \beta n_{t+1}^{\alpha-2}u^{\prime}(c_{t+1})\kappa n_{t+2}\ln(n_{t+1})\right.\\ &&\quad\quad\quad\left.- n_{t}^{\alpha-1}u^{\prime}(c_{t})\kappa\ln(n_{t})\right]d\alpha. \end{array} $$

The resource constraint requires that d n _t+1 = −γ d n _t ;we also impose a balanced growth path so that n _t+1 = g n _t , where$g = \left (\beta \gamma \right )^{\frac {1}{1-\alpha }}$, and c _t = c ≡ 1 − κ g. Thenwe rearrange to find:

$$\begin{array}{@{}rcl@{}} &&n_{t}^{\alpha-2}\left[\beta\alpha(\alpha-1)u(c)\left( 1 + \frac{\gamma}{g}\right) + 2\beta(\alpha-1)u^{\prime}(c)\kappa\left( 2\gamma + g\right)\right.\\ &&\quad\quad\quad\left.+ \beta u^{\prime\prime}(c)\kappa^{2}\left( g^{2} + \gamma^{2} + 4\gamma g\right)\right]dn_{t}\\ &=& \beta n_{t}^{\alpha-1}\ln(g)\left[\alpha u(c) + (g-\gamma)\kappa u^{\prime}(c)\right]d\alpha. \end{array} $$

Each of the terms in the square brackets on the left-hand side contains either (α − 1)or u ^″(c), both of whichare negative, so it is immediately apparent that the left-hand side is some negative number multipliedby d n _t .Therefore, $\frac {dn_{t}}{d\alpha } < 0$if and only if the square bracket on the right-hand side is positive; on the assumption that β γ > 1so that g > 1, weneed:

$$\kappa > \frac{\alpha u(c)}{(g-\gamma)u^{\prime}(c)} $$

which is exactly the condition for the resource constraint to be binding. □

If all the n _t are allowed to vary with α, the problem becomes much more complicated; it is intuitive that the same result should typically hold, but since the overall pattern of effects could be quite complex, we are unable to extend our theoretical result to a more general case; similarly, we cannot extend the result analytically beyond the setting of a balanced growth path.³⁰

Appendix B: Comparison of α = 0 and α = 1

In this appendix, we present a special case of our model in which we can compare results with α = 0 and α = 1. In each of the models presented so far, at least one of these values of α leads to degenerate results, so we now consider a setting in which utility from consumption is quadratic. Additionally, to simplify the analysis to come, which will rely on continuity of derivatives with respect to time, we focus on a continuous-time setting:

$$u(t) = 1 + \psi_{1}n(t) - \psi_{2}n(t)^{2} $$

where ψ ₁,ψ ₂ > 0.

In this case, the planner’s problem is:

$$\mathcal{L} = {\int}_{t=0}^{\infty}{e^{-rt}n(t)^{\alpha}(1+\psi_{1}n(t)-\psi_{2}n(t)^{2})} - \lambda\left( {\int}_{t=0}^{\infty}{\frac{n(t)}{T(t)}} - X\right) $$

where r is the instantaneous discount rate, and the first-order condition for n(t) is:

$$e^{-rt}n(t)^{\alpha-1}\left[\alpha (1+\psi_{1}n(t)-\psi_{2}n(t)^{2}) + n(t)(\psi_{1}-2\psi_{2}n(t))\right] = \frac{\lambda}{T(t)}. $$

Simplifying, normalizing T(0) = 1, and writing T(t) = e ^pt where p is the technological growth rate (to maintain consistent notation), we arrive at:

$$e^{(p-r)t}n(t)^{\alpha-1}\left[\alpha + (1+\alpha)\psi_{1}n(t) - (2+\alpha)\psi_{2}n(t)^{2}\right] = \lambda. $$

We then differentiate with respect to t:

$$\begin{array}{@{}rcl@{}} &&(p\,-\,r)\lambda + e^{(p-r)t}n(t)^{\alpha-2}\left[(\alpha\,-\,1)\alpha \,+\, \alpha(1\,+\,\alpha)\psi_{1}n(t) \,-\, (1\,+\,\alpha)(2\,+\,\alpha)\psi_{2}n(t)^{2}\right]\\ &&\times\frac{dn(t)}{dt} = 0 \end{array} $$

and therefore we have the following results for α = 0 and α = 1:

$$\frac{dn(t)}{dt}|_{\alpha=0} = \frac{\lambda_{0}(p-r)e^{-(p-r)t}}{2\psi_{2}} $$

$$\frac{dn(t)}{dt}|_{\alpha=1} = \frac{\lambda_{1}(p-r)e^{-(p-r)t}}{6\psi_{2}n(t)-2\psi_{1}} $$

where λ _α denotes the equilibrium value of λ for a given value of α.

We want to prove that, at any value of t for which n(t) is equal for α = 0 and α = 1, $\frac {dn(t)}{dt}|_{\alpha =1} > \frac {dn(t)}{dt}|_{\alpha =0}$; if so, then given continuity, it is impossible for n(t)| _α=1 to descend from above n(t)| _α=0 to below n(t)| _α=0 as t increases. If the resource constraint is binding in both cases—which we assume is the case— n(t)| _α=1 cannot be always above or always below n(t)| _α=0. Therefore, if $\frac {dn(t)}{dt}|_{\alpha =1} > \frac {dn(t)}{dt}|_{\alpha =0}$ when n(t)| _α=1 = n(t)| _α=0, we will have proved a single-crossing property, in which n(t)| _α=1 starts out below n(t)| _α=0 and then rises above n(t)| _α=0 after some critical t ^∗.

To prove this result, we need:

$$\frac{\lambda_{1}(p-r)e^{-(p-r)t}}{6\psi_{2}n(t)-2\psi_{1}} > \frac{\lambda_{0}(p-r)e^{-(p-r)t}}{2\psi_{2}}. $$

Assuming that p > r—which is analogous to our usual assumption that β γ > 1—and substituting for λ _α , we need:

$$\frac{1 + 2\psi_{1}n(t) - 3\psi_{2}n(t)^{2}}{3\psi_{2}n(t)-\psi_{1}} > \frac{\psi_{1} - 2\psi_{2}n(t)}{\psi_{2}}. $$

We further assume that 3ψ ₂ n(t) > ψ ₁, which is implicitly a condition on the parameters of the utility function: the increase in utility with n(t) cannot be too steep. Then the necessary condition is:

$$\psi_{2} + 2\psi_{1}\psi_{2}n(t) - 3{\psi_{2}^{2}}n(t)^{2} > 3\psi_{1}\psi_{2}n(t) - {\psi_{1}^{2}} - 6{\psi_{2}^{2}}n(t)^{2} + 2\psi_{1}\psi_{2}n(t) $$

which simplifies to:

$${\psi_{1}^{2}} + \psi_{2} > 3\psi_{2}n(t)\left( \psi_{1} - \psi_{2}n(t)\right). $$

It is straightforward to prove that this final condition is always satisfied. The right-hand side is maximized at $n(t) = \frac {\psi _{1}}{2\psi _{2}}$, at which value the right-hand side is equal to $\frac {3{\psi _{1}^{2}}}{4}$, which is less than ${\psi _{1}^{2}} + \psi _{2}$. Therefore, as long as the parameters are such that 3ψ ₂ n(t) > ψ ₁ at the optimum, and as long as p > r, so that the optimal population is growing over time, there will be a critical time t ^∗, before which n(t) is higher for α = 0, and after which n(t) is higher for α = 1. When the planner is total-utilitarian rather than average-utilitarian, they will desire a population that starts out lower but rises higher in the long-run.

Appendix C: Extensions of main model

This appendix contains four extensions of the main models in the paper. First, we consider a case where the planner has a choice over the optimal level of emissions; we show that the optimal population equation takes the same form, and that the simulation results are very similar to those from the dynasty model in the main text. Second, we add capital to the dynasty model with variable resource use, and show that increasing substitutability between capital and exhaustible resources in production weakens our result, by moving to an earlier date the point at which the effect of α on population becomes positive, but does not overturn the result in the cases we consider. We then examine optimal population when the resource is renewable; we find that making the resource renew faster weakens our result and eventually overturns it. Finally, we show that if larger populations induce innovations that improve technology in the future, our result can again be overturned.

In each case, the new feature added to the model amounts to a weakening of the exhaustibility of the resource, either directly or indirectly. In this way, this appendix shows the limit of our result, and demonstrates that the essential feature required is the exhaustibility of the resource and the inability of society to find satisfactory substitutes.

1.1 C.1 Variable resource use in dynasty model

In this appendix, the recursive dynasty model from Section 3.3 is extended to include a choice over the amount of resource used up in each period. To be precise, the planner chooses e _t , the amount of resource income obtained by each individual at time t, and the output function is linear in population size, so that per-capita income is f(e _t ), where f ^′(e) > 0 and f ^″(e) < 0. Thus, the amount of X used up by each generation is $\frac {n_{t}e_{t}}{T_{t}}$. The planner’s problem is defined by:

$$\mathcal{L} = {\sum}_{t=0}^{\infty}{\beta^{t}n_{t}^{\alpha}u\left( f(e_{t}) - \kappa\frac{n_{t+1}}{n_{t}}\right)} - \lambda\left( {\sum}_{t=1}^{\infty}{\frac{n_{t}e_{t}}{T_{t}}} - X\right). $$

where we continue to assume a constant growth rate γ for T. As noted in footnote 22, our numerical results are very similar if consumption is instead specified as $f(e_{t})\left (1 - \kappa \frac {n_{t+1}}{n_{t}}\right )$, where fertility costs are consistent with a time cost interpretation; the crossing point t ^∗ moves slightly earlier, but the qualitative results are the same.

The planner chooses e _t and n _t for all t ≥ 1, and we assume that both are fixed and normalized to one in period 0; differentiating with respect to e _t and n _t , and then using the derivatives for e _t+1 and n _t+1 to cancel out the λ, we find:

$$\begin{array}{@{}rcl@{}} \frac{e_{t+1}}{e_{t}}&&{}\left( \beta\alpha n_{t}^{\alpha-1}u(c_{t}) + \beta n_{t}^{\alpha-2}u^{\prime}(c_{t})\kappa n_{t+1} - n_{t-1}^{\alpha-1}u^{\prime}(c_{t-1})\kappa\right)\\ &&= \beta\gamma\left( \beta\alpha n_{t+1}^{\alpha-1}u(c_{t+1}) + \beta n_{t+1}^{\alpha-2}u^{\prime}(c_{t+1})\kappa n_{t+2} - n_{t}^{\alpha-1}u^{\prime}(c_{t})\kappa\right) \end{array} $$

$$n_{t}^{\alpha-1}u^{\prime}(c_{t})f^{\prime}(e_{t}) = \beta\gamma n_{t+1}^{\alpha-1}u^{\prime}(c_{t+1})f^{\prime}(e_{t+1}). $$

Here, the problem is even more complicated than in Section 3.3; a balanced growth path in which n _t+1 = g n _t and e _t+1 = e _t is feasible, and now for any value of α there is a constant e that satisfies the resource constraint, but there is no reason to think that any given combination of α and e would actually be optimal. Meanwhile, the analytical analysis is complicated by the fact that the optimal path of e _t is not necessarily fixed.

Thus, we focus on simulations, using linear utility and the same parameter values as those presented in Fig. 2, but with e _t chosen as well subject to a production function $f(e_{t}) = 2e_{t} - {e_{t}^{2}}$. The latter implies that e _t = 1, the implied value in Section 3.3, is the maximum feasible value, and thus that the choice of e _t will be in [0, 1]. Figure 5 presents the optimal population and growth paths, exactly as in Fig. 2, and the result is also qualitatively identical: a planner with a higher α—a more total-utilitarian planner—would prefer a lower population for approximately the first 100 generations. The population growth rate is once again significantly higher in the first few periods with α = 0.4, as the population level is corrected to a higher path, and then growth settles down to a value slightly lower than that with α = 0.6.

Fig. 5

Optimal population paths in dynasty model with variable resource use. Notes: All panels present results for α = {0.4,0.6}, from simulations in which β = 0.975, γ = 1.04, T ₀ = 1, X = 100, κ = 0.1811, and f(e _t ) = 2e _t − e t2. Panel (a) presents results from the first 140 periods, while (b) only presents the first 20 periods, as population growth rates remain flat after that point

Figure 6, meanwhile, presents the optimal paths for e _t , which settle down very quickly, with a high- α planner preferring a lower resource use per period. This is intuitive, as a planner who places more value on population size relative to individual utility would prefer to restrict consumption in order to make a larger population possible. However, even with this effect, the main result of the paper holds: it is quite possible, even likely, that a planner with a higher welfare weight on population size would prefer a smaller population now and long into the future.

Fig. 6

Optimal resource income paths in dynasty model. Notes: Results are presented for the first 140 periods for α = {0.4,0.6}, from simulations in which β = 0.975, γ = 1.04, T ₀ = 1, X = 100, κ = 0.1811, and f(e _t ) = 2e _t − e t2

1.2 C.2 Capital and exhaustible resources

We now add physical capital to the dynasty model, and allow the planner to choose both the level of resources used per period, e _t , and the level of capital investment. If m _t is the level of capital per person in period t and δ is the depreciation rate, investment per person in period t is m _t+1 − (1 − δ)m _t , which we assume can be made at a cost w(m _t+1 − (1 − δ)m _t )². Then, specify output per person as $f(e_{t},m_{t},n_{t}) = (2z_{t}(e_{t},m_{t}) - z_{t}(e_{t},m_{t})^{2})n_{t}^{\phi -1}$, where $z_{t}(e_{t},m_{t}) = \left (ae_{t}^{\frac {\sigma -1}{\sigma }} + (1-a)m_{t}^{\frac {\sigma -1}{\sigma }}\right )^{\frac {\sigma }{\sigma -1}}$ is a CES aggregate of resource and capital with elasticity of substitution σ; we assume diminishing marginal returns to population by assuming that ϕ < 1 (this simply helps with ensuring convergence of the solution algorithm). Since $2z_{t} - {z_{t}^{2}}$ takes a maximum at 1, this prevents the planner from accumulating ever-increasing amounts of economic resources per person. However, planners with different tastes may prefer different divisions of z _t into e _t and m _t , and the more substitutable capital and the exhaustible resource are—that is, the larger is σ—the easier it will be to cut back on resource use and use capital instead, thus weakening the exhaustible resource constraint.

The planner’s problem is summarized in the following lagrangian:

$$\mathcal{L} = {\sum}_{t=0}^{\infty}{\beta^{t}n_{t}^{\alpha}u\left( f(e_{t},m_{t},n_{t}) - \kappa\frac{n_{t+1}}{n_{t}} - w\left( m_{t+1} - (1-\delta)m_{t}\right)^{2}\right)} - \lambda\left( {\sum}_{t=1}^{\infty}{\frac{n_{t}e_{t}}{T_{t}}} - X\right) $$

and we simplify the analysis by assuming linear utility, as in the simulations throughout the paper; then the first-order conditions are as follows:

$$\frac{\partial\mathcal{L}}{\partial n_{t}} = \beta^{t}n_{t}^{\alpha-2}\left( \alpha n_{t}c_{t} - (1-\delta)f(e_{t},m_{t},n_{t})n_{t} + \kappa n_{t+1}\right) - \beta^{t-1}n_{t-1}^{\alpha-1}\kappa - \lambda\frac{e_{t}}{T_{t}} = 0 $$

$$\frac{\partial\mathcal{L}}{\partial e_{t}} = \beta^{t}n_{t}^{\alpha}f_{e_{t}} - \lambda\frac{n_{t}}{T_{t}} = 0 $$

$$\frac{\partial\mathcal{L}}{\partial m_{t}} = \beta^{t}n_{t}^{\alpha}\left( f_{m_{t}} + 2(1-\delta)w(m_{t+1} - (1-\delta)m_{t})\right) - \beta^{t-1}n_{t-1}^{\alpha}2w(m_{t} - (1-\delta)m_{t-1}) = 0 $$

where $f_{e_{t}}$ and $f_{m_{t}}$ are partial derivatives of output with respect to e _t and m _t . If ϕ < 1, the second-order conditions are not guaranteed to be satisfied, but we have confirmed that they are satisfied in each of our numerical analyses to follow.

The first-order conditions are analytically quite complicated, but they can be simulated to illustrate their meaning. We use the standard calibration from before, except that X = 50 (and the simulations only run for 200 periods rather than 1000), and that we assume a = 0.5, w = 1, δ = 0.1, and ϕ = 0.9. Figure 7 presents optimal population paths for α = 0.4 and α = 0.6, as usual, but for four different values of σ, from 0.2 to 0.8.

Fig. 7

Optimal population sizes in capital and resource model. Notes: All panels present results for the first 50 periods for α = {0.4,0.6}, from simulations in which β = 0.975, γ = 1.04, T ₀ = 1, X = 50, κ = 0.1811, a = 0.5, w = 1, δ = 0.1, and ϕ = 0.9

We can see that, as σ increases, optimal populations generally increase, because easier substitution to capital means that the exhaustible resource constraint is less binding. More importantly for our purposes, however, the time at which the effect of α on optimal population becomes positive moves earlier. Higher- α planners have a greater incentive to switch from resource e _t to capital m _t , to loosen the exhaustible resource constraint and allow larger populations; we find that e _t drops faster and to a lower level when α = 0.6, while m _t increases faster and to a higher level with α = 0.6 in all but the σ = 0.2 case.

However, in the cases we study, our basic result continues to hold: the initial optimal population is lower for α = 0.6. The reason is that, in order to build up a large capital stock, the α = 0.6 planner must spend a lot on capital investment initially, which leaves few resources for fertility; as σ gets large, the optimal population actually drops initially to enable the planner to accumulate capital. At levels of σ above 0.8, we have had severe difficulties in finding a solution, as the planner wants to lower n ₁ to a very low level—and at that level the cost of reproduction for period 2 is very high, making consumption in period 1 low and thus reducing the desire to have people in period 1. Thus, the problem appears to be unstable when substitution is sufficiently easy, but the reason is that high- α planners want to lower initial populations too much, so it is unlikely that a high- α planner will ever want a larger initial population than a lower- α planner.

Therefore, we conclude that allowing for substitution between capital and exhaustible resources can weaken our result, in that the crossing point at which $\frac {dn_{t}}{d\alpha }$ becomes positive moves earlier. It is possible that sufficiently strong substitution could, under certain parameter configurations, completely overturn our result, but this does not appear to happen in the case we study.

1.3 C.3 Renewable resource

Next, we consider a case in which the resource is not exhaustible, but rather renews at a slow rate. Denoting the remaining stock of the resource at time t as X _t , we assume that:

$$X_{t+1} = X_{t} - \frac{n_{t}}{T_{t}} + \eta(\bar{X}-X_{t}) $$

where T ₁ = 1 and T _t = γ ^t−1, and where we now use η to denote the rate of resource renewal, and where $\bar {X}$ is the starting value of the resource stock. The resource renews itself in each period based on the amount existing at the beginning of the period, and grows at a faster rate when the resource is more depleted; one could easily think of alternative specifications for resource renewal, but this specification gives us a single parameter η that describes how renewable the resource is, which we can vary in the subsequent analysis.

We pursue our analysis in the exogenous-utility model. The planner’s problem is most easily specified as follows:

$$\max_{n_{t},X_{t}} W = {\sum}_{t=1}^{\infty}{\beta^{t-1}n_{t}^{\alpha}u_{t}} \hspace{0.2cm} s.t. \hspace{0.2cm} X_{t+1} = X_{t} - \frac{n_{t}}{T_{t}} + \eta(\bar{X}-X_{t}) \& \hspace{0.2cm} \frac{n_{t}}{T_{t}} \leq X_{t}, \forall t. $$

Thus, we think about the planner’s problem as choosing both n _t and X _t subject to the resource renewal equation and the feasibility constraint that states that no more resource can be used up in a period than exists at the start of the period. The lagrangian expression is:

$$\mathcal{L} = {\sum}_{t=1}^{\infty}{\left[\beta^{t-1}n_{t}^{\alpha}u_{t} - \lambda_{t}\left( X_{t+1} - X_{t} + \frac{n_{t}}{T_{t}} - \eta(\bar{X}-X_{t})\right) - \mu_{t}\left( \frac{n_{t}}{T_{t}} - X_{t}\right)\right]} $$

and taking the partial derivatives while assuming u _t = 1 for all t:

$$\frac{\partial\mathcal{L}}{\partial n_{t}} = \beta^{t-1}\alpha n_{t}^{\alpha-1} - \frac{\lambda_{t}}{T_{t}} - \frac{\mu_{t}}{T_{t}} = 0 $$

$$\frac{\partial\mathcal{L}}{\partial X_{t}} = \lambda_{t}(1-\eta) - \lambda_{t-1} + \mu_{t} = 0. $$

Initially, the feasibility constraint will not be binding, unless the planner wishes to use up all of the available resource in the first period; we can check that the latter condition does not hold, and so we assume that μ _t = 0 up to some time t ^∗ at which the constraint begins to bind. This time t ^∗ could be at infinity, or it could be a finite time. However, if t ^∗ occurs in finite time, then once the feasibility constraint begins to bind, it must bind forever, and the resource stock must stay at the steady-state value of $\theta \equiv \frac {\eta \bar {X}}{1+\eta }$, which is used up and renewed in every subsequent period.

To understand this, consider that allowing X _t to increase from 𝜃 and converge to some higher value is clearly dominated by remaining at X _t = 𝜃: the allowable resource usage is lower in every period in the former case, and thus population and welfare are also lower. Meanwhile, an oscillatory deviation from X _t = 𝜃 is also welfare-reducing under the standard assumption that β γ ^α < 1. To prove this, suppose that we have reached a resource stock of 𝜃 at time t, and have to decide whether or not to remain at 𝜃 in period t + 1 and beyond. The baseline case is that we stay at X _t+1 = X _t+2 = 𝜃, and so forth to infinity. Against this, consider the welfare obtained from an increase in X _t+1 to 𝜃 + 𝜖, followed by a return to 𝜃 for X _t+2 and thereafter. In the former case, the population sizes are n _t = 𝜃 γ ^t−1 and n _t+1 = 𝜃 γ ^t, and the welfare over those two periods is:

$$W_{b} = \beta^{t-1}(\theta\gamma^{t-1})^{\alpha} + \beta^{t}(\theta\gamma^{t})^{\alpha}. $$

Meanwhile, in the oscillatory deviation, the population sizes are n _t = (𝜃 − 𝜖)γ ^t−1 and n _t+1 = (𝜃 + (1 − η)𝜖)γ ^t, and the welfare over those two periods is:

$$W_{d} = \beta^{t-1}\left( (\theta-\epsilon)\gamma^{t-1}\right)^{\alpha} + \beta^{t}((\theta + (1-\eta)\epsilon)\gamma^{t})^{\alpha}. $$

Therefore, the net welfare gain from the oscillation is:

$${\Delta} W \equiv W_{d} - W_{b} = \left( \beta\gamma^{\alpha}\right)^{t-1}\left[(\theta-\epsilon)^{\alpha} - \theta^{\alpha}\right] + \left( \beta\gamma^{\alpha}\right)^{t}\left[(\theta + (1-\eta)\epsilon)^{\alpha} - \theta^{\alpha}\right]. $$

Now, assume that β γ ^α < 1 as before; then a necessary (but not sufficient) condition for the welfare gain to be positive is:

$$(\theta-\epsilon)^{\alpha} + (\theta + (1-\eta)\epsilon)^{\alpha} > 2\theta^{\alpha}. $$

However, Jensen’s inequality tells us that, since f(x) = x ^α is an increasing and concave function:

$$(\theta-\epsilon)^{\alpha} + (\theta + (1-\eta)\epsilon)^{\alpha} < (\theta-\epsilon)^{\alpha} + (\theta + \epsilon)^{\alpha} < 2\theta^{\alpha}. $$

Therefore, the oscillation must reduce welfare, and the only possible optimum is one in which X _t remains at 𝜃 after t ^∗ if t ^∗ is reached in finite time.

Therefore, for the time up to t ^∗, we can solve the first-order conditions setting μ _t = 0, which leads us to the following condition:

$$\frac{n_{t+1}}{n_{t}} = ((1-\eta)\beta\gamma)^{\frac{1}{1-\alpha}}. $$

If η = 0, then of course we are back in the standard exhaustible-resource world of Section 3, and the population growth rate is $(\beta \gamma )^{\frac {1}{1-\alpha }}$. However, faster resource renewal reduces the population growth rate, at least until the feasibility constraint binds; this may seem counter-intuitive, but the reason is that larger populations in early periods, by depleting the resource, now lead to more resource renewal in future periods, so there is an added benefit of a larger population at the beginning. Thus, the starting population level will increase with η, while the growth rate declines, so that later population values may increase or decrease.

The effect of α on the optimal population levels and growth rate is the same as before up to t ^∗, except that (1 − η) is added; if (1 − η)β γ > 1, so that population growth is positive, an increase in α raises the optimal population growth rate, and thus lowers the optimal starting population. However, this condition is less likely to be satisfied when η increases, so that an increase in resource renewability will eventually overturn our result: with sufficiently high η, population growth will be negative (though from a high value) until t ^∗ is reached, and higher α will make that growth rate more negative, with a larger starting population.

This is illustrated in Fig. 8, which presents optimal population paths for the standard parameter values, with η = {0.001, 0.004, 0.008, 0.015}. In each case, t ^∗ is reached in finite time, as can be seen in the final three figures. In the first three figures, the population growth rate increases with α, at least up to the point at which the constraint begins to bind, and the initial optimal population is lower when α = 0.6. When η = 0.015, population starts from a high level and then descends slowly, but faster and from a higher level when α = 0.6, exactly as predicted above.

Fig. 8

Optimal population sizes with renewable resource. Notes: All panels present optimal population paths for α = {0.4,0.6}, from simulations in which β = 0.975, γ = 1.04, T ₁ = 1, $\bar {X} = 100$, and η = {0.001,0.004,0.008,0.015}. Panels (a) through (c) present the first 100 periods, while (d) presents the first 50 periods

1.4 C.4 Population-induced innovation

Finally, this appendix considers the case in which population at time t determines the technology level at time t + 1 (and perhaps beyond). This captures, in a reduced-form way, the idea that a larger population could provide more opportunities or impetus for innovation, and is one more way in which the exhaustibility of our resource could be weakened.

Making T _t dependent on past population values adds considerably to the complexity of our model, so to illustrate the impact we focus on a simple 2-period version of our exogenous-utility model. We assume that T ₁ = 1, but that T ₂ is an increasing function of n ₁: $T_{2} = \gamma An_{1}^{\delta }$, where δ > 0, and where A is a constant with respect to α that we will vary with δ (as explained later) to prevent changes in δ from directly affecting the relative technological efficiencies.

We can then set up the maximization problem:

$$\mathcal{L} = n_{1}^{\alpha}u_{1} + \beta n_{2}^{\alpha}u_{2} - \lambda\left( n_{1} + \frac{n_{2}}{\gamma An_{1}^{\delta}} - X\right) $$

and if u ₁ = u ₂ = 1, the first-order conditions are:

$$\alpha n_{1}^{\alpha-1} = \lambda\left( 1 - \frac{\delta n_{2}}{\gamma An_{1}^{\delta+1}}\right) $$

$$\beta\alpha n_{2}^{\alpha-1} = \frac{\lambda}{\gamma An_{1}^{\delta}}. $$

Substituting for λ, we have:

$$n_{1}^{\alpha-1} = \beta\gamma An_{1}^{\delta}n_{2}^{\alpha-1}\left( 1 - \frac{\delta n_{2}}{\gamma An_{1}^{\delta+1}}\right) $$

along with the resource constraint, $n_{1} + \frac {n_{2}}{\gamma An_{1}^{\delta }} = X$.

Even in a two-period setting, the resulting algebra is rather complicated, so as before we use numerical simulations to explore the effect of δ on our results. In Fig. 9, we present simulations for β = 0.975, γ = 1.04, and X = 1; for any value of α, we also set $A = ({n_{1}^{0}})^{-\delta }$, where ${n_{1}^{0}}$ is the optimal n ₁ when δ = 0 for that value of α. In this way, changing the value of δ will affect the technological parameter only by shifting n ₁; there will be no direct effect whereby raising δ directly raises (if n ₁ > 1) or lowers (if n ₁ < 1) the value of $\gamma n_{1}^{\delta }$.

Fig. 9

Optimal population sizes and growth rate with induced innovation. Notes: Panel (a) presents the optimal population levels n ₁ and n ₂, while panel (b) presents the optimal growth rate $\frac {n_{2}}{n_{1}}$, for α = {0.4,0.6}, from simulations in which β = 0.975, γ = 1.04, X = 1, and δ = [0,0.05]

At low values of δ, of course, our standard result holds: population growth is positive, and a higher α raises population growth further while lowering n ₁. However, as δ increases, the same result obtains as in the previous appendix: population growth eventually falls below zero, and our result is flipped: a higher α means a larger initial population and a lower growth rate. As with renewable resources, induced innovation raises the value of large populations at early times, as they help to make the resource constraint less binding in the future. Therefore, a more total-utilitarian planner who wants to increase population when it is already large may choose to expand the population more in earlier periods.

Appendix D: An alternative specification for average utilitarianism

In our main analysis, we treat average utilitarianism as a generational average, applying our α parameter to the size of each generation. While we describe in Section 3 that this is the standard specification in the recent literature on the economics of population, it is not the universal standard: Nerlove et al. (1982, 1985, 1986b) all find that total utilitarians prefer larger populations in a setting in which average utilitarianism evaluates the average utility across all individuals over time. Furthermore, some philosophical objections have been raised against the generational average specification as a way of modelling average utilitarianism; for example, (Dasgupta 1998) argues that such a specification is ad hoc and lacks philosophical foundations.³¹

Therefore, in this appendix, we demonstrate that similar results hold when we model average utilitarianism as an average across all time, rather than across a generation; we thank Marc Fleurbaey for suggesting this analysis. We will specify our general social welfare function as:

$$W = \frac{{\sum}_{t=0}^{\infty}{\beta^{t}n_{t}u_{t}}}{\left( {\sum}_{t=0}^{\infty}{\beta^{t}n_{t}}\right)^{1-\alpha}} $$

where α is once again the weight placed on population; if α = 0, the function expresses pure average utilitarianism, while if α = 1, the term on the bottom is equal to one and the social welfare function corresponds to total utilitarianism.³² We will evaluate the effect of marginal changes to α on our results, but in each case the same conclusion would hold if considering the extreme cases of average (α = 0) and total (α = 1) utilitarianism.

To keep the algebra simple, we will only present here the two-period exogenous utility case; exactly analogous results follow in the infinite-horizon and dynasty models when considering changes in n _t and n _t+1, and are available upon request. The planner’s problem is:

$$\mathcal{L} = \frac{n_{1}u_{1} + \beta n_{2}u_{2}}{\left( n_{1} + \beta n_{2}\right)^{1-\alpha}} - \lambda\left( \frac{n_{1}}{T_{1}} + \frac{n_{2}}{T_{2}} - X\right). $$

However, we need to make one minor modification to the model to prevent knife’s-edge solutions. If u _t is simply a constant (or if $u_{t} = 1 - \kappa \frac {n_{t+1}}{n_{t}}$ in the dynasty model), then an interior solution can only exist if β γ = 1, and then any interior solution is valid. Instead, we assume that output features diminishing marginal returns with respect to population size, so that total output is equal to $n_{t}^{\delta }$ for δ < 1, and we assume linear utility over per-capita consumption so that $u_{t} = n_{t}^{\delta -1}$. Then the planner’s problem becomes:

$$\mathcal{L} = \frac{n_{1}^{\delta} + \beta n_{2}^{\delta}}{\left( n_{1} + \beta n_{2}\right)^{1-\alpha}} - \lambda\left( \frac{n_{1}}{T_{1}} + \frac{n_{2}}{T_{2}} - X\right). $$

Differentiating with respect to n ₁ and n ₂, we find:

$$\frac{\partial\mathcal{L}}{\partial n_{1}} = \frac{\delta n_{1}^{\delta-1}\left( n_{1} + \beta n_{2}\right) - (1-\alpha)\left( n_{1}^{\delta} + \beta n_{2}^{\delta}\right)}{\left( n_{1} + \beta n_{2}\right)^{2-\alpha}} - \frac{\lambda}{T_{1}} $$

$$\frac{\partial\mathcal{L}}{\partial n_{2}} = \frac{\beta\delta n_{2}^{\delta-1}\left( n_{1} + \beta n_{2}\right) - \beta(1-\alpha)\left( n_{1}^{\delta} + \beta n_{2}^{\delta}\right)}{\left( n_{1} + \beta n_{2}\right)^{2-\alpha}} - \frac{\lambda}{T_{2}}. $$

At the optimum, of course, both of these derivatives are equal to zero. Concavity cannot be guaranteed for all parameter values, but it holds in simulations with values similar to those used in Section 3; therefore, we assume concavity of the value function with respect to n ₁ and n ₂, and that thus an interior optimum exists.

However, the expressions are sufficiently complicated that we are not going to try to solve for n ₁ and n ₂; as it happens, we do not need to in order to sign the effect of α on n ₁ and n ₂. Suppose that we start from some $\hat {\alpha }$; then we could solve the equations above for the optimal $n_{1}\left (\hat {\alpha }\right )$ and $n_{2}\left (\hat {\alpha }\right )$, and the associated $\lambda \left (\hat {\alpha }\right )$. Now assume that α increases marginally to $\tilde {\alpha } = \hat {\alpha } + \varepsilon$, where ε is vanishingly small. Further assume that we hold n ₁ and n ₂ fixed for now, and that λ adjusts with α to hold $\frac {\partial \mathcal {L}}{\partial n_{2}} = 0$, so that λ takes the value $\lambda _{2}(\tilde {\alpha })$ that would make the planner willing to hold n ₂ unchanged. This, however, will not generally be the value of λ that will make $\frac {\partial \mathcal {L}}{\partial n_{1}} = 0$, so we can evaluate whether the planner would like to increase or decrease n ₁. If $\frac {\partial \mathcal {L}}{\partial n_{2}} = 0$, then we know that:

$$\lambda_{2}(\tilde{\alpha}) = T_{2}\frac{\beta\delta n_{2}^{\delta-1}\left( n_{1} + \beta n_{2}\right) - \beta(1-\tilde{\alpha})\left( n_{1}^{\delta} + \beta n_{2}^{\delta}\right)}{\left( n_{1} + \beta n_{2}\right)^{2-\tilde{\alpha}}} $$

and assuming that T ₂ = γ and T ₁ = 1, this means:

$$\begin{array}{@{}rcl@{}} \frac{\partial\mathcal{L}}{\partial n_{1}} &=& \frac{\delta n_{1}^{\delta-1}\left( n_{1} + \beta n_{2}\right) - (1-\tilde{\alpha})\left( n_{1}^{\delta} + \beta n_{2}^{\delta}\right)}{\left( n_{1} + \beta n_{2}\right)^{2-\tilde{\alpha}}}\\ &&- \gamma\frac{\beta\delta n_{2}^{\delta-1}\left( n_{1} + \beta n_{2}\right) - \beta(1-\tilde{\alpha})\left( n_{1}^{\delta} + \beta n_{2}^{\delta}\right)}{\left( n_{1} + \beta n_{2}\right)^{2-\tilde{\alpha}}}\\ &=& \frac{\delta\left( n_{1} + \beta n_{2}\right)\left( n_{1}^{\delta-1} - \beta\gamma n_{2}^{\delta-1}\right) + (1-\tilde{\alpha})\left( n_{1}^{\delta} + \beta n_{2}^{\delta}\right)\left( \beta\gamma-1\right)}{\left( n_{1} + \beta n_{2}\right)^{2-\tilde{\alpha}}}. \end{array} $$

As stated, at $\hat {\alpha }$, we know that $\frac {\partial \mathcal {L}}{\partial n_{1}} = 0$ by definition, and thus the numerator must be equal to zero at $\hat {\alpha }$. Increasing α affects the bottom of the fraction, but without changing the sign, whereas it also reduces the top as long as β γ > 1. Therefore, if β γ > 1, $\frac {\partial \mathcal {L}}{\partial n_{1}}(\tilde {\alpha }) < 0$. If the equilibrium λ with $\tilde {\alpha }$ was such that both $\frac {\partial \mathcal {L}}{\partial n_{1}}$ and $\frac {\partial \mathcal {L}}{\partial n_{2}}$ took the same sign at $n_{1}(\hat {\alpha })$ and $n_{2}(\hat {\alpha })$, there would be no allocation consistent with constrained optimization, as concavity would imply that both n ₁ and n ₂ should be moved in the same direction, which is not feasible given the resource constraint. Therefore, the optimal λ must be such that $\frac {\partial \mathcal {L}}{\partial n_{1}}$ and $\frac {\partial \mathcal {L}}{\partial n_{2}}$ take opposite signs at $n_{1}(\hat {\alpha })$ and $n_{2}(\hat {\alpha })$ given $\tilde {\alpha }$, and given that both are decreasing in λ, this is only possible at a $\lambda ^{*} < \lambda _{2}(\tilde {\alpha })$ at which $\frac {\partial \mathcal {L}}{\partial n_{1}} < 0$ and $\frac {\partial \mathcal {L}}{\partial n_{2}} > 0$. Then, by concavity, it must be that n ₁ decreases with α and n ₂ increases.

Thus, we again find the standard result from the paper: if β γ > 1, then a planner with a larger value of α—a more total-utilitarian planner—will want to reduce population today to allow for a larger population tomorrow. This proves that our result holds in this alternative specification for average utilitarianism, in the two-period exogenous-utility case; as stated earlier, exactly analogous results follow from the infinite-horizon and dynasty models when we consider changes in n _t and n _t+1, holding all other n fixed. In this specification, an increase in α means that increases in total discounted population are weighted more strongly, as they do not count as heavily in the denominator; therefore, the planner would like to increase population in the most efficient way, and if β γ > 1, technology is improving faster than discounting and it is best to increase the future population at the expense of today.