A.1. Proof of Proposition 1

Conditional on mating, a woman will always prefer an α, since she gets a higher utility for any parental investment. This proves (i). As a result, the maximization problem of women is

$$\begin{eqnarray*} &&\max _{c,\upsilon,z}\ln c+\gamma \nu(p_{\alpha}\ln (\alpha z^{\psi})+(1-p_{\alpha})\ln (\beta z^{\psi})), \\ &&\quad s.t.\text{}c+\upsilon z \le Ah, \\ &&\quad 0 \le \upsilon \le n. \end{eqnarray*}$$

Writing the Lagrangian as

$$\begin{equation*} \ln c+\gamma \nu(p_{\alpha}\ln (\alpha z^{\psi})+(1-p_{\alpha})\ln (\beta z^{\psi}))+\lambda \left( Ah-c-\upsilon z\right) +\mu (n-\upsilon) +\varepsilon \upsilon, \end{equation*}$$

it is easy to see that the full fertility regime (μ ⩾ 0) prevails iff

(32)$$\begin{equation} a+\psi \ln z \ge \psi . \end{equation}$$

In this case, the solution is given by

(33)$$\begin{eqnarray} \upsilon &=&n \end{eqnarray}$$

(34)$$\begin{eqnarray} z &=&\frac{\gamma \psi}{1+\gamma n\psi}Ah; \end{eqnarray}$$

(35)$$\begin{eqnarray} c &=&\frac{1}{1+\gamma n\psi}Ah. \end{eqnarray}$$

Consequently, (32) holds iff

(36)$$\begin{equation} a+\psi \left( \ln A+\ln h\right) +\psi \ln (\gamma \psi) -\psi \ln (1+\gamma n\psi) \ge \psi . \end{equation}$$

Clearly, from this, the minimum level of human capital for a woman to be in this regime is $\underline{h},$ as defined in (1). Therefore, all women are willing to mate n times with an α man as long as they all have more human capital than $\underline{h}.$

Next, a beta offspring with a mother of human capital h has a human capital given by βz ^ψ. This is larger than $\underline{h}$ iff

(37)$$\begin{equation} \ln \beta +\psi \left( \ln A+\ln h\right) +\psi \ln (\gamma \psi) -\psi \ln (1+\gamma n\psi) \ge \ln \underline{h}. \end{equation}$$

If the distribution of human capital is above $\underline{h},$ a sufficient condition for the distribution of human capital among offsprings to have that same property is that (37) holds at $h=\underline{h}.$ Rearranging using (1), we see that this is equivalent to (6) . This proves that (v) holds if the two assumptions in the Proposition hold. To check that (i) and (ii) hold, we need to check that it is rational for alpha men to mate. If they do so, their utility is clearly given by

(38)$$\begin{equation} \bar{U}_{\alpha}^{\ast}(h^{\ast})=\ln A+\ln h^{\ast}+\frac{n}{p_{\alpha}}E\ln h^{\prime}. \end{equation}$$

This is because the proportion of alpha men in the total population of men is p _α, therefore each alpha man mates n/p _α times. For this to be larger than utility without mating, it must be that Elnh ⩾ 0. Clearly, the expected human capital of an offspring condition on the mother’s human capital h is given by

$$\begin{eqnarray*} E(\ln h^{\prime} &\mid &h)=a+\psi \ln z \\ &=&a+\psi \left( \ln A+\ln h\right) +\psi \ln (\gamma \psi) -\psi \ln (1+\gamma n\psi) \\ &=&\psi (\ln h-\ln \underline{h}+1) \ge 0. \end{eqnarray*}$$

This proves that Elnh ⩾ 0 and therefore that it is also in the interest of alpha men to mate. This proves claims (ii) and (iii). From the preceding formula we get that $E\ln h^{\prime}=\psi (E\ln h-\ln \underline{h}+1),$ which, once substituted into (38), proves (4). Equation (3) holds straightforwardly from (i). Equation (5) holds from substituting the optimal policies (33)–(35) into the woman’s utility function. This proves (iv).

A.2. Proof of Proposition 2

The household maximizes

$$\begin{equation*} \max _{c,c^{\ast},\upsilon,z}\theta (\ln c+\gamma \upsilon E\ln h^{\prime})+(1-\theta) (\ln c^{\ast}+\gamma \upsilon E\ln h^{\prime}), \end{equation*}$$

for some θ ∈ [0, 1]. The budget constraint for the couple is now

$$\begin{equation*} c+c^{\ast}+\nu z\le A(h+h^{\ast}). \end{equation*}$$

Furthermore

$$\begin{equation*} E\ln h^{\prime}=b+\psi \ln z. \end{equation*}$$

The Lagrangian is

$$\begin{eqnarray*} L &=&\theta (\ln c+\gamma \upsilon E\ln h^{\prime})+(1-\theta) (\ln c^{\ast}+\gamma \upsilon E\ln h^{\prime}) \\ &&+\;\lambda \left( A(h+h^{\ast})-\left( c+c^{\ast}+\nu z\right) \right) +\mu (n-\upsilon) +\varepsilon \upsilon . \end{eqnarray*}$$

Looking at the FOCs, a regime such that $\upsilon =n$ arises iff (8)–(35) holds and if μ ⩾ 0, i.e. $\partial L/\partial \upsilon \ge 0,$ or equivalently

$$\begin{equation*} \gamma (b+\psi \ln z) \ge \lambda z. \end{equation*}$$

From λ = θ/c and (8)–(35), one can check that this inequality is equivalent to $h+h^{\ast} \ge \underline{h}(\beta) .$ One can then prove (11)–(12) by simple substitution of the optimal policies into the utility function.

A.3. Proof of Proposition 3

Confronting (11) with (5), we see that for the woman to gain from marriage she must at least get a share θ_min of consumption, where

$$\begin{equation*} \theta _{\min}=e^{k}\left( 1+\frac{h^{\ast}}{h}\right) ^{-(1+\gamma n\psi) }. \end{equation*}$$

The match is efficient if the man’s utility U*_β(h, h*, θ), computed at θ = θ_min, is greater than the man’s outside option, which is given by (3). That defines the condition under which the marriage takes place:

$$\begin{equation*} 1-\theta _{\min}\ge \tilde{h}^{\gamma n\psi}h^{\ast}(h+h^{\ast})^{-(1+\gamma n\psi) }. \end{equation*}$$

Substituting the value of θ_min, we get Equation (13).

A.4. Proof of Proposition 4

Suppose it’s not. Then we can find two married couples, (h ₀, h*₁) and (h ₁, h*₀), such that h ₀ ⩽ h ₁ and h*₀ = h*(h ₁) ⩽ h*₁ = h*(h ₀). Let θ₀ = θ(h ₀) and θ₁ = θ(h ₁). For this assignment to be an equilibrium, condition (iii) in Definition 2 must hold. Let us apply it for h* = h*₁ and h = h ₁. Using (11), we see that

(39)$$\begin{equation} \ln \hat{\theta}(h_{1},h_{1}^{\ast})=\ln \theta _{1}-(1+\gamma n\psi) (\ln (h_{1}+h_{1}^{\ast})-\ln (h_{1}+h_{0}^{\ast})). \end{equation}$$

Using (12) and (17), we see that we must have

$$\begin{equation*} \ln (1-\hat{\theta}(h_{1},h_{1}^{\ast}))\le \ln (1-\theta _{0})+(1+\gamma n\psi) (\ln (h_{1}^{\ast}+h_{0})-\ln (h_{1}^{\ast}+h_{1})). \end{equation*}$$

Substituting (39), we get that the following inequality must hold:

$$\begin{equation*} (h_{1}+h_{1}^{\ast})^{1+\gamma n\psi}\le (1-\theta _{0})(h_{1}^{\ast}+h_{0})^{1+\gamma n\psi}+\theta _{1}(h_{1}+h_{0}^{\ast})^{1+\gamma n\psi}. \end{equation*}$$

If we now apply condition (iii) to h* = h*₀ and h = h ₀, we get a similar condition

$$\begin{equation*} (h_{0}+h_{0}^{\ast})^{1+\gamma n\psi}\le (1-\theta _{1})(h_{1}+h_{0}^{\ast})^{1+\gamma n\psi}+\theta _{0}(h_{1}^{\ast}+h_{0})^{1+\gamma n\psi}. \end{equation*}$$

Adding these two inequalities, we get that the following must hold

(40)$$\begin{equation} (h_{1}+h_{1}^{\ast})^{1+\gamma n\psi}+(h_{0}+h_{0}^{\ast})^{1+\gamma n\psi}\le (h_{1}+h_{0}^{\ast})^{1+\gamma n\psi}+(h_{1}^{\ast}+h_{0})^{1+\gamma n\psi}. \end{equation}$$

However, the strict convexity of the function ψ(x) = x ^{1 + γnψ} precludes it. Let $\mu =\frac{h_{1}-h_{0}}{h_{1}-h_{0}+h_{1}^{\ast}-h_{0}^{\ast}}\in [0,1].$ One has h ₁ + h*₀ = μ(h ₁ + h*₁) + (1 − μ)(h ₀ + h*₀), and h*₁ + h ₀ = (1 − μ)(h ₁ + h*₁) + μ(h ₀ + h*₀). Therefore, ψ(h*₁ + h ₀) ⩽ (1 − μ)ψ(h ₁ + h*₁) + μψ(h ₀ + h*₀) and ψ(h ₁ + h*₀) ⩽ μψ(h ₁ + h*₁) + (1 − μ)ψ(h ₀ + h*₀). Adding these two inequalities, we clearly contradict (40).

A.5. Proof of Proposition 5

The first step is to show that if a man is married, all men with greater human capital must also be married. To see this, consider a man with human capital h*₀, married with a woman with human capital h ₀. Let θ₀ = θ(h ₀) be her corresponding equilibrium output share. A man with human capital h*₁ ⩾ h ₀* can marry her if he gives her a share $\hat{\theta}(h_{0},h_{1}^{\ast})$ such that $U_{\beta}(h_{0},h_{1}^{\ast},\hat{\theta}(h_{0},h_{1}^{\ast}))=V(h_{0})=U_{\beta}(h_{0},h_{0}^{\ast},\theta _{0}),$ or equivalently, using (11),

(41)$$\begin{equation} \ln \hat{\theta}(h_{0},h_{1}^{\ast})=\ln \theta _{0}+(1+\gamma n\psi) \left[ \ln (h_{0}+h_{0}^{\ast})-\ln (h_{0}+h_{1}^{\ast})\right] . \end{equation}$$

If h*₁ is single, then he must not be better-off by marrying h ₀ and offering her an output share equal to $\hat{\theta} (h_{0},h_{1}^{\ast});$ otherwise, (17) would be violated. Therefore, we must have $U_{\beta}^{\ast}(h_{0},h_{1}^{\ast},\hat{\theta} (h_{0},h_{1}^{\ast}))\le V(h_{1}^{\ast})=\bar{U}_{\beta}^{\ast}(h_{1}^{\ast}),$ or equivalently, using (12) and (3),

(42)$$\begin{equation} \ln (1-\hat{\theta}(h_{0},h_{1}^{\ast}))\le \gamma n\psi \ln \tilde{h}+\ln h_{1}^{\ast}-(1+\gamma n\psi) \ln (h_{0}+h_{1}^{\ast}). \end{equation}$$

Substituting (41) into (42), we see that the following inequality must hold:

(43)$$\begin{equation} (h_{0}+h_{1}^{\ast})^{1+\gamma n\psi}-\theta _{0}(h_{0}+h_{0}^{\ast})^{1+\gamma n\psi}\le \tilde{h}^{\gamma n\psi}h_{1}^{\ast}. \end{equation}$$

At the same time, h*₀ must be better-off married with h ₀ than single, otherwise (16) would be violated. Using (12) and (3), this is equivalent to

(44)$$\begin{equation} (h_{0}+h_{0}^{\ast})^{1+\gamma n\psi}-\theta _{0}(h_{0}+h_{0}^{\ast})^{1+\gamma n\psi}\ge \tilde{h}^{\gamma n\psi}h_{0}^{\ast}. \end{equation}$$

Putting together (44) and (43), we see that the following inequality must hold:

(45)$$\begin{equation} (h_{0}+h_{0}^{\ast})^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{0}^{\ast}\ge (h_{0}+h_{1}^{\ast})^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{1}^{\ast}. \end{equation}$$

Observe that $\underline{h}(\beta) =\tilde{h}\exp (1-\ln (1+\gamma n\psi) /\gamma n\psi) \ge \tilde{h},$ and that the expression $(h_{0}+x)^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}x$ is strictly increasing in x over the relevant range provided $h_{\min} \ge \frac{1}{2}(1+$$\gamma n\psi) ^{-\gamma n\psi}\tilde{h},$ which is clearly implied by our assumption that $h_{\min} \ge \underline{h}(\beta) .$ Therefore, (45) cannot hold for h*₁ ⩾ h ₀*. Therefore, h*₁ must be married too. Consequently, it must be that $S^{\ast}=[\underline{h}^{\ast},h_{\max}].$

Next, we show that the inverse assignment function h*⁻¹() must be continuous over S*. Suppose it is not the case. Since it is monotonic, the set of its discontinuity points is at most countable. Then there exists some $h_{0}^{\ast}\in \mathring{S}$ such that h ₁ = lim_{h* → h*⁺0}h*⁻¹() ⩾ h*⁻¹(h*₀) = h ₀.Footnote ⁴⁰ Then, all women in (h ₀, h ₁) must be single. Furthermore, a man with human capital h*₀ must be indifferent between marrying a woman with human capital h ₀ or a woman with human capital (arbitrarily close to) h ₁, since he prefers the former but a man with an arbitrarily close human capital to his prefers the latter. Denoting θ₀ = θ(h ₀) and θ₁ = lim_{h → h ⁺1}θ(h), this can be written as

(46)$$\begin{equation} \ln (1-\theta _{1})+(1+\gamma n\psi) \ln (h_{0}^{\ast}+h_{1})=\ln (1-\theta _{0})+(1+\gamma n\psi) \ln (h_{0}^{\ast}+h_{0}). \end{equation}$$

Another equilibrium condition is that all women such that h ∈ (h ₀, h ₁) could not be better-off if they married h*₀. The woman’s output share that would leave him indifferent between marrying h ₀ or h ₁ and marrying h is $\hat{\theta}^{\ast}(h,h_{0}^{\ast})$ such that

(47)$$\begin{equation} \ln (1-\hat{\theta}^{\ast}(h,h_{0}^{\ast}))=(1+\gamma n\psi) \ln (h_{0}^{\ast}+h_{0})+\ln (1-\theta _{0})-(1+\gamma n\psi) \ln (h_{0}^{\ast}+h). \end{equation}$$

That a woman with h ∈ (h ₀, h ₁) prefers to be single than marrying h*₀ under these terms can be written as

(48)$$\begin{equation} k+(1+\gamma n\psi) \ln h\ge \ln \hat{\theta}^{\ast}(h,h_{0}^{\ast})+(1+\gamma n\psi) \ln (h_{0}^{\ast}+h). \end{equation}$$

Note that $\hat{\theta}^{\ast}(h_{0},h_{0}^{\ast})=\theta _{0}$ and $\hat{ \theta}^{\ast}(h_{1},h_{0}^{\ast})=\theta _{1}.$ Taking limits in (48) for h → h ₀ and h → h ₁ and noting that a woman with h = h ₀ or h arbitrarily close to h ₁ is married in equilibrium and thus not worse-off than single, we see that (48) must hold with equality at the bounds of (h ₀, h ₁), i.e.

(49)$$\begin{equation} k+(1+\gamma n\psi) \ln h_{0}=\ln \theta _{0}+(1+\gamma n\psi) \ln (h_{0}^{\ast}+h_{0}); \end{equation}$$

(50)$$\begin{equation} k+(1+\gamma n\psi) \ln h_{1}=\ln \theta _{1}+(1+\gamma n\psi) \ln (h_{0}^{\ast}+h_{1}). \end{equation}$$

Using (49) and (50) to eliminate θ₀ and θ₁ in (46), we get that

$$\begin{equation*} (h_{0}^{\ast}+h_{1})^{1+\gamma n\psi}-e^{k}h_{1}^{1+\gamma n\psi}=(h_{0}^{\ast}+h_{0})^{1+\gamma n\psi}-e^{k}h_{0}^{1+\gamma n\psi}. \end{equation*}$$

Let ϕ(h) be the function defined by ϕ(h) = (h*₀ + h)^{1 + γnψ} − e^kh ^{1 + γnψ}. It is easy to see that ϕ′(h) is positive and then negative as h goes from zero to infinity. Since ϕ(h ₀) = ϕ(h ₁), ϕ() must be hump-shaped between h ₀ and h ₁, implying that

$$\begin{equation*} \phi (h) \ge \phi (h_{0})=\phi (h_{1})\text{ for}h\in (h_{0},h_{1}). \end{equation*}$$

But, substituting (47) into (48), we see that we must also have

$$\begin{equation*} (h+h_{0}^{\ast})^{1+\gamma n\psi}-e^{k}h^{1+\gamma n\psi}\le (1-\theta _{0})(h_{0}^{\ast}+h_{0})^{1+\gamma n\psi}. \end{equation*}$$

Substituting again the value of θ₀ from (49), we see that this is equivalent to

$$\begin{equation*} (h+h_{0}^{\ast})^{1+\gamma n\psi}-e^{k}h^{1+\gamma n\psi}=\phi (h)\le \phi (h_{0})=(h_{0}+h_{0}^{\ast})^{1+\gamma n\psi}-e^{k}h_{0}^{1+\gamma n\psi}, \end{equation*}$$

which is clearly a contradiction. Therefore, the inverse assignment function must be continuous, implying that S is an interval.

Let $\underline{h}$ be the lower bound of S. It must be that $h^{\ast}( \underline{h})=\underline{h}^{\ast}.$ If $\underline{h}^{\ast}=h_{\min},$ then all men are married, so must all women, and one must have S = [h _min, h _max]. The equilibrium is then Victorian. Assume then that $ \underline{h}^{\ast} > h_{\min}.$ Assume $\underline{h} > h_{\min}.$ Then, all women such that $h \le \underline{h}$ are single. We can use similar steps as the ones used to derive (47)–(50) to show that $\underline{h}$ is just indifferent between being married and single, i.e.

$$\begin{equation*} k+(1+\gamma n\psi) \ln \underline{h}=\ln \theta (\underline{h})+(1+\gamma n\psi) \ln (\underline{h}^{\ast}+\underline{h}). \end{equation*}$$

By the same token, $\underline{h}^{\ast}$ is also indifferent between being married and single, that is

$$\begin{equation*} \ln (1-\theta (\underline{h}))+(1+\gamma n\psi) \ln (\underline{h}^{\ast}+ \underline{h})=\gamma n\psi \ln \tilde{h}+\ln \underline{h}^{\ast}. \end{equation*}$$

Putting these two conditions together, we see that the marriage viability condition (13) must be satisfied with equality at $h=\underline{h}$ and $h^{\ast}=\underline{h}^{\ast}.$ But this implies that it is satisfied strictly for any $h < \underline{h}$ and $h^{\ast}=\underline{h}^{\ast}.$ Therefore, a woman with human capital $h < \underline{h}$ can underbid $ \underline{h}$ to marry $\underline{h}^{\ast}$ and give him a positive surplus, meaning that condition (iii) in Definition 2 must be violated. Hence, it cannot be that $\underline{h} > h_{\min},$ implying that if $ \underline{h}^{\ast} > h_{\min}$ the equilibrium must be SATC.

A.6. Proof of Proposition 6

Assume (18) holds. We show that a Victorian assignment along with a sharing rule defined by (19) matches all equilibrium conditions in Definition 2. Let us start with condition (iii). Since all women are married, their reservation utility is given by

(51)$$\begin{equation} V(h)=\ln \text{}\theta (h)+(1+\gamma n\psi) (\ln A+\ln (h+h^{\ast}(h)))+\pi _{\beta}.\end{equation}$$

By marrying another man with human capital h* and get a fraction θ of consumption, their utility would be given by (11). Therefore, the consumption share that would make them indifferent between their marriage and this alternative marriage is given by

(52)$$\begin{equation} \ln \hat{\theta}(h,h^{\ast})=\ln \theta (h)+(1+\gamma n\psi) (\ln (h+h^{\ast}(h))-\ln (h+h^{\ast})). \end{equation}$$

The new husband utility is now

$$\begin{equation*} U_{\beta}^{\ast}(h,h^{\ast},\hat{\theta}(h,h^{\ast}))=\ln \text{}(1- \hat{\theta}(h,h^{\ast}))+(1+\gamma n\psi) (\ln A+\ln (h+h^{\ast}))+\pi _{\beta}. \end{equation*}$$

It must not exceed the utility he had in his assigned marriage

(53)$$\begin{equation} V^{\ast}(h^{\ast})=\ln (1-\theta (h^{\ast -1}(h^{\ast})))+(1+\gamma n\psi) (\ln A+\ln (h^{\ast -1}(h^{\ast})+h^{\ast}))+\pi _{\beta}. \end{equation}$$

Using (52) to eliminate $\hat{\theta}(h,h^{\ast}),$ and rearranging, we see that the condition $U_{\beta}^{\ast}(h,h^{\ast},\hat{\theta} (h,h^{\ast}))\le V^{\ast}(h^{\ast})$ is equivalent to

(54)$$\begin{equation} (h+h^{\ast})^{1+\gamma n\psi}\le \theta (h)(h+h^{\ast}(h))^{1+\gamma n\psi}+(1-\theta (h^{\ast -1}(h^{\ast})))(h^{\ast -1}(h^{\ast})+h^{\ast})^{1+\gamma n\psi}, \end{equation}$$

and, since the actual equilibrium assignment defines V*(h*), this will hold with equality for h* = h*(h).

In our candidate equilibrium, we have h*(h) = h. The preceding formula can simply be rewritten as

$$\begin{equation*} (h+h^{\ast})^{1+\gamma n\psi}\le \theta (h)(2h)^{1+\gamma n\psi}+(1-\theta (h^{\ast}))(2h^{\ast})^{1+\gamma n\psi}. \end{equation*}$$

Suppose now that we have $\theta (h)=\frac{1}{2}\left( 1+\lambda h^{-(1+\gamma n\psi) }\right),$ this boils down to

$$\begin{equation*} \left(\frac{h+h^{\ast}}{2}\right)^{1+\gamma n\psi}\le \frac{h^{1+\gamma n\psi}+h^{\ast}{}^{1+\gamma n\psi}}{2}, \end{equation*}$$

which is true by convexity. This proves that our candidate equilibrium satisfies (iii) in Definition 2.

Let us now check condition (i), i.e. that women are better-off married than single. Comparing (51) and (5), we see that the condition $ V(h)\ge \bar{U}_{\alpha}(h)$ is equivalent to

$$\begin{equation*} \theta (h)\ge e^{k}2^{-(1+\gamma n\psi) }. \end{equation*}$$

Given (19), this is equivalent to

$$\begin{equation*} \lambda \ge \max (\left( e^{k}2^{-\gamma n\psi}-1\right) h_{\min}^{1+\gamma n\psi},\left( e^{k}2^{-\gamma n\psi}-1\right) h_{\max}^{1+\gamma n\psi}). \end{equation*}$$

This defines the set of values of λ for which (i) in Definition 2 holds.

Turning now to condition (ii), using (53) and (3), and the fact that h*⁻¹(h*) = h*, the condition $V^{\ast}(h^{\ast})\ge \bar{U}_{\beta}^{\ast}(h^{\ast})$ is equivalent to

(55)$$\begin{equation} 1-\theta (h^{\ast})\ge 2^{-(1+\gamma n\psi) }\tilde{h}^{\gamma n\psi}h^{\ast -\gamma n\psi}. \end{equation}$$

If λ ⩾ 0 then the LHS goes up with h*; since the RHS falls with h*, then for this to hold for all h* it must hold for h* = h _min. We get the condition that

(56)$$\begin{equation} \lambda \le \left( 1-2^{-\gamma n\psi}\tilde{h}^{\gamma n\psi}h_{\min}^{-\gamma n\psi}\right) h_{\min}^{1+\gamma n\psi}. \end{equation}$$

Note that the RHS to this inequality is always positive.

If λ ⩽ 0 then the LHS of (55) is always greater than 0.5, and therefore always exceeds the RHS since $h^{\ast} \ge \underline{h}(\beta) \ge \tilde{h}.$

Summarizing all these findings, we see that there exist values of λ which satisfy (i) and (ii) if and only if (18) holds – which is always the case if e^k ⩽ 2^γnψ. The relevant interval of values of λ is then clearly defined by (20).

The preceding steps imply that if (18) holds, then a Victorian equilibrium exists. Furthermore, the claims of part B hold by construction. We now prove that those conditions are necessary.

We have seen that for the equilibrium condition (87) to hold (see footnote 30), it must be that (54) holds. We also know that (54) holds with equality at h* = h*(h). Thus, h*(h) must be a local extremum of the RHS of (54) minus its LHS, as a function of h*. Locally, this means that:

(57)$$\begin{eqnarray} &&(1+\gamma n\psi) (h+h^{\ast}(h))^{\gamma n\psi} \nonumber\\ &&\quad=(1-\theta (h))(1+\gamma n\psi) (h+h^{\ast}(h))^{\gamma n\psi}\times \left( 1+h^{\ast \prime}(h)^{-1}\right) \nonumber\\ &&\qquad-\;(h+h^{\ast}(h))^{1+\gamma n\psi}\frac{\theta ^{\prime}(h)}{h^{\ast \prime}(h)}. \end{eqnarray}$$

In a Victorian equilibrium h*(h) = h, and this simplifies to

$$\begin{equation*} \theta ^{\prime}(h)=\frac{1+\gamma n\psi}{2h}(1-2\theta (h)). \end{equation*}$$

The solution to this differential equation is given by (19), which is therefore a necessary condition for a Victorian equilibrium. We then have already seen that (20) is necessary for (i) and (ii) in Definition 2 to be satisfied, and that this interval of values of λ is nonempty if and only if (18) holds. This proves that (18) is necessary and that the conditions in B necessarily hold.

A.7. Proof of Proposition 7

A. Constructing the assignment for the SATC equilibrium

We now show that an SATC equilibrium can be constructed. For this, we look for a pair ($\underline h$$^{\ast},\bar{h})$ such that $\underline h$* ⩾ h _min, $\bar{h} \le h_{\max},$ and the assignment given by

(58)$$\begin{equation} h^{\ast}(h)=F^{-1}(F(h)+F(\underline{h}^{\ast})), \end{equation}$$

is an equilibrium one. Clearly, given $\bar{h},$ if we choose

(59)$$\begin{equation} \underline{h}^{\ast}=F^{-1}(1-F(\bar{h}))=\underline{h}^{\ast}(\bar{h}), \end{equation}$$

the candidate assignment will map $S=[h_{\min},\bar{h}]$ to S* = [$\underline h$*, h _max] and satisfy (14). Therefore, it is indeed an assignment:

• • We have proved that given any $\bar{h},$ the value of $\underline{h} ^{\ast}$ given by (59) and the h*() function defined by (58) are an assignment.

B. Checking that married people cannot underbid one another

Next, let us assume that the sharing function θ(h) satisfies (23). We show that (iii) in Definition 2 holds for h ∈ S and h* ∈ S*. As shown in the proof of Proposition 6, this is equivalent to (54). Substituting (23), we get that this is equivalent to

(60)$$\begin{equation} (h+h^{\ast})^{1+\gamma n\psi}\le (h^{\ast -1}(h^{\ast})+h^{\ast})^{1+\gamma n\psi}+(1+\gamma n\psi) \int _{h^{\ast -1}(h^{\ast})}^{h}(z+h^{\ast}(z))^{\gamma n\psi}dz,\end{equation}$$

for all h, h* ∈ S × S*. Clearly, equality holds for h* = h*(h). Furthermore, the derivative of the RHS of (60) with respect to h* is (1 + γnψ)(h*⁻¹(h*) + h*)^γnψ, while the derivative of the LHS is (1 + γnψ)(h + h*)^γnψ. Given that h*(h) is increasing, the former is clearly larger than the latter for h* ⩾ h*(h), and smaller for h* ⩽ h*(h). Consequently, the difference between the RHS and the LHS reaches it minimum at h* = h*(h); hence (60) holds.

• • We have proved that if θ(h) satisfies (23), then condition (iii) holds for (h, h*) ∈ S × S*.

C. Deriving the value-matching conditions at the frontier of S and S*

Next, we show that there exist values for $\mu,\bar{h}$ and $\underline h$* such that, in addition to (59), the two following conditions hold:

(61)$$\begin{equation} \bar{U}_{\alpha}(\bar{h})=U_{\beta}(\bar{h},h_{\max},\theta (\bar{h})); \end{equation}$$

(62)$$\begin{equation} \bar{U}_{\beta}^{\ast}(\underline{h}^{\ast})=U_{\beta}^{\ast}(h_{\min}, \underline{h}^{\ast},\theta (h_{\min}). \end{equation}$$

These two conditions mean that the reservation utilities V() and V*() do not jump as one crosses the boundaries of S and S*. Otherwise, the equilibrium conditions would be violated. Suppose, for example, that a woman such that h is marginally higher than $\bar{h}$ has a utility higher than $U_{\beta}(\bar{h},h_{\max},\theta (\bar{h}))$ by a discrete amount. Then, since the θ() function is continuous over S, women with h below $\bar{h}$ but arbitrarily close to it would be better-off being single, and condition (i) in Definition 2 would be violated. Suppose now that a woman with h marginally higher than $\bar{h}$ has a utility lower than $U_{\beta}(\bar{h},h_{\max},\theta (\bar{h}))$ by a discrete amount. Then $\hat{\theta}(h,h_{\max}) \le \theta (\bar{h}):$ since these women are arbitrarily close to $\bar{h},$ but have a discretely lower utility than the married women with $\bar{h},$ they can reach that same utility by marrying a man with h _max and get a lower fraction of the surplus. But the h _max man would then be better-off and this would violate condition (iii). Therefore, (61) must hold. A similar reasoning applies to (62). Using (5) and (11), we see that (61) is equivalent to

$$\begin{equation*} \ln \theta (\bar{h})=k+(1+\gamma n\psi) \ln \bar{h}-(1+\gamma n\psi) \ln ( \bar{h}+h_{\max}). \end{equation*}$$

Substituting (23), we see that this is equivalent to

(63)$$\begin{equation} \mu =e^{k}\bar{h}^{1+\gamma n\psi}-\left( 1+\gamma n\psi \right) \int _{h_{\min}}^{\bar{h}}(z+h^{\ast}(z))^{\gamma n\psi}dz=\mu _{H}(\bar{h}) . \end{equation}$$

Similarly, we can substitute (3) and (12) into (62) and get

$$\begin{equation*} \ln (1-\theta (h_{\min}))=-\gamma n\psi \ln A-\pi _{\beta}+\ln \underline{h}^{\ast}-(1+\gamma n\psi) \ln (h_{\min}+\underline{h}^{\ast}), \end{equation*}$$

or equivalently, given (23),

(64)$$\begin{equation} \mu =(h_{\min}+\underline{h}^{\ast})^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}\underline{h}^{\ast}=\mu _{L}(\bar{h}). \end{equation}$$

Equations (65) and (64) define a 2 × 2 system in $\bar{h}$ and μ, where $\underline{h}^{\ast}$ is implicitly treated as a function of $ \bar{h}$ defined by (59).

• • We have proved that μ and $\bar{h}$ must satisfy (63) and (64) in equilibrium.

D. Showing that there is a solution, for $\mu,\bar{h},\underline{h} ^{\ast}$which satisfies the value-matching conditions as well as condition (iii) in Definition 2 for pairs of singles

To prove that it has a solution, we use the intermediate value theorem. First, we show that μ_H(h _max) ⩾ μ_L(h _max). If $\bar{h} =h_{\max}$, then h*(h) = h. Therefore, μ_H(h _max) = e^kh ^{1 + γnψ}_max − 2^γnψ(h ^{1 + γnψ}_max − h ^{1 + γnψ}_min), and μ_L(h _max) = $2^{1+\gamma n\psi}h_{\min}^{1+\gamma n\psi}-\tilde{h} ^{\gamma n\psi}h_{\min}.$ Clearly, the condition $e^{k}h_{\max}^{1+\gamma n\psi}-2^{\gamma n\psi}\left( h_{\max}^{1+\gamma n\psi}-h_{\min}^{1+\gamma n\psi}\right) \ge 2^{1+\gamma n\psi}h_{\min}^{1+\gamma n\psi}- \tilde{h}^{\gamma n\psi}h_{\min}$ is equivalent to (18) being violated, which is true by assumption.

Next, let $\hat{h}$ be the minimum possible value of $\bar{h}$ such that condition (iii) in Definition 2 holds for $h \ge \bar{h}$ and $h^{\ast} \le \underline{h}^{\ast}$. The threshold $\hat{h}$ is such that a marriage between the least skilled single woman and the most skilled single man is barely viable, i.e.

(65)$$\begin{equation} (\hat{h}+\underline{h}^{\ast}(\hat{h}))^{1+\gamma n\psi}=\tilde{h}^{\gamma n\psi}\underline{h}^{\ast}(\hat{h})+e^{k}\hat{h}^{1+\gamma n\psi}. \end{equation}$$

Observe that (59) implies that $\underline{h}^{\ast \prime}() \le 0,$ while (65) states that $(\underline{h}^{\ast}(\hat{h}),\hat{h})$ lies on the upward sloping marriage viability frontier. Thus, (65) holds for at most one value of $\hat{h}.$ Furthermore, at $\hat{h}=h_{\max}$, we have that $\underline{h}^{\ast}(h_{\max})=h_{\min}$ and that

(66)$$\begin{equation} (h_{\max}+h_{\min})^{1+\gamma n\psi} < \tilde{h}^{\gamma n\psi}h_{\min}+e^{k}h_{\max}^{1+\gamma n\psi}. \end{equation}$$

To see this, note that since (18) is violated by assumption, we have that

$$\begin{equation*} e^{k}h_{\max}^{1+\gamma n\psi} > 2^{\gamma n\psi}h_{\min}^{1+\gamma n\psi}+2^{\gamma n\psi}h_{\max}^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}. \end{equation*}$$

Clearly, the inequality

$$\begin{equation*} 2^{\gamma n\psi}h_{\min}^{1+\gamma n\psi}+2^{\gamma n\psi}h_{\max}^{1+\gamma n\psi} > (h_{\max}+h_{\min})^{1+\gamma n\psi}, \end{equation*}$$

holds. This proves that (66) holds and therefore that the LHS of (65) is smaller than its RHS at $\hat{h}=h_{\max}.$ Next, for B small enough, at $\hat{h}=h_{\min},$ we have that $\underline{h}^{\ast}(h_{\min})=h_{\max}$ and that

(67)$$\begin{equation} (h_{\max}+h_{\min})^{1+\gamma n\psi} > \tilde{h}^{\gamma n\psi}h_{\max}+e^{k}h_{\min}^{1+\gamma n\psi}. \end{equation}$$

To see this, note that if (18) holds with equality, then

$$\begin{equation*} e^{k}h_{\max}^{1+\gamma n\psi}=2^{\gamma n\psi}h_{\min}^{1+\gamma n\psi}+2^{\gamma n\psi}h_{\max}^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}, \end{equation*}$$

or equivalently

$$\begin{eqnarray*} e^{k}-2^{\gamma n\psi} &=&2^{\gamma n\psi}\frac{h_{\min}^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}}{h_{\max}^{1+\gamma n\psi}} \\ & < &2^{\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}^{-\gamma n\psi}, \end{eqnarray*}$$

where the inequality comes from the fact that $h_{\min} > \underline{h}(\beta) \ge \tilde{h},$ hence the numerator in the fraction is ⩾ 0. This inequality is then equivalent to

(68)$$\begin{equation} e^{k}h_{\min}^{1+\gamma n\psi} \le 2^{\gamma n\psi +1}h_{\min}^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}. \end{equation}$$

Observe that the function $(x+h_{\min})^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}x$ is increasing with x for x ⩾ 0 since $h_{\min} > \tilde{h},$ which implies that (67) holds since (68) does. Since (67) holds if (18) holds with equality, by continuity it will also hold if it is marginally violated, i.e. if B is small enough.

Together, (66) and (67) imply that $\hat{h}$ exists and satisfies $h_{\min} \le \hat{h} \le h_{\max}.$

We now show that $\mu _{H}(\hat{h}) \le \mu _{L}(\hat{h}).$ Substituting (65) into (63), we see that this is equivalent to

$$\begin{equation*} (\hat{h}+\underline{h}^{\ast}(\hat{h}))^{1+\gamma n\psi}-\left( 1+\gamma n\psi \right) \int _{h_{\min}}^{\hat{h}}(z+h^{\ast}(z))^{\gamma n\psi}dz \le (h_{\min}+\underline{h}^{\ast}(\hat{h}))^{1+\gamma n\psi}. \end{equation*}$$

This inequality always holds.Footnote ⁴¹ Thus, by the intermediate value theorem, there exists a solution to (63)–(64) such that $ \hat{h} \le \bar{h} \le h_{\max}.$ Since $\bar{h} \ge \hat{h},$ by construction, no single woman in this solution wants to marry a single men; hence condition (iii) in Definition 2 holds for pairs of singles.

• • We have proved that there exists a pair $(\mu,\bar{h})$ such that (63)–(64) hold and that condition (iii) in Definition 2 holds for (h, h*) ∈ [h _min, h _max] − S × [h _min, h _max] − S*.

E. Checking that the constructed solution satisfies (iii) for a single woman and a married man

Another requirement is that condition (iii) hold for $h \ge \bar{h}$ and h* ∈ S*. Using (5) and (11), we must have

(69)$$\begin{equation} \ln \hat{\theta}(h,h^{\ast})=(1+\gamma n\psi) (\ln h-\ln (h+h^{\ast}))+k. \end{equation}$$

Using (12), we see that (17) is equivalent to

(70)$$\begin{eqnarray} &&\ln (1-\hat{\theta}(h,h^{\ast}))+(1+\gamma n\psi) (\ln A+\ln (h+h^{\ast}))+\pi _{\beta} \nonumber \\ &\le &V^{\ast}(h^{\ast}) \nonumber \\ &=&\ln (1-\theta (h^{\ast -1}(h^{\ast})))+(1+\gamma n\psi) (\ln A+\ln (h^{\ast}+h^{\ast -1}(h^{\ast})))+\pi _{\beta .} \end{eqnarray}$$

Substituting (69) and (23), we see that this is equivalent to

(71)$$\begin{eqnarray} \mu &\le &(h^{\ast}+h^{\ast -1}(h^{\ast}))^{1+\gamma n\psi}-\left( 1+\gamma n\psi \right) \int _{h_{\min}}^{h^{\ast -1}(h^{\ast})}(z+h^{\ast}(z))^{\gamma n\psi}dz \\ &&+e^{k}h^{1+\gamma n\psi}-(h+h^{\ast})^{1+\gamma n\psi} \nonumber \\ &=&\phi (h,h^{\ast}). \nonumber \end{eqnarray}$$

Since (63) holds by construction, we have that $\phi (\bar{h},h_{\max})=\mu .$ Furthermore,

$$\begin{equation*} \frac{\partial \phi}{\partial h}=\left( 1+\gamma n\psi \right) \left[ e^{k}h^{\gamma n\psi}-(h+h^{\ast})^{\gamma n\psi}\right] . \end{equation*}$$

Therefore, $\frac{\partial \phi}{\partial h}\ge 0$ if and only if $e^{k}\ge (1+\frac{h^{\ast}}{h})^{\gamma n\psi}.$ Let us assume that

(72)$$\begin{equation} e^{k}\ge (1+\frac{h_{\max}}{\bar{h}})^{\gamma n\psi}. \end{equation}$$

It must then be that $e^{k}\ge (1+\frac{h^{\ast}}{h})^{\gamma n\psi}$ for all h* ∈ S* and for all $h\ge \bar{h}.$ Consequently, $\phi (h,h^{\ast})\ge \phi (\bar{h},h^{\ast}).$

Furthermore, since $\bar{h}\in S,$ (17) is equivalent to (54) for $(\bar{h},h^{\ast}),$ and we already know that from part B of this proof that (54) holds for $(\bar{h},h^{\ast}).$ Using (23) and rearranging, we see that this is equivalent to

$$\begin{equation*} (\bar{h}+h^{\ast})^{1+\gamma n\psi}\le \left( 1+\gamma n\psi \right) \int _{h^{\ast -1}(h^{\ast})}^{\bar{h}}(z+h^{\ast}(z))^{\gamma n\psi}dz+(h^{\ast}+h^{\ast -1}(h^{\ast}))^{1+\gamma n\psi}; \end{equation*}$$

but inspection of (71) and making use of (63) shows that this condition is equivalent to $\phi (\bar{h},h^{\ast})\ge \mu .$ Therefore, ϕ(h, h*) ⩾ μ.

Hence, condition (72) is sufficient for (iii) in Definition 2 to hold for h* ∈ S* and $h \ge \bar{h}.$ Furthermore, if condition (18) holds with equality, we have that e^k ⩾ 2^γnψ,and in this limit case the solution to (63)–(64) is $\bar{h}=h_{\max}.$ Condition (72) then strictly holds. By continuity, if h _max is such that (18) is not violated by too much, i.e. B in (21) is not too large, then (72) will hold.

• • We have proved that we can choose B such that the values of μ and $\bar{h}$ constructed in D are such that condition (iii) in Definition 2 holds for (h, h*) ∈ [h _min, h _max] − S × S*.

F. Checking that the constructed solution satisfies (i)

The condition that married women are better-off than if they were single can be written

(73)$$\begin{equation} \ln \theta (h)+(1+\gamma n\psi) \ln (h+h^{\ast}(h))\ge (1+\gamma n\psi) \ln h+k,\forall h\le h^{\ast}, \end{equation}$$

or equivalently using the formula for θ(h):

(74)$$\begin{equation} \mu \ge e^{k}h^{1+\gamma n\psi}-(1+\gamma n\psi) \int _{h_{\min}}^{h}(z+h^{\ast}(z))^{\gamma n\psi}dz=\phi (h). \end{equation}$$

Again, (63) implies that it holds with equality at $h=\bar{h}.$ Furthermore, ϕ′(h) = (1 + γnψ)(e^kh ^γnψ − (h + h*(h))^γnψ). We have ϕ′(h) ⩾ 0 if and only if $h^{\ast}(h)/h \le e^{\frac{k}{\gamma n\psi}}-1.$ This is again true in the limit case where (18) holds with equality, since we then have h*(h) = h and $e^{\frac{k}{\gamma n\psi}} \ge 2.$ Therefore, in this limit equilibrium we have $\phi (h) \le \phi (\bar{h}).$ By continuity, this remains true if (18) is not violated by too much. Then (74) holds, and condition (i) in Definition 2 is satisfied.

• • We have proved that we can choose B such that, in addition to the properties spelled out above, condition (i) in Definition 2 holds.

G. Checking that the constructed solution satisfies (iii) for a single man and a married woman

We now prove that (iii) holds when $h^{\ast} \le \underline{h}^{\ast}$ and $h \le \bar{h}.$ Here, it is more convenient to use the alternative formulation defined in footnote 30. Using (3) and (12) allows us to compute $\hat{\theta}^{\ast}(h,h^{\ast}):$

(75)$$\begin{equation} \hat{\theta}^{\ast}(h,h^{\ast})=1-\frac{\tilde{h}^{\gamma n\psi}h^{\ast}}{(h+h^{\ast})^{1+\gamma n\psi}}. \end{equation}$$

Comparing (11) for $\theta =\hat{\theta}^{\ast}(h,h^{\ast})$ and for θ = θ(h) and h*(h) instead of h*, we see that the condition in footnote 30 holds if and only if

$$\begin{equation*} \hat{\theta}^{\ast}(h,h^{\ast})\le \theta (h)\frac{(h+h^{\ast}(h))^{1+\gamma n\psi}}{(h+h^{\ast})^{1+\gamma n\psi}}. \end{equation*}$$

Substituting (23) and (75), we get the following condition

(76)$$\begin{equation} (h+h^{\ast})^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h^{\ast}\le \mu +(1+\gamma n\psi) \int _{h_{\min}}^{h}(z+h^{\ast}(z))^{\gamma n\psi}dz. \end{equation}$$

Note that this holds with equality for h = h _min and $h^{\ast}= \underline{h}^{\ast},$ by virtue of (64). Next, note that the LHS is an increasing function of h*. Therefore, (76) holds for all $ h^{\ast} \le \underline{h}^{\ast}$ if and only if it holds for $h^{\ast}= \underline{h}^{\ast}.~$Next, note that the derivative of the RHS with respect to h is (1 + γnψ)(h + h*(h))^γnψ, while the derivative of the LHS with respect to h at $h^{\ast}=\underline{h} ^{\ast}$ is $(1+\gamma n\psi) (h+\underline{h}^{\ast})^{\gamma n\psi}.$ The former is clearly larger than the latter since $h^{\ast}(h)\ge \underline{h}^{\ast}.$ Therefore, the difference between the RHS of (76) and its LHS at $h^{\ast}=\underline{h}^{\ast}$ is an increasing function of h. Since (76) holds with equality for h = h _min and $h^{\ast}=\underline{h}^{\ast},$ it also holds for any h ⩾ h _min and $h^{\ast}=\underline{h}^{\ast}.$ As we have already seen, that in turn implies that it holds for any h ⩾ h _min and $h^{\ast} \le \underline{h} ^{\ast}.$ This completes the proof that (iii) holds for single men underbidders.

• • We have proved that the values of μ and $\bar{h}$ constructed in D are such that condition (iii) in Definition 2 holds for (h, h*) ∈ S × [h _min, h _max] − S*.

H. Proof that (ii) holds for the constructed solution

The last thing we have to check is that (ii) holds, that is, married men are better-off than if they were single. Denoting by h the wife of a married man and by h*(h) this man, we see that this is equivalent to

$$\begin{equation*} \ln A+\ln h^{\ast}(h)\le \ln (1-\theta (h))+(1+\gamma n\psi) \left( \ln A+\ln (h+h^{\ast}(h))\right) +\pi _{\beta}. \end{equation*}$$

Substituting in (23), we see that this is equivalent to

(77)$$\begin{equation} \mu \le (h+h^{\ast}(h))^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h^{\ast}(h)-(1+\gamma n\psi) \int _{h_{\min}}^{h}(z+h^{\ast}(z))^{\gamma n\psi}dz. \end{equation}$$

Again, this holds with equality for h = h _min, because of (64). Furthermore, the RHS’s derivative with respect to h is equal to $h^{\ast \prime}(h)\left[ (1+\gamma n\psi) (h+h^{\ast}(h))^{\gamma n\psi}-\tilde{h} ^{\gamma n\psi}\right],$ which is clearly positive since h*() is increasing and $h+h^{\ast}(h) \ge \underline{h}(\beta) \ge \tilde{h}.$ Therefore, the RHS of (77) is an increasing function of h and is always greater for h ⩾ h _min than for h = h _min, where it holds with equality. Hence, (77) always holds:

• • We have proved that the values of μ and $\bar{h}$ constructed in D are such that condition (ii) in Definition 2 holds.

This completes the proof of Proposition 7.

A.8. Proof of Proposition 8

The discussion in D in the proof of proposition 7 implies that we can always pick an equilibrium such that locally, the RHS of (63) as a function of $\underline{h}^{\ast}$ is flatter than that of (64). It is then clear that a rise in k shifts the RHS of (63) up, and has no effect on (64). Therefore, both $\underline{h}^{\ast}$ and μ go up, which proves claim (i). Similarly, a greater $\tilde{h}$ reduces the RHS of (64), with no effect on (63), so that $\underline{h} ^{\ast}$ goes up again while μ falls. This proves claim (ii).

Finally, note that h _max does not enter in (64) and that for a uniform distribution, (63) is equivalent to

(78)$$\begin{equation} \mu =e^{k}(h_{\max}-\delta) ^{1+\gamma n\psi}-2^{\gamma n\psi}\left[ (h_{\max}-\delta /2)^{1+\gamma n\psi}-(h_{\min}+\delta /2)^{1+\gamma n\psi}\right], \end{equation}$$

where $\delta =\underline{h}^{\ast}-h_{\min},$ and h*(h) = h + δ. The constructed equilibrium is such that (72) holds. This also implies that the RHS of (78) is increasing in h _max, holding $\underline{h}^{\ast}$ or equivalently δ constant. Consequently, a greater h _max raises the RHS of (63), so that the equilibrium values of μ and $\underline{h}^{\ast}$ go up. The proportion of married people is $1-\frac{\delta}{h_{\max}-h_{\min}},$ and it must go down. It falls iff $\frac{d\delta}{dh_{\max}} \ge \frac{\delta}{h_{\max}-h_{\min}},$ which is true if δ is small enough, which is true in the constructed equilibrium of Prop. 7.

A.9. Proof of Proposition 9

Since the formula for θ(h) is the same as in Proposition 6, it is clear that (54) is still satisfied by our candidate equilibrium and therefore that equilibrium condition (iii) in Definition 2 holds.

The utility of a beta woman outside marriage is now given by $\bar{U}_{\beta}^{\ast}(h)$; like beta men, they cannot have children and are therefore in a symmetrical situation. Using (51) and (3), we see that condition (i) in Definition 2 holds if and only if

$$\begin{equation*} \ln (1+\lambda h^{-(1+\gamma n\psi) })+\gamma n\psi \ln h\ge -\gamma n\psi \ln 2+\gamma n\psi \ln \tilde{h},\forall h\epsilon [h_{\min},h_{\max}]. \end{equation*}$$

This condition always holds for λ ⩾ 0 since $h \ge \tilde{h}.$ For λ ⩽ 0, the LHS is clearly an increasing function of h, so (i) holds provided the above holds for h = h _min, that is $\lambda \ge -\left( 1-2^{-\gamma n\psi}\tilde{h}^{\gamma n\psi}h_{\min}^{-\gamma n\psi}\right) h_{\min}^{1+\gamma n\psi},$ which defines the lower bound of (25).

For men, the proof that (ii) holds if $\lambda \le \left( 1-2^{-\gamma n\psi}\tilde{h}^{\gamma n\psi}h_{\min}^{-\gamma n\psi}\right) h_{\min}^{1+\gamma n\psi}$ is the same as in the proof of Proposition 6.

A.10. Proof of Proposition 10

We compare utility in the two types of equilibria for all agents. In all what follows, h*(h) refers to the assignment function in the SATC equilibrium, since in the Victorian equilibrium we can readily replace it by h.

We start with women who are married in the SATC equilibrium. Using (11) for both equilibria, we see that V_S(h) ⩾ V_V(h; λ) if and only if

(79)$$\begin{equation} \ln \theta _{S}(h)+(1+\gamma n\psi) \ln (h+h^{\ast}(h)) \ge \ln \theta _{V}(h;\lambda) +(1+\gamma n\psi) \ln (2h), \end{equation}$$

where θ_S(h) and θ_V(h; λ) are the appropriate shares, i.e.

$$\begin{eqnarray*} \theta _{V}(h,\lambda) &=&\frac{1}{2}(1+\lambda h^{-(1+\gamma n\psi) }), \\ \theta _{S}(h) &=&\left( h+h^{\ast}(h)\right) ^{-(1+\gamma n\psi) }\left[ (1+\gamma n\psi) \int _{h_{\min}}^{h}(z+h^{\ast}(z))^{\gamma n\psi}dz+\mu \right] . \end{eqnarray*}$$

Condition (79) is equivalent to

(80)$$\begin{equation} (1+\gamma n\psi) \int _{h_{\min}}^{h}(z+h^{\ast}(z))^{\gamma n\psi}dz+\mu \ge 2^{\gamma n\psi}(\lambda +h^{1+\gamma n\psi}). \end{equation}$$

Clearly, if it holds for the maximum possible value of λ, it must hold for any equilibrium value of λ. Hence, substituting $\lambda =\left( 1-2^{-\gamma n\psi}\tilde{h}^{\gamma n\psi}h_{\min}^{-\gamma n\psi}\right) h_{\min}^{1+\gamma n\psi}$ into (80), we get

(81)$$\begin{equation} \mu \ge 2^{\gamma n\psi}h_{\min}^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}+2^{\gamma n\psi}h^{1+\gamma n\psi}-(1+\gamma n\psi) \int _{h_{\min}}^{h}(z+h^{\ast}(z))^{\gamma n\psi}dz. \end{equation}$$

By differentiating the last two terms with respect to h, we see that together they are a decreasing function of h. Therefore the preceding inequality will hold for all h iff it holds for h _min, that is

$$\begin{equation*} \mu \ge 2^{\gamma n\psi +1}h_{\min}^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}. \end{equation*}$$

Subsituting (64), which must hold in any SATC equilibrium, we get

$$\begin{equation*} (h_{\min}+\underline{h}^{\ast})^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi} \underline{h}^{\ast} \ge 2^{\gamma n\psi +1}h_{\min}^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}. \end{equation*}$$

Noting that there is equality at $\underline{h}^{\ast}=h_{\min}$ and that the LHS is increasing with $\underline{h}^{\ast},$ we conclude that this always holds. Consequently, (81) holds for all $h \le \bar{h}.$ Therefore, (80) holds for all λ and $h \le \bar{h}.$ Hence married beta women have a greater utility under the SATC equilibrium than under the Victorian one.

We next consider single women. Comparing (5) and (11), we get that V_S(h) ⩾ V_V(h; λ) iff

$$\begin{equation*} (1+\gamma n\psi) \ln h+\pi _{\alpha} \ge \ln \theta _{V}(h)+(1+\gamma n\psi) \ln (2h), \end{equation*}$$

or equivalently

(82)$$\begin{equation} e^{k}2^{-\gamma n\psi} \ge 1+\lambda h^{-(1+\gamma n\psi) }. \end{equation}$$

Again, if this inequality holds for the maximum equilibrium value of λ, it will hold for any of them. Furthermore, at this maximum λ, which is positive, the RHS is decreasing in h, so that the inequality holds for all $h\ge \bar{h}$ iff it holds for $h=\bar{h}.$ Substituting both $\lambda =\left( 1-2^{-\gamma n\psi}\tilde{h}^{\gamma n\psi}h_{\min}^{-\gamma n\psi}\right) h_{\min}^{1+\gamma n\psi}$ and $h= \bar{h}$ into (82) we get, after rearranging:

$$\begin{equation*} e^{k}\bar{h}^{1+\gamma n\psi} \ge 2^{\gamma n\psi}\bar{h}^{1+\gamma n\psi}+h_{\min}^{1+\gamma n\psi}2^{\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}. \end{equation*}$$

Substituting (63) and then (64), this is equivalent to

(83)$$\begin{eqnarray} &&(h_{\min}+\underline{h}^{\ast})^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}\underline{h}^{\ast}+(1+\gamma n\psi) \int _{h_{\min}}^{\bar{h}}(z+h^{\ast}(z))^{\gamma n\psi}dz \nonumber\\ &&\quad \ge 2^{\gamma n\psi}\bar{h}^{1+\gamma n\psi}+h_{\min}^{1+\gamma n\psi}2^{\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}. \end{eqnarray}$$

Since $(h_{\min}+\underline{h}^{\ast})^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}\underline{h}^{\ast} \ge h_{\min}^{1+\gamma n\psi}2^{1+\gamma n\psi}- \tilde{h}^{\gamma n\psi}h_{\min},$ a sufficient condition for (83) to hold is

$$\begin{eqnarray*} &&h_{\min}^{1+\gamma n\psi}2^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}+(1+\gamma n\psi) \int _{h_{\min}}^{\bar{h}}(z+h^{\ast}(z))^{\gamma n\psi}dz \nonumber\\ &&\quad \ge 2^{\gamma n\psi}\bar{h}^{1+\gamma n\psi}+h_{\min}^{1+\gamma n\psi}2^{\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}, \end{eqnarray*}$$

that is

$$\begin{eqnarray*} &&(1+\gamma n\psi) \int _{h_{\min}}^{\bar{h}}(z+h^{\ast}(z))^{\gamma n\psi}dz \nonumber\\ &&\quad \ge 2^{\gamma n\psi}(\bar{h}^{1+\gamma n\psi}-h_{\min}^{1+\gamma n\psi}). \end{eqnarray*}$$

This inequality holds since the RHS equals the LHS for $\bar{h}=h_{\min},$ while differentiation shows that the LHS increases faster with $\bar{h}$ than the RHS. Therefore, (82) holds for the highest λ and for $\bar{h},$ i.e. for any equilibrium λ and all $h\ge \bar{h}.$ Thus single women prefer the SATC equilibrium as well.

We now turn to married men. Using the same steps as for married women, we get that they prefer the Victorian equilibrium iff

$$\begin{equation*} (1+\gamma n\psi) \int _{h_{\min}}^{h^{\ast -1}(h^{\ast})}(z+h^{\ast}(z))^{\gamma n\psi}dz+\mu \ge 2^{\gamma n\psi}(\lambda -h^{\ast 1+\gamma n\psi})+(h^{\ast -1}(h^{\ast})+h^{\ast})^{1+\gamma n\psi}. \end{equation*}$$

Again, this holds for all equilibrium values of λ if it holds for the largest one, $\lambda =\left( 1-2^{-\gamma n\psi}\tilde{h}^{\gamma n\psi}h_{\min}^{-\gamma n\psi}\right) h_{\min}^{1+\gamma n\psi}.$ Substituting, we get

(84)$$\begin{equation} \mu \ge 2^{\gamma n\psi}h_{\min}^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}+\phi (h^{\ast -1}(h^{\ast})), \end{equation}$$

where ϕ(h) ≡ (h + h*(h))^{1 + γnψ} − 2^γnψh*(h)^{1 + γnψ} − (1 + γnψ)∫^h_{h min}(z + h*(z))^γnψdz. Now, note that ϕ′(h) = h*′(h)(1 + γnψ)[(h + h*(h))^γnψ − 2^γnψh*(h)^γnψ] ⩽ 0. Thus, (84) holds for all $h^{\ast}\ge \underline{h}^{\ast}$ iff it holds at $h^{\ast}=\underline{h}^{\ast},$ that is

$$\begin{equation*} \mu \ge 2^{\gamma n\psi}h_{\min}^{1+\gamma n\psi}-\tilde{h}^{\gamma n\psi}h_{\min}+(h_{\min}+\underline{h}^{\ast})^{1+\gamma n\psi}-2^{\gamma n\psi}\underline{h}^{\ast 1+\gamma n\psi}. \end{equation*}$$

Substituting (64), we see that this is equivalent to

(85)$$\begin{equation} 2^{\gamma n\psi}\left( \underline{h}^{\ast 1+\gamma n\psi}-h_{\min}^{1+\gamma n\psi}\right) \ge \tilde{h}^{\gamma n\psi}(\underline{h}^{\ast}-h_{\min}). \end{equation}$$

As the RHS equates the LHS for $\underline{h}^{\ast}=h_{\min},$ and as the LHS increases more with $\underline{h}^{\ast}$ than the RHS, this inequality clearly holds, which proves that married beta men are better-off in the sexually repressed Victorian equilibrium than in the SATC equilibrium regardless of λ.

We finally consider the case of single men. Using (3) and (12) and substituting $\theta =\theta _{V}(h^{\ast},\lambda) =\frac{1}{2} (1+\lambda h^{\ast -(1+\gamma n\psi) }),$ then rearranging, we see that they prefer the Victorian equilibrium iff

$$\begin{equation*} \lambda \le h^{\ast 1+\gamma n\psi}-2^{-\gamma n\psi}\tilde{h}^{\gamma n\psi}h^{\ast}. \end{equation*}$$

This holds for the maximum $\lambda,\left( 1-2^{-\gamma n\psi}\tilde{h} ^{\gamma n\psi}h_{\min}^{-\gamma n\psi}\right) h_{\min}^{1+\gamma n\psi},$ iff

$$\begin{equation*} h_{\min}^{1+\gamma n\psi}-2^{-\gamma n\psi}\tilde{h}^{\gamma n\psi}h_{\min} \le h^{\ast 1+\gamma n\psi}-2^{-\gamma n\psi}\tilde{h}^{\gamma n\psi}h^{\ast}. \end{equation*}$$

This is clearly satisfied since the RHS grows with h* and is equal to the LHS at h* = h _min.

Therefore single men also prefer the Victorian outcome. This completes the proof of Proposition 10.

A.11. Proof of Proposition 11

First, note that a Victorian equilibrium among the alphas always exists. This is because the alpha’s marriage problem is identical to the beta’s except that k = 0 in this case: women marrying an alpha would access the same genetic material if they were single instead. Since Proposition 6 holds for k = 0, such an equilibrium exists.

Next, assume that at date t, all alpha agents have a human capital level lower than h^LR _{max, α}, and all beta agents have a human capital lower than h^LR _{max, β}.

Clearly, for generation t + 1, the human capital level of an alpha cannot exceed that of the alpha children of a couple such that the man is alpha and both spouses have human capital h^LR _{max, α}. This can be written as

$$\begin{equation*} \ln h\le \psi \ln 2+\ln \alpha +\psi \ln \frac{\gamma \psi A}{1+n\gamma \psi}+\psi \ln h_{\max,\alpha}^{LR}. \end{equation*}$$

By construction, the RHS is equal to h^LR _{max, α}. Therefore, all the alphas in generation t + 1 have a human capital which cannot exceed h^LR _{max, α}. As for the betas of that generation, they cannot do better than the beta children of an alpha couple with human capital h^LR _{max, α}:

$$\begin{equation*} \ln h\le \psi \ln 2+\ln \beta +\psi \ln \frac{\gamma \psi A}{1+n\gamma \psi}+\psi \ln h_{\max,\alpha}^{LR}=\ln h_{\max,\beta}^{LR} \le \ln h_{\max,\alpha}^{LR}. \end{equation*}$$

Therefore, the property that h ⩽ h^LR _{max, β} for all the betas and h ⩽ h^LR _{max, α} for all the alphas will remain true across all generations, regardless of how the betas mate, as long as the conditions of Proposition 2 hold.

Assume that at date t, all agents have a human capital level larger than h^LR _{min, β}. We know that (18) is more likely to hold, the smaller h _max and the larger h _min. By assumption, (18) holds for h _min = h^LR _{min, β} and h _max = h^LR _{max, β}. By assumption, the distribution of the beta’s human capital at t is such that h^LR _{min, β} ⩽ h _{min, t} ⩽ h _{max, t} ⩽ h^LR _{max, β}. Therefore, (18) holds. Also, by virtue of (28), so does the condition $h_{\min} \ge \underline{h}(\beta) $ and therefore the conditions of Proposition 2. Hence, there exists a marriage market equilibrium at t which is Victorian for the betas. Furthermore, in generation t + 1, the lowest human capital level of a beta cannot exceed that of a beta offspring of a beta couple with human capital h^LR _{min, β}:

$$\begin{equation*} \ln h\ge \psi \ln 2+\ln \beta +\psi \ln \frac{\gamma \psi A}{1+n\gamma \psi}+\psi \ln h_{\min,\beta}^{LR}=\ln h_{\min,\beta}^{LR}. \end{equation*}$$

Therefore, the property that h ⩾ h^LR _{min, β} still holds among the betas (and, a fortiori, among the alphasFootnote ⁴²) of generation t + 1, implying that a Victorian equilibrium also exists for them. By induction, h ⩾ h^LR _{min, β} for all generations and a Victorian equilibrium exists at all dates. This proves claim (i) in Proposition 11.

Claim (ii) derives straightforwardly from the fact that all agents marry, so that a fraction ρ_t of children have alpha fathers.

Claim (iii) derives straightforwardly from aggregating log human capital among all offsprings of types alpha and beta.

A.12. Proof of Proposition 12

The proof is straightforward from (27), (29), (30) and from noting that the expected human capital of the betas evolves according to

$$\begin{equation*} E\ln h_{t+1,\beta}=\psi (\ln 2+1+E\ln h_{t,\beta}-\underline{h}(\beta) ). \end{equation*}$$

Finally note that (31) is equivalent to

(86)$$\begin{equation} e^{k}\le 2^{\gamma n\psi}, \end{equation}$$

and compare with (26).

A.13. Proof of Proposition 13

The required inequality holds for married women, since they get a higher quality beta husband in the SATC assignment than in the Victorian one. Let us thus focus on unmarried ones. In the Victorian assignment, their average offspring log human capital would be equal to

$$\begin{equation*} E_{V}\ln h^{\prime}(h)=b+\psi \ln (2h)+\ln \frac{A\gamma \psi}{1+\gamma n\psi}. \end{equation*}$$

In the SATC equilibrium, we get

$$\begin{equation*} E_{\text{SATC}}\ln h^{\prime}(h)=a+\psi \ln h+\ln \frac{A\gamma \psi}{1+\gamma n\psi}. \end{equation*}$$

The inequality holds iff

$$\begin{equation*} \psi \ln 2 \le (p_{\alpha}-p_{\beta})\ln \frac{\alpha}{\beta}. \end{equation*}$$

This condition is equivalent to e^k ⩾ 2^γnψ, i.e to (86) being violated. We can show that this is a necessary condition for an SATC equilibrium to exist. Going back to the proof of Proposition 5, part E, we can see that (72) is a necessary condition for an SATC equilibrium. If it does not hold, we can show that women slightly above $\bar{h}$ can profitably underbid the women at $h=\bar{h},$ by just inverting the reasoning in part E and (72) clearly implies that e^k ⩾ 2^γnψ.