行政院國家科學委員會專題研究計畫 成果報告
為什麼總是夫比妻大?一個演化上的解釋
計畫類別： 個別型計畫
計畫編號： NSC91-2415-H-002-018-
執行期間： 91 年 08 月 01 日至 92 年 07 月 31 日
執行單位： 國立臺灣大學經濟學系暨研究所
計畫主持人： 黃貞穎
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中 華 民 國 92 年 10 月 31 日
On Spousal Age Gap: An Evolutionary
Explanation
Chen-Ying Huang and Wei-Torng Juang
∗October, 2003
Abstract
We set up an evolutionary model to address the influence of demo-graphics on the demand and supply condition in the marriage market, which in turn has an implication on the social norm whereby men marry younger women. If each couple produces more than two babies and the (male-to-female) sex ratio is greater than one, when the revision dynam-ics satisfy a Darwinian type property, we show that in any mixed steady state, marriages where husbands are older than their wives must exist. Al-though husbands of these marriages delay entering the marriage market, they do have the advantage of choosing their wives from a larger cohort because the population is growing. This implies a greater probability of getting married. The cost and benefit of their postponing marriages must cancel out in a steady state. We further lay out a dynamic under which starting from any initial state, marriages where husbands are older than wives must survive in the long run, if the sex ratio is sufficiently large relative to the reproduction rate and the revision rate.
JEL Classification Numbers: J12, D19
Keywords: spousal age gap, Darwinian type property, revision
∗Huang: Department of Economics, National Taiwan University. Juang: Institute of Economics, Academia Sinica and King’s College, Cambridge. Correspondence: Wei-Torng Juang, King’s College, Cambridge CB2 1ST, UK. Tel: ++44 1223 335225. E-mail: wtj1v@econ.cam.ac.uk.
1
Introduction
The pattern of men marrying women younger than themselves is universal and persistent. According to United Nations (1990), between 1950 and 1985, men who got married before the age of 50 stayed single longer than women in every country surveyed. Most previous research explains the reason why men marry later than women from two perspectives. The first perspective is socio-economic. It is argued that some desirable characteristics of men require time to become known or developed. For instance, Bergstrom and Bagnoli (1993) suggest that the economic roles played by men are more varied than those played by women, and consequently more desirable men choose to marry later under the belief that time helps make their quality known to women. More recently, Edlund (1999) proposes that when son preferences exist, more parents opt for the technology which increases the probability that male babies will be born. This results in a latent deficit of women and makes men, especially those from less sterling families, marry later because they need time to climb up the social ladder so that they can compete with men from better families in the marriage market. The second perspective is biological. Research along this line argues that since women are fecund only for a limited period of time, natural selection will favor men who prefer younger women, since younger women have more reproductive years ahead than older ones. Siow (1998), for instance, indicates how fecundity may interact with other markets to affect gender roles.
Despite of the difference in these papers, to render tractable implications, most models assume constant population size which can be doubtful. Moreover, different models result in different predictions. For example, Bergstrom and Bagnoli (1993) suggest that more desirable men choose to marry later while Edlund (1999) proposes the opposite so that men, especially those from less sterling families, marry later. Fecundity consideration undoubtedly plays an important role in forming the spousal gap, but fecundity alone cannot fully explain diverse spousal age gaps across countries. A quick glance at the spousal age gap of the first marriage across countries suggests that the sex ratio or more
generally the supply and demand condition in the marriage market may play a role in determining the spousal age gap.1
We therefore take a different view in this paper. We invoke neither the story where men use time to signal (or develop) their desirability nor the differential fecundity between men and women. Instead, we attempt to provide an evolu-tionary story to account for the spousal age gap by focusing on the demand and supply in the marriage market. The gist of our story goes like this. Two demographic facts might affect the demand and supply in the marriage market and therefore help explain the spousal age gap. For one, the natural population sex ratio of men to women is about 1.05, which is greater than one (Edlund (1999)). In countries where son preferences are strong like China, Korea, India and Bangladesh, the ratio increases considerably to the range of 1.1 to 1.5 (Tu-liapurkar, Li and Feldman (1995)). Second, the population is generally growing. For instance, over the past 40 years, the population has been growing at 2% to 3% annually over all the countries surveyed except for three (United Nations (1992)).
Suppose, for simplicity, that there are three types of men so that a man could look for his wife from a cohort of women younger than himself, of the same age or older than himself. Since the sex ratio is greater than one, it follows that women are typically in shortage and that men need to carefully consider which cohort of women to target in order to get married. Consider two men of the same age where the first man looks for a younger wife while the second man does not. Then the first man (compared to the second man) has to wait longer for his targeted bride to enter the marriage market. The result is that the first man becomes a father, thus passing down his type, at an older age. His children therefore might be born at the same time as the grandchildren of the
1For instance, in Table B1 of Bergstrom and Bagnoli (1993), we find that the average spousal age gap in Africa is 6.1, compared to 2.9 in North America, Oceania, and Europe. We note that in some countries of Africa, the mean number of wives per husband is greater than one because of polygamy. The practice of polygamy certainly exacerbates the shortage of women. Readers can also refer to Figure 1 in Edlund (1999) or United Nations (1990).
second man, for the latter does not look for a younger wife, hence saving on the waiting time before getting married. Inherent here is a disadvantage on the first man as ceteris paribus, he fails to reproduce as soon as the second man. When each couple produces more than two babies, the number of the grandchildren is typically larger than that of the children. Therefore, the first man’s offspring may be outnumbered by that of the second man. We term this the “grandchildren effect.” Yet, the first man, compared to the second man, finds his wife from a younger cohort of women. A younger cohort is typically larger in number than an older cohort when the population is growing. If men are in oversupply, this makes the oversupply faced by the first man less serious. His probability of getting a wife is thus higher, implying a higher probability of his having children, which is an advantage. We term this the ”probability effect.” Finally, consider the following social learning mechanism. Since men need to compete to find a wife, it is reasonable to assume that the higher the probability of getting a wife, the less severe the competition faced by men in the marriage market. When this is the case, men may learn to revise to a type that entails less competition. This results in an additional advantage to the first man because the second man may consider revising into his type. We term this the ”revision effect.” In a steady state where men targeting different cohorts of women coexist, the grandchildren effect, the probability effect and the revision effect must cancel each other out.
This simple way of decomposing the total effect of marrying later into the grandchildren effect, the probability effect, and the revision effect is helpful. Suppose we call a steady state mixed if more than one type coexist. By evalu-ating these three effects, we can argue that there is no steady state where men looking for older wives and those looking for wives of their own age coexist. This is because men of neither type need to wait to find a wife2. They therefore
give birth to their children at the same time if they indeed get married. The grandchildren effect therefore does not distinguish these two types. On the other
2Recall that only men who look for wives younger than themselves need to wait for their brides to enter the marriage market.
hand, both the probability effect and the revision effect work to the advantage of men who look for wives of their own age. This is because their targeted co-hort is younger and a younger coco-hort is larger when the population is growing.3
Therefore, men looking for wives of their own age will outnumber those looking for older wives.4 By contrast, there can be a steady state where men looking
for younger wives and those looking for wives of their own age coexist.5 This
is because the grandchildren effect now works to the advantage of men looking for wives of their own age since they save on waiting time for their targeted brides to enter the marriage market. However, the other two effects now work to the advantage of men looking for younger wives because their targeted cohort is larger due to population growth. We have just argued that men looking for older wives cannot coexist with those looking for wives of their age, therefore in any mixed steady state, men looking for younger wives must always exist. This provides the first support for men marrying later than women.
Turning to the dynamic aspect of the story, we will prove that under certain circumstances, women looking for husbands of their own age are in overdemand and in the long run, they drive out women looking for younger husbands. This is because if the latter can have any advantage compared to the former, it must be due to the probability effect and revision effect since the targeted cohort of the latter is younger and thus larger. However, the overdemand of the former implies that the probability of their getting a husband has reached 1, the highest possible. Since both the probability and the revision effect depend positively on the probability of getting married, these two effects cannot work against the former. When the population is growing, the grandchildren effect works
3Recall the men looking for older wives target a cohort born earlier. When the population is growing, it is smaller.
4One may then wonder the similar logic would apply to women so that there is no mixed steady state where women looking for older husbands and those looking for husbands of their age coexist. This is not true because even though women looking for older husbands are targeting an older cohort, their targeted cohort could outnumber them due to the uneven sex ratio.
5By similar logic, there can be a steady state where men looking for wives younger than themselves and men looking for wives older than themselves coexist.
for the former but against the latter. Therefore, the latter must be driven out by the former. It can also be shown that when the sex ratio is sufficiently large relative to the reproduction rate, women looking for older husbands are constantly in overdemand. This is because even though their husbands are born earlier than they are (and a growing population implies that those born earlier are less numerous), if the sex ratio is large enough, their potential husbands are still more than them. Based on the same reasoning as why women looking for husbands of their own age survive, they must survive in the long run, too. This provides another support for women marrying earlier than men.
We wish to emphasize that all the arguments in the paper depend only on the demand and supply in the marriage market. Whenever possible, we leave the revision effect (or revision dynamic) as abstract as possible because presumably the revision dynamic is affected by the socio-economic forces. In this paper, we hope to point out a purely demographic force, different from socio-economic influences, that may explain the spousal age gap. Therefore, we either make the revision dynamic abstract or discuss the case where the revision effect is small. This will become clear in the following sections. The rest of the paper is organized as follows. We set up the basic model in Section 2. The steady states are solved in Section 3. The dynamic support for them is provided in Section 4. We provide some discussion about the underlying assumptions of the model in Section 5.
2
The Model
Consider a situation where people live for two periods and they mature when born. The sole purpose of their life is to get married to reproduce. Suppose that there are three types of people: those who marry spouses older than themselves, those who marry spouses of their own age and those who marry spouses younger than themselves.6 The first type, therefore, marries at one year of age while his
6We assume that people marry as soon as possible. This can be justified if, faced with the same set of partners, people tend to marry early.
(her resp.) spouse is two years of age. Likewise, the third type waits until he (she resp.) is two years of age to marry his (her resp.) spouse of one year of age. The second type, by assumption, marries at one year of age when his (her resp.) spouse is also one year of age.
Successful marriages bear offspring in the next period. Newborns inherit their parents’ types. In the absence of other influences, a male baby, there-fore, copies his father’s type, whereas a female baby copies her mother’s type, presumably because parents are their closest role models. To account for influ-ences other than the parental influence, newborns may reevaluate their parents’ choices by talking to other newborns. As a result, if different types perform dif-ferently (in terms of how easily they get a spouse), a better performing type will grow relatively to a worse performing type. This is much in the spirit of Dar-win’s survival of the fittest and is not uncommon in the literature of learning.7
In the following, we impose a minimal structure on this type revision dynamic whenever possible.8 Additional structure needed for more elaborate results will
be provided in due course.
2.1
Parental Influence
Formally, suppose time t is discrete, so that t ∈ Z, the set of integers. Let Mt
denote the number of men born in period t and Wtdenote that of women born
in period t. There are three possible marriages in every period: a marriage in which the husband is older than the wife (type O), a marriage in which the husband is of the same age as the wife (type S), and a marriage in which the husband is younger than the wife (type Y).
Let pi
t(qit, 1−pit−qitresp.) be the proportion of Mtwhose parents’ marriages
are of type Y (S, O resp.). Let ui
t(vit, 1 − uit− vitresp.) be the proportion of Wt
whose parents’ marriages are of type O (S, Y resp.). The superscript i stands
7For instance, Kandori, Mailath and Rob (1993) explicitly assume a deterministic dynamic that has a ”Darwinian” property.
8This is because we want to abstract from the socio-economic forces because presumably the revision dynamic is shaped by the socio-economic influences.
for inheritance. Since newborns may change their inherited types after talking to other newborns, we need some notations for the proportion of newborns of each type subsequent to type revision. Let pt (qt, 1 − pt− qt resp.) be the
proportion of Mt who decides to look for a type Y (S, O resp.) marriage after
type revision. Let ut(vt, 1 − ut− vtresp.) be the proportion of Wtwho decides
to look for a type O (S, Y resp.) marriage after type revision.
For convenience, we call a man looking for a type O marriage a type O man while a woman looking for a type O marriage a type O woman. We similarly define types S and Y men (women resp.). Suppose the matching process is effi-cient,9 then the number of type O marriages in period t, denoted by O(t), is the minimum between type O men who are born in period t−1 and wait till period t to look for younger wives and type O women who are born in period t and decide to look for older husbands. Thus, O(t) = min{(1 − pt−1− qt−1)Mt−1, utWt}.
Similarly, the number of type S marriages in period t, denoted by S(t), is min{qtMt, vtWt}, while the number of type Y marriages in period t, denoted by
Y(t), is min{ptMt, (1 − ut−1− vt−1)Wt−1}. The matching process is depicted
in Figure 1 for ease of remembrance.
To capture the notion that the population is growing, it is assumed that each couple, on average, produces (1 + r)(1 + a) babies: (1 + r) of them are female and a(1 + r) are male. Accordingly, r > 0 is a parameter that reflects the reproduction rate. A higher r implies that the population is reproducing faster. On the other hand, a > 1 is a parameter that reflects the sex ratio of the society. A larger a implies a higher sex ratio which may be due to stronger
9This is to abstract from the friction in the marriage market and helps ensure that the population is growing. If the marriage market is not frictionless, then even if every couple produces more than two babies, the population might be shrinking because there is always a non-negligible fraction of them who cannot find partners to marry. Empirically, the majority of men and women do marry. For instance, Becker (1991) noted that in the United States in 1975, only about 4.6 percent of women and 6.3 percent of men aged 45-54 had never married. Even though the market is not frictionless, the whole population may still be increasing as long as every couple produces sufficiently many babies on average. This will not affect our results qualitatively.
son preferences.10
The efficient matching process, the uneven sex ratio and a reproduction rate greater than one together imply the following population equations:
Mt+1 = a(1 + r)(O(t) + S(t) + Y(t)) and (1)
Wt+1 = (1 + r)(O(t) + S(t) + Y(t)).
That is, there are O(t)+S(t)+Y(t) couples getting married in period t. Each cou-ple, on average, produces a(1 + r) male babies and (1 + r) female babies. Hence, the number of male babies born in period t + 1 is a(1 + r)(O(t)+S(t)+Y(t)). That of female babies born in period t + 1 is (1 + r)(O(t)+S(t)+Y(t)). Notice that Mt= aWtfor all t by assumption.
Without type revision, newborns simply copy their parents’ types. There-fore,
pit+1 = 1 − uit+1− vt+1i = Y(t)
O(t) + S(t) + Y(t), (2) qt+1i = vt+1i = S(t)
O(t) + S(t) + Y(t) and 1 − pit+1− qt+1i = uit+1=
O(t) O(t) + S(t) + Y(t).
For instance, among all the successful marriages in period t, Y(t) of them are of type Y. Since the reproduction rate and the sex ratio are constant, the propor-tion of male babies whose parents’ marriages are of type Y, is just that of type Y marriages, O (t)+S(t)+Y(t)Y(t) . Therefore, pit+1 = O (t)+S(t)+Y(t)Y(t) . The proportion of their sisters among all female babies is the same or pit+1= 1 − uit+1− vit+1.
2.2
Imposing a minimal structure on the revision dynamic
To make our model more realistic, we assume that newborns copy their par-ents’ types, but they may reevaluate the performance of their inherited types. Suppose the performance of a type is summarized by its probability of getting married. For example, the success of type O men in period t is(1−p O (t)
t−1−qt−1)Mt−1.
This is because there are (1 − pt−1− qt−1)Mt−1 type O men but only O(t) type
O marriages in the marriage market of period t. Thus, (1 O (t)
−pt−1−qt−1)Mt−1 is
the average probability that a type O man successfully marries a younger wife in period t. Presumably, this reflects the demand and supply situation in the marriage market of period t. When it is large (small resp.), it is easy (difficult resp.) for a type O man to find a wife younger than himself. Simply put, it summarizes the degree of short-term success of this type. If newborns either are myopic or are mostly affected by the ”fresh” memory of the most recent marriage market conditions, this probability is the most important factor in determining newborns’ revision decision in period t + 1.
Let Rm(O(t)) (Rm(S(t)), Rm(Y(t)) resp.) stand for the performance of type
O (S, Y resp.) men in period t. Similarly, let Rw(O(t)) (Rw(S(t)), Rw(Y(t))
resp.) stand for the performance of type O (S, Y resp.) women in period t.11
Formally, Rm(O(t)) = O (t) (1−pt−1−qt−1)Mt−1 if 1 − pt−1− qt−1> 0, 0 if 1 − pt−1− qt−1= 0, (3) Rm(S(t)) = S(t) qtMt if qt> 0, 0 if qt= 0, Rm(Y(t)) = Y(t) ptMt if pt> 0, 0 if pt= 0, and Rw(O(t)) = O (t) utWt if ut> 0, 0 if ut= 0, (4) Rw(S(t)) = S(t) vtWt if vt> 0, 0 if vt= 0, Rw(Y(t)) = Y(t) (1−ut−1−vt−1)Wt−1 if 1 − ut−1− vt−1> 0, 0 if 1 − ut−1− vt−1= 0.
Now, suppose in addition to parental influence, newborns revise their types
based on the performance of their types in the previous period. We treat the composition of inherited types as a base and assume that some newborns of a worse performing type switch to a better performing one so that a better performing type will be better represented. As an illustrating example, suppose Rm(Y (t)) > Rm(S(t)) > 0, so that type Y is performing better than type
S for men in period t. Then we want the revision dynamic to satisfy that
pt+1
qt+1 >
pi t+1
qi
t+1. In words, since type Y is performing better than type S in the last
period, some newborns inheriting type S may switch to type Y. Therefore, the relative proportion of type Y to S after revision (pt+1
qt+1) is higher than that before
revision (p
i t+1
qi
t+1), when only parental influence is taken into account. Notice how
the revision dynamic is left as abstract as possible.
Formally, we assume the revision dynamic to satisfy the following “Dar-winian” property:
Let l(j(t + 1)) denote the proportion of newborns of type j ∈ {O,S,Y} for sex l ∈ {m, w} (where m stands for men and w for women) in period t + 1. For instance, m(Y(t + 1)) is the proportion of newborns of type Y for men, using our notation earlier, it is simply pt+1. Let li(j(t + 1)) denote the inherited
proportion of newborns of type j for sex l in period t+1. Similarly, mi(Y(t+1))
is simply pi t+1.
• When marriages of types j and k for sex l occur in period t, or Rl(j(t))Rl(k(t)) >
0, then, sign(l(j(t + 1)) l(k(t + 1) − li(j(t + 1)) li(k(t + 1))) (5) = sign(Rl(j(t)) − Rl(k(t))).
To make (5) well defined, it is required that l(j(t + 1)), l(k(t + 1)) > 0 in this case.
• When marriages of type j for sex l do not occur in period t so that Rl(j(t)) =
0, then
This dynamic in (5) says that if for sex l, type j is doing better than type k in period t, then the ratio of type j relative to type k after revision is higher than that before revision. This reflects the fact that some newborns inheriting type k may switch to type j. The process of switching is left as abstract as possible so as to accommodate various possible processes.
3
The Steady States
We will now solve for the steady states. Although (5) is left very abstract, its “Darwinian” property is enough to let us discuss the forms of the steady states. In this respect, the result in this section is quite strong since we do not need to pin down the specific dynamic in the revision process at all. Let St ≡ (pt, qt, ut, vt) be the composition of types for both sexes in period t. A
steady state consists of a quadruplet (p, q, u, v), where {St, Mt, Wt}t∈Z evolves
according to (1), (2), (5), (6) and if
St= (p, q, u, v) for some t ∈ Z,
then
St0 = (p, q, u, v) for all t0> t.
In words, a steady state is a composition of types that is independent of the evolution of time.
We now solve for the steady states step by step. For the ease of exposition, we discuss its intuition before each lemma and relegate the proof to the Appendix. The first lemma shows that if two types of the same sex enter the marriage market at the same age, their performance must be the same. This is because the grandchildren effect does not distinguish types entering the marriage market at the same age since they give birth at the same time. In a steady state where the composition of types remains fixed, the probability effect has to cancel out the revision effect. However, these two effects both depend positively on the performance, and as a result, they must perform equally well. Recall that types
Y and S for men, and types O and S for women all enter the market at one year old.
Lemma 1 In a steady state where marriages of types Y and S for men occur or pq > 0, Rm(Y(t)) = Rm(S(t)) for all sufficiently large t.
Similarly, in a steady state where marriages of types O and S for women occur or uv > 0, Rw(O(t)) = Rw(S(t)) for all sufficiently large t.
The second lemma shows that when the population is growing, a type that delays entering the marriage market must perform better than a type that does not. This is because if the population is growing, the grandchildren effect works against the type that delays marriage, so the probability effect and the revision effect must work in favor of it in a steady state, indicative of a better perfor-mance. On the other hand, if the population is shrinking, the grandchildren effect works for the type that delays marriage, and hence its performance must be worse.
Lemma 2 In a steady state where marriages of types O and Y occur, for all sufficiently large t,
sign( Mt
Mt−1 − 1) = sign(Rm(O(t)) − Rm
(Y(t))) = sign(Rw(Y(t)) − Rw(O(t))) .
In a steady state where marriages of types O and S for men occur, for all sufficiently large t,
sign( Mt
Mt−1 − 1) = sign(Rm(O(t)) − Rm
(S(t))).
In a steady state where marriages of types Y and S for women occur, for all sufficiently large t,
sign( Mt
Mt−1 − 1) = sign(R
w(Y(t)) − Rw(S(t))).
The third lemma shows that if two types of the same sex differ in their time of entering the marriage market, they must differ in performance in every pe-riod. This is because suppose to the contradiction that these two types perform
equally well in period t, then by Lemma 2 the population in period t is as large as that in period t − 1. But this contradicts the positive reproduction rate r. Lemma 3 In a steady state where marriages of types O and Y occur, Rm(O(t)) 6=
Rm(Y(t)) and Rw(Y(t)) 6= Rw(O(t)) for all sufficiently large t.
In a steady state where marriages of types O and S for men occur , Rm(O(t)) 6=
Rm(S(t)) for all sufficiently large t.
In a steady state where marriages of types Y and S for women occur, Rw(Y(t)) 6=
Rw(S(t)) for all sufficiently large t.
Let a mixed steady state be one where marriages of at least two types coexist. We have shown in Lemma 3 that in a mixed steady state, it is impossible that the population growth Mt
Mt−1 is one. The fourth lemma goes a step further. It
shows that in a mixed steady state, if the population growth Mt
Mt−1 is less than
1 in period t, then in the next period, the population growth Mt+1
Mt must be
greater than one, to be consistent with the positive reproduction rate r. Lemma 4 In a mixed steady state, for all sufficiently large t, if Mt
Mt−1 is less
than 1, then Mt+1
Mt must be greater than one.
The fifth lemma shows that types Y and S cannot coexist in a steady state. We have shown that the population growth cannot be one in Lemma 3. Suppose to the contrary that they did coexist, then by Lemma 4 there would have to be a period t in which the population growth is greater than one. In this period, women of type Y must perform better than those of type S by Lemma 2. This implies that men of type S are in overdemand,12 a contradiction to the sex ratio
being greater than one. One can also “roughly” read the inference this way. If women of type Y perform better than those of type S, then men of type S must be in overdemand. In other words, men of type S perform weakly better than those of type Y. Since the former and the latter enter the marriage market at the same time, the grandchildren effect does not differentiate them. If men of type S indeed perform better than those of type Y, then the probability effect
and the revision effect both work for men of type S. Hence we cannot be in a steady state. This interpretation corresponds better to the intuition we provide in the Introduction.
Lemma 5 Types Y and S cannot coexist in any steady state.
By Lemma 5 we have shown that in any mixed steady state, type O must exist. Yet we can show more. The sixth lemma shows that in any mixed steady state, to be consistent with the positive reproduction rate r, the population growth is greater than one in every period.
Lemma 6 In any mixed steady state, the population growth Mt+1
Mt is greater
than 1 for all sufficiently large t. In particular, in a steady state where only marriages of types O and Y occur, Mt+1
Mt−1 > a(1 + r) and
Mt+1
Mt < 1 + r for all
sufficiently large t. In a steady state where only marriages of types O and S occur, Mt+1
Mt = 1 + r for all sufficiently large t.
Let a pure steady state be one where marriages of only one type exist. The pure steady states are easy to solve. Note that among all pure steady states, in the one where only type Y exists, the population grows the slowest. It becomes 1 +r times bigger every other period since women reproduce at two years of age. This is no good news when reproduction means everything. In fact, it implies that type Y will be driven out in the long run as we will see in Section 4. As for type O, as summarized in the seventh lemma, we will see that whether the sex ratio a is greater than the reproduction rate 1 + r crucially determines whether a pure type O steady state can reproduce as fast as a pure type S steady state. We will see that the same theme reappears in the convergence result in Section 4. We summarize all the pure steady states in the seventh lemma.
Lemma 7 In a steady state where only marriages of type S occur, Mt+1
Mt = 1 + r
for all sufficiently large t.
In a steady state where only marriages of type O occur: if a ≥ 1 + r, then
Mt+1
Mt = 1 + r for all sufficiently large t; if a < 1 + r, then
Mt+1
Mt−1 = a(1 + r) for
In a steady state where only marriages of type Y occur, Mt+1
Mt−1 = 1 + r for all
sufficiently large t.
We now summarize all the lemmas into the main result of this paper. Proposition 8 In a mixed steady state, type O must exist. Moreover, the pop-ulation growth (Mt+1
Mt ) is higher when it coexists with type S as opposed to type
Y.
In a pure steady state, the population growth is lowest when only type Y exists. When the sex ratio is large enough so that a ≥ 1 + r, the population growth is highest either when only type O exists or when only type S exists.
We wish to point out that all the results up to now are very general because the revision process in (5) is left so abstract. In particular, it could even be the case that the specific revision process is different from period to period. Nevertheless, all the results still hold. Note further that Proposition 8 says type Y does not fare too well as far as its growth is concerned. In a world where reproduction means everything, this is not good news because type Y reproduces too slowly. We now turn to provide a dynamic support for this view in the following section.
4
Dynamic Support
The analysis above shows that in any mixed steady state, type O must exist. Moreover, type Y does not grow fast enough no matter in a pure steady state or in a mixed one. One may wonder, logically, that starting from any initial state where all three types exist, will type Y eventually go extinct? We now turn to answer this.
In order to study the long run convergence behavior, we need to impose more structure on the revision dynamic. We assume that every male (female resp.) newborn knows the performance of his (her resp.) type in the previous period. He (she resp.) presumably gets this information from his (her resp.) parent of
the same sex. For instance, a male type O baby born in period t + 1 knows Rm(O(t)), presumably from his father’s marriage market experience. His female
siblings know Rw(O(t)), presumably from his mother’s experience. Suppose
further that every newborn has probability α to ask another newborn of the same sex about the performance of the latter’s type. If the former learns that the performance of the latter’s type is strictly better than that of his type, then the former switches to the latter’s type. Otherwise, the former sticks to his type. This, in essence, captures the spirit of “Darwinism.” Such a set-up is quite common in the literature. For example, Norman (1976), Ellison and Fudenberg (1995), Borger and Sarin (1997) and Juang (2001) all assume this, to name a few.
Formally, the revision dynamic of type O for men is
1 − pt+1− qt+1 = (1 − pit+1− qit+1)[1 − α + α(1 − pit+1− qt+1i ) (7)
+αqt+1i (IRm(O (t))≥Rm(S(t))+ IRm(O (t))>Rm(S(t)))
+αpit+1(IRm(O (t))≥Rm(Y(t))+ IRm(O (t))>Rm(Y(t)))],
where IA is an indicator function such that IA = 1 if event A is true and
0 otherwise. The revision dynamic of each other type for men or women is similarly defined. We now explain equation (7). A proportion (1 − pi
t+1− qt+1i )
of male babies inherits their fathers to be of type O. With probability 1 − α, they do not reevaluate but simply follow their fathers. This explains the first term in the square bracket. With probability α, they reevaluate. When this occurs, with probability (1 − pi
t+1− qt+1i ), they talk to another newborn of
type O and thus stick to their type as no new information about other types is conveyed. This explains the second term in the square bracket. With probability qi
t+1 (pit+1 resp.), they talk to another newborn of type S (Y resp.). If the
performance of their own type is weakly better than that of type S (Y resp.), i.e. IRm(O (t))≥Rm(S(t)) = 1 (IRm(O (t))≥Rm(Y(t)) = 1 resp.), they stick to their
type. This explains the third term (the fifth term resp.) in the square bracket. For the remaining terms, it can be reasoned the other way around. There is a proportion qi
resp.). When these newborns reevaluate (with probability α), with probability (1 − pi
t+1− qit+1), they talk to a newborn of type O. If the performance of
their own type is strictly worse than that of type O, i.e. IRm(O (t))>Rm(S(t))= 1
(IRm(O (t))>Rm(Y(t)) = 1 resp.), they switch to type O. This explains the fourth
(the sixth resp.) term in the square bracket. Note that the dynamic in (7) satisfies (5) as long as 1 > α > 0.
We discuss the long run convergence of the dynamic defined by (7). The question we pose is, starting from any interior initial state13 when all three
types coexist, will any type become extinct as time goes by? We study the case in which the type revision is small.14
We first discuss the intuition for convergence. When the type revision para-meter α is small, type S must survive in the long run. This is because women of type S must be in overdemand since men of type S are almost an a multiple of them given parental dynamic dominates.15 If women of type S are in
overde-mand, implying that the probability of their getting married is one, neither the probability effect nor the revision effect can work against them. Since type S women do not delay entering the marriage market, they must reproduce at the highest possible speed and become at least 1 + r times larger in number in every period. On the other hand, type Y must vanish in the long run. This is because even if women of type Y are in overdemand, they reproduce at a slower speed due to their delay in entering the marriage market. Thus, they become about 1 + r times larger in every other period. Finally, as in Proposition 8, the long run behavior of type O varies. When the sex ratio a is much greater than 1 + r, women of type O must be in overdemand because even though men of type O
1 3An interior initial state consists of the proportions of each type in periods -1 and 0. That is, (p−1M−1, q−1M−1, (1 − p−1 − q−1)M−1), (p0M0, q0M0, (1 − p0 − q0)M0), (u−1W−1, v−1W−1, (1 − u−1− v−1)W−1) and (u0W0, v0W0, (1 − u0− v0)W0). Note that every element is strictly positive.
1 4When type revision is large, the long run behavior depends crucially on the initial state. For instance, when α becomes one, in period 1, almost all will revise into the best performing type of period 0.
1 5We concentrate on women since the sex ratio a > 1 hints at the shortage of women. Therefore it is women who determine the population growth.
delay entering the marriage market, the high sex ratio makes them still in over-supply relative to women of type O. Therefore, women of type O must survive for the same reason as those of type S. On the other hand, when the sex ratio is much lower than 1 + r, women of type O are in oversupply. Since they do not perform as well as those of type S but enter the marriage market at the same time as type S, they eventually vanish.
The following lemma provides a sufficient condition under which type S must survive in the long run. In proving this, we incidentally prove that type Y must vanish.
Lemma 9 When α is sufficiently small or a is sufficiently large so that a ≥
1+α
1−α, type S must survive and type Y must vanish in the long run.
When Lemma 9 holds and a is relatively large compared with 1 + r, we can show that type O must survive. On the other hand, when Lemma 9 holds and a is relatively small compared with 1 + r, we can show that type O must vanish. The intuition is, when a is relative large compared with 1 + r, women of type O are in overdemand. For the same reason as why type S survives in Lemma 9, type O survives too. This corresponds well to the intuition in Lemma 7. We formalize this by providing a sufficient condition in the following lemma. Lemma 10 When a ≥ (1 + r)(1+α)1−α2, type O must survive in the long run.
When a ≥ 1+α
1−α and a ≤ (1 + r) 1−α
1+α, type O must vanish in the long run.
5
Discussion
We have pointed out that the uneven sex ratio and the positive population growth can affect the demand and supply in the marriage market, which in turn may account for the spousal age gap. We will now make some remarks regarding the assumptions underlying the model. These remarks may shed light on how one can run empirical tests concerning the validity of our story.
First, one crucial element of our story is that the sex ratio is greater than one. Yet how biased the sex ratio is toward men should be carefully examined when
turning to an empirical analysis. An estimate of the ratio of male population to female population is not adequate because the longer expectancy of women will bias the ratio downward. Conceptually, what we really need is the estimate of the sex ratio at marriageable ages.
Second, some may question whether the population growth is indeed positive in most recent years. Like all evolutionary stories, we read our story as one that long enough time is essential to let the evolutionary force make its mark. Therefore, we feel that a few years of negative population growth, if indeed is the case in some countries, is not enough to invalidate our assumption of positive population growth. In other words, it takes time to form or deform a social norm. For completeness, interested readers can nevertheless infer that if the reproduction rate r is negative or 1 > 1 + r > 0, Lemmas 1 to 3 still hold. To be consistent with the negative reproduction rate, Lemma 4 should be changed to that, if Mt
Mt−1 is greater than one, then
Mt+1
Mt must be less than one in
a mixed steady state. Now it is not necessary that type Y cannot coexist with type S. Recall that before, the probability effect and the revision effect work against men of type Y because their targeted cohort is older and thus smaller. Now since the population is shrinking, an older cohort is larger. Therefore the probability effect and the revision effect can work for men of type Y. We provide a steady state where types Y and S coexist in the following example.
Example 11 Consider the following parameters. If a = 1.02, 1 + r = 0.75, α = 29 in equation (7), then there is a mixed steady state in which types Y and S coexist or p = 0.1, q = 0.9, u = 0, v = 0.92. In this steady state, the population growth Mt+1
Mt is 0.765. Men of types S and Y are both in overdemand.
One important intuition can be gained from this example. When the repro-duction rate is positive, typically, men are in oversupply because of the uneven sex ratio. Now since the reproduction rate is negative, men could be in overde-mand in the market. Take type Y men born in period t as an example. The uneven sex ratio typically will make them larger in number than type Y women born in period t. However, since the population is shrinking, type Y women
born in period t − 1 is also larger in number than type Y women born in period t. Since type Y marriages are between type Y men born in period t and type Y women born in period t − 1, a priori, one cannot definitely say that type Y men will be in oversupply. Rightly because of this reason, type Y men may survive now while they hardly have this chance when the population is growing. We do not want to make too much out of this finding. Yet this suggests that the social norm where men marry earlier than women can be prevalent only when the population is declining.
Third, even if a society’s sex ratio is greater than one and the population is growing as discussed above, some may still wonder, since the sex ratio and the population growth rate are not very far from one, whether these small deviations from being one can really matter. We hasten to emphasize that it is not the absolute magnitudes of a and 1 + r that determine which type may survive. Instead, it is the relative magnitude that matters. A quick glance of Lemma 10 tells us that it is whether a is sufficiently large compared to 1 + r that matters. By Lemma 6, we also know that if types O and S coexist, since the population growth is 1 + r, men of type O must be in oversupply because every woman must get married with probability one to generate this population growth. This implicitly implies that a cannot be too small compared to 1 + r because if it were, then type O men born in period t will be outnumbered by type O women born in period t + 1 and thus the former cannot be in oversupply.
Fourth, it might be argued that the story we describe is mechanical and that it may not capture the whole picture underlying the spousal age gap. We fully recognize the importance of other theories in the literature, no matter whether the focus is socio-economic or biological. However, our paper is best read as an attempt at pointing out a mechanical force, which relies entirely on the demographics, to help explain the age gap after controlling other plausible factors. In doing so, we put emphasis on the effect of the positive population growth and the uneven sex ratio on the demand and supply in the marriage market. In this respect, we view our approach as a complement to that of Bergstrom and Bagnoli (1993), Edlund (1999) or Siow (1998). The empirical
relevance of our theory remains to be tested.16 Appendix
Proof of Lemma 1. We consider the case for men. The case for women is analogous. Suppose to the contrary that there exists a steady state where pq > 0 but Rm(Y(t)) 6= Rm(S(t)) for some t sufficiently large. Consider Rm(Y(t)) >
Rm(S(t)) (the case where Rm(Y(t)) < Rm(S(t)) can be analogously argued).
By (3),
Y(t) = Rm(Y(t))pMt and
S(t) = Rm(S(t))qMt.
Because of parental influence specified in (2), pi t+1 qi t+1 =Y(t) S(t) > p q. By (5), Rm(Y(t)) > Rm(S(t)) implies that
p q > pit+1 qi t+1 , which is a contradiction. ¥
Proof of Lemma 2. We prove that ( Mt
Mt−1− 1) and (Rm(O(t)) − Rm(Y(t))
have the same sign in steady states where types O and Y exist for all t sufficiently large. The proofs for all other cases are analogous.
Note that
sign(Rm(O(t)) − Rm(Y(t)))
= sign(1 − pt+1− qt+1 pt+1 − 1 − pi t+1− qit+1 pi t+1 ),
1 6We do make a very rough attempt at regressing the spousal age gap on the sex ratio and the population growth across countries. Our primary finding does indicate that the spousal age gap is larger for countries with higher sex ratio or where polygamy is still practiced. Notice that the practice of polygamy essentially exacerbates the shortage of women and should play the same role as the uneven sex ratio in our story. However since we haven’t controlled other socio-economic factors, we will leave a rigorous empirical test for future research.
by (5). By (2), 1 − pit+1− qt+1i pi t+1 = O(t) Y(t) = Rm(O(t))(1 − pt−1− qt−1)Mt−1 Rm(Y(t))ptMt , where the second equality follows by (3). Therefore,
sign(1 − pt+1− qt+1 pt+1 − 1 − pi t+1− qt+1i pit+1 ) = sign(1 − p − q p − Rm(O(t))(1 − p − q)Mt−1 Rm(Y(t))pMt ) since we are in a steady state. Hence
sign(Rm(O(t)) − Rm(Y(t)))
= sign(1 −Rm(O(t))Mt−1 Rm(Y(t))Mt ), or sign( Mt Mt−1 − 1) = sign(Rm(O(t)) − Rm (Y(t))). ¥
Proof of Lemma 3. We prove the case where marriages of types O and Y occur. The proofs for other cases are analogous.
Suppose to the contrary that either Rm(O(t)) = Rm(Y(t)) or Rw(Y(t)) =
Rw(O(t)) for some t sufficiently large. By Lemma 2, MMt−1t = 1. Therefore,
Rm(O(t)) = Rm(Y(t)) and Rw(Y(t)) = Rw(O(t)).
Since by Lemma 1, types Y and S for men perform equally well and so do types O and S for women. Therefore, all types for each sex perform equally well in period t. This, together with Mt
Mt−1 = 1, implies that men of each type
is a times of women of that type.17 Therefore, 1 − p − q = u, p = 1 − u − v
and women are in overdemand in period t. Hence Wt+1= (1 + r)(uWt+ vWt+
(1 − u − v)Wt−1) = (1 + r)Wtor MMt+1t = 1 + r. By Lemma 2, this implies that
1 7The performance of each type is the same in period t and thus (1−p−q)Mt−1
uWt = qMt vWt = pMt (1−u−v)Wt−1. Since Mt
Mt−1 = 1, and Mk = aWk for all k, we have (1 − u − v)Wt−1 = (1 − u − v)Wt and (1 − p − q)Mt−1 = (1 − p − q)Mt. Therefore, (1−p−q)MuW t t = qMt vWt = pMt (1−u−v)Wt = Mt Wt = a.
Rm(O(t + 1)) > Rm(Y(t + 1)) and that Rw(Y(t + 1)) > Rw(O(t + 1)). By (5)
the first inequality implies that 1 − p − q p > 1 − pit+2− qit+2 pi t+2 , while the second inequality implies that
1 − u − v u > 1 − uit+2− vt+2i ui t+2 . However, this is impossible because 1−pi
t+2−qt+2i = uit+2and pit+2= 1−uit+2−
vi
t+2by (2). ¥
Proof of Lemma 4. We prove the case where marriages of all three types coexist. The proofs for other cases are analogous.
By Lemma 2, Mt
Mt−1 < 1 implies that Rm(O(t)) < Rm(Y(t)) and that
Rw(Y(t)) < Rw(O(t)). Therefore, Rm(Y(t)) = Rw(O(t)) = 1.18 By Lemma
1, types Y and S for men and types O and S for women perform equally well, thus Rm(Y(t)) = Rm(S(t)) = Rw(O(t)) = Rw(S(t)) = 1. Hence by (3) and (4),
O(t) = uWt, Y(t) = pMtand S(t) = qMt= vWt.
By (2),
uit+1= u u + v + pa.
Since women of type Y perform the worst in period t, by (5), some of them must switch over to types O and S, therefore, u > uit+1. This implies that
u+v +pa > 1. We thus have Wt+1= (1+r)(uWt+vWt+pMt). The population
growth in the next period is, therefore, Mt+1
Mt = (u + v + pa)(1 + r) > 1 + r. ¥
Proof of Lemma 5. Suppose to the contrary that there is a steady state in which types Y and S coexist. Since by Lemma 3 the population growth cannot be one, there must be a period t in which it is greater than one by Lemma 4. By Lemma 2 Rw(Y(t)) > Rw(S(t)), and therefore Rm(S(t)) = 1. Women of
type S perform worse than those of type Y (and if there exists type O women,
1 8Note that R
m(O(t)) < 1 implies that men of type O are in oversupply and hence women of type O get married with probability one. Similarly, Rw(Y(t)) < 1 implies that men of type Y get married with probability one.
by Lemma 1, they must perform equally well as those of type S). Thus some of them must revise to type Y by (5). Therefore, v < vi
t+1. However, men of type
S get married with probability one, so none of them can possibly revise to other types, i.e., q ≥ qi
t+1. Since qit+1= vt+1i by (2) and Mt= aWt, thus qMt> vWt,
contradicting Rm(S(t)) = 1. ¥
Proof of Lemma 6. By Lemma 5, there are only two possible mixed steady states: either types O and Y coexist or types O and S coexist. Moreover, by Lemma 3, types must perform differently and hence Mt
Mt−1 6= 1 for all t sufficiently
large. We discuss the two possible mixed steady states separately. When only types O and Y exist: Either (i) Mt
Mt−1 > 1 or (ii)
Mt
Mt−1 < 1.
(i) If Mt
Mt−1 > 1, then Rm(O(t)) > Rm(Y(t)) and Rw(Y(t)) > Rw(O(t))
by Lemma 2. Therefore, Rm(O(t)) = Rw(Y(t)) = 1 or O(t) = (1 − p)Mt−1,
Y(t) = (1 − u)Wt−1. Hence 1 − pit+1=
(1−p)a
(1−p)a+1−u by (2). Because Rm(O(t)) >
Rm(Y(t)), by (5), 1−pp > 1−pit+1
pi
t+1 or 1 − u > pa. By (1) Wt+1 = (1 + r)((1 −
p)Mt−1+ (1 − u)Wt−1). The endogenous population growth is then MMt−1t+1 =
((1 − p)a + 1 − u)(1 + r) > a(1 + r). Note also that Rw(O(t)) < 1 implies that
O(t) < uWt. Since Y(t) = (1 − u)Wt−1 and WWt−1t =
Mt
Mt−1 > 1, therefore Y(t) <
(1 − u)Wt. Hence Wt+1= (1 + r)(O(t)+Y(t)) < (1 + r)Wt, or MMt+1t < 1 + r.
(ii) If Mt
Mt−1 < 1, then Rm(O(t)) < Rm(Y(t)) and Rw(Y(t)) < Rw(O(t)) by
Lemma 2. Thus, Rm(Y(t)) = Rw(O(t)) = 1 or O(t) = uWt, Y(t) = pMt. Hence
uit+1 = pa+uu by (2). Because Rw(O(t)) > Rw(Y(t)), by (5), u 1−u >
uit+1
1−ui
t+1 or
pa > 1−u. This, however, is impossible because by Lemma 4, in the next period
Mt+1
Mt > 1, but this implies that 1 − u > pa as we just argued in (i). Hence, (ii)
can be ruled out.
When only types O and S exist: Either (iii) Mt
Mt−1 > 1 or (iv)
Mt
Mt−1 < 1. In
(iii), by Lemma 2, Rm(O(t)) > Rm(S(t)). Therefore, Rw(S(t)) = 1. By Lemma
1, Rw(S(t)) = Rw(O(t)) = 1. In (iv), by Lemma 2, Rm(O(t)) < Rm(S(t)).
Therefore, Rw(O(t)) = 1. By Lemma 1, Rw(O(t)) = Rw(S(t)) = 1. Thus
in both (iii), (iv), O(t) = uWt, S(t) = vWt where u + v = 1. Therefore,
Wt+1= (1 + r)(uWt+ vWt) or MMt+1t = 1 + r. ¥
We nevertheless provide some intuition. The steady state of pure type S is obvious in that women get married with probability one on account of the sex ratio a being greater than one. It follows, then, that everyone of them reproduces 1 + r female babies.
The result concerning the steady state of type O requires more explanation. Note that Wt is at most (1 + r)Wt−1. When a ≥ 1 + r, Wt≤ (1 + r)Wt−1 ≤
Mt−1. Thus, women always get married with probability one, implying that
the population growth is 1 + r. By contrast, when a < 1 + r, if women always get married with probability one, then Wt = (1 + r)Wt−1 > Mt−1, which is
a contradiction since men should then be in overdemand. It can be argued that in a steady state, men get married with probability one, implying that Mt−1(1 + r) = Wt+1 or MMt+1t−1 = a(1 + r).
As for the result concerning the steady state of type Y, note that men cannot always get married with probability one for otherwise, Mt = a(1 + r)Mt−1 >
Wt−1, which is a contradiction. It can be shown that women get married with
probability one so that Wt+1= (1 + r)Wt−1 or MMt+1t−1 = 1 + r. ¥
Proof of Lemma 9. By (7), qt> (1−α)qitbecause at least 1−α proportion
will stick to its inherited type. (1 + α)vi
t> vtbecause at most no one will switch
out but α proportion of the other two types will switch in. Since qi
t = vit by
(2), qtMt > a(11+α−α)vtWt ≥ vtWt for all t ≥ 1. Hence, women of type S are
in overdemand from period 1 irrespective of the initial states. Since women of type S get married with probability 1, the share of women of type S relative to that of type O, or vt
ut, cannot decrease over time by (7). This is because both
types enter the marriage market at one year old and the performance of type S cannot be higher.
To prove that type Y must vanish, observe that since women of type S are always in overdemand, so vt+1≥ vit+1and S(t + 1) = vt+1Wt+1≥ vt+1i Wt+1=
vtWt(1+r). Thus, they become at least 1+r times larger every period. Because
Y (t + 1) = min{pt+1Mt+1, (1 − ut− vt)Wt}, so Y (t + 1) ≤ (1 − ut− vt)Wt. Therefore, vt(1+r) 1−ut−vt ≤ S(t+1) Y (t+1) = vi t+2 1−ui t+2−vt+2i ≤ vt+2
1−ut+2−vt+2, where the last
Thus, the relative ratio of type S to Y becomes bigger and bigger. Since we have shown that the relative ratio of type S to O cannot decrease over time, type S must survive and type Y must vanish in the long run. In fact, as long as women of type S are always in overdemand, the details of the revision dynamic in (7) are not important for this part of the proof. It is enough that the revision dynamic satisfies (5). ¥
Proof of Lemma 10. For the first part of the lemma, by (7), 1 − pt−
qt > (1 − α)(1 − pit− qit) because at least 1 − α proportion will stick to its
inherited type. (1 + α)ui
t > ut because at most no one will switch out but
α proportion of the other two types will switch in. Since 1 − pit− qti = uit
by (2), (1 − pt− qt)Mt > a(11+α−α)utWt for all t ≥ 1. Moreover, ut+1Wt+1 <
(1 + r)(1 + α)utWt since at most all women of type O get married in period
t and at most α percent of the other two types switching into type O. Hence (1 − pt− qt)Mt> a(1+α)1−α2(1+r)ut+1Wt+1≥ ut+1Wt+1. Hence, women of type O
are in overdemand from period 2 on. By a similar argument as that in Lemma 9, type O must survive.
For the second part of the lemma, note that if for some period t ≥ 1, women of type O are in overdemand or (1−pt−1−qt−1)Mt−1≥ utWt, then they must be
in oversupply in the next period or (1 − pt− qt)Mt< ut+1Wt+1. To prove this,
note that 1 − pt− qt< (1 + α)(1 − pit− qit) because at most no one switches out
and 1 + α proportion of the other two types switch in. (1 −α)uit< utbecause at
least 1 − α proportion sticks to its inherited type. Since 1 − pi
t− qit= uitby (2),
(1 − pt− qt)Mt< a1+α1−αutWt. Observe also that ut+1Wt+1≥ (1 + r)utWtsince,
by assumption, women of type O are in overdemand in t. Combining the last two inequalities, we get (1 − pt− qt)Mt< a(1−α)(1+r)1+α ut+1Wt+1≤ ut+1Wt+1. This
means that women of type O are strictly in oversupply in t + 1. Since women of type S are in overdemand in every period and women of types O and S marry at the same time, this implies that women of type S must grow in proportion relative to those of type O, or ut+1
vt+1 <
ut
vt. Thus, the ratio
ut
vt must decrease
at least every other period. Note that type Y can be made infinitely small by Lemma 9. Moreover, every other period α proportion of type O switches out to
type S. Therefore type O must vanish as t becomes large. ¥
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