Pierre André Chiappori (Columbia) "Family Economics" - Cemmap

11. Marriage, Divorce, Children 493
commitment to his ex-wife raises the consumption of the step child and
decreases λf. Hence, ∂λm
∂λf
∂σ ≤ 0 and ∂σ ≤ 0. But in a symmetric equilibrium
all couples make the same choices and ∂λm ∂λf
∂σ = ∂σ , which would imply that
∂ 2 Vf
∂σ2 ≥ 0. Therefore, there is no interior symmetric equilibrium, and the
only symmetric equilibria are such that all couples must be at one of the
boundaries, σ =0or σ = wf −c∗−(1−p)σ0 . Notice that the upper boundary
is not determined by the budget constraint but by the requirement that
the mother is just indifferent between remarriage and remaining single.
Suppose that all other couples set σ0 = wf −c ∗
2−p .Thenbysetting
σ = wf − c ∗ − (1 − p)σ 0 = wf − c ∗
(11.50)
2 − p
the father can guarantee that the mother is just indifferent between marriage
and remaining single. This is seen by noting that the mother participation
constraint becomes
(1+α)[wmhm+ wf − c∗ 2 − p +c∗−c]+γ(1−hm)+g(c) ≥ γ+g(c ∗ )+(1+α) wf − c∗ 2 − p
(11.51)
But
Max{(1
+ α)[wmhm + c
c,hm
∗ − c]+γ(1 − hm)+g(c)}
≤ Max{ γhm +(1+α)(c ∗ − c)+γ(1 − hm)+g(c)}
c,hm
= γ + g(c ∗ ). (11.52)
Therefore, (11.51) must hold as equality.
It remains to show that σ = wf −c ∗
2−p is indeed an optimal choice, given that
others maintain σ0 = wf −c ∗
∂Vf
2−p . For a marginally lower σ we have that ∂σ
1
approaches ∞ because λm = 1+α−g0 (c) approaches −∞ as c approaches c∗ .
For marginally higher σ we have that ∂Vf
∂σ approaches −∞, because when
the mother chooses not to marry the father suffers a discrete loss since he
pays the mother σ with certainty. Thus, σ = wf −c ∗
2−p is a local maximum.
However, it need not be a global maximum. In particular, for a small p, it
is always the case that ∂Vf
∂σ < 0, because the father bears the costs with
high probability and the benefits with a low probability, so that σ =0is
also a local maximum. However, in contrast to the selection of σ = wf −c ∗
2−p ,
the selection of σ =0is a local maximum only if p is small. The difference
arises because at low levels of σ, c < c∗ so that λm and λf are finite and
∂Vf
∂σ must change sign from negative to positive as p rises from zero to one.
We have noted in the text that if remarried mothers work part time
then ∂Vf
∂σ is independent of σ0 . However, if the mother does not work or
1
works full time so that λm = 1+α−g0 (c) , then an increase in σ0 will raise c,