Proof of Lemma 2. We prove the result by developing a sufficient condition for the LHS and RHS of (A5) to satisfy single crossing w.r.t. L0, with the LHS crossing the RHS from above. 23Recall that the consumer does not pay any interest when she defaults.
24To see why, notice that
G(0.5, 0.5) = 0.5(G1(0.5, 0.5) + G2(0.5, 0.5)) = G1(0.5, 0.5) < G1((y − B + L)/(2θxy), 0.5),
where the last inequality is due to the concavity of G w.r.t. each of its arguments and the fact that for L to be binding we must have (y − B + L)/(2θxy) < 1/2.
We do so by differentiating both sides w.r.t. L0. We need to consider two cases: for low L0 both lines may be utilized, while as L0 grows eventually only the second round line will be used. When both lines are used the LHS is given by
c0|0 = G(y − B + (1 − R/2)ˆb0|0− L0R/2, y − (1 + R/2)ˆb0|0− L0R/2),
where ˆb0|0 denotes total borrowing. If L is not binding, by the envelope theorem we have that
∂c0|0
∂L0 =
R
2 G1(c1,0|0, c2,0|0) + G2(c1,0|0, c2,0|0) .
Since R(G1+ G2) = 2(G1− G2) by the FOC for borrowing we must have that
∂c0|0
∂L0 ≤ G1(c1,0|0, c2,0|0) − G2(c1,0|0, c2,0|0) < G1(y − B, y) − G2(y − B, y),
where the last inequality comes from the concavity of G.
When L is binding, L0 must be binding too so the effect of increasing L0 is similar to the next case.
When only the second round line is used by the consumer, increasing L only affects c0|0
whenever L0 is binding. Accordingly,
∂c0|0
∂L0 = G1(c1,0|0, c2,0|0) − G2(c1,0|0, c2,0|0) < G1(y − B, y) − G2(y − B, y).
Thus, we just need to show that the rate at which consumption under default grows with L0 is higher than G1(y − B, y) − G2(y − B, y). This is guaranteed by Assumption 1, since,
as shown in the above proofs, this rate is at least G(0.5, 0.5) = G1(y, y). Continuity and
differentiability are a direct consequence of the differentiability of G, the continuity of c0|0
and the fact that ∂c0|0
∂L0 has the same expression regardless whether the borrower is constrained
or not.
Finally, it is easy to see why Lmax is decreasing in L. As long as L + L0 remains fixed,
raising L lowers c0|0 whenever L0 is fully utilized while the RHS of (A5) does not change.
Thus, for (A5) to hold L + L0 must go down, since increasing L + L0 would increase the RHS more than the LHS by the above argument. If L0 is not fully utilized after increasing L while keeping L + L0 fixed then both sides of (A5) remain constant and Lmax does not change.
Proof of Lemma 3. First, notice that Lmax is bounded above by (1 − θ0)y. To see why notice
that the highest c0|0 is associated with full smoothing at R = 0. Thus, from (A5) we have
that
(2y − B)G(0.5, 0.5) = ((1 + θ0)y − B + Lmax(0, L, 0))G(q0|0D , 1 − qD0|0),
which yields, for all L,
Lmax(0, L, 0) = (1 − θ0)y
G(0.5, 0.5) G(qD
0|0, 1 − q0|0D )
≤ (1 − θ)y
Given this, default losses when κ = x are bounded above by p(1 − θ0)y. In addition, at R = 0
and L = Lmax(0, L, 0) we have that b0|0 = min{B/2, Lmax(0, L, 0)}, which is greater than
zero as long as B > 0 and θ0 < 1. But then, since Lmax and b0|0 are continuous,25 we can find
R > 0 such that interest revenue, given by Rb0|0 is strictly positive. This, in turn, implies
that we can find p > 0 such that pL ≤ p(1 − θ0)y ≤ Rb0|0, i.e. L is strictly feasible at all
p < p.
Proof of Lemma 4. We obtain the sufficient condition by finding a lower bound on Lmax.
Notice that Lmax satisfies
G(y − B − (1 − R/2)b0|0, y − (1 + R/2)b0|0) = ((1 + θ0)y − B + Lmax)G(q0|0D , 1 − q D 0|0).
The LHS is minimized by setting b0|0 = 0 and G(qD0|0, 1 − q0|0D ) = G(0.5, 0.5), which yields
Lmax ≥
G(y − B, y)
G(0.5, 0.5) − (1 + θ0)y + B.
Proof of Proposition 2. We proceed by computing the slope of the RFCF using
dcL dcH = dcL dL dcH dL .
By the argument in the text, if L < Lmax is not binding we have that L0 = 0. Accordingly, cH
and cL now equal aggregated consumption on the repayment and default paths, respectively.
25Continuity of b
0|0 is straightforward when it is given by the FOC (A4). If borrowing constraints are binding
Given this, for all cL at which L is not binding, we have that
dcL
dL = G(0.5, 0.5).
In order to calculate dcH
dL we use the zero-profit condition of the first round line, given by
EIR = (1 − p)RbH = pL, where bH denotes the borrowing level when κ = 0. In addition,
since L is not binding the first order condition pinning down bH is satisfied with equality.
Therefore, we can apply the envelope theorem and obtain
dcH dL = p dcH dEIR = p ∂G(y − B + (1 − R/2)bH, y − (1 + R/2)bH) ∂R dR dEIR = −pbH G1(qH, 1 − qH) + G2(qH, 1 − qH) 2 dR dEIR,
where qH denotes the fraction of first period consumption over total consumption under κ = 0
and the last equality uses the fact that Gj is homogeneous of degree zero for j = 1, 2. Now,
since bH is decreasing in R and EIR is increasing in R on the frontier,26 we must have
0 ≤ dEIRdR < (1 − p)bH and thus
dcH dL < − p 1 − p G1(qH, 1 − qH) + G2(qH, 1 − qH) 2 .
Notice that G1(qH, 1 − qH) + G2(qH, 1 − qH) is decreasing in qH for qH ≤ 1/2.27 Thus,
G1(qH, 1 − qH) + G2(qH, 1 − qH)
2 >
G1(0.5, 0.5) + G2(0.5, 0.5)
2 = G(0.5, 0.5),
where the last equality comes from applying Euler’s theorem. But this implies that the slope of the frontier is flatter than the AFTL:
dcL dcH = dcL dL dcH dL > −1 − p p 2G(0.5, 0.5) (G1(qH, 1 − qH) + G2(qH, 1 − qH)) > −1 − p p .
26Otherwise, the planner could lower the interest rate and increase revenues, making the consumer strictly
better off.
27Differentiating this expression w.r.t. q
H we get G11− G22, where Gii denotes the second partial derivative
w.r.t to G’s ith argument. It is straightforward to check that G11(c1, c2) < G22(c1, c2) < 0 if c2 > c1 ≥ 0,
Proof of Proposition 4. Notice that at L = Lmax consumption under entry and no-entry
(λ = 0, 1) coincide, and so certainty equivalents equal aggregated consumption given κ. Also, since L + L0 = Lmax on the PFCF, from the definition of Lmax we have that consumption
under repayment equals consumption under default for κ = 0. Hence, to check whether cL> cH, we just need to compare consumption under default for different values of κ, which
boils down to comparing θ0 and θx.
Proof of Lemma 8. I directly follows from the argument in the text.
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