Internet Appendix for A Pyrrhic Victory? Bank Bailouts and Sovereign Credit Risk

Internet Appendix for A Pyrrhic Victory? Bank Bailouts and Sovereign Credit Risk This Internet Appendix serves as a companion to the paper A Pyrrhic Victory? Bank Bailouts and Sovereign Credit Risk. It
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Internet Appendix for A Pyrrhic Victory? Bank Bailouts and Sovereign Credit Risk This Internet Appendix serves as a companion to the paper A Pyrrhic Victory? Bank Bailouts and Sovereign Credit Risk. It contains the proofs to the propositions and other results that were not included in the main text in order to conserve space. IA.1 Derivations IA.1.1 Proof of Lemma 1 Use (6) to substitute for w s in the financial sector s first-order condition and then take the derivative with respect to the transfer T 0 :. d 2 f(k 0, s 0 ) ds 0 dp solv p ds 2 solv + w s c (s 0 ) ds 0 = 0 0 dt ( 0 ) ds 0 dp solv d 2 f(k 0, s 0 ) = w s / p solv c (s 0 ) ds 2 0 (IA.1) Since dp solv / = p(a 1 ), this term is positive so long as A 1 is in the support of à 1 and the transfer increases the probability of solvency by decreasing the solvency threshold A 1. Hence the numerator of the right hand side in the second line is negative. That the denominator is also negative follows from the concavity of f and the convexity of c. This establishes that the right side is positive and hence ds 0 / 0. IA.1.2 A Candidate for V (K) based on f(k, s) Consider the frictionless counterpart to our setting, with p solv = 1. In a dynamic setting, the expression for V would reflect the value of future production of the non-financial sector as a function of its future capital, K. For simplicity, consider one extra period of output. The case of more than one future period should be similar as it is the sum of multiple one-period output. The output of the additional period is given by max s f(k, s). It is natural then to let V (K) = max f(k, s) w s s s with w s determined by the financial sector s first-order condition. With f(k, s) = αk 1 ϑ s ϑ, this implies that V (K) = (1 ϑ)αk 1 ϑ s ϑ where s is the optimal choice of s. Let c(s) = 1 m sm for m 2. Then the first-order condition of the financial sector implies that w s = s m 1 and the first-order condition of the non-financial sector implies that: ϑαk 1 ϑ s ϑ 1 = w s = s m 1 Solving for s, substituting into the expression above for V (K), and simplifying gives: s = (ϑα) 1 m ϑ K 1 ϑ m ϑ V (K) = (1 ϑ)α m m ϑ K γ where γ = (1 ϑ) 1 ϑ m Hence, V (K) has the power form that is used in the paper. Moreover,for m 2 (which is assumed), γ 1. 2 IA.1.3 Properties of Expected Tax Revenue: T For the assumed parametric forms, we obtained the following results: T = θ 0 γ γ 1 γ (1 θ0 ) γ 1 γ dt = γ γ 1 γ (1 θ0 ) γ γ 1 γ θ0 dθ 0 1 γ γ γ 1 γ (1 θ0 ) γ 1 γ 1 = T θ d 2 T = 2 γ dθ0 2 1 γ γ γ 1 γ (1 θ0 ) γ 1 γ 1 + θ 0 1 θ 0 ( 1 γ 1 γ 1 θ 0 ( ) γ γ 1 γ 1 1 γ γ γ 1 γ (1 θ0 ) γ 1 γ 1 The second line shows that dt /dθ 0 0 on [0, θ0 max ) and dt /dθ 0 0 on (θ0 max, 1) where = 1. It is zero at θ max and at 1 (where T = 0). θ max 0 solves: γ 1 γ θ max 0 1 θ max 0 The third line implies that d 2 T /dθ 2 0 0 on [0, θ max 0 ] so T is increasing but concave on this region. To see this, note that the third line can be rewritten as: d 2 T dθ 2 0 = ( 2 + γ 1 γ γ θ 0 1 γ 1 θ 0 θ 0 θ ) 0 γ 1 θ 0 1 θ 0 1 γ γ γ 1 γ (1 θ0 ) γ 1 γ 1 We know that 1 + 0 on [0, θ0 max ] and so, on this region, the leading term in parenthesis is negative. Since the remaining terms are positive, d 2 T /dθ0 2 0 in this region. θ 0 ) IA.1.4 The Government s First-Order Condition From (3) we obtain the following first order condition of the government for the tax rate, θ 0 : [ ] f(k0, s 0 ) c ds0 dt (s 0 ) + [V (K 1 ) 1] dk 1 = 0 (IA.2) s 0 dt dθ 0 dθ 0 Note that the derivatives of s 0 and T here are total derivatives, since the government s choices are subject to the equilibrium choices of the financial and non-financial sectors. As shown above, dt /dθ 0 is positive and decreasing (towards zero), but remains positive, on [0, θ max 0 ]. Therefore, the mapping from tax level (θ 0 ) to the marginal rate of transformation of taxes into tax revenue (dt /dθ 0 ), is invertible on this region. A high tax rate corresponds to a low marginal rate of transformation of taxes into tax revenue and vice versa. Note that the optimal tax rate must be in the region [0, θ max 0 ], since any further increase in θ 0 beyond θ max 0 reduces tax revenue and investment. Hence, we can limit the consideration of 3 the optimal tax rate to this region. Since dt /dθ 0 is positive and the mapping from θ 0 to T is invertible in this region, we can instead consider the government s first order condition with respect to T, which turns out to be more intuitive for analyzing the government s problem. Dividing (IA.2) through by dt /dθ 0, and rewriting (dk 1 /dθ 0 )/(dt /dθ 0 ) = dk 1 /dt we obtain this alternative first-order condition: [ ] f(k0, s 0 ) c ds0 (s 0 ) + [V (K 1 ) 1] dk 1 s 0 dt = 0 (IA.3) where the term /dt, which equals 1 under a no-default government policy, is omitted from the expression. IA.1.5 Under-Investment Loss Due to Taxes We want to obtain an expression for the second term in (8), the transfer version of the government s first-order condition: [V (K 1 ) 1] dk 1 dθ 0 dt dθ 0 The first-order condition for investment of the non-financial sector, (7), and the parametric form for V imply that: V (K 1 ) 1 = θ 0 V (K 1 ) = θ 0 γk γ 1 Substituting in the parametric form also gives: dk 1 dθ 0 = 1 1 θ 0 1 γ 1 K 1 Moreover, from (7) we can solve for the equilibrium K 1 as a function of θ 0 : K 1 = γ 1 1 γ (1 θ0 ) 1 1 γ We can obtain the numerator to our fraction of interest by multiplying the expressions for 4 the two terms together: [V (K 1 ) 1] dk 1 dθ 0 = θ 0 γ (1 θ 0 )(γ 1) Kγ = θ 0 1 θ 0 γ γ 1 γ γ 1 γ (1 θ0 ) γ 1 γ = T θ 0 θ 0 1 θ 0 γ γ 1 where the second line follows by substituting in the expression for K 0 and the third line follows by substituting in the expression for T. Appendix IA.1.3 derives dt /dθ 0. Dividing the expression for the numerator by the expression for dt /dθ 0 shows that the marginal loss per transfer is given by: θ 0 1 θ 0 dl dt = [V (K 1 ) 1] dk1 dθ 0 = dt dθ 0 1 θ 0 1 θ 0 γ 1 γ γ 1 γ From this it is clear that dl/dt as θ 0 θ max (since at θ max the denominator is 0). Additionally, we have: d 2 L dt 2 = d2 L dθ 0 dt dθ 0 dt 0 for θ 0 [0, θ max ). Hence, the marginal loss to the economy is increasing in magnitude (getting worse) as the tax rate increases up to θ max and expected tax revenue rises to T max. In other words, marginal tax revenues becomes increasingly expensive to raise as the marginal loss to the economy from underinvestment rises in the tax rate/level of tax revenues. IA.1.6 Proof of Proposition 1A Substituting (6) into (5) and solving, we obtain the equilibrium level of s 0 (note that we refer to the equilibrium level of s 0 also as s 0, an abuse of notation intended to reduce clutter): s 0 = ( ) 1 ϑα m ϑ 1 ϑ K m ϑ β 0 p 1 m ϑ solv 5 Now substitute this into the expression for dg/dt to get: dg dt = f(k 0, s 0 ) s (1 p solv ) ds 0 = 1 m ϑ Taking derivative again with respect to T shows that: dt 2 ( ) ϑ m ϑ 1 p ϑ ( m ϑ 1 dpsolv solv (1 p solv ) dp solv ) 2 + p ϑ p ϑ m ϑ 2 solv where /dt = 1 is omitted. ( ) m ϑαk 1 ϑ m ϑ 0 β ϑ m ϑ 1 solv (1 p solv ) d2 p solv dt0 2 ϑ m ϑ p m ϑ 1 solv (1 p solv ) dp solv Since the second term in the above expression is always negative, a sufficient condition to ensure that /dt 2 0 is to ensure that the first and third terms in the above expression are non-positive. The condition: m 2ϑ 0 ensures that the first term is non-positive. The third term is negative if the slope of the probability density of à 1 at A 1 is non-positive. Letting Ã1 take a uniform distribution sets this term to zero. 1 Since we have shown that both G and L are concave in T, the government s problem is concave in T. Furthermore, the optimum tax revenue, ˆT, must correspond to a tax rate ˆθ θ max, because the first-order condition is negative at θ max. To see that this is the case, note that dl/dt as θ θ max while dg/dt is finite for p solv 0. IA Impact of L 1 and N D on T Let x = L 1 or N D. Rewriting (8) using the gain and loss notation as dg/dt + dl/dt = 0 and then taking the derivative with respect to x gives: dxdt + d2 L dxdt = 0 (IA.4) 1 Using an exponential distribution would also be sufficient. For the log-normal distribution, this term will be negative for a range of values below a cutoff. 6 Using the Implicit Function Theorem, the two terms on the right side evaluate to the following: dxdt = d dp solv d 2 L dxdt = d2 L dt dt 2 dx ( ) { ( dg psolv T0 dt dt T 0 T dx + T ) 0 + p } solv x x Substituting into (IA.4) and combining the terms multiplying dt /dx yields: dt dx [ d dp solv ( ) ] dg psolv T 0 dt T 0 T + d2 L = d ( ) { dg psolv T 0 dt 2 dp solv dt T 0 x + p } solv x (IA.5) Note for the left-hand side term in parenthesis: d dp solv ( ) dg psolv T 0 dt T 0 T + d2 L dt = 2 d2 G dt + d2 L 0 2 dt 2 For x = N D : p solv T 0 T 0 x + p solv x = p solv T 0 (k A 1) 0 since T 0 / N D = 1 and p solv / N D = ( p solv / T 0 )k A. For x = L 1 : p solv T 0 T 0 x = 0 and p solv x 0 so for either value of x, the term in braces on the right side is negative. Finally, the intermediate steps in the proof of the concavity of G in T show that d dp solv ( ) dg 0 dt Combining these results shows that dt /dx 0 for x = L 1 or N D. IA Impact of N D on T 0 To show how T 0 changes with N D, begin by using the result above for T. In particular, letting x = N D in (IA.5) and simplifying the right-side expression using p solv T 0 + p solv T 0 = x x 7 p solv T 0 (k A 1) and /( dt ) = /dt 2 gives: dt dn D [ ] dt + d2 L = (1 k 2 dt 2 A ) d2 G dt 2 dt = (1 k A) d2g dt 2 0 dt 1 k dn D + d2 L A dn dt 2 dt 2 D Since T 0 = T N D, = dt 1 1 k A dn D dn D dn d Moreover, this shows that T 0 + k A N D, the gross transfer to the financial sector, is decreasing in N D. IA.1.7 Proof of Proposition 1B The tradeoff involved in default is the loss of the deadweight cost D, versus the benefit of a larger transfer with reduced underinvestment made possible by diluting pre-existing debt. The net benefit of this tradeoff can be written as follows: ˆT0 def knd ˆT 0 no def d G ˆT def dl + ˆT no def dt dt D (IA.6) where the first integral is the gain due to increasing the (gross) transfer, while the second integral is the reduction in underinvestment loss due to reducing tax revenue. Note that dg/ here is evaluated at the no-default values. If (IA.6) is positive, it is optimal for the sovereign to choose default, while if it is negative then no-default is optimal. To prove point (1), take the derivative of (IA.6) with respect to L 1 and simplify the resulting expression to obtain: ˆT0 def ka N D ˆT 0 no def ( ) d dg 0 dl 1 The intermediate steps in Appendix IA.1.4 show that the derivative in the integrand is positive. As shown in Appendix IA.1.6.2, the gross transfer is decreasing in N D, so T def 0 no def k A N D + T0 and hence the integral is positive. 8 To prove the statement for N D, take the derivative of (IA.6) with respect to N D. Sim- plifying the derivative at the upper integration boundary gives k A dg/ def ˆT0 ka while N D from the lower boundary we get we get dg/ no def ˆT0. The remaining part of the derivative is: ˆT0 def ka N D ˆT 0 no def d dn D ( ) dg def ˆT0 ka N D = k A d ( ) dg no def ˆT 0 ( dg = k A dg ) def no def ˆT0 KA N D ˆT0 Combining the three parts of the derivatives gives: (1 k A )dg/ ˆT0 no def 0. To show that the benefit of defaulting is convex in N D, take a second derivative to obtain: (1 k A )/dt0 2 no def no def ˆT0 dt0 /dn D 0. Finally, taking the derivative with respect to k A, we obtain (dg/ )N D 0 at the upper integration boundary and 0 at the lower boundary. In the interior we obtain ˆT0 def ka N D ˆT 0 no def d dk A ( ) dg def ˆT0 ka N D = N D ˆT 0 no def ( ) d dg 0 so the derivative is negative. IA.1.8 Optimal Tax Revenue Under Uncertainty Since N T = (T N D /H)H and P 0 is given by (10) under uncertainty, T 0 can be written in terms of T and H as follows: T 0 = N T P 0 = (T N [ D H )E 0 min (H, R )] V. (IA.7) As earlier, the first order condition for the government s choice of T is given by: dg dt + dl dt = 0 Whereas under certainty /dt =1, this is no longer the case. Taking the derivative of T 0 in (IA.7) with respect to T (while holding H constant) and then using (9) to substitute into 9 the resulting expression gives /dt = P 0 H. Therefore, the first-order condition for T is: dg HP 0 + dl T 0 dt = 0 (IA.8) with T 0 given in (IA.7). The loss due to underinvestment, L, is the same as under certainty. Recall that it is concave, with the magnitude of the marginal loss, dl/dt, increasing in T. Similarly, dg/, the gain to the economy from the increased provision of financial services, remains the same with uncertainty and is decreasing in T 0. However, the rate at which T 0 increases in T is now HP 0 rather than 1. Note that this rate is a constant in T, as P 0 is only a function of H, and is less than 1. 2 Finally, the second-order condition for T holds T 2 0 (HP 0 ) 2 + d2 L dt 2 0 as G and L are concave and HP 0 is a function only of H. IA.1.9 Optimal Probability of Default Under Uncertainty Changing H affects two components of the government s objective. As can be seen from (IA.7), increasing H changes T 0. Unlike the case with T, however, increasing H does not have any effect on investment. Instead, the cost associated with increasing H is that it increases the probability of default, and so also the expected deadweight cost. The firstorder condition for H shows this tradeoff: dg dh D dp def dh = 0 (IA.9) From (10), it is clear that dp def /dh 0 and we can think of choosing H exactly as choosing the probability of default. The effect on T 0 = P 0 N T is less immediately clear, since increasing H increases N T, but decreases P 0. However, (IA.7) shows that /dh 0. To see this we break up T 0 into two terms based on (IA.7) and consider their derivatives: d (T N D H )/dh = N D H 0 2 d E 0 [min (H, R )] V /dh = (1 p def ) 0 2 To see this, note that HP 0 = E 0 [min (H, R )] V E 0 [ R V ] = 1. (IA.10) (IA.11) 10 Demonstrating the equivalence in the second line is straightforward, as shown in Appendix IA We refer to (IA.10) as increasing the dilution of existing bondholders claim, since the increase in H reduces the share of tax revenues that goes to the holders of the existing debt, N D. We refer to (IA.11) as reducing either the default buffer or precautionary taxation, since by increasing H, it increases the probability that R V H, in which case the government defaults. Hence, (IA.10) and (IA.11) show that increasing H (while holding T constant) increases T 0. It immediately follows that dg/dh 0 and there is a benefit to increasing H. Substituting in for /dh, the first-order condition becomes: dg ( ND [ H E 2 0 min (H, R )] V + (T N ) D H )(1 p def) D dp def dh = 0 Appendix IA.1.10 also shows that as H increases, raising it further becomes decreasingly effective at increasing T 0 : d 2 T 0 dh = 2N H D xp 2 H 3 RV (x)dx (T N D 0 H )p (H) 0 RV where p RV (x) denotes the probability density of RV evaluated at x. In other words, T 0 is concave in H. Together with the concavity of G in T 0, this implies that G is concave in H, e.g., /dh 2. 3 The implication is that while increasing H provides a benefit to the government by increasing the transfer through dilution and reduction of precautionary taxation, the marginal benefit is decreasing. Meanwhile, the government bears a cost for increasing H; the resulting increased likelihood of default increases the expected deadweight cost of default. We assume that at the optimal choice of H, d 2 p def /d 2 H 0. 3 Note that in the first-order conditions, we have assumed that the government takes into account the (negative) impact of higher H on prices. Thus, we have NOT treated the government here as a price-taker. If we instead treat the government as a price-taker, the resulting conditions are simpler: /dh = P 0 T (as dp 0 /dh is omitted due to the price-taking assumption) and the first-order condition is: dg/ (P 0 T ) Dd p def /dh = 0. In this case, concavity of G in H still holds because G is concave in T 0. 11 IA.1.10 To derive d E 0 [min Uncertainty Calculations (H, R V )] /dh, rewrite the expectation as: E 0 [min (H, R )] H V = x p RV (x)dx + H 0 H p RV (x)dx Now taking the derivative with respect to H, one obtains: d E 0 [min (H, R )] V /dh =Hp RV (H) Hp RV (H) + = H p RV (x)dx =(1 p def ) H p RV (x)dx The first line is just the derivative, while the last line follows by definition of p def. Using this result we have that: dh = N [ D H E 2 0 min (H, R )] ( V + T N ) D (1 p def ) H Substituting in the expression above for E 0 [min (H, R )] V, taking the derivative with respect to T 0, and simplifying gives: [ d 2 T H 0 dh = 2N D 2 H 3 0 x p RV (x)dx + H + N ( D H (1 p def) T N D 2 H = 2N [ H ] D x p H 3 RV (x)dx 0 H ] p RV (x)dx + N D H (1 p def) 2 ) p RV (H) ( T N D H ) p RV (H) Since (T N D /H) = N T /H 0, it is clear that d 2 T 0 /dh 2 0. IA.1.11 Proof of Proposition 2 The starting point are the first-order conditions for T and for H, given by (IA.8) and (IA.9), respectively. Substituting out dg and rearranging gives the relation 12 dl dt dh = HP 0D dp def dh = 0 (IA.12) Differentiating with respect to L 1 gives on the left-hand side: d2 L dt dt 2 dl 1 dh dl dt and on the right-hand side: (1 p def )D dp def dh Combining the terms in dt dl 1 d 2 L dt 2 dh dl d 2 T 0 dt dt dh d 2 T 0 dt dh dh + HP 0 D d2 p def dl 1 dh 2 gives: dt dl d 2 T 0 dh dl 1 dt dh 2 dl 1 dh dl 1 and it is not difficult to see that each term has a positive sign. Combining the terms in dh dl 1 gives: dl dt d 2 T 0 dh + (1 p def)d dp def 2 dh + HP 0D d2 p def dh 2 ( and again each term is positive. Thus, we see that at the optimal values, sgn ( dh sgn dl 1 ). It remains to show that both of these signs are indeed positive. ) dt dl 1 To that end, let V represent the objective function of the government with the first-order conditions given by (IA.8) and (IA.9). Let X = [T, H] be the vector of the two controls. Then the first order conditions can be written as just dv/dx = 0. Differentiating this with respect to L 1 then gives dv dl 1 dx + d2 V dx = 0. dx 2 dl 1 By assumption, the optimal X is internal and so d 2 V/dX 2 is negative definite. Isolating = 13 dx/dl 1 then gives ( ) dx d 2 1 V dv = dl 1 dx 2 dl 1 dx. Premultiplying by dv T dl 1 dx we obtain dv T dx = dv ( ) T d 2 1 V dv dl 1 dx dl 1 dl 1 dx dx 2 dl 1 dx 0 where the sign follows since the Hessian is negative definite. Since dl 1 dt 0 it is straightforward to see that dv dl 1 dx 0, i.e., both terms in the vector are positive. Hence, we must have that dx/dl 1 0 as well since both terms in this vector are of the same sign. Similar steps prove the result for ϑ. IA.1.12 Proposition 3 Below we derive the return on financial sector equity, debt, and the sovereign bond. A complication created by the guarantee is that the number of outstanding sovereign bonds is state contingent, since it depends on the realization of à 1. Let N G (Ã1) denote the number of new bonds issued towards the guarantee. This means there will also be a different price for sovereign bonds contingent on the realization of à 1. Hence, P 0 will now depend on Ã1, as will T 0. This state-contingency is implicit below but will be omitted to avoid excessive notation. Assume that Ã1 U[A min, A max ] and consider two types of shocks. The first is a shock to the value of the risky asset held by the financial sector. This shock changes the mean of à 1 by shifting the support of Ã1 by an amount da. Thus, à 1 remains uniformly distributed with the same dispersion, but a different mean. The second shock affects the sovereign bond price by changing the expected growth rate of future output by dr. For R V uniformly distributed this corresponds to a dr shift in its support. 14 From the model we have that the value of financial sector equity is given by E = Amax A 1 (Ã1 + T 0 L 1 )p(ã1)dã1 where p(ã1) is the uniform probability density. Calculating the change in E in
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