Online Appendix I. 1 1+r ]}, Bψ = {ψ : Y E A S S}, B W = +(1 s)[1 m (1,0) (b, e, a, ψ (0,a ) (e, a, s); q, ψ, W )]}, (29) exp( U(d,a ) (i, x; q)

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Download "Online Appendix I. 1 1+r ]}, Bψ = {ψ : Y E A S S}, B W = +(1 s)[1 m (1,0) (b, e, a, ψ (0,a ) (e, a, s); q, ψ, W )]}, (29) exp( U(d,a ) (i, x; q)"

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1 Online Appendix I Appendix D Additional Existence Proofs Denote B q = {q : A E A S [0, +r ]}, Bψ = {ψ : Y E A S S}, B W = {W : I E A S R}. I slightly abuse the notation by defining B q (L q ) the subset of B q in which q is uniformly Lipschitz continuous in s with condition L q, i.e., B q (L q ) = {q B q : q(a, e, a, s) q(a, e, a, s ) < L q s s, (a, e, a), s, s }. Similarly, B ψ (L ψ ) = {ψ B ψ : ψ (d,a ) (a, e, a, s) ψ (d,a ) (a, e, a, s ) < L ψ s s, (d, a ), (a, e, a), s, s }. B W (L ψ ) = {W B W : W (i, e, a, s) W (i, e, a, s ) < L W s s, (i, e, a), s, s }. Define mapping T = (T, T, T ) : B q B ψ B W B q B ψ B W as following: T q(a, e, a, s; q, ψ, W ) = + r {s[ m (,0) (g, e, a, ψ (0,a ) (e, a, s); q, ψ, W )] e E +( s)[ m (,0) (b, e, a, ψ (0,a ) (e, a, s); q, ψ, W )]}, (9) T ψ (d,a ) (e, a, s; q, ψ, W ) = ξ (d,a ) (x; q, ψ, W )Ω(g g) + ( ξ (d,a ) (x; q, ψ, W ))Ω(g b), (0) T W (i, e, a, s; q, ψ, W ) = γ C + log[ where, m (d,a ) (i, e, a, s; q, ψ, W ) = (d,a ) M(x;q) U(d,a ) (i,e,a,s;q) U( d, ( d, ã ) M(e,a,s;q) U(d,a ) (i, x; q) + β i EW (i, e, a, ψ (d,a ) (e, a, s)) a ) (i,e,a,s;q) + β i EW (i, e, a, ψ (d,a ) (e, a, s))], (), for (d, a ) M(e, a, s; q), + β i EW (i, e, ã, ψ ( d,ã ) (e, a, s)))) 0, for (d, a ) / M(e, a, s; q); () odd (d,a ) (e, a, s; q, ψ, W ) = U(,0) (b,e,a,s) β b EW (b, e, a, ψ (d,a ) (e, a, s))) β gew (g, e, a, ψ (d,a ) (e, a, s))) + β b EW (b, e, a, ψ (d,a ) (e, a, s))) U(,0) (g,e,a,s) + β gew (g, e, a, ψ (d,a ) (e, a, s))) ( d, a ) M(e,a,s;q) U( d, a ) (i,e,a,s;q) ( d, a ) M(e,a,s;q) U( d, a ) (i,e,a,s;q) ( d, a ) M(e,a,s;q) U( d, a ) (e,a,s,x;q) ( d, a ) M(e,a,s;q) U( d, a ) (e,a,s,x;q) + β gew (g, e, ã, ψ ( d, a ) (e, a, s)) + β b EW (b, e, ã, ψ ( d, a ) (e, a, s))) + β gew (g, e, ã, ψ ( d, a ) (e, a, s)) + β b EW (b, e, ã, ψ ( d, a ) (e, a, s))) (), for (d, a ) = (0, a ), for (d, a ) = (, 0) 78

2 ξ (d,a ) (e, a, s; q, ψ, W ) = + s s odd (d,a ) (e, a, s; q, ψ, W ) (4) The goal is to find large enough, s.t. (L q, L ψ, L W ), the image of T is contained in B q (L q ) B ψ (L ψ ) B W (L W ). Lemma 6. >, > 0, the function h(c) = c ) over c (0, ˉc] satisfies following properties: (i) lim c 0+ h(c) = 0; (ii) lim c 0+ h (c) = 0; (iii) h (c), c (0, ˉc]; (iv) h(c) ˉc. Proof. (i) is obvious. For (ii), consider Consider the transformation t = c which by L Hospital s rule is equal to 0. For (iii), consider h (c) = c c )., then lim c 0+ h (c) = lim t t) (( ), t + h (c) = c c ) + c c ) = c c )(c + c ). Therefore, the maximum of h (c) is attained at c = ( ), and h (c ) = ( ) ) = By Mean Value Theorem, we get (iv).. Now I show that B W can be chosen a set of functions with bounded value. Lemma 7. W ( ) and W ( ) s.t. if W ( ) W (i, e, a, s) W ( ), i, e, a, s, then W ( ) T W (i, e, a, s; q, ψ, W ) W ( ), i, e, a, s; q, ψ. Moreover, W ( ) can be chosen that is increasing in ; W ( ) can be chosen to be W that does not depend on. 79

3 Proof. Since default is always an option, Therefore, U(d,a ) (e, a, s, x; q) + β i EW (i, e, a, ψ (d,a ) (e, a, s))) (d,a ) M(e,a,s) [φ i (c (,0) (e, a, s) v(e))] ) β i W ( )). T W (i, e, a, s; q, ψ, W ) γ C + [φ i (c (,0) (e, a, s) v(e))] + β i W ( ). Note I have used the property the consumption upon default c (,0) (e, a, s) does not depend q. Thus fix i it suffices to set the lower bound s.t. W ( ) = γ C + [φ i (c (,0) (e, a, s) v(e))] + β i W ( ). Taking the lower across i we thus have W ( ) = min i {b,g} Since >, we have W ( ) is increasing in. (γ C + [φ i (c (,0) (e, a, s) v(e))] ). β i Next, since >, U (d,a ) (i, e, a, s; q) < 0, (i, e, a, s). Therefore, T W (i, e, a, s; q, ψ, W ) γ C + log(n A + ) + β i W ( ), where N A is the cardinality of set A. Therefore, it suffices to set which does not depend on. W ( ) = max i b,g β i (γ C + log(n A + )), Next I establish Lipschitz condition for T W. I first show the following intermediate function is Lipschitz continuous. From now on denote L q, L ψ, L W as Lipschitz conditions of q, ψ, W, respectively. The underlying assumptions in each following lemma is that q, ψ, W are Lipschitz continuous (note it does not mean I have imposed the image of T to be Lipschitz). 80

4 Consider function μ (d,a ) (i, e, a, s; q) defined as μ (d,a ) (i, e, a, s; q) = c (d,a ) (e, a, s; q) () ), for (d, a ) M(e, a, s; q), 0, for (d, a ) / M(e, a, s; q). Lemma 8. μ (d,a ) (i, e, a, s; q) μ (d,a ) (i, e, a, s ; q) L q C μ s s, (d, a ), i, e, a, s, where C μ is a constant that does not depend on. Further, μ (d,a ) (i, e, a, s; q) for some constant C μ 0 that does not depend on. Proof. Case, If (d, a ) M(e, a, s; q) and (d, a ) M(e, a, s ; q), using Lemma 6 and Mean Value Theorem, μ (d,a ) (i, e, a, s; q) μ (d,a ) (i, e, a, s ; q) C μ 0 c (d,a ) (e, a, s; q) c (d,a ) (e, a, s ; q) max a A a L q s s. Case, suppose (d, a ) M(e, a, s; q) and (d, a ) / M(e, a, s ; q). a = 0. Now I claim Notice in this case, c (d,a ) (e, a, s; q) s s a L q Suppose not, then we have s s < c(d,a ) (e, a, s; q) a L q, which implies c (d,a ) (e, a, s ; q) c (d,a ) (e, a, s; q) L q a s s < c (d,a ) (e, a, s; q), which implies c (d,a ) (e, a, s ; q) > 0 that contradicts (d, a ) / M(e, a, s ; q). Therefore, μ (d,a ) (i, e, a, s; q) μ (d,a ) (i, e, a, s ; q) = μ (d,a ) (i, e, a, s; q) c (d,a ) (e, a, s; q) 0 max a A a L q s s 8

5 Setting C μ = maxa A a we have proved the first part. For the second part, apply (iv) of Lemma 6 we have μ (d,a ) (i, e, a, s; q) where ˉc can be any constant greater than the maximum total consumption attainable, which can be set as ˉc = we max n(e max )( τ) a min + r + a max + Tr v(n(e max )). Setting C μ 0 = ˉc, we have proved the second part. We review the following properties about Lipschitz continuity. Lemma 9. Suppose functions f (x), g(x) are defined on closed interval [x, x] and are bounded in sup norm f and g, and f (x), g(x) are Lipschitz continuous in x with condition L f and L g, respectively. Then the following are true: () L α f α L f, α R; () L f +g L f + L g ; () L f g L f g + L g f ; (4) For function h that the compound evaluation h( f ) is defined, suppose h( ) is Lipschitz continuous with condition L h, then L h( f ) L h L f. If h is differentiable with bounded derivative h, then L h( f ) h L f. Lemma 0. T W (i, e, a, s ; q, ψ, W ) T W (i, e, a, s; q, ψ, W ) < L T W s s, where L T W = [C T W ( )L q + C T W ( )L W L ψ ], where C T W ( ) > 0 is decreasing in, and C T W ( ) > 0 is decreasing in. Proof. First, default is always an option so the sum of exponential value is bounded ˉc 8

6 below, denoted by LB( ): U(d,a ) (i, e, a, s; q) + β i EW (i, e, a, ψ (d,a ) (e, a, s))) (d,a ) M(e,a,s;q) min [Tr + ( τ)we min n(e min ) n(e min )] + β i {b,g} i W ( )) = LB( ). Notice that LB( ) is increasing in since > and W ( ) is increasing in (Lemma 7). Denote L ( f ) as the Lipschitz condition for the function f w.r.t. s. Notice can be written as U(d,a ) (i, e, a, s; q) + β i EW (i, e, a, ψ (d,a ) (e, a, s)) (d,a ) M(x;q) μ (d,a ) (i, e, a, s; q) β i EW (i, e, a, ψ (d,a ) (e, a, s))), (d,a ) Y because μ (d,a ) (i, e, a, s; q) assigns 0 to (d, a ) / M(e, a, s; q). Using Lemma 9, L (β i EW (i, e, a, ψ (d,a ) (e, a, s)))) β i β i W )L W L ψ. And using Lemma 8, L (μ (d,a ) (i, e, a, s; q)) C μ L q, μ (d,a ) (i, e, a, s; q) C μ 0, 8

7 Therefore, using Lemma 9, C μ 0 Therefore, Therefore, L [ L (μ (d,a ) (i, e, a, s; q) β i EW (i, e, a, ψ (d,a ) (i, e, a, s)))) L (β i EW (i, e, a, ψ (d,a ) (e, a, s)))) + β i W )L (μ (d,a ) (i, e, a, s; q)) (d,a ) M(x;q) U(d,a ) (i, e, a, s; q) β i C μ 0 β i W )L W L ψ + C μ L q. β g C μ 0 β g W )L W L ψ + C μ L q. + β i EW (i, e, a, ψ (d,a ) (i, e, a, s)))] [β g C μ 0 β g W )L W L ψ + C μ L q ](N A + ). Setting L [T W (i, e, a, s; q, ψ, W ) = L [log( U(d,a ) (i, e, a, s; q) + β i EW (i, e, a, ψ (d,a ) (e, a, x))))] (d,a ) M(x;q) LB( ) [β gc μ 0 β g W )L W L ψ + C μ L q ](N A + ) C T W ( ) = LB( ) β gc μ 0 β gw )(N A + ) C T W ( ) = LB( ) Cμ (N A + ). Since LB ( ) > 0 and is increasing in, we have proved the results. Lemma. m defined in Equation is Lipschitz continuous in s with condition L m = [C m ( )L q + C m ( )L W L ψ ], where C m ( ) > 0 is decreasing in, and C m ( ) > 0 is decreasing in. 84

8 Proof. Notice m (d,a ) (i, e, a, s; q, ψ, W ) = μ (d,a ) (e, a, s; q) β i EW (i, e, a, ψ (d,a ) (e, a, s)) ( d, ã ) Y μ( d,ã ) (e, a, s; q) β i EW (i, e, ã, ψ ( d,ã ) (e, a, s))). As shown in Lemma 0, L [μ (d,a ) (i, e, a, s; q) β i EW (i, e, a, ψ (d,a ) (e, a, s)))] β g C μ 0 β g W )L W L ψ + C μ L q. L [ ( d,ã ) Y μ ( d,ã ) (e, a, s; q) β i EW (i, e, ã, ψ ( d,ã ) (e, a, s)))] [β g C μ 0 β g W )L W L ψ + C μ L q ](N A + ) Apply Lemma 9, μ ( d,ã ) (e, a, s; q) β i EW (i, e, ã, ψ ( d,ã ) (e, a, s))) LB( ) ( d,ã ) Y L (m (d,a ) (i, e, a, s; q, ψ, W )) LB ( ) [β gc μ 0 β g W )L W L ψ + C μ L q ](N A + ) β g W ) +[β g C μ 0 β g W )L W L ψ + C μ L q ] LB( ) = [ LB ( ) (N A + ) β g W )C μ 0 + LB( ) Cμ 0 ]β g β g W L W L ψ Setting and C m( ) = [ LB ( ) (N A + ) β g W ) + LB( ) ]Cμ +[ LB ( ) (N A + ) β g W ) + LB( ) ]Cμ C m( ) = [ LB ( ) (N A + ) β g W ) + LB( ) ]Cμ 0 β g β g W ). Since LB( ) is increasing in, we have proved the result. Similarly we can show L q 85

9 Lemma. odd defined in Equation is Lipschitz continuous in s with condition L odd = [C odd ( )L q + C odd ( )L W L ψ ], where C odd ( ) > 0 is decreasing in, and C odd ( ) > 0 is decreasing in. Lemma. ξ defined in Equation 4 is Lipschitz continuous in s with condition L ξ = [C ξ ( )L q + C ξ ( )L W L ψ ] + C ξ, where C ξ ( ) > 0 is decreasing in, C ξ ( ) > 0 is decreasing in, and C ξ depend on. > 0 does not Proof. Notice 0 < s s s <, s s s s and d s s ds = s [ s, s ] Using Lemma 9, L [ s s odd (d,a ) (e, a, s; q, ψ, W )] s + L odd( s ) L [ + s s odd (d,a ) (e, a, s; q, ψ, W ) ] s + L odd( s ) s + ( s )( [C odd ( )L q + C odd ( )L W L ψ ]). Then we set C ξ ( ) = ( s )Codd ( ) C ξ ( ) = ( s )Codd ( ) 86

10 C ξ = s Lemma 4. T ψ defined in Equation 0 is Lipschitz continuous in s with condition L T ψ = [C T ψ ( )L q + C T ψ ( )L W L ψ ] + C T ψ, where C T ψ ( ) > 0 is decreasing in, C T ψ ( ) > 0 is decreasing in, and C T ψ > 0 does not depend on. Proof. Apply Lemma 9, L T ψ (Ω(g g) + Ω(g b))l ξ (Ω(g g) + Ω(g b))[ (C ξ ( )L q + C ξ ( )L W L ψ ) + C ξ ]. Lemma 5. Tq defined in Equation 9 is Lipschitz continuous in s with condition L T q = [C T q ( )L q L ψ + C T q ( )L W L ψ] + C T q, where C T q ( ) > 0 is decreasing in, C T q ( ) > 0 is decreasing in, and C T q > 0 does not depend on. Proof. Apply Lemma 9, L T q + r (sl ml ψ + + ( s)l m L ψ + ) + r + ( + s s)l ψ[ [C m( )L q + C m( )L W L ψ ]]. Lemma 6. L q, L ψ, L W, (L q, L ψ, L W ) s.t. > (L q, L ψ, L W ), L T W < L W. Proof. From Lemma 0, L T W = [C T W ( )L q + C T W ( )L W L ψ ]. 87

11 Choose (L q, L ψ, L W ) s.t. C T W ( )L ψ < / and This can be done since lim. C T W ( )L q < L W Then we have > (L q, L ψ, L W ), L T W < L W. = 0 and C T W ( ) and C T W ( ) are decreasing in Lemma 7. L q, L ψ C T ψ, L W, (L q, L ψ, L W ) s.t. > (L q, L ψ, L W ), L T ψ < C T ψ. Proof. From Lemma 4, L T q = [C T q ( )L q L ψ + C T q ( )L W L ψ] + C T q. L q, L ψ < C T ψ, L W, choose (L q, L ψ, L W ) s.t. C T q. This can be done since lim in. Similarly, we can establish [C T q ( )L q L ψ + C T q ( )L W Lψ ] < = 0 and C T ψ ( ) and C T ψ ( ) are decreasing Lemma 8. L q C T q, L ψ, L W, (L q, L ψ, L W ) s.t. > (L q, L ψ, L W ), L T q < C T q. Through the above three lemmas, we have the following Lemma 9. For L q = C T q, L ψ = C T ψ, L W =, s.t. >, L T q C T q, L T ψ C T ψ, L T W. Proof. Choose (L q, L ψ, L W ), (L q, L ψ, L W ), (L q, L ψ, L W ) from Lemma 8, Lemma 7, and Lemma 6, respectively. Choose = max{ (L q, L ψ, L W ), (L q, L ψ, L W ), (L q, L ψ, L W )}. Then the result follows Lemma 8, 7, and 6. Up to now we have established that by choosing large enough, T maps B q (L q ) B ψ (L ψ ) B W (L W ) to itself. Now we prove the mapping T is continuous. Denote as sup norm. 88

12 Lemma 0. The intermediate function μ (d,a ) (i, e, a, s; q) defined in Lemma 8 is continuous in q with sup norm for q B q (L q ). Proof. Consider μ (d,a ) (i, e, a, s; q) μ (d,a ) (i, e, a, s; q ) under different cases. Case : (d, a ) M(e, a, s; q), and (d, a ) M(e, a, s; q ), by Mean Value Theorem μ (d,a ) (i, e, a, s; q) μ (d,a ) (i, e, a, s; q ) c (d,a ) (e, a, s; q) c (d,a ) (e, a, s; q ) max a A a q q Case : (d, a ) M(e, a, s; q), and (d, a ) / M(e, a, s; q ), Notice in this case, a = 0. Now I claim c (d,a ) (e, a, s; q) q(a, e, a, s) q (a, e, a, s) a Suppose not, then we have which implies q(a, e, a, s) q (a, e, a, s) < c(d,a ) (e, a, s; q) a, c (d,a ) (e, a, s; q ) c (d,a ) (e, a, s; q) q(a, e, a, s) q (a, e, a, s) a < c (d,a ) (e, a, s; q), which implies c (d,a ) (e, a, s; q ) > 0 that contradicts (d, a ) / M(e, a, s; q ). Therefore, μ (d,a ) (i, e, a, s; q) μ (d,a ) (i, e, a, s; q ) = μ (d,a ) (i, e, a, s; q) 0 c (d,a ) (e, a, s; q) 0 q(a, e, a, s) q (a, e, a, s) a max a A a q q. Similarly for case where (d, a ) / M(e, a, s; q), and (d, a ) M(e, a, s; q ). And for 89

13 case 4 (d, a ) / M(e, a, s; q) and (d, a ) / M(e, a, s; q ) we have μ (d,a ) (i, e, a, s; q) μ (d,a ) (i, e, a, s; q ) = 0. Summarize over all four cases we have μ (d,a ) (i, e, a, s; q) μ (d,a ) (i, e, a, s; q ) Therefore μ (d,a ) (i, e, a, s; q) is continuous in q with sup norm. max a A a q q. Lemma. T W (i, e, a, s; q, ψ, W ) is continuous in (q, ψ, W ) with sup norm on B q (L q ) B ψ (L ψ ) B W (L W ). Proof. Write T W (i, e, a, s; q, ψ, W ) as T W (i, e, a, s; q, ψ, W ) = γ C + log( (d,a ) Y μ (d,a ) (i, e, a, s; q) β i EW (i, e, a, ψ (d,a ) (e, a, s)))) Since (d,a ) Y μ (d,a ) (i, e, a, s; q) β i EW (i, e, a, ψ (d,a ) (e, a, s))) is uniformly bounded positive below (by LB( ) established in Lemma 0), μ (d,a ) (i, e, a, s; q) is continuous in q, W (i, e, a, s) is Lipschitz continuous in s, therefore, T W (i, e, a, s; q, ψ, W ) is continuous in (q, ψ, W ). Lemma. m (d,a ) (i, e, a, s; q, ψ, W ) defined in Equation is continuous in (q, ψ, W ) with sup norm on B q (L q ) B ψ (L ψ ) B W (L W ). Proof. Write m (d,a ) (i, e, a, s; q, ψ, W ) as following m (d,a ) (i, e, a, s; q, ψ, W ) = μ (d,a ) (e, a, s; q) β i EW (i, e, a, ψ (d,a ) (e, a, s)) ( d, ã ) Y μ( d,ã ) (e, a, s; q) β i EW (i, e, ã, ψ ( d,ã ) (e, a, s))). Note the following properties. () W is uniformly bounded above and below. () W is Lipschitz in s. () μ is continuous in q. (4) ( d, ã ) Y μ( d,ã ) (e, a, s; q) β i EW (i, e, ã, ψ ( d,ã ) (e, a, s))) is uniformly bounded above and uniformly positively bounded below (established in the proof of Lemma 7). Therefore, we have m is continuous in (q, ψ, W ). We next establish Lemma. T q(a, e, a, s; q, ψ, W ) is continuous in (q, ψ, W ) with sup norm on B q (L q ) B ψ (L ψ ) B W (L W ). 90

14 Proof. Consider (q n, ψ n, W n ) (q, ψ, W ) with sup norm, we have max m(d,a ) (i, e, a, ψ (0,a ) (e, a, s); q, ψ, W ) m (d,a ) (i, e, a, ψ (0,a ) i,e,a n (e, a, s); q n, ψ n, W n ),e,a,s max m(d,a ) (i, e, a, ψ (0,a ) (e, a, s); q, ψ, W ) m (d,a ) (i, e, a, ψ (0,a ) (e, a, s); q n, ψ n, W n ) i,e,a,e,a,s + max i,e,a,e,a,s m(d,a ) (i, e, a, ψ (0,a ) (e, a, s); q n, ψ n, W n ) m (d,a ) (i, e, a, ψ (0,a ) n (e, a, s); q n, ψ n, W n ) First, lim n max m(d,a ) (i, e, a, ψ (0,a ) (e, a, s); q, ψ, W ) m (d,a ) (i, e, a, ψ (0,a ) (e, a, s); q n, ψ n, W n ) = 0 i,e,a,e,a,s since m( ; q, ψ, W ) is continuous in (q, ψ, W ). Second, m (d,a ) (i, e, a, ψ (0,a ) (e, a, s); q n, ψ n, W n ) m (d,a ) (i, e, a, ψ (0,a ) n (e, a, s); q n, ψ n, W n ) L m ψ ψ n since m (d,a ) (i, e, a, s; q, ψ, W ) is uniformly Lipschitz continuous in s with condition L m, as established in Lemma. lim n Therefore, max m(d,a ) (i, e, a, ψ (0,a ) (e, a, s); q, ψ, W ) m (d,a ) (i, e, a, ψ (0,a ) i,e,a n (e, a, s); q n, ψ n, W n ) = 0,e,a,s Therefore, consider T q( ; q, ψ, W ) T q(a, e, a, s; q, ψ, W ) = + r {s[ m (,0) (g, e, a, ψ (0,a ) (e, a, s); q, ψ, W )] e E +( s)[ m (,0) (b, e, a, ψ (0,a ) (e, a, s); q, ψ, W )]}, lim max T q(a, e, a, s; q, ψ, W ) T q(a, e, a, s; q n, ψ n, W n ) = 0 n a,e,a,s Similarly we can prove Lemma 4. T ψ(a, e, a, s; q, ψ, W ) is continuous in (q, ψ, W ) with sup norm on B q (L q ) B ψ (L ψ ) B W (L W ). 9

15 We next apply Schauder fixed point theorem on the mapping T. Lemma 5. s.t. >, T defined in Equation 9, 0, and has fixed points. Proof. First, from Lemma 7, for large enough, B W can be chosen the set of functions uniformly bounded by W () and W. (Notice W ( ) is increasing in ). Fix L q = C T q, L ψ = C T ψ, L W =. From Lemma 9, s.t. >, L T q C T q, L T ψ C T ψ, L T W. Now denote T : B q (C T q ) B ψ (C T ψ ) B W () B q B ψ B W as mapping that agrees with T on the subset B q (C T q ) B ψ (C T ψ ) B W (). Then T(B q (C T q ) B ψ (C T ψ ) B W ()) B q (C T q ) B ψ (C T ψ ) B W (). And from Lemma,, and 4, T is continuous. By Arzelà Ascoli theorem, the set of bounded Lipschitz functions B q (C T q ) B ψ (C T ψ ) B W () is compact. All conditions applying Schauder fixed point theorem are satisfied. Therefore, T has fixed points, which are also fixed points of T. Now given q, ψ, W and the induced policy function m( ; q, ψ, W ), I establish the existence of stationary probability measure. Denote M Φ as the set of probability measure Φ over (I E A S, P(I) P(E) P(A) B(S)), where P( ) is the power set of underlying discrete sets, and B(S) is the Borel algebra of [s, s]. Denote T 4 : M Φ M Φ defined as following: T 4 Φ(i, e, a, S ; Φ) = +(a = 0) i,e,a Ω(i i)γ(e e) m (0,a ) (i, e, a, s; q, ψ, W )(ψ (0,a ) (e, a, s) S )Φ(i, e, a, ds) a Ω(i i)γ(e e)m (,0) (i, e, a, s; q, ψ, W )(ψ (,0) (e, a, s) S )Φ(i, e, a, ds), i,e,a We prove T 4 is continuous in Φ under the weak topology. Similar proof strategies have been used in Cao (06). We first have the following: Lemma 6. f Lipschitz continuous (with condition L f ) and sequence of Φ n that is weakly converging to Φ lim f (s )T 4 Φ(i, e, a, ds ; Φ n ) = f (s )T 4 Φ(i, e, a, ds ; Φ) n s S s S depends on (r, τ, Tr) and other model parameters. i, e, a, S B(S) (5) 9

16 i, e, a, S B(S). Proof. Consider = lim f (s )[ n s S = lim [ n +(a = 0) +(a = 0) s i,e,a = lim [ n s i,e,a lim f (s )T 4 Φ(i, e, a, ds ; Φ n ) n s S Ω(i i)γ(e e) m (0,a ) (i, e, a, s; q, ψ, W )(ψ (0,a ) (e, a, s) ds )Φ n (i, e, a, ds) a s i,e,a Ω(i i)γ(e e)m (,0) (i, e, a, s; q, ψ, W )(ψ (,0) (e, a, s) ds )Φ n (i, e, a, ds)] Ω(i i)γ(e e) a m (0,a ) (i, e, a, s; q, ψ, W ) s i,e,a +(a = 0) i,e,a f (s )(ψ (0,a ) (e, a, s) ds )Φ n (i, e, a, ds) s S Ω(i i)γ(e e)m (,0) (i, e, a, s; q, ψ, W ) f (s )(ψ (,0) (e, a, s) ds )Φ n (i, e, a, ds)] s S {s:ψ (0,a ) (e,a,s) S } i,e,a Ω(i i)γ(e e) m (0,a ) (i, e, a, s; q, ψ, W ) f (ψ (0,a ) (e, a, s))φ n (i, e, a, ds) a {s:ψ (0,a ) (e,a,s)} S Ω(i i)γ(e e)m (,0) (i, e, a, s; q, ψ, W ) f (ψ (0,a ) (e, a, s))φ n (i, e, a, ds)] The second equal changes order of integration since the inner integral is finite (m is positive bounded below ). Since f, m, ψ are all bounded and Lipschitz in s ( f is bounded because f is Lipschitz and defined on a compact set), by definition of weak convergence the above = i,e,a +(a = 0) i,e,a {s:ψ (0,a ) (e,a,s) S } {s:ψ (0,a ) (e,a,s) S } Ω(i i)γ(e e) m (0,a ) (i, e, a, s; q, ψ, W ) f (ψ (0,a ) (e, a, s))φ(i, e, a, ds) a Ω(i i)γ(e e)m (,0) (i, e, a, s; q, ψ, W ) f (ψ (0,a ) (e, a, s))φ(i, e, a, ds) = f (s )T 4 Φ(i, e, a, ds ; Φ) s S Therefore, T 4 ( ; Φ) is continuous in Φ with weak topology, we have Lemma 7. T 4 has fixed points. Proof. Since S is convex and compact, M Φ is a convex and compact set. And since T 4 is continuous. The conditions of Schauder fixed point theorem are satisfied. T 4 has fixed points. Now I construct the remaining parts of equilibrium out of q, ψ, W, Φ. 9

17 Proof for Proposition, restated in Lemma 8 Lemma 8. For any r > 0, τ [0, ), and Tr 0, (r, τ, Tr) s.t. > (r, τ, Tr), a stationary equilibrium SCE(r, τ, Tr) exists. Proof. Given (r, τ, Tr), w = z η η. Using Lemma 5, we have, s.t. >, (q, ψ, W ) that is a fixed point of operator T. Then pick and fix (q, ψ, W ), using Lemma 7, we have Φ that is a fixed point of operator T 4. Now pick and fix a Φ. Construct the following (sequentially): W(i, e, a, s) = W (i, e, a, s) i = ˉr B = [ π = ˉπ = a m (0,a ) (i, e, a, s; q, ψ, W )q(a, e, a, s)a Φ(di, de, da, ds)]( + r) n(e) = v (( τ t )w t e) N = Γ(e)n(e)e e E Pro f it = zn wn G = B + Pro f it + τwn Tr B + r Y = zn Then one verifies all the equilibrium conditions are satisfied. 94

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