Asset Bubbles and Monetary Policy Figures and Appendix

C(%) I(%) 0 0 −0.2 −1 −0.4 −2 −0.6 −3 −0.8 5 10 15 20 −4 Y (%) 0 −0.5 −1 5 10 15 20 5 K(%) N (%) 0 10 15 20 15 20 15 20 R(%) 0 −0.2 20 15 −1 −0.4 10 −2 −0.6 −0.8 −1.5 5 10 15 20 −3 5 5 10 15 20 0 5 p h (%) L/P (%) Cutoffs(%) 2 10 0 100 −5 50 * ε ε** 1 0 −1 5 10 15 (Π − 1) ∗ 100 20 −10 5 10 15 (Π − 1) ∗ 100 20 0 5 10 (Π − 1) ∗ 100 Figure 1: Comparative statics for the bubbly steady state. Parameter values are given in Section 5.1. Ct It Yt 1 4 2 0.5 2 1 0 0 5 10 15 0 0 5 Nt 10 15 0 4 0 −0.5 2 5 10 15 −1 0 5 10 15 0 1 5 −0.5 0 0 5 10 15 −1 0 5 10 15 −5 1 1 −0.05 0 0 0 5 10 Quarters 15 −1 0 5 10 Quarters 15 10 15 0 5 10 15 ε∗∗ t 0 −0.1 5 ε∗t real rate 10 gt 0 0 0 p ht Rt −1 5 Lt+1/Pt 0 0 0 Πt 1 −1 Taylor Weak Strong 15 −1 0 5 10 15 Quarters Figure 2: Impulse responses to a positive 1% technology shock. All vertical axes are measured in percentage. The solid, dashed, and dot dashed lines represent responses under the Taylor rule in (42), the weak inflation targeting rule in (43) with φπ = 1.1, and the strong inflation targeting rule in (43) with φπ = 2. In all these rules, φp = 0. Ct It 0.5 Taylor Weak Strong 0 0 5 10 15 Yt 2 1 0 0.5 −2 0 5 Nt 10 15 0 2 0 0 0 5 10 15 −0.2 0 5 10 15 −2 0.5 2 0 0 0 0 5 10 15 −0.5 0 5 10 15 −2 0.5 0.5 0 0 0 0 5 10 Quarters 15 −0.5 0 5 10 Quarters 10 15 0 5 10 15 ε∗∗ t 0.5 −0.5 5 ε∗t real rate 15 gt 0.2 −0.2 0 p ht Rt 10 Lt+1/P t 0.2 0 5 Πt 2 −2 0 15 −0.5 0 5 10 15 Quarters Figure 3: Impulse responses to a negative 0.25% shock to the nominal interest rate. All vertical axes are measured in percentage. The solid, dashed, and dot dashed lines represent responses under the Taylor rule in (42), the weak inflation targeting rule in (43) with φπ = 1.1, and the strong inflation targeting rule in (43) with φπ = 2. In all these rules, φp = 0. Ct It 0.01 0 −0.01 Yt 0.2 0.04 0 0 5 10 15 −0.2 0.02 0 5 10 15 0 0 0 0 −0.05 5 10 15 −0.02 0 5 10 15 −0.1 4 0.1 0 2 0 5 15 x 10 2 0 0 0 0 5 5 5 10 15 Quarters 10 15 −0.1 0 5 ε∗t real rate −3 4 10 0.02 0.01 0.01 0 5 15 10 15 10 15 Quarters 10 15 ε∗∗ t 0.02 0 10 gt 0.02 0 0 p ht Rt −0.02 5 Lt+1/Pt 0.02 0 0 Πt Nt 0.1 −0.1 Taylor Weak Strong 0 0 5 10 15 Quarters Figure 4: Impulse responses to a positive standard-deviation (0.613%) shock to the sentiment. All vertical axes are measured in percentage. The solid, dashed, and dot dashed lines represent responses under the Taylor rule in (42), the weak inflation targeting rule in (43) with φπ = 1.1, and the strong inflation targeting rule in (43) with φπ = 2. In all these rules, φp = 0. Ct It 1 Resp No resp Flexible 0.5 0 0 5 10 15 Yt 4 2 2 1 0 0 5 Nt 10 15 0 0 5 10 15 Lt+1/P t Πt 0.5 0 4 0 −0.2 2 −0.4 −0.5 0 5 10 15 0 5 10 15 0 1 5 −0.5 0 0 5 10 15 −1 0 5 10 15 −5 1 1 −0.1 0 0 0 5 10 Quarters 15 −1 0 5 10 Quarters 15 5 10 15 ε∗∗ t 0 −0.2 0 ε∗t real rate 10 gt 0 0 5 p ht Rt −1 0 15 −1 0 5 10 15 Quarters Figure 5: Impulse responses to a positive 1% TFP shock under the interest rate rule in (42). All vertical axes are measured in percentage. The solid lines represent the case with the optimal weight on asset bubbles given in Table 1. The dashed lines represent the case with φp = 0. The dot dashed lines represent the responses in the flexible price equilibrium Ct Resp No resp Flexible 0 −0.1 0 5 Yt It 0.1 10 15 0.5 0.2 0 0.1 −0.5 0 5 Nt 10 15 0 0.5 0 0 0 5 10 15 −0.1 0 5 10 15 −0.5 4 0.5 0 2 0 0 5 10 15 0 0 5 10 15 −0.5 0.1 0.1 0 0 0 0 5 10 15 Quarters −0.1 0 5 10 15 0 5 10 15 Quarters 10 15 ε∗∗ t 0.02 −0.02 5 ε∗t real rate 15 gt 0.05 −0.05 0 p ht Rt 10 Lt+1/P t 0.1 0 5 Πt 0.5 −0.5 0 −0.1 0 5 10 15 Quarters Figure 6: Impulse responses to a positive standard deviation (0.613%) sentiment shock under the interest rate rule in (42). All vertical axes are measured in percentage. The solid lines represent the case with the optimal weight on asset bubbles given in Table 1. The dashed lines represent the case with φp = 0. The dot dashed lines represent the responses in the flexible price equilibrium Inflation volatility: TFP Bubble volatility: TFP 1.4 1.4 1.3 1.3 1.2 1.2 % % 1.1 1 1.1 0.9 0.8 1 0.7 −0.2 0 0.2 0.4 0.9 −0.2 0.6 0 0.2 0.4 0.6 φp φp Inflation volatility: Sentiment Bubble volatility: Sentiment 2 6.5 1.5 6 % 7 % 2.5 1 5.5 0.5 5 0 −0.2 0 0.2 φp 0.4 0.6 4.5 −0.2 0 0.2 0.4 0.6 φp Figure 7: The inflation and output volatilities as functions of φp under the Taylor-type rule in (42) when only sentiment shocks or TFP shocks are turned on. Appendix A Proofs Proof of Proposition 1: We conjecture that the value function takes the following form    ∞ (A.1) Vt Kjt , Sjt , Ljt , Hj,t|t−k k=0 , εjt = φkt (εjt ) Kjt + φst (εjt ) Sjt − φlt (εjt ) Ljt + ∞ X φht|t−k (εjt ) Hj,t|t−k , k=0 where φit (εjt ) , i ∈ {k, s, l, h} , satisfy qtk 1 Pt Rlt 1 Rt Pt = = = pht|t−k = Z Λt+1 βEt φkt+1 (ε) dF (ε) , Λt Z Λt+1 βEt φst+1 (ε) dF (ε) , Λt Z Λt+1 βEt φlt+1 (ε) dF (ε) , Λt Z Λt+1 βEt φht+1|t−k (ε) dF (ε) . Λt (A.2) (A.3) δh . By convention, we set Hj,t|t = 1−δ h Substituting (21), (24), and the above conjecture into the Bellman equation (26), we obtain    ∞ Vt Kjt , Sjt , Ljt , Hj,t|t−k k=0 , εjt = max Ijt ,Sjt+1 ,Ljt+1 ,Hjt+1 + (1 − δh ) ∞ X k=0 Rkt Kjt − Ijt + pht|t−k Hj,t|t−k − ∞ X 1 (Sjt Rt−1 − Sjt+1 + Ljt+1 − Ljt Rlt−1 ) Pt pht|t−k Hj,t+1|t−k k=0 ∞ X 1 Rlt 1 Sjt+1 − Ljt+1 + pht|t−k Hj,t+1|t−k Pt Rt Pt k=0   R R t−1 lt−1 Sjt − Ljt = max Rkt + qtk (1 − δ) Kjt + Ijt ,Sjt+1 ,Ljt+1 Pt Pt   ∞   X 1 Rlt h k + (1 − δh ) pt|t−k Hj,t|t−k + qt εjt − 1 Ijt − − 1 Ljt+1 . Pt Rt +qtk ((1 − δ)Kjt + εjt Ijt ) + (A.4) k=0 Since Ijt ≥ 0 and Djt ≥ 0, it follows from (A.4) that Ijt = 0 if εjt < 1/qtk ≡ ε∗t , but the firm makes as much investment as possible so that Djt = 0 if εjt > ε∗t . Therefore, when εjt ≥ 1/qtk ≡ ε∗t , it follows from (24) that Sjt Rt−1 − Sjt+1 + Ljt+1 − Ljt Rlt−1 Pt ∞ ∞ X X h + (1 − δh ) pt|t−k Hj,t|t−k − pht|t−k Hj,t+1|t−k . Ijt = Rkt Kjt + k=0 k=0 (A.5) Using this investment rule, we can simplify (A.4) for εjt < ε∗t as    ∞ Vt Kjt , Sjt , Ljt , Hj,t|t−k k=0 , εjt   Rlt−1 Rt−1 k = max Rkt + qt (1 − δ) Kjt + Sjt − Ljt Ljt+1 Pt Pt   ∞ X 1 Rlt h + (1 − δh ) − 1 Ljt+1 . pt|t−k Hj,t|t−k − Pt Rt (A.6) k=0 Given Rlt > Rt , we have Ljt+1 = 0. Therefore, when εjt < ε∗t , equation (A.6) becomes    ∞ Vt Kjt , Sjt , Ljt , Hj,t|t−k k=0 , εjt =  (A.7) ∞  X Rt−1 Rlt−1 Rkt + qtk (1 − δ) Kjt + Sjt − Ljt + (1 − δh ) pht|t−k Hj,t|t−k . Pt Pt k=0 Matching coefficients in (A.1) and (A.7), we have φkt (εjt ) = Rkt + qtk (1 − δ) , φht|t−k (εjt ) = (1 − δh ) pht|t−k , Rt−1 Rlt−1 , φlt (εjt ) = , Pt Pt φst (εjt ) = for εjt < ε∗t . Next consider the case of εjt ≥ ε∗t = 1/qtk . Substituting (A.5) into (A.4) yields    ∞ Vt Kjt , Sjt , Ljt , Hj,t|t−k k=0 , εjt = ∞  X Rt−1 Rlt−1 Rkt + qtk (1 − δ) Kjt + pht|t−k Hj,t|t−k Sjt − Ljt + (1 − δh ) Sjt+1 ,Ljt+1 ,Hjt+1 Pt Pt k=0      Sjt Rt−1 − Sjt+1 + Ljt+1 − Ljt Rlt−1 1 Rlt k + qt εjt − 1 Rkt Kjt + − − 1 Ljt+1 Pt Pt Rt " # ∞ ∞   X X + qtk εjt − 1 (1 − δh ) pht|t−k Hj,t|t−k − pht|t−k Hj,t+1|t−k max  k=0 k=0      R t−1 = max Rkt + qtk εjt − 1 Rkt + qtk (1 − δ) Kjt + 1 + qtk εjt − 1 Sjt Sjt+1 ,Ljt+1 ,Hjt+1 Pt ∞    R    X lt−1 k k − 1 + qt εjt − 1 Ljt + 1 + qt εjt − 1 (1 − δh ) pht|t−k Hj,t|t−k Pt k=0   ∞     X 1 Rlt k k k − qt εjt Ljt+1 − qt εjt − 1 pht|t−k Hj,t+1|t−k . (A.8) − qt εjt − 1 Sjt+1 − Pt Rt   k=0 Thus we must have Sjt+1 = 0 and Hj,t+1|t−k = 0, k = 0, 1, 2, ..., Moreover, using (23) gives Ljt+1 = Pt ( 0, µKjt , if εjt < ε∗∗ t , otherwise where we define ε∗∗ t ≡ Rlt Rlt ∗ ε . = k Rt t Rt qt ∗ Since Rlt > Rt , ε∗∗ t > εt . Using the preceding decision rules, we can derive that    ∞ Vt Kjt , Sjt , Ljt , Hj,t|t−k k=0 , εjt        R t−1 = Rkt + qtk εjt − 1 Rkt + qtk (1 − δ) Kjt + 1 + qtk εjt − 1 Sjt Pt ∞    R    X lt−1 − 1 + qtk εjt − 1 Ljt + 1 + qtk εjt − 1 (1 − δh ) pht|t−k Hj,t|t−k Pt k=0 +qtk µKjt max {εjt − ε∗∗ t , 0} , (A.9) for εjt ≥ ε∗t . Matching coefficients in (A.1) and (A.9) yields   k φkt (εjt ) = Rkt + qtk εjt − 1 Rkt + qtk (1 − δ) + max {εjt − ε∗∗ t , 0} qt µ    εjt k −1 Rkt + qtk (1 − δ) + max {εjt − ε∗∗ = 1+ t , 0} qt µ, ε∗t     R εjt Rt−1 t−1 = 1+ , −1 = 1+ −1 ∗ Pt εt Pt       R εjt Rlt−1 lt−1 l k φt (εjt ) = 1 + qt εjt − 1 = 1+ −1 , ∗ Pt εt Pt      εjt φht (εjt )  = 1 + qtk εjt − 1 pht|t−k = 1 + − 1 pht|t−k , 1 − δh ε∗t φst (εjt )   qtk εjt for εjt ≥ ε∗t . Combining the above two cases, we deduce that, for any εjt ∈ (εmin , εmax ), we have    εjt k k φt (εjt ) = 1 + max − 1, 0 Rkt + qtk (1 − δ) + max {εjt − ε∗∗ t , 0} qt µ, ε∗t  εjt Rt−1 = 1 + max − 1, 0 , ∗ εt Pt    εjt Rlt−1 l φt (εjt ) = 1 + max − 1, 0 , ∗ εt Pt    φht (εjt ) εjt = 1 + max − 1, 0 pht|t−k . 1 − δh ε∗t φst (εjt )   Substituting the preceding four equations into (A.2)-(A.3) yields (29), (30), (31) in the proposition. Q.E.D. Define the function Z εmax Π G(z) = max(ε, z)dF (ε) − z = 0, z ∈ [εmin , εmax ] . β εmin Proof of Lemma 1: Π Π Note that G0 (z) = F (z) − Π β < 0, G(εmin ) = E [ε] − β εmin > 0, and G(εmax ) = εmax − β εmax < 0 by assumption 1. The intermediate value theorem ensures that there exists a unique solution for ε ∈ (εmin , εmax ) in the equation in the lemma. Q.E.D. In any steady state, we can use (18) and (56) to derive that Proof of Proposition 2: ε ∗∗ λ ∗ Rl ∗ 1 − R(ε∗ ) ∗ ε − = ε = ε = R 1−λ λβ Π R εmax εmin max(ε∗ , ε)dF (ε) 1−λ . (A.10) By assumption 1 and λ ∈ (0, 1) , we can show that ∗ 1 − λβ dε∗∗ Π F (ε ) = > 0. dε∗ 1−λ Thus ε∗∗ = ε (ε∗ ) increases with ε∗ . We need the following result. Lemma 2 For a sufficiently small µ, Rk (ε∗ ) decreases with ε∗ for ε∗ ∈ (εmin , εmax ). A sufficient condition is 1 β 0<µ< max 1−F (ε(ε∗ )) R εmax ε∗ ∈[εmin ,εmax ] F (ε∗ ) εmin −1+δ max(ε∗ , ε)dF (ε) 1− λβ F (ε∗ ) Π 1−λ + R εmax ε(ε∗ ) . (A.11) (ε − ε (ε∗ )) dF (ε) Proof. Using (58), we can compute that dRk (ε∗ ) dε∗ = = i h R ∗ , ε)dF (ε) dε∗∗ − 1 − 1 + δ − µ εmax (ε − ε∗∗ ) dF (ε) F (ε∗ ) max(ε ∗ ∗∗ dε β ε εmin hR i2 εmax ∗ εmin max(ε , ε)dF (ε) i h R εmax R εmax 1− λβ F (ε∗ ) 1 ∗∗ ) dF (ε) F (ε∗ ) Π (ε − ε µ (1 − F (ε∗∗ )) εmin max(ε∗ , ε)dF (ε) − − 1 + δ − µ ∗∗ 1−λ β ε . hR i2 εmax ∗ εmin max(ε , ε)dF (ε) µ [1 − F (ε∗∗ )] R εmax If µ = 0, dRk (ε∗ ) = − hR dε∗ εmax εmin 1 β −1+δ max(ε∗ , ε)dF (ε) i2 < 0. Since Rk (ε∗ ) is a continuous function, the above inequality is also true for a sufficiently small µ. Define the function   Z εmax Z εmax λµ 1 − F (ε (ε∗ )) H (ε∗ ) = Rk (ε∗ ) + εdF (ε) + µ εdF (ε) , 1−λ Π ε∗ ε(ε∗ ) (A.12) for ε∗ ∈ [ε, εmax ] . Since ε (ε∗ ) increases with ε∗ , it follows from the preceding lemma that H (ε∗ ) decreases with ε∗ . Note that H (εmax ) = 0 and   Z εmax Z εmax λµ 1 − F (ε (ε)) εdF (ε) + µ εdF (ε) > δ, H (ε) = Rk (ε) + 1−λ Π ε ε(ε) given the following assumption Rε δ − Rk (ε) ε max εdF (ε) . µ> R R εmax λ 1−F (ε(ε)) εmax εdF (ε) εdF (ε) + 1−λ Π ε ε(ε) (A.13) By the intermediate value theorem, there exists a unique solution for ε∗f ∈ (ε, εmax ) in the equation   H (ε∗ ) = δ. Since ε∗f > ε, it follows Lemma 1 and (56) that Rf = R ε∗f > R (ε) = 1. Finally, the additional assumptions on µ not explicitly specified in the proposition are (A.11) and (A.13). Q.E.D. Proof of Proposition 3: Equation (58) gives Rkf = Rk (ε∗f ). Since Rkf = αYf /Kf , we can determine Yf /Kf . We now use the following procedure to derive other bubbleless steady-state variables. Equation (60) determines mf /Kf . Equation (51) gives  

If mf (A.14) = Rkf + 1 − F (ε∗f ) + µ 1 − F (ε∗∗ f ) . Kf ΠKf Thus we can derive If /Yf = (If /Kf ) / (Yf /Kf ) , and hence Cf /Yf = 1 − If /Yf . Using equation (12), we can derive that = wf Yf 1 = (1 − α) , Cf Nf Cf and thus Nf = 1 − α Yf . Cf Now we use Rkf = αAKfα−1 Nf1−α to solve for Kf and then use the ratios derived above to determine Yf , mf , If , and Cf . Consider the function H defined in (A.12). It decreases with Π. Since the solution for ε∗f is given by the intersection of the downward sloping curve H (ε∗ ) and the line δ, ε∗f decreases with Π.   Since Rkf = αKfα−1 Nf1−α = αYf /Kf = Rk ε∗f , it follows from Lemma 2 that Kf /Nf decreases with Π, but Yf /Kf increases with Π. The rest of results are straightforward to prove and hence are omitted. Q.E.D. Proof of Proposition 4: the function We first prove sufficiency. Suppose that condition (61) holds. Define   Z εmax   ε ∗ − 1 dF (ε) (A.15) G (ε ) = βs 1 + ε∗ ε∗ for ε∗ ∈ [ε∗f , εmax ). Since R (ε∗ ) increases with ε∗ , it follows from (57) that ε (ε∗ ) and ε (ε∗ ) /ε∗   increase with ε∗ . Thus G (ε∗ ) decreases with ε∗ . Moreover, G (εmax ) = β < 1 and G ε∗f < 1 by (61). By the intermediate value theorem, there exists a unique solution εb∗ ∈ (ε∗f , ε∗max) to (62).     Since Rf = R ε∗f > 1 and R (ε∗ ) increases with ε∗ , we have R (ε∗b ) > R ε∗f > 1. Once the bubbly steady-state cutoff ε∗b is determined, other bubbly steady-state values can be easily solved. ∗ ∗ In particular, we use equation (57) to derive ε∗∗ b = ε(εb ). Equation (58) gives Rkb = Rk (εb ). Since Rkb = αYb /Kb , we can determine Yb /Kb . The steady-state version of equation (52) implies that R εmax   εdF (ε) h ε(ε∗b ) m δ 1 p b R εmax = Rk (ε∗b ) + + µ R εmax , (A.16) + Kb Π Kb εdF (ε) εdF (ε) ε∗ ε∗ b b where we use equation (49) to derive mb λ = µ (1 − F (ε∗∗ b )) . Kb 1−λ In turn, we can solve for ph /Kb as h iR R εmax εmax ∗ ) + µ λ (1 − F (ε (ε∗ ))) δ − R (ε εdF (ε) − µ εdF (ε) ∗ k b b Π 1−λ ε ε(ε∗b ) ph b R εmax . = Kb εdF (ε) ε∗ b We can show that the numerator on the right-hand side of the equation above is an increasing function of ε∗b when µ is sufficiently small. Since that expression is equal to zero when ε∗b is replaced with ε∗f (see (59)), it follows from ε∗b > ε∗f that ph /Kb > 0. Using (51), we can solve for Ib /Kb and hence Ib /Yb = (Ib /Kb ) / (Yb /Kb ) . It follows from the resources constraint that Cb /Yb = 1 − Ib /Yb . Using (12) and wb = (1 − α) Yb /Nb , we can show that Nb = 1 − α Yb . Cb Using Rkb = αKbα−1 Nb1−α , we can solve for Kb . We can then determine other equilibrium variables using the ratios derived above. We next prove necessity. Suppose that a bubbly steady state exists. In a bubbly steady state ph > 0 and in a bubbleless steady state ph = 0. Since (A.16) holds for both bubbly and bubbleless steady states, we have   Z εmax Z εmax

λ µ ∗ ∗ δ = Rk (εf ) + εdF (ε) + µ εdF (ε) 1 − F ε(εf ) 1−λΠ ε∗f ε(ε∗f )   Z εmax Z εmax λ µ ∗ ∗ > Rk (εb ) + (1 − F (ε (εb ))) εdF (ε) + µ εdF (ε) . 1−λΠ ε∗b ε(ε∗b ) When µ is sufficient small, Lemma 2 implies that Rk (ε∗ ) decreases with ε∗ . We have also shown that ε (ε∗ ) increases with ε∗ . Thus we deduce that  Z εmax  Z εmax λ µ Rk (ε∗ ) + (1 − F (ε (ε∗ ))) εdF (ε) + µ εdF (ε) 1−λΠ ε∗ ε(ε∗ ) decreases with ε∗ . The preceding inequality then implies that ε∗b > ε∗f . We have shown that G (ε∗ ) ∗ defined in (A.15) decreases with ε∗ . Since   equation (48) implies that G (εb ) = 1 in a bubbly steady state, it follows from ε∗b > ε∗f that G ε∗f > 1. Q.E.D. In the bubbly steady state, G (ε∗b ) = 1 where G is defined in (A.15). Proof of Proposition 5: Since we have Z εmax ε∗∗ ε (ε∗b ) Rlb b = = > 1, ∗ ∗ εb εb Rb  ε(ε∗b ) ε (ε∗b ) ε − ε∗b ε∗b  Thus εmax Z βs 1 +  ε∗b This implies that Z εmax  ε∗b Z dF (ε) < εmax  ε∗b  ε − 1 dF (ε) . ε∗b !  ε − 1 dF (ε) > G (ε∗b ) = 1. ε∗b  1 ε − 1 dF (ε) > − 1. ∗ εb βs By (56), we have 1 < R (ε∗b ) =  β 1+ Π R εmax  ε ε∗b ε∗b − 1 dF (ε) Thus Π> Moreover, since ε∗b > ε∗f , < Π 1 = sΠ. 1 β 1 + βs −1 1 > 1. s Π > sΠ > R(ε∗b ) > R(ε∗f ). We then obtain the desired result. B  Q.E.D. Log-linearized System The log-linearized equilibrium system can be described by 17 equations for 17 variables {Ĉt , Iˆt , K̂t , Ŷt , N̂t , Π̂t , p̂wt , ŵt , p̂ht , q̂tk , R̂kt , ε̂∗t , ε̂∗∗ t , R̂lt , m̂t , ĝt , R̂t }. • Aggregate demand Rβ Ĉt = Et Ĉt+1 − (R̂t − Et Π̂t+1 ) + Π εmax ε∗ ε dF (ε) Et ε̂∗t+1 , ε∗ (B.1) m Rk K (1 − F (ε∗ )) (R̂kt + K̂t ) + (1 − F (ε∗ )) (m̂t − Π̂t ) (B.2) I Π µK ph + (1 − F (ε∗∗ )) K̂t + (1 − F (ε∗ )) p̂ht I I  m  f (ε∗ )ε∗ ∗ f (ε∗∗ )ε∗∗ ∗∗ − Rk K + ε̂t − µK ε̂t , Π I I Z εmax   = (1 − δ)K̂t + Rk εdF (ε) K̂t + R̂kt (B.3) ε∗ Z εmax Z εmax   Z εmax m ph +µ εdF (ε) K̂t + εdF (ε) m̂t − Π̂t + εdF (ε) p̂ht ΠK ε∗ K ε∗ ε∗∗   h m p − Rk + + (ε∗ )2 f (ε∗ )ε̂∗t − µ (ε∗∗ )2 f (ε∗∗ )ε̂∗∗ t , ΠK K Iˆt = K̂t+1 Z Ŷt = • Aggregate supply C I Ĉt + Iˆt . Y Y (B.4) 1 Π̂t = (1 − ξ)(1 − βξ)p̂wt + βEt Π̂t+1 , ξ (B.5) Ŷt = Ât + αK̂t + (1 − α)N̂t , (B.6) ŵt = p̂wt + Ât + αK̂t − αN̂t , (B.7) Ĉt = ŵt . (B.8) • Asset prices p̂ht q̂tk = = Et p̂ht+1   + Ĉt − Et Ĉt+1 + ŝt − sβ Z εmax ε∗ ε dF (ε) Et ε̂∗t+1 , ε∗  R Π k k Ĉt − Et Ĉt+1 + k Et R̂kt+1 + β (1 − δ) Et q̂t+1 q R  Z εmax Z ∗∗ ∗∗ ∗∗ ∗∗ (ε − ε ) dF (ε) + Rk −βµε (1 − F (ε )) Et ε̂t+1 − β µ (B.9)  ε∗∗ R̂kt = p̂wt + Ât + (α − 1)K̂t + (1 − α)N̂t . • Lending rate and two cutoffs ∗ R̂lt − R̂t = ε̂∗∗ t − ε̂t , ε̂∗t = −q̂tk , ∗ ε̂∗∗ t = ε̂t + • Money demand and supply • Interest rate rule m̂t+1 = k̂t − (B.10) εmax ε∗  εdF (ε) Et ε̂∗t+1 , (B.11) (B.12) (B.13) λ/R R̂t . 1 − λ/R (B.14) ε∗∗ f (ε∗∗ ) ∗∗ ε̂ , 1 − F (ε∗∗ ) t (B.15) m̂t+1 = m̂t − Π̂t + ĝt . (B.16) R̂t = φπ Π̂t + φy Ŷt + φp p̂ht + v̂t , (B.17) R̂t = φπ Et Π̂t+1 + φp p̂ht + v̂t , (B.18) or • Exogenous shocks Ât = ρa Ât−1 + εat , v̂t = ρv v̂t−1 + εvt , ŝt = ρs ŝt−1 + εst .