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货币理论与政策解决方案手册，第 4 版，麻省理工学院出版社 2017 年*

  卡尔·沃尔什

  2017 年 3 月

  内容

  1 引言 ...1

2 第 2 章：Money-in-the-Utility 函数......2

3 第 3 章：货币和交易......15

4 第 4 章：货币与公共财政......35

5 第 5 章：信息和投资组合的刚性......46

6 第 6 章：自由裁量权政策和时间不一致......52

7 第七章：名义价格和工资刚性 ...82

8 第 8 章：新凯恩斯主义货币经济学 ...98

9 第 9 章：开放经济中的货币政策 ...127

10 第 10 章：金融市场和货币政策 ...146

11 第 11 章：有效的下限和资产负债表政策......163

12 第 12 章：货币政策操作程序 ...175

  1 引言


本手册包含《货币理论》和《政治 i c y , 4 t h i c y , 4 t h icy,4^(th)i c y, 4^{t h} 》版（剑桥，马萨诸塞州：麻省理工学院出版社，2017 年）中所有章末问题的解决方案。请向 walshc@ucsc.edu 报告任何错误或建议的更正。拼写错误和更正将发布在 http://people.ucsc.edu/~ walshc/mtp4e/。

敦促用户检查 http://people.ucsc.edu/~walshc/mtp4e/before 分配问题中发布的更正。对于与这些练习相关的 Matlab 程序，应发送至 walshc@ucsc.edu。


2 第 2 章：Money-in-the-Utility 函数


  1. 第 2.2 节的 MIU 模型意味着货币和消费之间的边际替代率被设定为等于 i t / ( 1 + i t ) i t / 1 + i t i_(t)//(1+i_(t))i_{t} /\left(1+i_{t}\right) （见 （2.12））。该模型假设代理人带着资源进入时期 t t tt ，并使用这些资源 ω t ω t omega_(t)\omega_{t} 来购买资本、消费、名义债券和货币。这些货币持有的实际价值在期间 t t tt 产生了效用。相反，假设在 period t t tt 中选择的货币持有量在 period t + 1 t + 1 t+1t+1 之前不会产生效用。效用和 ∑ β i U ( c t + i , M t + i / P t + i ) ∑ β i U c t + i , M t + i / P t + i sumbeta^(i)U(c_(t+i),M_(t+i)//P_(t+i))\sum \beta^{i} U\left(c_{t+i}, M_{t+i} / P_{t+i}\right) 以前一样，但预算限制采用
ω t = c t + M t + 1 P t + b t + k t ω t = c t + M t + 1 P t + b t + k t omega_(t)=c_(t)+(M_(t+1))/(P_(t))+b_(t)+k_(t)\omega_{t}=c_{t}+\frac{M_{t+1}}{P_{t}}+b_{t}+k_{t}

和家庭选择 c t , k t , b t c t , k t , b t c_(t),k_(t),b_(t)c_{t}, k_{t}, b_{t} ，并在 M t + 1 M t + 1 M_(t+1)M_{t+1} 时期 t t tt 。家庭的真正财富 ω t ω t omega_(t)\omega_{t} 是由
ω t = f ( k t − 1 ) + ( 1 − δ ) k t − 1 + ( 1 + r t − 1 ) b t − 1 + m t ω t = f k t − 1 + ( 1 − δ ) k t − 1 + 1 + r t − 1 b t − 1 + m t omega_(t)=f(k_(t-1))+(1-delta)k_(t-1)+(1+r_(t-1))b_(t-1)+m_(t)\omega_{t}=f\left(k_{t-1}\right)+(1-\delta) k_{t-1}+\left(1+r_{t-1}\right) b_{t-1}+m_{t}

推导住户选择的一阶条件， M t + 1 M t + 1 M_(t+1)M_{t+1} 并证明
U m ( c t + 1 , m t + 1 ) U c ( c t + 1 , m t + 1 ) = i t U m c t + 1 , m t + 1 U c c t + 1 , m t + 1 = i t (U_(m)(c_(t+1),m_(t+1)))/(U_(c)(c_(t+1),m_(t+1)))=i_(t)\frac{U_{m}\left(c_{t+1}, m_{t+1}\right)}{U_{c}\left(c_{t+1}, m_{t+1}\right)}=i_{t}

（由 Kevin Salyer 建议。设 value 函数为
V ( ω t , m t ) = max { U ( c t , m t ) + β V ( ω t + 1 , m t + 1 ) } V ω t , m t = max U c t , m t + β V ω t + 1 , m t + 1 V(omega_(t),m_(t))=max{U(c_(t),m_(t))+beta V(omega_(t+1),m_(t+1))}V\left(\omega_{t}, m_{t}\right)=\max \left\{U\left(c_{t}, m_{t}\right)+\beta V\left(\omega_{t+1}, m_{t+1}\right)\right\}
  受制于
ω t + 1 = f ( k t ) + ( 1 − δ ) k t + ( 1 + r t ) b t + m t + 1 ω t + 1 = f k t + ( 1 − δ ) k t + 1 + r t b t + m t + 1 omega_(t+1)=f(k_(t))+(1-delta)k_(t)+(1+r_(t))b_(t)+m_(t+1)\omega_{t+1}=f\left(k_{t}\right)+(1-\delta) k_{t}+\left(1+r_{t}\right) b_{t}+m_{t+1}
  和
ω t − c t − m t + 1 ( 1 + π t + 1 ) − b t − k t = 0 ω t − c t − m t + 1 1 + π t + 1 − b t − k t = 0 omega_(t)-c_(t)-m_(t+1)(1+pi_(t+1))-b_(t)-k_(t)=0\omega_{t}-c_{t}-m_{t+1}\left(1+\pi_{t+1}\right)-b_{t}-k_{t}=0

在这种设置中， r t r t r_(t)r_{t} 是债券的实际回报。让我们 λ t λ t lambda_(t)\lambda_{t} 表示最后一个约束的拉格朗日。 c t , m t + 1 , b t c t , m t + 1 , b t c_(t),m_(t+1),b_(t)c_{t}, m_{t+1}, b_{t} 的一阶条件 和 k t k t k_(t)k_{t} 加上包络定理得到
U c ( c t , m t ) = λ t β V m ( ω t + 1 , m t + 1 ) + β V ω ( ω t + 1 , m t + 1 ) = λ t ( 1 + π t + 1 ) β ( 1 + r t ) V ω ( ω t + 1 , m t + 1 ) = λ t V ω ( ω t , m t ) = λ t V m ( ω t , m t ) = U m ( c t , m t ) . U c c t , m t = λ t β V m ω t + 1 , m t + 1 + β V ω ω t + 1 , m t + 1 = λ t 1 + π t + 1 β 1 + r t V ω ω t + 1 , m t + 1 = λ t V ω ω t , m t = λ t V m ω t , m t = U m c t , m t . {:[U_(c)(c_(t),m_(t))=lambda_(t)],[betaV_(m)(omega_(t+1),m_(t+1))+betaV_(omega)(omega_(t+1),m_(t+1))=lambda_(t)(1+pi_(t+1))],[beta(1+r_(t))V_(omega)(omega_(t+1),m_(t+1))=lambda_(t)],[V_(omega)(omega_(t),m_(t))=lambda_(t)],[V_(m)(omega_(t),m_(t))=U_(m)(c_(t),m_(t)).]:}\begin{gathered} U_{c}\left(c_{t}, m_{t}\right)=\lambda_{t} \\ \beta V_{m}\left(\omega_{t+1}, m_{t+1}\right)+\beta V_{\omega}\left(\omega_{t+1}, m_{t+1}\right)=\lambda_{t}\left(1+\pi_{t+1}\right) \\ \beta\left(1+r_{t}\right) V_{\omega}\left(\omega_{t+1}, m_{t+1}\right)=\lambda_{t} \\ V_{\omega}\left(\omega_{t}, m_{t}\right)=\lambda_{t} \\ V_{m}\left(\omega_{t}, m_{t}\right)=U_{m}\left(c_{t}, m_{t}\right) . \end{gathered}

现在梳理其中的第二个和第三个以获得
β V m ( ω t + 1 , m t + 1 ) + β V ω ( ω t + 1 , m t + 1 ) = β ( 1 + π t + 1 ) ( 1 + r t ) V ω ( ω t + 1 , m t + 1 ) β V m ω t + 1 , m t + 1 + β V ω ω t + 1 , m t + 1 = β 1 + π t + 1 1 + r t V ω ω t + 1 , m t + 1 betaV_(m)(omega_(t+1),m_(t+1))+betaV_(omega)(omega_(t+1),m_(t+1))=beta(1+pi_(t+1))(1+r_(t))V_(omega)(omega_(t+1),m_(t+1))\beta V_{m}\left(\omega_{t+1}, m_{t+1}\right)+\beta V_{\omega}\left(\omega_{t+1}, m_{t+1}\right)=\beta\left(1+\pi_{t+1}\right)\left(1+r_{t}\right) V_{\omega}\left(\omega_{t+1}, m_{t+1}\right)
  或
β V m ( ω t + 1 , m t + 1 ) = [ ( 1 + π t + 1 ) ( 1 + r t ) − 1 ] β V ω ( ω t + 1 , m t + 1 ) = i t β V ω ( ω t + 1 , m t + 1 ) . β V m ω t + 1 , m t + 1 = 1 + π t + 1 1 + r t − 1 β V ω ω t + 1 , m t + 1 = i t β V ω ω t + 1 , m t + 1 . {:[betaV_(m)(omega_(t+1),m_(t+1))=[(1+pi_(t+1))(1+r_(t))-1]betaV_(omega)(omega_(t+1),m_(t+1))],[=i_(t)betaV_(omega)(omega_(t+1),m_(t+1)).]:}\begin{aligned} \beta V_{m}\left(\omega_{t+1}, m_{t+1}\right) & =\left[\left(1+\pi_{t+1}\right)\left(1+r_{t}\right)-1\right] \beta V_{\omega}\left(\omega_{t+1}, m_{t+1}\right) \\ & =i_{t} \beta V_{\omega}\left(\omega_{t+1}, m_{t+1}\right) . \end{aligned}
  因此
V m ( ω t + 1 , m t + 1 ) V ω ( ω t + 1 , m t + 1 ) = i t V m ω t + 1 , m t + 1 V ω ω t + 1 , m t + 1 = i t (V_(m)(omega_(t+1),m_(t+1)))/(V_(omega)(omega_(t+1),m_(t+1)))=i_(t)\frac{V_{m}\left(\omega_{t+1}, m_{t+1}\right)}{V_{\omega}\left(\omega_{t+1}, m_{t+1}\right)}=i_{t}

但是最后一个一阶条件暗示 V m ( ω t + 1 , m t + 1 ) = U m ( c t + 1 , m t + 1 ) V m ω t + 1 , m t + 1 = U m c t + 1 , m t + 1 V_(m)(omega_(t+1),m_(t+1))=U_(m)(c_(t+1),m_(t+1))V_{m}\left(\omega_{t+1}, m_{t+1}\right)=U_{m}\left(c_{t+1}, m_{t+1}\right) ，而第一个和第四个产生 V ω ( ω t + 1 , m t + 1 ) = λ t + 1 = U c ( c t + 1 , m t + 1 ) V ω ω t + 1 , m t + 1 = λ t + 1 = U c c t + 1 , m t + 1 V_(omega)(omega_(t+1),m_(t+1))=lambda_(t+1)=U_(c)(c_(t+1),m_(t+1))V_{\omega}\left(\omega_{t+1}, m_{t+1}\right)=\lambda_{t+1}=U_{c}\left(c_{t+1}, m_{t+1}\right) 。因此
U m ( c t + 1 , m t + 1 ) U c ( c t + 1 , m t + 1 ) = i t U m c t + 1 , m t + 1 U c c t + 1 , m t + 1 = i t (U_(m)(c_(t+1),m_(t+1)))/(U_(c)(c_(t+1),m_(t+1)))=i_(t)\frac{U_{m}\left(c_{t+1}, m_{t+1}\right)}{U_{c}\left(c_{t+1}, m_{t+1}\right)}=i_{t}

在第 2.2 节考虑的案例中，当时 t t tt 持有的现金在当时 t t tt 产生了效用，但机会成本在于本可以持有的债券的利息收入损失。由于这笔利息支付发生在 t + 1 t + 1 t+1t+1 ，因此必须将其贴现回去，以与当时持有货币的边际价值进行比较 t t tt 。

2. （Carlstrom 和 Fuerst （2001）。假设代表住户的效用取决于消费和可用于消费的实际货币余额水平。设 A t / P t A t / P t A_(t)//P_(t)A_{t} / P_{t} 为进入效用函数的真实货币存量。如果忽略资本，则家庭的目标是在预算约束下实现最大化 ∑ β i U ( c t + i , A t + i / P t + i ) ∑ β i U c t + i , A t + i / P t + i sumbeta^(i)U(c_(t+i),A_(t+i)//P_(t+i))\sum \beta^{i} U\left(c_{t+i}, A_{t+i} / P_{t+i}\right)
Y t + M t − 5 P t + τ t + ( 1 + i t − 1 ) B t 尺 1 P t = C t + M t P t + B t P t , Y t + M t − 5 P t + τ t + 1 + i t − 1 B t 尺 1 P t = C t + M t P t + B t P t , Y_(t)+(M_(t-5))/(P_(t))+tau_(t)+((1+i_(t-1))B_(t尺1))/(P_(t))=C_(t)+(M_(t))/(P_(t))+(B_(t))/(P_(t)),Y_{t}+\frac{M_{t-5}}{P_{t}}+\tau_{t}+\frac{\left(1+i_{t-1}\right) B_{t 尺 1}}{P_{t}}=C_{t}+\frac{M_{t}}{P_{t}}+\frac{B_{t}}{P_{t}},尺

其中收入 Y t Y t Y_(t)Y_{t} 被视为一个外生过程。假设产生效用的货币存量是购买债券之后但在收到收入或购买消费品之前持有的货币的实际价值：
A t P t = M t − 1 P t + τ t + ( 1 + i t − 1 ) B t − 1 P t − B t P t A t P t = M t − 1 P t + τ t + 1 + i t − 1 B t − 1 P t − B t P t (A_(t))/(P_(t))=(M_(t-1))/(P_(t))+tau_(t)+((1+i_(t-1))B_(t-1))/(P_(t))-(B_(t))/(P_(t))\frac{A_{t}}{P_{t}}=\frac{M_{t-1}}{P_{t}}+\tau_{t}+\frac{\left(1+i_{t-1}\right) B_{t-1}}{P_{t}}-\frac{B_{t}}{P_{t}}

（a） 推导 B t B t B_(t)B_{t} 和 的 A t A t A_(t)A_{t} 一阶条件。设 value 函数为
V ( z t ) = max { U ( c t , z t − b t ) + β V ( z t + 1 } , V z t = max U c t , z t − b t + β V z t + 1 , V(z_(t))=max{U(c_(t),z_(t)-b_(t))+beta V(z_(t+1)},:}V\left(z_{t}\right)=\max \left\{U\left(c_{t}, z_{t}-b_{t}\right)+\beta V\left(z_{t+1}\right\},\right.
  哪里
z t + 1 = M t P t + 1 + τ t + 1 + ( 1 + i t ) B t P t + 1 = ( P t P t + 1 ) m t + τ t + 1 + ( 1 + i t ) ( P t P t + 1 ) b t z t + 1 = M t P t + 1 + τ t + 1 + 1 + i t B t P t + 1 = P t P t + 1 m t + τ t + 1 + 1 + i t P t P t + 1 b t {:[z_(t+1)=(M_(t))/(P_(t+1))+tau_(t+1)+((1+i_(t))B_(t))/(P_(t+1))],[=((P_(t))/(P_(t+1)))m_(t)+tau_(t+1)+(1+i_(t))((P_(t))/(P_(t+1)))b_(t)]:}\begin{aligned} z_{t+1} & =\frac{M_{t}}{P_{t+1}}+\tau_{t+1}+\frac{\left(1+i_{t}\right) B_{t}}{P_{t+1}} \\ & =\left(\frac{P_{t}}{P_{t+1}}\right) m_{t}+\tau_{t+1}+\left(1+i_{t}\right)\left(\frac{P_{t}}{P_{t+1}}\right) b_{t} \end{aligned}
  和
Y t + z t − c t − m t − b t = 0 Y t + z t − c t − m t − b t = 0 Y_(t)+z_(t)-c_(t)-m_(t)-b_(t)=0Y_{t}+z_{t}-c_{t}-m_{t}-b_{t}=0

在时间 t t tt ，家庭选择 c t , m t , b t c t , m t , b t c_(t),m_(t),b_(t)c_{t}, m_{t}, b_{t} 。让我们 λ t λ t lambda_(t)\lambda_{t} 表示最后一个约束的拉格朗日。一阶条件加上包络定理得到
U c ( c t , z t − b t ) = λ t β ( P t P t + 1 ) V z ( z t + 1 ) = λ t − U a ( c t , a t ) + β ( 1 + i t ) ( P t P t + 1 ) V z ( z t + 1 ) = λ t V z ( z t ) = U a ( c t , z t − b t ) + λ t U c c t , z t − b t = λ t β P t P t + 1 V z z t + 1 = λ t − U a c t , a t + β 1 + i t P t P t + 1 V z z t + 1 = λ t V z z t = U a c t , z t − b t + λ t {:[U_(c)(c_(t),z_(t)-b_(t))=lambda_(t)],[beta((P_(t))/(P_(t+1)))V_(z)(z_(t+1))=lambda_(t)],[-U_(a)(c_(t),a_(t))+beta(1+i_(t))((P_(t))/(P_(t+1)))V_(z)(z_(t+1))=lambda_(t)],[V_(z)(z_(t))=U_(a)(c_(t),z_(t)-b_(t))+lambda_(t)]:}\begin{gathered} U_{c}\left(c_{t}, z_{t}-b_{t}\right)=\lambda_{t} \\ \beta\left(\frac{P_{t}}{P_{t+1}}\right) V_{z}\left(z_{t+1}\right)=\lambda_{t} \\ -U_{a}\left(c_{t}, a_{t}\right)+\beta\left(1+i_{t}\right)\left(\frac{P_{t}}{P_{t+1}}\right) V_{z}\left(z_{t+1}\right)=\lambda_{t} \\ V_{z}\left(z_{t}\right)=U_{a}\left(c_{t}, z_{t}-b_{t}\right)+\lambda_{t} \end{gathered}

现在将其中的第二个和第三个组合在一起以消除 value 函数，可以得到 − U a ( c t , z t − b t ) + ( 1 + i t ) λ t = λ t − U a c t , z t − b t + 1 + i t λ t = λ t -U_(a)(c_(t),z_(t)-b_(t))+(1+i_(t))lambda_(t)=lambda_(t)-U_{a}\left(c_{t}, z_{t}-b_{t}\right)+\left(1+i_{t}\right) \lambda_{t}=\lambda_{t} ，或
U a ( c t , a t ) = i t λ t . U a c t , a t = i t λ t . U_(a)(c_(t),a_(t))=i_(t)lambda_(t).U_{a}\left(c_{t}, a_{t}\right)=i_{t} \lambda_{t} .

在第四个方程 中使用它， V z ( z t ) = U a ( c t , a t ) + λ t = λ t ( 1 + i t ) V z z t = U a c t , a t + λ t = λ t 1 + i t V_(z)(z_(t))=U_(a)(c_(t),a_(t))+lambda_(t)=lambda_(t)(1+i_(t))V_{z}\left(z_{t}\right)=U_{a}\left(c_{t}, a_{t}\right)+\lambda_{t}=\lambda_{t}\left(1+i_{t}\right) .现在更新这一时期并使用它来消除 V z ( z t + 1 ) V z z t + 1 V_(z)(z_(t+1))V_{z}\left(z_{t+1}\right) 第二个 F O C F O C FOCF O C 时期，我们只剩下
U c ( c t , a t ) = λ t λ t = β ( 1 + i t + 1 ) ( P t P t + 1 ) λ t + 1 = β ( 1 + i t + 1 1 + π t + 1 ) λ t + 1 U c c t , a t = λ t λ t = β 1 + i t + 1 P t P t + 1 λ t + 1 = β 1 + i t + 1 1 + π t + 1 λ t + 1 {:[U_(c)(c_(t),a_(t))=lambda_(t)],[lambda_(t)=beta(1+i_(t+1))((P_(t))/(P_(t+1)))lambda_(t+1)=beta((1+i_(t+1))/(1+pi_(t+1)))lambda_(t+1)]:}\begin{gathered} U_{c}\left(c_{t}, a_{t}\right)=\lambda_{t} \\ \lambda_{t}=\beta\left(1+i_{t+1}\right)\left(\frac{P_{t}}{P_{t+1}}\right) \lambda_{t+1}=\beta\left(\frac{1+i_{t+1}}{1+\pi_{t+1}}\right) \lambda_{t+1} \end{gathered}

（乙）这些条件跟经文里的条件有什么不同？在第 2.2 节的基本 MIU 模型中，方程 （2.10） 和 （1） 相同。使用 （2.7） 和 （2.10），方程 （2.6） 可以写成（设置 n = 0 n = 0 n=0n=0 后）
TESTB λ t I ∄ β ( 1 + i t 1 + π t + 1 ) λ t + 1 TESTB λ t I ∄ β 1 + i t 1 + π t + 1 λ t + 1 TESTBlambda_(t)I∄beta((1+i_(t))/(1+pi_(t+1)))lambda_(t+1)\mathrm{TESTB} \lambda_{t} I \nexists \beta\left(\frac{1+i_{t}}{1+\pi_{t+1}}\right) \lambda_{t+1}
which differs from (2) in that i t i t i_(t)i_{t} rather than i t + 1 i t + 1 i_(t+1)i_{t+1} appears. This difference in timing is related to the dependence of utility on M t / P t M t / P t M_(t)//P_(t)M_{t} / P_{t} in the model of the text and on A t / P t = A t / P t = A_(t)//P_(t)=A_{t} / P_{t}= M t − 1 / P t + τ t + [ ( 1 + i t − 1 ) B t − 1 − B t ] / P t M t − 1 / P t + τ t + 1 + i t − 1 B t − 1 − B t / P t M_(t-1)//P_(t)+tau_(t)+[(1+i_(t-1))B_(t-1)-B_(t)]//P_(t)M_{t-1} / P_{t}+\tau_{t}+\left[\left(1+i_{t-1}\right) B_{t-1}-B_{t}\right] / P_{t} in the model of this problem. In the basic MIU model of section 2.2, the household can reduce current consumption at a cost of U c ( c t , m t ) = λ t U c c t , m t = λ t U_(c)(c_(t),m_(t))=lambda_(t)U_{c}\left(c_{t}, m_{t}\right)=\lambda_{t} and purchase bonds instead, yielding a real payoff of ( 1 + i t ) / ( 1 + π t + 1 ) 1 + i t / 1 + π t + 1 (1+i_(t))//(1+pi_(t+1))\left(1+i_{t}\right) /\left(1+\pi_{t+1}\right) in period t + 1 t + 1 t+1t+1. This gross interest income could be spend on consumption, but it did not serve to augment the period t + 1 t + 1 t+1t+1 holdings of money that enter the utility function. In the model of this problem, the same option to lower current consumption to purchase bonds is open, but purchasing bonds has an addition cost since purchasing additional bonds reduces A t / P t A t / P t A_(t)//P_(t)A_{t} / P_{t} which reduces current utility by U a ( c t , z t − b t ) = i t λ t U a c t , z t − b t = i t λ t U_(a)(c_(t),z_(t)-b_(t))=i_(t)lambda_(t)U_{a}\left(c_{t}, z_{t}-b_{t}\right)=i_{t} \lambda_{t} in period t t tt. Thus, the utility cost of purchasing a bond is U c ( c t , z t − b t ) + U a ( c t , z t − b t ) = ( 1 + i t ) λ t U c c t , z t − b t + U a c t , z t − b t = 1 + i t λ t U_(c)(c_(t),z_(t)-b_(t))+U_(a)(c_(t),z_(t)-b_(t))=(1+i_(t))lambda_(t)U_{c}\left(c_{t}, z_{t}-b_{t}\right)+U_{a}\left(c_{t}, z_{t}-b_{t}\right)=\left(1+i_{t}\right) \lambda_{t}. The bond pays off 1 + i t ) / ( 1 + π t + 1 ) 1 + i t / 1 + π t + 1 {:1+i_(t))//(1+pi_(t+1))\left.1+i_{t}\right) /\left(1+\pi_{t+1}\right) in period t + 1 t + 1 t+1t+1 and this return, since it is paid out before the goods market opens, yields the marginal utility of A t + 1 / P t + 1 A t + 1 / P t + 1 A_(t+1)//P_(t+1)A_{t+1} / P_{t+1} (equal to i t + 1 λ t + 1 i t + 1 λ t + 1 i_(t+1)lambda_(t+1)i_{t+1} \lambda_{t+1} ) plus the marginal utility of consumption ( λ t + 1 ) λ t + 1 (lambda_(t+1))\left(\lambda_{t+1}\right). That is, the gross interest income from the bond augments the real stock of money that enters time t t tt utility and can be spend on period t t tt consumption. Hence, 
( 1 + i t ) λ t = β ( 1 + i t 1 + π t + 1 ) ( 1 + i t + 1 ) λ t + 1 ⇒ λ t = β ( 1 + i t + 1 1 + π t + 1 ) λ t + 1 1 + i t λ t = β 1 + i t 1 + π t + 1 1 + i t + 1 λ t + 1 ⇒ λ t = β 1 + i t + 1 1 + π t + 1 λ t + 1 (1+i_(t))lambda_(t)=beta((1+i_(t))/(1+pi_(t+1)))(1+i_(t+1))lambda_(t+1)=>lambda_(t)=beta((1+i_(t+1))/(1+pi_(t+1)))lambda_(t+1)\left(1+i_{t}\right) \lambda_{t}=\beta\left(\frac{1+i_{t}}{1+\pi_{t+1}}\right)\left(1+i_{t+1}\right) \lambda_{t+1} \Rightarrow \lambda_{t}=\beta\left(\frac{1+i_{t+1}}{1+\pi_{t+1}}\right) \lambda_{t+1}