The Optimal Choice of Monetary Instruments Š The Poole …cm.de.iscte.pt/Poole_model.pdfThe Optimal...

The Optimal Choice of Monetary Instruments� The Poole Model �

Vivaldo M. Mendes

ISCTE � Lisbon University Institute

06 November 2013

(Vivaldo M. Mendes) The Poole Model 06 November 2013 1 / 27

Summary

1 Tools, targets and goals2 A simple model with demand shocks (the Poole model)3 Extending the model: monetary base as potential instrument4 Poole�s model with supply shocks (not covered, lack of time)


Tools, targets and goals




The temporal steps of monetary policy

1 Policy tools2 Operating targets (instruments)3 Intermediate targets4 Final goals



The temporal steps of monetary policy

1 The CB uses (controls) operating targets in order to a¤ect theintermediate targets.

2 Which operating target is better:Money supply? Short term interestrates?

3 Answer: it depends upon the kind of shocks that a¤ect the economy:

1 Aggregate demand shocks2 Money demand shocks3 Supply shocks

4 Major result: relative variances of macroeconomic shocks matter foroptimal choice of instrument

William Poole (1970). "Optimal choice of monetary policy instrumentsin a simple stochastic macro model," Sta¤ Studies 57, Federal ReserveBoard, Washington.



The Taylor rule

A famous example of an operating target rule



The ECB example


What is the New Keynesian Model?

ECB key interest rates

(Vivaldo M. Mendes) November 2018 3 / 3

Only demand shocks

A simple model with demand shocks


Only demand shocks

The basic equations of the model

1 Consider an economy with an IS�LM model with RationalExpectations

2 The IS function tells us that

yt = �b(it � Etπt+1) + zt (1)

Positive shocks on the IS (zt) boost output at given interest rates.3 The LM curve tells us that real money demand (mt � pt) isdependent upon output and the interest rate

mt � pt| {z }log values

= yt � ait + vt (2)

4 Positive shocks on the LM (vt) boost money demand at given interestrates and output.


Only demand shocks

Basic assumptions

1 In the old IS�LM framework, prices and expected in�ation are kept asconstant. So

Pt = 1, then ln Pt � pt = 0Etπt+1 = 0

2 The policy maker cannot observe the shocks. It can set either thenominal interest rate (i) or the money supply (m).

3 The problem of the policy maker is to choose between these twoinstruments to minimize the variance of output

min E[yt]2, or min E[yt � y�]2

4 What is best as instrument? Money supply (m) or (i)?


vmpm

Realce

Only demand shocks

Solving the model

1 With Etπt+1 = 0, and pt = 0, The IS and LM eq can be written as

yt = �bit + zt

mt = yt � ait + vt

2 If there were no shocks, the solutions would be (no subscripts forsimplicity)

y� = �bi� =b

a+ bm�

and

i� = � 1a+ b

m�

3 The objective of the CB is to minimize

min E[yt � y�]2


Only demand shocks

The CB controls (pegs) the interest rate

1 The objective of the CB is

min E[yt � y�]2

2 Therefore, the CB sets i = i�, and output is given by

y = �bi� + z= y� + z

3 Then, it is immediate to see that

E[y� y�]2i� = Var(z)| {z }σ2

z


Only demand shocks

The CB controls (pegs) the money stock (m)

1 Now the CB instead pegs the money stock (m = m�).2 From the IS we know that

y = �bi+ z3 And from the LM we know that

i =1a(y�m+ v)

4 Then we have

y = �ba(y�m� + v) + z

...

= y� � ba+ b

v+a

a+ bz


E[y� y�]2m� =�

ba+ b

�2

σ2v +

�a

a+ b

�2

σ2z


Only demand shocks

Comparing the instruments

The comparison of both instruments is easy

E[y� y�]2i� = E[y� y�]2m�

σ2z =

�b

a+ b

�2

σ2v +

�a

a+ b

�2

σ2z

(a+ b)2σ2z = b2σ2

v + a2σ2z

...(2ab+ b2)σ2

z = b2σ2v

σ2z =

�b

2a+ b

�| {z }

φ<1

σ2v


Only demand shocks

Comparing the instruments: summary

E[y� y�]2i� = E[y� y�]2m�

σ2z =

�b

2a+ b

�| {z }

φ<1

σ2v

1 Hence, choose an interest rate targeting procedure whenever there is

1 relatively high money demand volatility2 relatively low aggregate demand volatility

2 In general targeting the interest rate is preferable if IS shocks aresmall, money demand is not sensitive to the interest rates (a is small),output is sensitive to the interest rate in IS (b is large):


Monetary base as potential instrument

Extending the model:

monetary base as potential instrument


vmpm

Caixa de texto

From now onwards, not covered this year


Monetary base as an instrument

1 Central banks control the money base, but not M (for example, M3)as in the model

2 Extend the model with endogenous money supply (M):

mt = ht + eit +ωt (3)

ht is the the monetary base.3 Notice that the money multiplier is given by (expressed in logs)

mt � ht = eit +ωt

ωtis mean-zero money multiplier shock.4 Money multiplier is increasing in interest rate (banks want to lendmore/consumers want to hold less cash)



Solving the model

1 The three fundamental equations are now

yt = �bit + zt (IS)

mt = yt � ait + vt (LM)

mt = ht + eit +ωt (M Multiplier)

2 The logic now is

y depends on ii depends on mm depends on h

3 The objective is to express y, i, m as determined only by h (which iscontrolled by the CB)



Solving the model (cont.)

1 If there were no shocks, the solutions would be (no subscripts forsimplicity, see Appendix 1 for details)

y� = �bi� =b

a+ b+ eh� (4)

i� = � 1a+ b+ e

h�

m� = h� + ei� =a+ b

a+ b+ eh�

2 The objective of the CB is to minimize

min E[yt � y�]2



The CB controls (pegs) the interest rate

1 The same results as above2 The solution is totally given by the IS function.3 The CB sets i = i�, and output is given by

y = �bi� + z= y� + z


E[y� y�]2i� = Var(z)| {z }σ2

z



The CB controls (pegs) the monetary base (h)

1 Now the CB pegs the monetary base (h = h�).2 Then, the Money Multiplier eq. comes as

m = h� + ei+ω

3 But from the LM eq. we know that

m = y� ai+ v

4 Therefore, by equating both previous eq. we get

h� + ei+ω = y� ai+ v

5 From where we can obtain

i =1

a+ e(y� h� + v�ω)



The CB controls (pegs) the monetary base (h)

1 Now, from the equilibrium in the money market we know that

i =1

a+ e(y� h� + v�ω)

2 Then, we should use the IS function to solve the entire problem

y = �bi+ z

= � ba+ e

(y� h� + v�ω) + z (5)

3 Solving for y, leads to (see Appendix 2 for details)

y = y� � ba+ b+ e

v+b

a+ b+ eω+

a+ ea+ b+ e

z (6)


E[y� y�]2h� =�

ba+ b+ e

�2 �σ2

v + σ2ω

�+

�a+ e

a+ b+ e

�2

σ2z



Comparing the instruments

1 Now it is easy to see which instrument is better

E[y� y�]2i� = E[y� y�]2h�

σ2z =

�b

a+ b+ e

�2 �σ2

v + σ2ω

�+

�a+ e

a+ b+ e

�2

σ2z

(a+ b+ e)2σ2z = (b)2

�σ2

v + σ2ω

�+ (a+ e)2 σ2

z

...

σ2z =

�b

2 (a+ e) + b

�| {z }

φ<1

�σ2

v + σ2ω

�

2 Hence, choose an interest rate targeting procedure whenever there is1 relatively high money demand volatility2 relatively high money multiplier volatility3 relatively low aggregate demand volatility


Introducing supply shocks

Extending the model:

introducing supply shocks� No time available for covering this point . Read pages 18-23 in thebibliography if you are curious about this point. But reading is not

compulsory�



Appendix 1

1 The 3 eq. with no shocks are

y = �bi (IS)

m = yt � ai (LM)

m = h+ ei (MM)

2 Using the LM and the MM eq., we get

i =1

a+ e(y� h)

3 Now insert this result into the IS eq.

y = �bi =b

a+ b+ eh

4 Then, it is easy to write back m and i also as functions of h

i = � 1a+ b+ e

h , m = h+ ei =a+ b

a+ b+ eh.



Appendix 2

1 How to get from eq. (5) to eq. (6)? From eq. (5), we have

y = � ba+ e

(y� h� + v�ω) + z

2 Solving for y, leads to

y(1+b

a+ e) =

ba+ e

h� � ba+ e

v+b

a+ eω+ z

y =b

a+ b+ eh� � b

a+ b+ ev+

ba+ b+ e

ω+a+ e

a+ b+ ez

3 Now, using the result in eq. (4)

y� =b

a+ b+ eh�

4 We arrive at eq. (6)

y = y� � ba+ b+ e

v+b

a+ b+ eω+

a+ ea+ b+ e

z(Vivaldo M. Mendes) The Poole Model 06 November 2013 27 / 27

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