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3. Rational Expectations Solution Technique

Tutorial 14

1. Consider the following model of a closed economy where symbols have their usualmeanings. y, p and m are in natural logarithms. E is the expectations operator and ε is arandom disturbance term with the usual properties.

yt = 20 + 0.25(pt − Et−1pt )

yt = 30 + 0.25(mt − pt )

mt = 5 + ε t

14.1

14.2

14.3

(a) Solve for

(i) the full-employment equilibrium value of y,(ii) the expected price level.

(b) If ε t = −2, what is (i) the price level? (ii) the output level?

2. Consider the following macroeconomic model where m, p and y are the logarithms ofmoney supply, the price level and output respectively. (14.4) is a money demand function,(14.5) is a variant of the Phillips curve and (14.6) is the authorities money supply rule. E isthe expectations operator and ε is a ‘white noise’ error term.

mt = pt + yt

pt = Et−1pt + 3(yt − 50)

mt = 55 − 0.4(yt−1 − 50) + ε t

14.4

14.5

14.6

(a) Obtain a solution for yt.

(b) Obtain a solution for (i) pt and (ii) its variance around expected price level.

(c) Can the authorities stabilise output in this model?

(d) Solve for pt using Lucas method of undetermined coefficients.

Solution

1(a) Note that the basic method will suffice for all Rational Expectations models inwhich there are expectations (at any date in the past) of current events only. The methodinvolves three steps:

1. Solve the model, treating expectations as exogenous.

2. Take the expected value of this solution at the date of the expectations, and solve forthe expectations.

3. Substitute the expectations solutions into the solution in 1, and obtain the completesolution.

Substitute equations (14.1) and (14.3) in (14.2) to get the reduced form:

⇒ 20 + 0.25(pt − Et−1pt ) = 30 + 0.25(5 + ε t − pt )

0.25ε t + 11.25 = 0.5pt − 0.25Et−1pt 14.7

In order to get Et−1pt run the expectations operator along equations 14.1-14.3:

Et−1yt = 20 + 0.25(Et−1pt − Et−1pt ) ≡ 20

Et−1yt = 30 + 0.25(Et−1mt − Et−1pt )

Et−1mt = 5

Substituting Et−1mt = 5 and Et−1yt = 20 in the second expression yields solution forEt−1pt.

Et−1pt = 45 14.8

The full-employment equilibrium value of y∗ can be found out by looking at equation(14.1). Note that in equilibrium pt − Et−1pt = 0. It follows that y∗ = 20.

Substituting (14.8) in (14.7) yields:

⇒ 0.25ε t + 11.25 = 0.5pt − 0.25(45)

pt = 45 + 0.5ε t 14.9

1(b(i)) If ε t = −2, substituting in (14.9) gives the price level:

pt = 45 + 0.5(−2) ≡ 44 14.10

1(b(ii)) Substituting (14.10) and (14.8) in (14.1) yields solution for output:

yt = 20 + 0.25(44 − 45) ≡ 19.75

2(a) Substitute equations (14.5) and (14.6) in (14.4) to get the reduced form:

55 − 0.4(yt−1 − 50) + ε t = Et−1pt + 3(yt − 50) + yt 14.11

In order to get Et−1pt run the expectations operator along equations 14.4-14.6:

Et−1mt = Et−1pt + Et−1yt

Et−1 pt = Et−1pt + 3(Et−1yt − 50)

Et−1mt = 55 − 0.4(yt−1 − 50) + Et−1ε t (Note Et−1ε t = 0)

Substituting Et−1mt = 55 − 0.4(yt−1 − 50) and Et−1yt = 50 in the first expression yieldssolution for Et−1pt.

Et−1pt = 25 − 0.4yt−1 14.12

Substituting (14.12) in (14.11) yields solution for output:

⇒ 55 − 0.4(yt−1 − 50) + ε t = 25 − 0.4yt−1 + 3(yt − 50) + yt

yt = 50 + 14(ε t ) 14.13

2 (b(i)) Substituting (14.12) and (14.13) in (14.5) yields solution for the price level:

⇒ pt = 25 − 0.4yt−1 + 3 50 + 14(ε t ) − 50

Substituting for yt−1(= 50 + 14(ε t−1 )) results in:

pt = 5 + 0.75ε t − 0.1ε t−1

2(b(ii)) Since we have the solution for pt and Et−1pt it is easy to compute the varianceof the price level.

Definition The variance of X will be denoted by σ2 whether X is a discrete or acontinuous type of random variable. It is useful to note thatσ2 = E[X − µ]2 = E[X − 2µX + µ2 ]; and since E is a linear operator,σ2 = E(X2) − 2µE(X) + µ2 = E(X2) − µ2.

E[pt − Et−1pt ]2 = E 34ε t

2= 9

16σ2

2(c) The authorities cannot stabilise output in this model as output is invariant to themoney supply rule. This can be proved by multiplying the money supply equation by (0)instead of (-0.4). The variance of output turns out to be the same ( 1

16 σ2) irrespective of the

rule in place. This is because the money supply rule is incorporated into agents’expectations at (t-1) and cannot cause any surprises.

2(d) Lucas method of undetermined coefficients

Substitute equations (14.5) and (14.6) in (14.4) to get the reduced form:

55 − 0.43

(pt−1 − Et−2pt−1 ) + ε t = pt + 50 + 13(pt − Et−1pt )

To apply the Lucas method write the solution for the endogenous variables (instead ofbeing written in terms of the constants and the errors) in terms of the ‘state variables,’ i.e.,current and past values of the exogenous variables (including the error terms of the modelequations) and past values of the endogenous variables. Thus write the solution for pt (onwhich we focus here) as

pt = π0 + π1ε t + π2ε t−1 + π3pt−1 + π4yt−1

By implicaton

Et−1pt = π0 + π2ε t−1 + π3pt−1 + π4yt−1

pt−1 = π0 + π1ε t−1 + π2ε t−2 + π3pt−2 + π4yt−2

Et−2pt−1 = π0 + π2ε t−2 + π3pt−2 + π4yt−2

Substituting these conjectures in the reduced form equation and collecting terms in ε t,ε t−1......

(cons tan ts) : 55 = π0 + 50 ∴ π0 = 5

(ε t ) : 1 = π1 + 13π1 ∴ π1 = 3

4

(ε t−1 ) : − 0.43

π1 = π2 ∴ π2 = −0.1

(pt−1 ) : 0 = π3

(yt−1 ) : 0 = π4

Substituting the π i

s in the conjecture solution for pt yields:

pt = 5 + 0.75ε t − 0.1ε t−1

Tutorial 15

Consider the following model. Fiscal policy will be held constant and monetary policywill be the only policy variable affecting demand for output. For expositional purposes theincome velocity of money is also held constant. With these assumptions, the aggregatedemand for output can be written in logs as:

mt +_v= pt + yt 15.1

The above equation is the equation of exchange in logs. The model is complete with theintroduction of the aggregate supply equation and a money supply rule.

yt = y∗ + α(pt − Et−1pt )

mt = βyt−1 + ε t

15.2

15.3

(a) Given that agents form expectations rationally, use the basic method to obtain asolution for (i) yt and (ii) pt.

(b) If expectations are backward looking i.e., pte − pt−1

e = γ(pt−1 − pt−1e ), where

0 γ 1, use the basic method to obtain a solution for (i) yt and (ii) pt. Is there any scopein this model for the policy authorities to influence output through systematic stabilisationpolicy?

Solution

(a) Substituting equations (15.2) and (15.3) in (15.1) yields the reduced form:

βyt−1 + ε t +_v= pt + y∗ + α(pt − Et−1pt ) 15.4

In order to solve for Et−1pt using the basic method:


2. Take the expected value of this solution at the date of the expectations, and solve forthe expectations.


Et−1mt +_v = Et−1pt + Et−1yt

Et−1yt = y∗ + α(Et−1pt − Et−1pt ) ≡ y∗

Et−1mt = βyt−1 + Et−1ε t ≡ βyt−1

15.5

15.6

15.7

∴ Et−1pt = βyt−1 +_v −y∗

Substituting the solution for Et−1pt in (15.4) yields solution for the price level:

⇒ βyt−1 + ε t +_v= pt + y∗ + α(pt − βyt−1 −

_v +y∗ )

⇒ βyt−1(1 + α) + ε t +_v −y∗ + α

_v −αy∗ = (1 + α)pt

∴ pt = βyt−1 − y∗ +_v + 1

1 + α ε t

Substituting the solutions for pt and Et−1pt in (15.2) yields:

⇒ yt = y∗ + α βyt−1 − y∗ +_v + 1

1 + α ε t − (βyt−1 +_v −y∗ )

yt = y∗ + 11 + α ε t

The solution for yt consist of an expected part (y∗) and an unexpected part (functionsof ε t). Rational expectations has incorporated anything known at t-1 with implications for yat time t into the expected part, so that the unexpected part is purely unpredictable.

(b) If expectations are backward looking i.e., pte − pt−1

e = γ(pt−1 − pt−1e ), where

0 γ 1:

⇒ pte = γpt−1 − γpt−1

e + pt−1e

⇒ pte = γpt−1 + (1 − γ)[γpt−2 − γpt−2

e + pt−2e ]

⇒ pte = γpt−1 + γ(1 − γ)pt−2 − γ(1 − γ)pt−2

e + (1 − γ)pt−2e

⇒ pte = γpt−1 + γ(1 − γ)pt−2 + (1 − γ)2pt−2

e

By continuous forward substitution for pt−2e , pt−3

e , ....

pte = Et−1pt = γ∑

i=0

∞

(1 − γ) ipt−1+i 15.8


⇒ βyt−1 + ε t +_v= pt + y∗ + α pt − γ∑

i=0

∞

(1 − γ) ipt−1+i

pt = 11 + α βyt−1 +

_v −y∗ + αγ∑

i=0

∞

(1 − γ) ipt−1+i + ε t

Substituting pt and Et−1pt in (15.2) yields solution for output:

yt = y∗ + α 11 + α

βyt−1 +_v −y∗ + αγ∑ i=0

∞ (1 − γ) ipt−1+i

+ε t − γ∑ i=0∞ (1 − γ) ipt−1+i

After collecting terms in γ∑ i=0∞ (1 − γ) ipt−1+i and resorting to some algebraic

manipulation we get:

yt = y∗ + α1 + α βyt−1 +

_v −y∗ + ε t − γ∑

i=0

∞

(1 − γ) ipt−1+i

Note that when expectations formation is backward looking (adaptive/error learning)the monetary policy or feedback rule (β -the feedback parameter) enters the solution foroutput i.e., authorities can stabilise output in this model.

Tutorial 16

1. Consider the following model where symbols have their usual meanings:

mt = pt + yt

mt =_m −β(yt−1 − y∗ ) + ε t

yt = y∗ + q pt − 12(Et−1pt + Et−2pt )

16.1

16.2

16.3

(a) Using the Muth method, obtain a solution for (i) yt and (ii) pt.

(b) Given that agents form expectations rationally, is there any scope in this model forthe policy authorities to influence output through systematic stabilisation policy?

Solution

1(a) Substitute (16.2) and (16.3) in (16.1) yields:

_m −β(yt−1 − y∗ ) + ε t = pt + y∗ + q pt − 1

2(Et−1pt + Et−2pt ) 16.12

It follows from (16.3) that:

yt−1 = y∗ + q pt−1 − 12(Et−2pt−1 + Et−3pt−1 ) 16.13


_m −β y∗ + q pt−1 − 1

2(Et−2pt−1 + Et−3pt−1 ) − y∗ + ε t = pt + y∗ +

q pt − 12(Et−1pt + Et−2pt )

or

_m −β(0.5q[(pt−1 − Et−2pt−1) + 0.5(pt−1 − Et−3pt−1)]) + ε t = pt + y∗ +

0.5q[(pt − Et−1pt) + (pt − Et−2pt)] 16.14

The Muth Method of Undetermined Coefficients:

In a stochastic linear economic system, such as the model we are currently workingwith, a variable can always be written as an infinite moving-average process in a randomerror. This is Wold’s decomposition theorem. We thus write a random variable yt as:

yt =−y +∑

i=0

∞

π iε t−i

or

yt =−y +π0ε t + π1ε t−1 + π2ε t−2 + ....

where−y = the mean of the series, π i = constant parameters, and ε = a normally

distributed error with a mean of 0, constant variance σε2 and zero covariance. The same is

true for the variable pt in our model. The Wold’s decomposition theorem forms the basisfor the Muth method of undetermined coefficients. Using the Muth method we can writethe solution for

pt =−p +∑

i=0

∞

π iε t−i

Ignoring the constants and expanding results give:

pt = π0ε t + π1ε t−1 + π2ε t−2 + ....

pt−1 = π0ε t−1 + π1ε t−2 + π2ε t−3 + ....

Et−2pt−1 = π1ε t−2 + π2ε t−3 + ....

Et−1pt = π1ε t−1 + π2ε t−2 + ....

Et−2pt = π2ε t−2 + ....

Et−3pt−1 = π2ε t−3 + ....

Substituting in (16.14) yields:

⇒_m −0.5βq(π0ε t−1 + π0ε t−1 + π1ε t−2) + ε t =

π0ε t + π1ε t−1 + π2ε t−2 + .... + y∗ + 0.5q[π0ε t + π0ε t + π1ε t−1 ]

or

_m −βqπ0ε t−1 − 0.5βqπ1ε t−2 + ε t =

π0ε t + π1ε t−1 + π2ε t−2 + .... + y∗ + qπ0ε t + 0.5qπ1ε t−1 16.15

We need to now evaluate π0 and π1 etc i.e., the undetermined coefficients. For this wecollect terms in ε t, ε t−1, ε t−2......

ε t : 1 = π0 + qπ0 ⇒ 1(1 + q)

ε t−1 : −βqπ0 = π1 + 0.5qπ1

⇒ −βq 1(1 + q)

= π1 + 0.5qπ1

∴ π1 = −βq(1 + q)(1 + 0.5q)

ε t−2 : −0.5βqπ1 = π2

∴ π2 = −0.5βq(−βq)(1 + q)(1 + 0.5q)

ε t−3 : 0 = π3

We know from (16.3) that

yt = y∗ + q pt − 12(Et−1pt + Et−2pt ) = y∗ + 0.5q[π0ε t + π0ε t + π1ε t−1 ]

∴ yt = y∗ + q1 + q

ε t −0.5βq2

(1 + q)(1 + 0.5q)ε t−1

1(b) Since β the parameter in the money supply rule enters the solution for output,stabilisation (monetary policy in this case) policy is effective. This illustration shows thatin the case of private agents not being able to respond to new information by changing theirwage-price decisions as quickly as the monetary authorities can change any of their control,then scope once again emerges for systematic stabilisation policy to have real effects.

Tutorial 17

1. Consider the following model of a closed economy where symbols have their usualmeanings. y, p and m are in natural logs. E is the expectations operator and ε is a randomdisturbance term with the usual properties.

yt = −αrt

yt = y∗ + γ(pt − Et−1pt )

yt = y∗ + γ(pt − Et−1pt )

mt =_m +ε t

Rt = rt + Et−1pt+1 − Et−1pt

17.1

17.2

17.3

17.4

17.5

(a) Derive the solution for output using the Muth method of undetermined coefficientsand comment on the response of output/unemployment to interest rate changes?

(b) Derive the solution for the price level using McCallum’s “minimal state variable”criterion?

Solution

1(a) Substituting equation (17.5) the Fisher equation in (17.1) the IS curve yields:

⇒ yt = −α(Rt − Et−1pt+1 + Et−1pt )

Rt = 1α −yt + αEt−1pt+1 − αEt−1pt 17.6

Substituting equations (17.6) and (17.4) in (17.3) yields:

⇒_m +ε t = pt + yt − ψ 1

α −yt + αEt−1pt+1 − αEt−1pt

Substituting for yt results in the reduced form equation:

_m +ε t = pt + 1 + ψ

α [γ((pt − Et−1pt) + y∗ )] − ψα [αEt−1pt+1 − αEt−1pt ] 17.7

Using the Muth method we can write the solution for pt as:

pt =−p +∑

i=0

∞

π iε t−i

Ignoring the constants and expanding results in:

pt = π0ε t + π1ε t−1 + π2ε t−2 + ....

pt+1 = π0ε t+1 + π1ε t + π2ε t−1 + ....

Et−1pt+1 = π2ε t−1 + π3ε t−2 + ....

Et−1pt = π1ε t−1 + π2ε t−2 + ....

Substituting all this in (17.7) yields:

_m +ε t = π0ε t + π1ε t−1 + π2ε t−2 + .... + 17.8

1 + ψα γ[π0ε t + y∗ ] − ψ

α απ2ε t−1 +

π3ε t−2 + ....− α

π1ε t−1 +π2ε t−2 + ....

We need to now evaluate π0 and π1 i.e., the undetermined coefficients. For this wecollect terms in ε t, ε t−1, ε t−2......

ε t : 1 = π0 + 1 + ψα γπ0

∴ π0 = 11 + γ(1 + ψ

α )

Note that we do not need ε t−1,ε t−2 etc for computing the solution for yt. Substitute thesolution for π0 in (17.2) to get solution for output.

yt = y∗ + γ(pt − Et−1pt ) ≡ y∗ + γ(π0ε t )

∴ yt = y∗ + γ1 + γ(1 + ψ

α )ε t

Note that the parameter for real interest rate α enters the solution for output. It impliesthat a rise in the interest rate shifts the aggregate supply curve upwards increasing output.Intuitively, a rise in r increases the attractiveness of working today relative to workingtomorrow. Thus, there is an increase in employment and output. The response of laboursupply to the interest rate is known as the intertemporal substitution in labour supply(Lucas and Rapping, 1969). Equilibrium business cycle theory (which we analyse later)uses the intertemporal substitution of labour to explain why employment and outputfluctuates over the business cycle.

1(b) McCallum’s MSV Criterion

The MSV criterion (see McCallum (1989(a))) is designed to yield a single bubble-freesolution by construction. Its definition begins by limiting solutions to those that are linearfunctions - analogous to our model - of a minimal set of “state variables,” i.e.,predetermined or exogenous determinants of current endogenous variables. Thus thesolution values involve variables that only appear in the model’s structural equations - no“extraneous” state variables are included, unlike the Lucas method discussed earlier.

It follows that the relevant determinants of pt include only ε t and a constant, and weconjecture that the solution is of the form:

pt = φ0 + φ1ε t

To find φ0 and φ1, we note that if our conjecture solution for pt is true, then

Et−1pt = φ0

Et−1pt+1 = φ0

Substituting these conjectures in the reduced-form equation (17.7) yields:

_m +ε t = φ0 + φ1ε t + 1 + ψ

α [γ((φ1ε t) + y∗ )] − ψα [αφ0 − αφ0 ]

The following conditions must pertain to the φ s:

(cons tan ts) :_m = φ0 + γ 1 + ψ

α y∗

(ε t ) : 1 = φ1 + γ 1 + ψα φ1

Now these two conditions - obtained by equating coefficients on both sides of thereduced form are just what is needed, for they permit us to solve for the two unknowns(“undetermined coefficients”) - the φ s. Thus we have,

∴ φ0 =_m −γ 1 + ψ

α y∗

∴ φ1 = 11 + γ(1 + ψ

α )

Therefore, the solution for pt is (substituting φ s in the conjecture solution for the pricelevel):

pt =_m −γ 1 + ψ

α y∗ + 11 + γ(1 + ψ

α )ε t

Tutorial 18

1. Consider the following model of a closed economy where symbols have their usualmeanings. E is the expectations operator and ε is a random disturbance term with the usualproperties.

yt = −α(Rt − Et−1pt+1 + Et−1pt ) + βGt − ψTt

yt = y∗ + δ(pt − Et−1pt )

Gt =_G +ε t

Tt = λyt

18.1

18.2

18.3

18.4

where β and ψ are government expenditure and tax elasticities respectively.

(a) Compute the solution for yt using the Muth method and show that fiscal policy

affects output? If so why?(b) Compute the variance of output and explain the role of automatic stabilisers? What

happens if we have a high tax elasticity?

Solution

(a) Substituting equations 18.2-18.4 in (18.1) yields the reduced form:

⇒ y∗ + δ(pt − Et−1pt ) = −α(Rt − Et−1pt+1 + Et−1pt ) +

β_G +ε t − ψλyt

or

y∗ + δ(pt − Et−1pt ) = −α(Rt − Et−1pt+1 + Et−1pt ) + β_G +ε t −

ψλ(y∗ + δ(pt − Et−1pt ))


pt =−p +∑

i=0

∞

π iε t−i

Ignoring the constants and expanding results in :

pt = π0ε t + π1ε t−1 + π2ε t−2 + ....

pt+1 = π0ε t+1 + π1ε t + π2ε t−1 + ....

Et−1pt+1 = π2ε t−1 + π3ε t−2 + ....

Et−1pt = π1ε t−1 + π2ε t−2 + ....

Plugging in these values in the reduced form equation yields:

y∗ + δ(π0ε t ) = −αRt − (π2ε t−1 + π3ε t−2 + ....) +

(π1ε t−1 + π2ε t−2 + ....)+

β_G +ε t − ψλ(y∗ + δ(π0ε t ))


ε t : δπ0 = β − ψλδπ0

∴ π0 = βδ + ψλδ

ε t−1 : 0 = απ2 − απ1

ε t−2 : 0 = απ3 − απ2

∴ π1 = π2 = π3

Substituting the value for π0 in the Phillips curve equation (18.2) yields solution foroutput:

⇒ yt = y∗ + δ(pt − Et−1pt ) ≡ y∗ + δ(π0ε t )

∴ yt = y∗ + δβδ + ψλδ ε t ≡ y∗ + β

1 + ψλ ε t

Et−1yt = y∗ (since Et−1ε t = 0)

Note that both the fiscal instruments (β and ψ are government expenditure and taxelasticities respectively) enter the solution for output. In order to explain why fiscal policyaffects output in this model see solution for Et−1pt.

Et−1pt = π1ε t−1 + π2ε t−2 + ....

Note that π0 (which contains β and ψ) does not enter the solution for the expected pricelevel. In other words the fiscal instruments are not taken into account when agents’ formexpectations at t-1 and hence can cause surprises.

b) The variance of output is given by:

⇒ σy2 = E(yt − Et−1yt )2 = E y∗ + β

1 + ψλ ε t − y∗2

∴ σy2 = β

1 + ψλ

2

σ2

Automatic Stabiliser:

By an automatic stabiliser we mean a mechanism in which a variable (say, taxliabilities) responds to current income levels, and therefore provide an automatic andimmediate adjustment to current disturbances. Automatic stabilisers are designed to reduce

the lags associated with stabilisation policy. The long and variable lags associated withmonetary and fiscal policy certainly make macroeconomic management more difficult.Automatic stabilisers are policies that help stimulate or depress the economy whennecessary without any deliberate shift in policy stance. In this particular model ψ is theelasticity of spending to temporary variations in tax liabilities. Note that the solution foroutput is not independent of the automatic stabiliser. A high tax elasticity reduces thevariance of output.

Tutorial 19

1. Consider the following model of a closed economy where symbols have their usualmeanings. Note that there is persistent shock to output in this model. E is the expectationsoperator and ε is a random disturbance term with the usual properties.

yt = −α(Rt − Et−1pt+1 + Et−1pt ) + µf(yt−1 − y∗ )

yt = y∗ + β(pt − Et−1pt ) + γ(yt−1 − y∗ )

mt = pt + yt − cRt + vt

mt =_m +µm(yt−1 − y∗ ) + ut

19.1

19.2

19.3

19.4

where µf(yt−1 − y∗ ) and µm(yt−1 − y∗ ) represent fiscal and monetary feedback responserespectively.

1) Using the Muth method, obtain a solution for yt.

2) Is there a scope in this model for policy authorities to influence output?

2. Solve the following model for pt and yt.

mt = pt + yt − α[Et−1pt+1 − Et−1pt ]

yt = y∗ + β(pt − Et−1pt ) + λ yt−1 − y∗

mt =_m +ε t

19.11

19.12

19.13

Solve by (a) Sargent’s forward root method (b) difference equation method and (c)

Muth method.

Solution

1) Note that we can transform (19.2) using the lag operator:

⇒ yt = y∗ + β(pt − Et−1pt ) + γ(Lyt − y∗ )

⇒ yt = y∗ + β(pt − Et−1pt ) + γL(yt − y∗ )

⇒ (1 − γL)(yt − y∗ ) = β(pt − Et−1pt )

yt = y∗ + β(pt − Et−1pt )(1 − γL)

19.5

From equation (19.1) one can get the solution for the nominal interest rate:

Rt = 1α − yt + αEt−1pt+1 − αEt−1pt + µf(yt−1 − y∗ ) 19.6

Substituting (19.6) in (19.3) and equating the resulting equation with (19.4) gives:

⇒_m +µm(yt−1 − y∗ ) + ut = pt +

yt − c 1α

− yt + αEt−1pt+1 −αEt−1pt + µf(yt−1 − y∗ )

+ vt

⇒_m +µ(yt−1 − y∗ ) + ut = pt + 1 + c

α yt − c(Et−1pt+1 − Et−1pt ) + vt

where µ = [µfcα + µm ].

Substituting the solution for output in (19.5) into the reduced form equation yields:

⇒_m +µ y∗ + β(pt−1 − Et−2pt−1 )

(1 − γL)− y∗ + ut = pt +

1 + cα y∗ + β(pt − Et−1pt )

(1 − γL)− c(Et−1pt+1 − Et−1pt ) + vt

or_m (1 − γ) + µβ(pt−1 − Et−2pt−1 ) + wt(1 − γL) = pt(1 − γL) +

1 + cα (1 − γ)y∗

+ 1 + cα β(pt − Et−1pt ) − c(1 − γL)(Et−1pt+1 − Et−1pt )

where wt = ut − vt.


pt =−p +∑

i=0

∞

π iε t−i

We need to now evaluate π0 and π1 i.e., the undetermined coefficients. For this wecollect terms in wt, wt−1, wt−2......

wt : 1 = π0 + β 1 + cα π0

∴ π0 = 11 + β(1 + c

α )

The identities in the other errors are irrelevant for our purpose here as it is clear bylooking at the solution for output. Substituting the solution for π0 in (19.5) yields solutionfor output.

2) After substituting for π0 in (19.5) we see that π0 does not depend either on µm or µf.Thus it is clear that systematic monetary policy does not influence the variance of output inthis model inspite of persistence in the aggregate supply curve.

2(a) Sargent’s forward operator method

Take expectations at t-1 to find Et−1pt from (19.11-13):

_m= Et−1pt(1 + α) − αEt−1pt+1 + y∗ + λβπ0ε t−1

1 − λL 19.14

Note that pt − Et−1pt = π0ε t and yt−1 − y∗ = βπ0εt−1

1−λL.

Hence,

_m= (1 + α) 1 − α

1 + α B−1 Et−1pt + y∗ + λβπ0ε t−1

1 − λL 19.15

where the operator B is defined by B−1(Ext+j Ω t ) = Ext+j+1 Ω t, where Ω t is the

information set at time ‘t’, in other words B instructs to lag the variable while leaving thedate of expectations unchanged.

∴ Et−1pt =_m −y∗

(1 + α) 1 − α1+α

− λβπ0ε t−1

(1 − λL)(1 + α) 1 − α1+α B−1

19.16

Note the last term (using the summation of an infinite series) can be expressed as:

λβπ0ε t−1

(1 − λL)(1 + α) 1 − α1+α B−1

= λβπ0

1 + α(1 − λ)ε t−1

1 − λL

Substituting this in (19.16) yields:

Et−1pt =_m −y∗ − λβπ0ε t−1

(1 − λL)(1 + α(1 − λ)) 19.17

It follows that:

Et−1pt+1 =_m −y∗ − Et−1

λβπ0ε t

(1 − λL)(1 + α(1 − λ))

Et−1pt+1 =_m −y∗ − λβπ0

1 + α(1 − λ)Et−1(ε t + λε t−1 + λ2ε t−2 + ......)

Et−1pt+1 =_m −y∗ − λ2βπ0

1 + α(1 − λ)ε t−1

1 − λL 19.18

Note that the reduced form equation of this model can be expressed as:_m +ε t = (pt − Et−1pt ) + Et−1pt − α(Et−1pt+1 − Et−1pt ) + y∗

+ β(pt − Et−1pt ) +λβπ0ε t−1

1 − λL 19.19

Substituting (19.17) for Et−1pt and (19.18) for Et−1pt+1 in (19.19) and multiplyingthroughout by 1 − λL yields:

ε t(1 − π0 − βπ0 ) − λε t−1(1 − π0 − βπ0 ) = 0

Using the method of undetermined coefficients yields:

(ε t ) : 1 − π0 − βπ0 = 0 ∴ π0 = 11 + β

(ε t−1 ) : Terms in ε t−1 also yield this.

Thus the solutions for pt and yt can be expressed as:

pt = 11 + β ε t +

_m −y∗ − λβπ0ε t−1

(1 − λL)(1 + α(1 − λ))

pt =_m −y∗ + 1

1 + β ε t − λ(1 + α(1 − λ))

(yt−1 − y∗ )

yt = y∗ + β1 + β ε t + λ(yt−1 − y∗ )

2(b) Difference equation method

Take expectations at t-1 to find Et−1pt from (19.11-13). Rearrange by multiplyingthroughout by 1 − λL to get:

(1 − λ)(_m −y∗ ) = −αEt−1pt+1 + αλEt−2pt + (1 + α)Et−1pt

− λ(1 + α)Et−2pt−1 + λπ0ε t−1

Take expectations at t-2:

(1 − λ)(_m −y∗ ) = −αEt−2pt+1 + (αλ + 1 + α)Et−2pt

− λ(1 + α)Et−2pt−1

Therefore

−(1 − λ)α (

_m −y∗ ) = Et−2pt+1 − λ + 1 + α

α Et−2pt

+ λ 1 + αα Et−2pt−1

or

−(1 − λ)α (

_m −y∗ ) = pt+1

e − λ + 1 + αα pt

e + λ 1 + αα pt−1

e

where pt+1e is pt+1 expected at t-2. Note that this is a second-order difference equation.

Thus the characteristic equation can be expressed as:

pt+ie = A1λ i+1 + A2

1 + αα

i+1+

_m −y∗ (i ≥ 1)

where A1 + A2 = pt−1e −

_m −y∗

A1λ + A21 + αα = pt

e −_m −y∗

If we appeal to stability requirement so that A2 = 0, we have

A1 = pt−1e − (

_m −y∗ ) and

pt+ie =

_m −y∗ + (pt−1

e − (_m −y∗ ))λ i+1 (i ≥ 0)

Now note that

Et−1pt+1 =_m −y∗ + λ(Et−1pt − (

_m −y∗ ))

Since

Et−2pt =_m −y∗ + λ(Et−2pt−1 − (

_m −y∗ ))

Substituting for Et−1pt+1 into (19.14) yields:

_m= Et−1pt(1 + α) − α

_m −y∗ + λ(Et−1pt − (

_m −y∗ )) + y∗ + λβπ0ε t−1

1 − λL

or

Et−1pt =_m −y∗ − λβπ0ε t−1

(1 − λL)(1 + α(1 − λ))

Rest of the solution follows from what we had done with Sargent’s forward operatormethod, especially from (19.17) onwards.

2(c) Muth method

This is in practice the easiest for this model.

Let pt =−p +∑ i=0

∞ π iε t−i. Substitute directly in (19.19) after multiplying (19.19) by

1 − λL to get:

(1 − λ)_m +ε t − λε t−1

=−p +∑

i=0

∞

π iε t−i − λ−p +∑

i=1

∞

π i−1ε t−i − α ∑i=1

∞

π i+1ε t−i − ∑i=1

∞

π iε t−i

+ αλ ∑i=2

∞

π iε t−i − ∑i=2

∞

π i−1ε t−i + (1 − λ)y∗ + βπ0ε t

Using the method of undetermined coefficients yields:

Collecting terms in:

(cons tan ts) : (1 − λ)_m= (1 − λ)

_p + (1 − λ)y∗ ∴

_p=

_m −y∗

(ε t ) : 1 = π0 + β π0 ∴ π0 = 11 + β

(ε t−1 ) : −λ = π1 − λπ0 − α(π2 − π1 )

ε t−i, i ≥ 2 : 0 = −απ i+1 + (1 + α + αλ)π i − λ(1 + α)π i−1

or

0 = π i+1 − 1 + αα + λ π i + λ 1 + α

α π i−1

This is a difference equation in π i, initial values (i = 2) . Thus the characteristicequation can be expressed as:

π i = A1λ i−1 + A21 + αα

i−1+

_m −y∗ (i ≥ 3) where

π1 = A1 + A2

π2 = A1λ + A21 + αα

Appeal to stability requirement to set A2 = 0, hence

π1 = A1; π2 = λπ1

Substitute π2 = λπ1 in (ε t−1 ) : −λ = π1 − λπ0 − α(λπ1 − π1 ) to get

π1 =λ −β

1+β

(1 + α(1 − λ))

Thus the solutions for pt and yt is:

pt =_m −y∗ + 1

1 + β ε t −λβ

(1 + β)(1 + α(1 − λ))(ε t−1 + λε t−2 + ......)

pt =_m −y∗ + 1

1 + β ε t − λ(1 + α(1 − λ))

(yt−1 − y∗ )

yt = y∗ + β1 + β ε t + λ(yt−1 − y∗ )

Tutorial 20

Consider the following model in natural logarithms.

_m −pt = α0 + α1yt + α2Rt

yt = y∗ + δ(pt − Et−1pt ) 20.1

20.2

Rt = (pt−1 − pt−2 ) + γ(yt−1 − y∗ ) +

β[(pt−1 − pt−2 ) − 2.5] + 2.5 + ε t 20.3

where γ = β = 0.5 is the weight placed on the deviation of output and inflation ratefrom target. Equation (20.3) is a simplified version of a backward looking Taylor rule (seeTaylor, 1993) which captures the spirit of recent research in macroeconomics.

Rt = Nominal interest rate

pt−1 − pt−2 = inflation rate in t-1

y∗ = potential real GDP

_m = constant money supply growth

ε t = white noise error

The policy rule in equation (20.3) has the feature that the nominal rate rises if inflationincreases above a target of 2.5% (set by the Chancellor for the MPC in 1997) or if realGDP rises above potential GDP. If both the inflation rate and the real GDP are on target,then the nominal rate would equal 5% or 2.5% in real terms.

1. Solve the model using the Muth method. Can authorities stabilise output in thismodel if they use a Taylor rule instead of an explicit money supply rule?

2. Would our conclusions change if we had a New Keynesian Phillips curve of the formyt = y∗ + 0.5q[(pt − Et−1pt ) + (pt − Et−2pt )] instead?

Solution

1) Substituting equations (20.3) and (20.2) in (20.1) yields the reduced form equation:

⇒_m −pt = α0 + α1(y∗ + δ(pt − Et−1pt )) +

α2 (pt−1 − pt−2 ) + γ(yt−1 − y∗ ) + β[(pt−1 − pt−2 ) − 2.5] + 2.5 + ε t

or_m −pt = α0 + α1(y∗ + δ(pt − Et−1pt )) +

α2(pt−1 − pt−2 ) + γ(y∗ + δ(pt−1 − Et−2pt−1 ) − y∗ ) +

β[(pt−1 − pt−2 ) − 2.5] + 2.5 + ε t


pt =−p +∑

i=0

∞

π iε t−i


pt = π0ε t + π1ε t−1 + π2ε t−2 + ....

pt−1 = π0ε t−1 + π1ε t−2 + π2ε t−3 + ....

Et−2pt−1 = π1ε t−2 + π2ε t−3 + ....

Et−1pt = π1ε t−1 + π2ε t−2 + ....

pt−2 = π0ε t−2 + π1ε t−3 + π2ε t−4 + ....

Et−3pt−1 = π2ε t−3 + ....


_m −(π0ε t + π1ε t−1 + π2ε t−2 + ....) = α0 + α1(y∗ + δ(π0ε t )) +

α2 (1 + β) π0ε t−1 + π1ε t−2 + π2ε t−3 + .... −π0ε t−2 + π1ε t−3 +

π2ε t−4 + ....+

α2γδπ0ε t−1 − α2β(2.5) + α2(2.5) + α2ε t


ε t : − π0 = α1δπ0 + α2

∴ π0 = − α2

1 + α1δ

ε t−1 : −π1 = α2(1 + β)π0 + α2γδπ0

∴ π1 = (α2 )2

1 + α1δ(1 + β + γδ)

The identities in the other errors are irrelevant for computing the solution for output.Substituting the solution for π0 in (20.2) yields solution for output:

yt = y∗ + δ(π0ε t ) ≡ y∗ − δ α2

1 + α1δε t

Note that the parameters of the interest rate rule do not enter the solution for output.Therefore authorities cannot stabilise output in this model. However, if one substitutes forπ0, π1,... in the price level equation pt =

−p +∑ i=0

∞ π iε t−i it is clear that β and γ(parameters in the interest rate rule) enter π1 (see above). It is clear that the interest raterule does affect the solution for the price level.

2) Suppose we had a New Keynesian Phillips curve, our reduced form equationbecomes:

_m −pt = α0 + α1(y∗ + 0.5q[(pt − Et−1pt ) + (pt − Et−2pt )]) +

α2(pt−1 − pt−2 ) + γ(0.5q[(pt−1 − Et−2pt−1 ) + (pt−1 − Et−3pt−1 )]) +

β[(pt−1 − pt−2 ) − 2.5] + 2.5 + ε t


_m −(π0ε t + π1ε t−1 + π2ε t−2 + ....) = α0 + α1

y∗ + qπ0ε t +0.5qπ1ε t−1

+

α2 (1 + β) π0ε t−1 + π1ε t−2 + π2ε t−3 + .... −π0ε t−2 + π1ε t−3 +

π2ε t−4 + ....+

α2γqπ0ε t−1 + α2γ0.5qπ1ε t−2 − α2β(2.5) + α2(2.5) + α2ε t


ε t : − π0 = α1qπ0 + α2

∴ π0 = − α2

1 + α1q

ε t−1 : −π1 = α10.5qπ1 + α2(1 + β)π0 + α2γqπ0

∴ π1 =(α2 )2(1 + β + γq)

(1 + α1q)(1 + α10.5q)

Note that we do not need ε t−2,ε t−3 etc for computing the solution for yt. Now substitutethe solution for π0 and π1 in the New Keynesian Phillips curve to get solution for output.

yt = y∗ + qπ0ε t + 0.5qπ1ε t−1

Note that the parameters of the Taylor rule β and γ enter the coefficient π1and hence thesolution for output. Thus, it is clear that short-run non-neutrality depends on what sort ofaggregate supply curve is in place and not on the way monetary policy (moneysupply/interest rate rule) is conducted.

Tutorial 21

Consider a version of Cagan (1956) model of hyperinflation. All symbols areinterpreted as being deviations from equilibrium. γ is the policy parameter describing thedirection and degree of intervention. For example, a value of γ 0 means authorities seekto offset the higher inflationary pressures by engaging in monetary contraction.

mt − pt = −β[Etpt+1 − pt ] + ε t

mt = −γpt

21.1

21.2

1. Comment on the solution path for pt by using the Muth method?

2. What happens when we impose the terminal condition π1 = 0 on the solution pathfor the price level?

Solution

1) Substituting equation (21.2) in (21.1) yields:

⇒ −γpt − pt = −β[Etpt+1 − pt ] + ε t

⇒ −pt(γ + 1 + β) − ε t = −β(Etpt+1 )

β(Etpt+1 ) = (1 + β + γ)pt + ε t


pt =−p +∑

i=0

∞

π iε t−i


pt = π0ε t + π1ε t−1 + π2ε t−2 + ....

pt+1 = π0ε t+1 + π1ε t + π2ε t−1 + ....

Etpt+1 = π1ε t + π2ε t−1 + ....


β(π1ε t + π2ε t−1 + ....) = (1 + β + γ)π0ε t + π1ε t−1 +π2ε t−2 + ....

+ ε t


ε t : βπ1 = (1 + β + γ)π0 + 1

∴ π0 = βπ1 − 1(1 + β + γ)

ε t−1 : βπ2 = (1 + β + γ)π1

∴ π1 = βπ2

(1 + β + γ)

or

π2 =(1 + β + γ)

β π1

ε t−2 : βπ3 = (1 + β + γ)π2

∴ π2 = βπ3

(1 + β + γ)

or

π3 =(1 + β + γ)

β π2

Similarly,

π i+1 =(1 + β + γ)

β π i i = 1,2..........

Therefore, the solution for pt is:

⇒ pt =_p +π0ε t + π1ε t−1 + π2ε t−2 + ....

pt =_p + βπ1 − 1

(1 + β + γ)ε t + π1

ε t−1 +(1+β+γ)

β ε t−2 +

(1+β+γ)β

2ε t−3 + ...

where π1 is undetermined. This model clearly has the saddle path property i.e., there isan infinity of paths/multiplicity of arbitrary solutions exist. One way to resolve this issue isto impose a terminal condition π1 = 0 on the solution path in order to rule out a bubble.

Tutorial 22

Consider the following model (a modified version of Taylor (1977)) where symbolshave their usual meanings.

yt = −β1[Rt − (Et−1pt+1 − Et−1pt )] + β2(mt − pt ) + u1t

mt − pt = yt − α1Rt + α2(mt − pt ) + u2t

yt = γ0 + γ1(mt − pt ) + u3t

mt = m

22.1

22.2

22.3

22.4

1. Determine the solution for pt using the Muth method and comment on the solution?

2. What sort of terminal condition is required to get a stable solution?

Solution

1) Our first objective is to get the reduced form equation. Note that we can get Rt fromequation (22.2).

⇒ −α1Rt = (1 − α2 )(mt − pt ) − yt − u2t

Rt = 1α1

[yt − (1 − α2 )(mt − pt ) + u2t ] 22.5

Substituting (22.5), (22.3) and (22.4) in (22.1) yields the reduced form equation:

⇒ γ0 + γ1(m − pt ) + u3t =

− β1

1α1

(γ0 + γ1(m − pt ) + u3t − (1 − α2 )(m − pt ) + u2t ) −(Et−1pt+1 − Et−1pt )

+

β2(m − pt ) + u1t

Collecting terms in (m − pt ) yields:

⇒ γ0 + γ1 +β1α1

γ1 −β1α1

(1 − α2 ) − β2 (m − pt ) + u3t =

− β1α1

[γ0 + u3t + u2t ] +

β1Et−1pt+1 − β1Et−1pt + u1t

or

Et−1pt+1 = Et−1pt + δ1pt + δ0 + ut

where

δ1 = (1 − α2 )α1

+ β2

β1− γ1

1α1

+ 1β1

δ0 = γ01α1

+ 1β1

− δ1m

ut = − 1β1

u1t + 1α1

u2t + 1α1

+ 1β1

u3t


pt =−p +∑

i=0

∞

π iut−i


pt = π0ut + π1ut−1 + π2ut−2 + ....

Et−1pt = π1ut−1 + π2ut−2 + π3ut−3 + ....

pt+1 = π0ut+1 + π1ut + π2ut−1 + ....

Et−1pt+1 = π2ut−1 + π3ut−2 + ....


⇒ π2ut−1 + π3ut−2 + .... = π1ut−1 + π2ut−2 + π3ut−3 + .... +

δ1(π0ut + π1ut−1 + π2ut−2 + ....) + δ0 + ut

We need to now evaluate π0 and π1 i.e., the undetermined coefficients. For this wecollect terms in ut, ut−1, ut−2......

ut : 0 = δ1π0 + 1

∴ π0 = − 1δ1

ut−1 : π2 = π1 + δ1π1

∴ π1 = π2

(1 + δ1 )

or

π2 = (1 + δ1 )π1

ut−2 : π3 = π2 + δ1π2

∴ π2 = π3

(1 + δ1 )

or

π3 = (1 + δ1 )π2

Similarly,

π i+1 = (1 + δ1 )π i i = 1,2..........

Therefore, the solution for pt is:

⇒ pt =_p +π0ut + π1ut−1 + π2ut−2 + ....

pt = − δ0

δ1− 1

δ1ut + π1 ∑

i=0

∞

(1 + δ1 ) iut−i−1

where π1 remains undetermined.

2) This model therefore has the saddlepath property i.e., there is an infinity of paths allbut one unstable. This immediately raises the question of which solution to choose. Taylorresolves this by proposing the criteria that variance of the price level be minimised. Thisimplies π1 = 0, so that the solution for pt reduces to:

pt = − δ0 + ut

δ1

Here, the terminal condition (π1 = 0) will select the most stable solution by, in effect,ruling out the root with the largest modulus.

Tutorial 23

1. Consider the following model (with future expectations) where symbols have theirusual meanings.

mt = pt + yt − α[Et−1pt+1 − Et−1pt ]

yt = y∗ + β(pt − Et−1pt )

mt =_m +ε t

23.1

23.2

23.3

Determine the solution for pt and yt using the method of forward substitution andcomment on the solution?

2. Consider the following model where symbols have their usual meanings.

mt = pt + yt

yt = y∗ + βpt − 0.5(Et−1pt + Et−2pt) + λ(yt−1 − y∗)

mt =_m +µ(yt−1 − y∗) + ε t

23.11

23.12

23.13

Determine the solution for pt and yt using the Muth method of undeterminedcoefficients?

Solution

1) Substituting (23.3) and (23.2) in (23.1) yields the reduced form equation:

_m +ε t = pt + y∗ + β(pt − Et−1pt ) − α[Et−1pt+1 − Et−1pt ]

In order to solve for Et−1pt using the basic method:


2. Take the expected value of this solution at the date of the expectations, and solve for

the expectations.


Et−1mt = Et−1pt + Et−1yt − α[Et−1pt+1 − Et−1pt ]Et−1yt = y∗ + β(Et−1pt − Et−1pt ) ≡ y∗

Et−1mt =_m +Et−1ε t ≡

_m

23.4

23.5

23.6

Substituting (23.5) and (23.6) in (23.4) yields:

Et−1pt =_m −y∗

1 + α + α1 + α Et−1pt+1 23.7

Note that this is not the solution for Et−1pt because Et−1pt+1 is not solved out; we haveshifted the problem into the future. To solve for Et−1pt+1 we lead the model by one period:

Etpt+1 =_m −y∗

1 + α + α1 + α Etpt+2 23.8

Leading it further by one period yields:

Et+1pt+2 =_m −y∗

1 + α + α1 + α Et+1pt+3 23.9

− − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − −

Taking expectations of (23.8) and (23.9) at time t-1 yields:

Et−1Etpt+1 =_m −y∗

1 + α + α1 + α Et−1Etpt+2

Et−1Et+1pt+2 =_m −y∗

1 + α + α1 + α Et−1Et+1pt+3

23.10

23.11

At this point we can invoke the Law of Iterated Expectations i.e.,

Et−1Etpt+1 = Et−1pt+1

Et−1Et+1pt+2 = Et−1pt+2

Hence, we can express (23.10) as follows:

Et−1pt+1 =_m −y∗

1 + α + α1 + α Et−1pt+2

Substituting successively (forward) for Et−1pt+2, Et−1pt+3 and so on results in:

⇒ Et−1pt+1 =_m −y∗

1 + α + α1 + α

_m −y∗

1 + α + α1 + α Et−1pt+3

⇒ Et−1pt+1 =_m −y∗

1 + α + α1 + α

_m −y∗

1 + α + α1 + α

2

×_m −y∗

1 + α + α1 + α

3Et−1pt+4

or

Et−1pt+1 = 11 + α ∑

i=0

N−1α

1 + αi(_m −y∗ ) + α

1 + αN

Et−1pt+N+1 23.12

As N → ∞ and applying the stability condition to Et−1pt+i,so that Et−1pt+N+1 → 0 as N → ∞.

Thus we can write (23.12) as:

⇒ Et−1pt+1 = 11 + α ∑

i=0

∞α

1 + αi(_m −y∗ )

Et−1pt+1 =_m −y∗

1 + α1 + α

1+α + α1+α

2 +α

1+α3 + .....

(Summation of an infinite series)

Et−1pt+1 =_m −y∗

1 + α1

1 − α1+α

≡_m −y∗

By the same argument but starting from (23.7) we get:

⇒ Et−1pt =_m −y∗

1 + α + α1 + α

_m −y∗

1 + α + α1 + α Et−1pt+2

⇒ Et−1pt =_m −y∗

1 + α + α1 + α

_m −y∗

1 + α + α1 + α

2_m −y∗

1 + α

+ α1 + α

3Et−1pt+3

or

Et−1pt =_m −y∗

1 + α ∑i=0

N−1α

1 + αi+ α

1 + αN

Et−1pt+N+1

∴ Et−1pt =_m −y∗

Note that the paths for events can be unstable. Our model here implies that all paths forprices, except that for which Et−1pt =

_m −y∗, explode monotonically. Prices would be

propelled into either ever-deepening hyperdeflation or ever-accelerating hyperinflation,even though money supply is held rigid. So we need to impose a terminal condition forselecting a unique stable path. In this model we use the method of forward substitution inorder to ensure convergence. Substituting Et−1pt and Et−1pt+1 in the reduced form equationyields solution for the price level:

⇒_m +ε t = pt + y∗ + β(pt − (

_m −y∗ )) − α[(

_m −y∗ ) − (

_m −y∗ )]

or

pt =_m −y∗ + 1

1 + β ε t

Substituting the solution for pt and Et−1pt in the Phillips curve equation (23.2) yieldssolution for output.

yt = y∗ + β1 + β ε t

2) Note that (23.12) can be expressed as

yt − y∗ = β(π0ε t + 0.5π1ε t−1 )1 − λL

The reduced form equation can be expressed as:

_m + µβ(π0ε t−1 + 0.5π1ε t−2 )

1 − λL+ ε t = ∑

i=0

∞

π iε t−i +_p +y∗ + β(π0ε t + 0.5π1ε t−1 )

1 − λL

Multiplying throughout by 1 − λL yields:

(1 − λ)(_m −y∗ ) + µβ(π0ε t−1 + 0.5π1ε t−2 ) + ε t − λε t−1

= (1 − λ)_p +∑

i=0

∞

π iε t−i − λ∑i=1

∞

π i−1ε t−i + β(π0ε t + 0.5π1ε t−1 )

Collecting terms in:

(cons tan ts) : (1 − λ)(_m −y∗ ) = (1 − λ)

_p ∴

_p=

_m −y∗

(ε t ) : 1 = π0 + β π0 ∴ π0 = 11 + β

(ε t−1 ) : µβπ0 − λ = π1 − λπ0 + β(0.5π1 )

∴ π1 = µβ + λ(1 − 1 + β)(1 + 0.5β)(1 + β)

(ε t−2 ) : 0.5µβπ1 = π2 − λπ1 ∴ π2 = (0.5µβ + λ)π1

ε t−i, i ≥ 3 : 0 = π i − λπ i−1

Therefore the solution for pt and yt can be expressed as:

pt =_m −y∗ + 1

1 + β ε t +µβ + λ(1 − 1 + β)(1 + 0.5β)(1 + β)

ε t−1 +(0.5µβ + λ)π1ε t−2

1 − λL

yt = y∗ +β

1+β ε t + 0.5β µβ+λ(1−1+β)(1+0.5β)(1+β)

ε t−1

1 − λL

Tutorial 24

Lucas Critique of Econometric policy evaluation:

Consider the following model where symbols have their usual meaning.

yt = β[pt − 0.5(Et−1pt + Et−2pt )]

mt = pt + yt

mt = µyt−1 + ε t

24.1

24.2

24.3

1. What is the solution for yt in terms of current and past ε t?

2. What is optimal µ?

3. What is the solution for yt in terms of current and lagged mt and yt?

4. If this was treated as the basis for calculating optimal mt, what would this be?

5. If this money supply (mt ) rule was implemented, what would then be the solution foroutput in terms of current and lagged ε t?

Solution

1) Substituting equations (24.1) and (24.3) in (24.2) yields the reduced form equation:

µβ(pt−1 − 0.5(Et−2pt−1 + Et−3pt−1 )) + ε t = pt + βpt − 0.5

(Et−1pt + Et−2pt ) 24.4


pt =−p +∑

i=0

∞

π iε t−i


pt = π0ε t + π1ε t−1 + π2ε t−2 + ....

Et−1pt = π1ε t−1 + π2ε t−2 + ....

Et−2pt = π2ε t−2 + π3ε t−3 + ....

pt−1 = π0ε t−1 + π1ε t−2 + π2ε t−3 + ....

Et−2pt−1 = π1ε t−2 + π2ε t−3 + ....

Et−3pt−1 = π2ε t−3 + π3ε t−4 + ....

Plugging in these values in the reduced form equation and collecting terms in ε t, ε t−1,

ε t−2......

ε t : 1 = π0 + βπ0

∴ π0 = 11 + β

ε t−1 : µβπ0 = π1 + β[π1 − 0.5π1 ]

∴ π1 = µβπ0

1 + 0.5β ≡ µβ(1 + β)(1 + 0.5β)

ε t−2 : µβ(π1 − 0.5π1 ) = π2 + β[π2 − 0.5(π2 + π2 )]

∴ π2 = 12µβπ1

π3 = π4 = π5 = 0

The identities in the other errors are irrelevant for computing the solution for output.Substituting the solution for π0 and π1 in the New Keynesian Phillips curve yields solutionfor output in terms of current and past ε t:

⇒ yt = β[pt − 0.5(Et−1pt + Et−2pt )] ≡ βπ0ε t + 0.5βπ1ε t−1

yt = β 11 + β ε t + 0.5β µβ

(1 + β)(1 + 0.5β)ε t−1 24.5

2) In order to compute the optimal value of µ take variance of yt:

Var(yt ) = E(yt − Et−1yt )2 = Eβ 1

1+β ε t + 0.5β

× µβ(1+β)(1+0.5β)

ε t−1

2

where the cross products (covariance matrix) are zero as we assume no serialcorrelation and no heteroscedasticity. Taking the expectations operator inside yields:

Var(yt ) = σ2 β1 + β

2

+ σ2 0.5µβ2

(1 + β)(1 + 0.5β)

2

If we set the value of µ = 0 (optimal policy response) then the variance of outputreduces to:

Var(yt ) = σ2 β1 + β

2

= β2σ2π0

3) Note that from equation (24.5) we know that:

yt =β

1 + β ε t +0.5µβ2

(1 + β)(1 + 0.5β)ε t−1

Similarly from the money supply rule equation (24.3) we know that:

ε t = mt − µyt−1

ε t−1 = mt−1 − µyt−2

24.6

24.7

Substituting for ε t and ε t−1 in equation (24.5) yields:

yt =β

1 + β(mt − µyt−1 ) +

0.5µβ2

(1 + β)(1 + 0.5β)(mt−1 − µyt−2 )

Thus the solution for yt in terms of current and lagged mt and yt is:

yt = σ1mt + σ2mt−1 + σ3yt−1 + σ4yt−2 24.8

where

σ1 = β1 + β

σ2 = 0.5µβ2

(1 + β)(1 + 0.5β)

σ3 = −µβ1 + β = −µσ1

σ4 = −0.5µ2β2

(1 + β)(1 + 0.5β)= −µσ2

4) From equation (24.8) we can calculate optimal money supply rule by setting yt = 0:

mt =−(σ2mt−1 + σ3yt−1 + σ4yt−2 )

σ1 24.9

According to our relationship in equation (24.5), this feedback rule for money supplyought to deliver zero fluctuations in output. Thus our expression for money supply from(24.9) after substituting for σi

s give:

mt =−0.5µβ

(1 + 0.5β)mt−1 + µyt−1 +

0.5µ2β(1 + 0.5β)

yt−2 + ε t

Using the lag operartor L we can simplify the above expression as follows:

mt = µyt−1 + ε t

1 + 0.5µβ1+0.5β L

24.10

This money supply rule ought to deliver zero fluctuations in output.

5) Substituting equations (24.1) and (24.10) in (24.2) yields the reduced form solutionfor output (with a new money supply rule) in terms of current and lagged errors. Thus thereduced form equation can be written as:

µβ(pt−1 − 0.5(Et−2pt−1 + Et−3pt−1 )) +

µβapt−2 −

0.5(Et−3pt−2 + Et−4pt−2 )+ ε t =

pt + β[pt − 0.5(Et−1pt + Et−2pt )] + apt−1 +

aβpt−1

−0.5(Et−2pt−1 + Et−3pt−1 )

where

a = 0.5µβ1 + 0.5β


pt =−p +∑

i=0

∞

π iε t−i


pt = π0ε t + π1ε t−1 + π2ε t−2 + ....

Et−1pt = π1ε t−1 + π2ε t−2 + ....

Et−2pt = π2ε t−2 + π3ε t−3 + ....

pt−1 = π0ε t−1 + π1ε t−2 + π2ε t−3 + ....

Et−2pt−1 = π1ε t−2 + π2ε t−3 + ....

Et−3pt−1 = π2ε t−3 + π3ε t−4 + ....

pt−2 = π0ε t−2 + π1ε t−3 + π2ε t−4 + ....

Et−3pt−2 = π1ε t−3 + π2ε t−4 + ....

Et−4pt−2 = π2ε t−4 + π3ε t−5 + ....

Plugging in these values in the reduced form equation and collecting terms in ε t, ε t−1,

ε t−2......

ε t : 1 = π0 + βπ0

∴ π0 = 11 + β

ε t−1 : µβπ0 = π1 + 0.5βπ1 + aπ0 + aβπ0

∴ π1 = µβ − a − aβ(1 + β)(1 + 0.5β)

The identities in the other errors are irrelevant for computing the solution for output.

Substituting the solution for a, π0 and π1 in the New Keynesian Phillips curve yieldssolution for output in terms of current and past ε t:

yt = β (1 + β)−1ε t + 0.25µβ (1 + β)(1 + 0.5β)2 −1ε t−1

Note that the feedback rule for money supply mt = µyt−1 + εt

1+ 0.5µβ1+0.5β L

ought to have

delivered zero fluctuations in output. However, it has delivered fluctuations nearly as largeas the feedback rule. It was Lucas (1976) who first pointed out that if expectations areformed rationally, then unless the estimated model equations are genuinely structural, theimplications drawn from such models may be seriously flawed. Where we went wrong wasthat we used the reduced form to calculate the optimal rule. But that reduced form itselfdepends on the monetary rule because the rule influences (rational) expectations and hencebehaviour of agents.

3. rational expectations solution technique - patrick · pdf filesolve the model, treating...

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