GR5215 2019
Ron Miller
Assignment 1; due September 24 in class
1) This problem asks you to analyze a typical Principles of Economics supplydemand system, but using the tools we used in class. The market for hot coffee is
perfectly competitive. Quantity demanded is given by: 𝐷(𝑃, 𝑃% ,T), where P is the
price of coffee, PC is the price of tea, which is a substitute in consumption for coffee
and T is the outside temperature. Assume that people want to consume fewer hot
drinks when temperatures are high. Quantity supplied is given by 𝑆(𝑃, 𝑃' ), where PI
is the price of labor (the wage), which is an input to coffee production. Equilibrium
is given by Q=S=D.
a) What are the endogenous variables in this model? What are the exogenous
variables?
b) What are the signs of the partial derivatives of S and D? For each, provide an
explanation.
c) Linearize the system and express it in matrix form.
d) Find the partial derivatives of each of the endogenous variables with respect to
each of the exogenous variables and evaluate their signs. (Hint: you can do this by
inverting the matrix, but I recommend doing it by substitution, just as you would
solve a two-equation linear system with high school mathematics.)
2) Consider the four-equation, fixed-wage model used in class.
a) Linearize the system and express it in matrix form.
b) Provide algebraic expressions for the partial derivatives of Y and P with respect
to M and G and evaluate their signs.
c) We did not discuss in class what might happen if the (fixed) wage increases.
Provide algebraic expressions for the partial derivatives of each of the endogenous
* . Evaluate their signs and provide economic intuition for
variables with respect to 𝑊
your results. Feel free to use an undergraduate macro textbook to help you on the
interpretations. (Hint: again, repeated substitution is the way to go here.)
3) Inflationary Finance
Consider a static Keynesian model with only the IS and LM relations:
(1)
y = c(y - t) + i(R- πe )+ g
(IS)
where t is real tax receipts, R is the nominal interest rate and πe is the expected
inflation rate, and all other symbols have their usual meanings and functions have
their usual slopes.
(2)
M/P = L(R,y)
(LM)
Why would there be a real I interest rate in the IS relation and a nominal interest
rate in the LM relation? Feel free to consult and intermediate macro textbook.
Consider two possible funding regimes for an increase in g without increases in t.
Here, only R and y are endogenous.
1
GR5215 2019
Ron Miller
Regime I: Bond Financing
(3a) πe = 0
a) In this regime, the government issues bonds to cover the increase in government
spending. That increase does not show up anywhere in this model (which was
always a criticism of the IS/LM framework!). Under Regime I, evaluate the effect of a
fiscal expansion (increase in g) on output by linearizing the model and calculating
the relevant partial derivative.
Regime II: Inflation Financing
(3b) g - t = πe L(R,y)
i.e. the expected inflation rate is equal to the rate that would be required to fund the
deficit by printing money. Note that we do not assume that the money supply
actually changes when there is fiscal expansion, simply that agents expect it to
expand in the future.
b) Repeat the analysis of part (a) under the inflationary finance regime and compare
the answers. Briefly give economic intuition for the results. Do you think these
conclusions are economically reasonable? Why or why not? (Hint: the expected rate
of inflation becomes an endogenous variable under this regime.)
2
Eddie Shore & Joe Saia
PS 1 Solutions
Let us know in office hours if you think you found any mistakes or typos. When solving a system of
equations we mixed and matched matrix inversion and repeated substitution to try and give a feeling for
both. On the exam its fine to do either or as long as you show your work. Writing out the equations you
are using and the signs of any partial derivatives in functions that are given, and then showing each step
of your work is best practice and helps show that you do know the core ideas, even if you make an algebra
mistake along the way.
Updates
1. The signs for the partials at the end of Q1 were orginally flipped for a few of them.
1
First we’ll outline the equations
S=D
(1)
Q=S
(2)
D = D(P, Pc , T )
(3)
S = S(P, Pl )
(4)
Equations (1) and (2) are the equilibrium conditions for this model while (3) and (4) are just functional
definitions. We could have also written (2) as Q = D, as we have the condtion D = S, but once we have (1)
and (2) we get Q = D for “free” and including it in the model does not actuall impose any more restrictions
on the equilibrium values in the model.
a)
The endogenous variables are as usual P and Q while the exogenous variables are Pc , T and Pl .
b)
The signs of the partial derivatives are as follows:
D1 (P, Pc , T ) < 0
Demand curves slope down
D2 (P, Pc , T ) > 0
When tea is more expensive, people substitute into coffee
D3 (P, Pc , T ) < 0
When it’s hot, people drink less coffee
S1 (P, Pl ) > 0
Supply curves slope up
S2 (P, Pl ) < 0
When input costs go up, producers reduce production
c)
For this section I will write D(P, Ps , T ) as simply D and S(P, PM ) as S. I will denote partial derivatives
as in the following form D1 = D1 (P, Ps , T ). I’ll linearize the system by taking the total derivates of each
equation.
S1 dP + S2 dPl = D1 dP + D2 dPc + D3 dT
dQ = S1 dP + S2 dPl
3
(5)
(6)
Eddie Shore & Joe Saia
PS 1 Solutions
Rearranging equations and converting to matrix notation we get:
S1 − D1
−S1
0 dP
−S2
=
1 dQ
S2
D2
0


dPl
D3 
dPc 
0
dT
(7)
d)
To get the partial derivatives of the endogenous variables with respect to the exogenous variables we simply
use the implicit function theorem using the matrix equation in the previous section. Note that S1 − D1 > 0.


dPl
D3 
dPc 
0
dT


dPl
dP
1
0
−S2 D2 D3 
∝
dPc 
dQ
S1 S1 − D1
S2
0
0
dT


dPl
dP
−S2
D2
D3 
∝
dPc 
dQ
−S2 D1 S1 D2 S1 D3
dT


dPl
dP
+ + − 
∝
dPc 
dQ
− + −
dT
S − D1
dP
= 1
−S1
dQ
0
1
−1 −S2
S2
D2
0
The partial derivatives can then be read off of each element of the matrix. The calculus gives us the
usual results that things that increase the quantity demanded increase both quantity and price while things
that decrease the quantity supplied reduce quantity but increase price.
Alternative Solution
Instead of inverting the matrix of coefficients one could use algebraic substitution, we’ll do so here.
S1 dP + S2 dPl = D1 dP + D2 dPc + D3 dT
(8)
dQ = S1 dP + S2 dPl
(9)
(S1 − D1 )dP = D2 dPc + D3 dT − S2 dPl
D2 dPc + D3 dT − S2 dPl
dP =
(S1 − D1 )
(10)
Rearrangging the first equation:
(11)
Plugging this solution into 9
dQ = S1
D2 dPc + D3 dT − S2 dPl
+ S2 dPl
(S1 − D1 )
(12)
At this point we have dP and dQ as functions of the exogenous differentials. By looking at the individual
partial derivates we know that S1 − D1 > 0. This gives us the same results as doing the proper matrix
4
Eddie Shore & Joe Saia
PS 1 Solutions
inversion. Mathematically we have implicitly done this matrix inversion which is why we got the same
answer, up to a constant which comes from the fact that when we did the matrix inversion we did not carry
through the deteriment. At this point we could write the solution for dP and dQ in matrix notation and
carry on as before. Either method is fine, but usually the substitution method is easier.
2
The four-equation, fixed-wage model from class consists of the following four equations:
Y = C(Y − T ) + I(Y, i) + G
M = P Y L(i)
(IS)
(LM)
Y = F (K, N )
(Production Function)
W = P FN (K, N )
(Labor Demand)
The variables of the model are (Y, T, i, G, P, M, K, N, W ). Slide 4 helpfully tells us which variables in the
classical model are exogenous and endogenous (slide 3 notes that K is exogenous as well). However, we now
assume (slide 7) that W is also fixed. Thus, the list of endogenous (ỹ) and exogenous (x) variables are:
 
T
 
Y
G
 
i
 

ỹ = 
x
=
M 
P 
 
K 
N
W
a)
You will now notice that whether you follow the method of total differentiation or just apply the implicit
function directly, you will have to invert a 4 × 4 matrix—if you have done this, then great. In these
solutions, we will follow recitation note 2 and reduce the number of endogenous variables. First, note that
(Production Function) allows us to replace Y everywhere else. Furthermore, (LM) allows us to replace P
everywhere else (with P = Y M
L(i) ). Plugging these two equations into (IS) and (Labor Demand) gives:
F (K, N ) = C(F (K, N ) − T ) + I(F (K, N ), i) + G
W =
M FN (K, N )
F (K, N )L(i)
(13)
(14)
Our new endogenous variables (y) are now:
N
y=
.
i
Let’s rearrange these slightly and write them as a function:
F (K, N ) − C(F (K, N ) − T ) − I(F (K, N ), i) − G
E(y, x) =
W F (K, N )L(i) − M FN (K, N )
Before we proceed, we define a few notational shortcuts, for the sake of space:
F = F (K, N )
I1 = I1 (F (K, N ), i)
FN = FN (K, N )
I2 = I2 (F (K, N ), i)
FK = FK (K, N ) FN N = FN N (K, N ) FN K = FN K (K, N )
L = L(i)
L1 = L1 (i)
where the subscripts were discussed extensively in recitation notes 2.
5
C1 = C1 (F (K, N ) − T )
Eddie Shore & Joe Saia
PS 1 Solutions
Now, let’s calculate Ey :


Ey = 
∂ E1
∂N
∂ E2
∂N
∂ E1
∂i
∂ E2
∂i



=
=⇒ Ey−1 =
FN (1 − C1 − I1 )
W LFN − M FN N
−I2
W F L1
1
W F L1
·
det(Ey ) M FN N − W LFN
I2
FN (1 − C1 − I1 )
where
det(Ey ) = FN (1 − C1 − I1 )W F L1 + I2 W LFN − M FN N .
Next, we need to calculate Ex :


Ex = 
=
∂ E1
∂T
∂ E1
∂G
∂ E1
∂M
∂ E1
∂K
∂ E1
∂W
2
2
2
2
2
∂E
∂T
C1
0
∂E
∂G
∂E
∂M
−1
0
0 −FN
∂E
∂K
∂E
∂W



FK (1 − C1 − I1 )
W FK L − M FN K
0
FL
Thus, the implicit function theorem gives us that
 dN
dN
dG
dN
dM
dN
dK
dN
dW
di
di
dT
dG
−Ey−1 Ex
di
dM
di
dK
di
dW
y0 (x) ≡ 
=
dT


1
W F L1
I2
C1 −1
0
FK (1 − C1 − I1 )
·
0
0 −FN W FK L − M FN K
det(Ey ) M FN N − W LFN FN (1 − C1 − I1 )

T
W F L1 C1
C1 (M FN N − W LFN )






−W F L1
−(M FN N − W LFN )








2


1
−I2 FN
−FN (1 − C1 − I1 )


·
=−

det(Ey ) 

W F L F (1 − C − I ) F (1 − C − I )(M F

1 K
1
1
K
1
1
N N − W LFN ) 



 +I2 (W FK L − M FN K ) +FN (1 − C1 − I1 )(W FK L − M FN K )




I2 F L
F LFN (1 − C1 − I1 )

T
d N/d T
d i/d T
 d N/d G d i/d G 




= d N/d M d i/d M 
d N

 /d K d i/d K 
d N/d W
d i/d W
=−
where AT means the transpose of matrix A.
6
0
FL
(15)
Eddie Shore & Joe Saia
PS 1 Solutions
b)
dY dY dP
dP
Next, we set out to to find dM
, dG , dM , and dG
. (Recitation note 2 discusses how to do this.) Let’s start
with Y . Let’s take the total derivative of (Production Function):
dY = FN dN + FK dK
• For
dY
dM ,
(16)
divide (16) by dM :
dY
dN
dK
= FN
+ FK
.
dM
dM
dM
Exogenous variables have no affect on one another, so
dK
dM
= 0. We can get
dN
dM
from equation (15):
dN
dY
= FN
dM
dM
FN
=−
(−I2 FN )
det(Ey )
=−
FN
(−I2 FN ).
FN (1 − C1 − I1 )W F L1 + I2 W LFN − M FN N
It will prove helpful to first sign the determinent of Ey :


W |{z}
W |{z}
det(Ey ) = FN (1 − C1 − I1 ) |{z}
F L1 + I2 |{z}
L FN − |{z}
M FN N  < 0.
|{z} |
|{z} |{z}
|{z}
| {z }
{z
}
+
+
+
+
−
−
+
+
+
+
−
So, combining this with the fact that FN > 0 and I2 < 0, we have
dY
>0
dM
• For
dY
dG ,
divide (16) by dG:
dN
dK
dY
= FN
+ FK
dG
dG
dG
FN
(−W F L1 )
FN (1 − C1 − I1 )W F L1 + I2 W LFN − M FN N | {z }
|
{z
}
+
=−
+
>0
Next, moving to P , totally differentiate (LM):
dM = Y LdP + P LdY + P Y L1 di
1
P
P L1
dM − dY −
di
=⇒ dP =
YL
Y
L
7
(17)
Eddie Shore & Joe Saia
• For
dP
dM ,
PS 1 Solutions
divide equation (17) by dM (remember: Y = F , and W = P FN ):
dP
1
P dY
P L1 di
=
−
−
dM
Y L Y dM
L dM
1
P
FN
P L1
−FN2 (1 − C1 − I1 )
=
−
−
(−I2 FN ) −
−
YL Y
det(Ey )
L
det(Ey )
P FN2
I2
L1 (1 − C1 − I1 )
1
−
+
=
Y L det(Ey ) Y
L
1
P FN2
=
1−
[I2 L + Y L1 (1 − C1 − I1 )]
YL
det(Ey )
1
I2 LP FN2 + Y L1 (1 − C1 − I1 )P FN2
=
1−
YL
FN (1 − C1 − I1 )W Y L1 + I2 (W LFN − M FN N )
1
Y L1 (1 − C1 − I1 )P FN2 + I2 LP FN2
=
1−
YL
Y L1 (1 − C1 − I1 )P FN2 + I2 (P LFN2 − M FN N )









φ
1 


1−
=

YL
φ − I2 M FN N 
| {z } 


>0


|
{z
}
Between (0,1)
>0
where we defined φ ≡ Y L1 (1 − C1 − I1 )P FN2 + I2 LP FN2 < 0 so we could see what was going on with
the fraction.
• For
dP
dG ,
divide equation (17) by dG:
dP
P dY
P L1 di
=−
−
dG
Y dG
L dG
P FN W F L1
P L1 M FN N − W LFN
=−
−
Y
det(Ey )
L
det(Ey )
|{z}
{z
} | {z } |
{z
}
|
+
−
+
+
P P Y L1
P L1 M FN N − P L
=−
−
Y det(Ey )
L
det(Ey )
−P L1
M FN N − P L
P+
=
det(Ey )
L
−P L1 LP + M FN N − P L
=
det(Ey )
L
−P L1 M FN N
=
det(Ey )
L
| {z } | {z }
−
bummer
−
> 0.
c)
Next, we want to find and such the partial derivatives and signs of the endogenous variables—N, i, Y and
P —with respect to W .
8
Eddie Shore & Joe Saia
• For
dN
dW
PS 1 Solutions
, we can read directly from (15):
dN
−1
I2 F L
=
det(Ey ) |{z} |{z} |{z}
dW
| {z } − + +
+
<0
• For
di
dW
, we can read directly from (15):
di
−1
F |{z}
L FN (1 − C1 − I1 )
=
|{z}
|{z} |
{z
}
det(E
dW
y)
| {z } + + +
+
+
>0
• For
dY
dW
, divide (16) by dW :
dY
dN
dK
= FN
+ FK
dW
dW
dW
−1
= FN
I2 F L
det(Ey )
<0
• For
dP
dW
, divide (17) by dW :
dP
1 dM
P dY
P L1 di
=
−
−
dW
Y L dW
Y dW
L dW
P dY
P L1 di
=−
−
Y dW
L dW
P
P L1
−1
−1
=−
I2 F L −
F LFN (1 − C1 − I1 )
FN
Y
det(Ey )
L
det(Ey )
|{z}
|
{z
} | {z } |
{z
}
+
−
−
+
>0
Since the real wage rises, firms want to hire fewer workers so N decreases. Since N decreases, production Y
falls. Then since Y falls the only way that lower output can be achieved in the IS equation is if i increases to
generate lower investment. Since both output falls and the interest rate increases, the public demand fewer
real money balances so prices must rise to equate the money market.
3
Writting out our equations we have:
Y = C(Y − T ) + I(R − π e ) + G
M
= L(R, Y )
P
The signs of our partial derivatives for our defined functions are:
C1 > 0
I1 < 0
L1 < 0
L2 > 0
9
(18)
(19)
Eddie Shore & Joe Saia
PS 1 Solutions
a)
As stated in the question, we’re assuming that π e = 0 and that no matter what and the government can
increase G by issuing bonds and that this moves no other exogenous variables. Linearizing the model we
have:
dY = C1 dY + I1 dR − C1 dT + dG
dM −
(20)
M
dP = L2 dY + L1 dR
P2
(21)
Rearranging we get
(1 − C1 )dY − I1 dR = −C1 dT + dG
M
L2 dY + L1 dR = dM − 2 dP
P
(22)
(23)
dT

1 0
0 
dG


M 
0 1 −P2
dM 
dP

1 − C1
L2
−I1
L1
dY
−C1
=
dR
0

(24)

dT

0
0 
 dG 
M 
dM 
1 −P2
dP


dT

0
0 
 dG 
M 
dM 
1 −P2
dP


dT
 dG 
−I1 PM2


M
− P 2 (1 − C1 ) dM 
dP


dT
 dG 
I1 PM2


M
dM 
2 (1 − C1 )

dY
1 − C1
=
dR
L2
−1 −I1
−C1
L1
0
1
0
dY
L1
∝−
dR
−L2
I1
1 − C1
−C1
0
1
0
dY
−C1 L1
∝−
L2 C1
dR
dY
C 1 L1
∝
dR
−L2 C1
L1
−L2
−L1
L2
I1
1 − C1
−I1
−(1 − C1 )
(25)
(26)
(27)
(28)
P
dP
dT

+ − 
dG


− + dM 
dP

dY
− +
∝
dR
− +

(29)
The negative sign in 26 comes from the fact that the determinant is negative. From the last line we can
read off the signs of all the partial derivatives.
b)
Now we are adding the assumption that G − T = π e L(R, Y ). This makes the expected inflation rate an
endgenous variable. Modifying the analysis above:
10
Eddie Shore & Joe Saia
PS 1 Solutions
dY = C1 dY + I1 dR − I1 dπ e − C1 dT + dG
(30)
M
dM − 2 dP = L2 dY + L1 dR
P
dπ e + L2 dY + L1 dR = dG − dT
(31)
(32)
Rearranging we get
(1 − C1 )dY − I1 dR + I1 dπ e = −C1 dT + dG
M
L2 dY + L1 dR = dM − 2 dP
P
dπ e + L2 dY + L1 dR = dG − dT
(33)
(34)
(35)
Now we note that with some substitution we can solve for the change in expected inflation
dπ e = dG − dT − dM +
M
dP
P2
(36)
Using this result we can get
(1 − C1 )dY − I1 dR + I1 (dG − dT − dM +
M
dP ) = −C1 dT + dG
P2
(1 − C1 )dY − I1 dR = (I1 − C1 )dT + (1 − I1 )dG + I1 dM − I1
L2 dY + L1 dR = dM −
M
dP
P2
M
dP
P2
(37)
(38)
(39)

dT
 dG 


dM 
dP

1 − C1
L2
−I1 dY
I − C1
= 1
0
L1
dR
1 − I1
0
I1
1
M
−I1 P 2
− PM2
(40)

dT

−I1 P 2 
 dG 
M

−P2
dM 
dP


dT

−I1 PM2 
 dG 
M

−P2
dM 
dP

dY
1 − C1
=
dR
L2
dY
L1
∝−
dR
−L2
−1 −I1
I1 − C 1
0
L1
1 − I1
0
I1
1
I1 − C1
0
1 − I1
0
I1
1
I1
1 − C1
M
(41)
(42)


dT
 dG 
dY
L1 (I1 − C1 )
L1 (1 − I1 )
I1 (L1 − 1)
−I1 PM2 (1 + L1 )


∝−
M
−L2 (I1 − C1 ) −L2 (1 − I1 ) −L2 I1 + (1 − C1 ) − P 2 (L2 I1 − (1 − C1 )) dM 
dR
dP
(43)


dT

dY
− + + X 
dG


∝
(44)

dR
− + − − dM 
dP
Going through each exogenous variables, allowing only one to change at a time:
11
Eddie Shore & Joe Saia
PS 1 Solutions
• T: An increase in T causes expected inflation to go down, which lowers the nominal interest rate. This
increases desired investment raising output. The LM equation tells us that if output increases then
the real interest rate increases as well. This means the increase in investment will increase the real
interest rate, holding everything else constant, attenuating the effect from the decrease in expected
inflation. The increase in T also lowers consumption which lowers Y. From the math we know that
contractionary effects win out and Y and R decrease.
• G: The same logic as T applies, except in reverse.
• M: An increase in M leads to a decrease in expected inflation. This decrease in expected inflation
lowers the real interest rate holding the nominal rate constant, which leads to an increase in output.
The nominal interest rate also decreases due to increase in the real money supply.
• P: An increase in P leads to a decrease in the real money supply. In order for real money demand to
adjust, we need for the real interest rate to increase and output to decrease. The increase in P also
increases expected inflation which lowers the nominal interest rate. These two offsetting affects mean
that the nominal interest rate may increase or decrease which means that investment may increase or
decrease in the IS equation. This means that the response of output is ambiguous but we know that
the nominal interest rate decreases.
12