Structure of JF Maths 2005

SummerPaper

Stats (50%) & Maths (50%)

JF Maths:

Section B

Answer Three out of Four Long Questions

*N.B. Unlike previous years there are no short questions.
Section B

Three out of Four Long Questions

1. Matrices

Course Manual: 2.5 and 12.1

Jacques (edition 3): Chapter 7.1- 7.2

2. Unconstrained Optimisation

Course Manual8.6-9.4

Jacques (edition 3): Chapter5.4

3. Constrained Optimisation

Course Manual10.1 – 10.2

Jacques (edition 3):Chapter5.5 & 5.6

4. Integration

Course Manual7.1-7.4

Jacques (edition 3): Chapter6.1-6.2

1. Matrices

Method 1

Solve the above equations directly, substituting expression for C in eq. (2) into eq. (1)

Thus,Y = a + bY+I+G

Solve for Y as:

Y – bY = a + I + G

Y(1 – b) = a + I + G

Thus,

Substitute this value for Y into eq. (2) and solve for C:

Method 2

Now solve the same problem using matrix algebra:

  • Rewrite (1) and (2) with endogenous variables, C and Y, on left hand side

From eq. 1: Y - C = I + G

From eq. 2:-bY + C = a

  • Now write this in matrix notation:

orA.X= B

  • We can solve for the endogenous variables X, by calculating the inverse of the A matrix and multiplying by B:

Since AX=B  X=A-1B

  • To invert the 2 X 2 A matrix, recall the steps from earlier in the lecture

If A = , then A –1 =

  • In this case, where

the determinant of A is :

|A| = 1.1 – [– 1.– b] = 1 – b

Cofactor Matrix:

Transpose Cofactor Matrix:

The inverse is:

  • so X=A-1B

where and

Thus, multiplying A-1B gives,

These are the solutions for the endogenous variables, C and Y, just as we derived using method 1.

Method 3: Using Cramers Rule

In the example above, where

  • Replace column 1 of A with the elements of vector B
  • Calculate the determinant of this as:
    |A1| = (I + G )(1) – ( –1)( a) = I + G + a
  • We saw earlier that the determinant of A is

| A | = 1– b

  • Therefore the solution using Cramers rule is:
  • Replace column 2 of A with the elements of vector b
  • Calculate the determinant of this as:
    |A2|=(1)(a) – (I+G)(– b) = a+b(I+G)
  • We saw earlier that the determinant of A is

| A | = 1– b

  • Solution using Cramers rule is:

2. Unconstrained Optimisation


Condition / Y = f(X) / Y = f(X,Z)
NeccesarySo required that……. / dY = 0
fX = 0 / dY = 0
fX = 0 AND
fZ = 0
Sufficient
For Min
So required that ….. / d2Y > 0
fXX >0 / d2Y > 0
fXX > 0 AND
fXX fZZ – fXZ fZX >0
Sufficient
For Max
So required that ….. / d2Y < 0
fXX < 0 / d2Y < 0
fXX < 0 AND fXX fZZ – fXZ fZX >0

Supply and Demand Equations of a good are given, respectively, as

P- t= 8 + QS

P = 80 – 3QD

A tax t per unit, imposed on suppliers, is being considered. At what value of t does the government maximise tax revenue in market equilibrium?

Set Supply equal to Demand

In equilibrium, QD = QS

Q + 8 + t = 80 – 3Q

Solve for Q

Qe = 18 – ¼ t

Tax Rev. T = t.Qe = t(18 – ¼ t)

MAX T(t) = 18t – ¼ t2

t*

  • First Order Condition for max:

dT/dt = 18 – ½ t = 0

t* = 36

  • Second Order Condition for max:

d2T/dt2 = -½ < 0 (Max)

Results

t* = 36

Qe = 18 – ¼ t*= 9

T = t*.Qe = 18t* – ¼ t*2 = 324

Pe = Qe + 8 + t*= 53

If t = 0, then

Qe = 18

Pe = Qe + 8 = 26

Is the full burden of the tax passed on to consumers?

Ex-ante (no tax) Pe = 26

Ex-post (t* =36) Pe = 53

Yet the tax is t*= 36 but the price increase is only 27 (75% paid by consumer).

Theory question on what determines the burden of the tax.

3. Constrained Optimisation

Max U = f (X1 X2) =

[ X1* X2* ]

s.t.

g (X1 X2)= P1X1 + P2X2- M = 0

Step 1: Define the Lagrangean Function L (objective function +  constraint)

Max L = f(X1 X2 ) + g(X1 X2)

[X1* X2* *]

Max L = X12X2 + (M – P1X1– P2X2 )

[X1* X2* *]

OR L = X21X2 – (P1X1 + P2X2 –M)

Step 2: First Order Conditions

Set dL= L1.dX1 + L2.dX2 = 0

1. L1 = 2X1X2 –  P1 = 0 eq1

2. L2 = X12 –  P2 = 0 eq2

Set g(X1X2) = 0

3. L = M – P1X1 – P2X2 = 0 eq3

Step 3: Solve the system of equations

Solving equations 1 & 2:

 2X1P2X2 = P1X12

 2P2X2 = P1X1

And substituting into eq 3

P1X1 + P2X2 – M= 0

2P2X2+ P2X2 – M= 0

from eq 3:

Substituting in for X2:

(again, note that rearranging reveals that P1X! = 2/3 M and P2X2 = 1/3 M . 2/3 of income spent on good 1, and 1/3 on good2)

Step 4: Second Order Condition

d2L = L11.dX21 + L12.dX1dX2 + L21.dX2dX1 + L22 dX22

s.t. g1.dX1 + g2.dX2= 0

or dX2 = -(g1/ g2).dX1

N. D. for a Max

d2L=[L11. g22 - 2L12.g1.g2+ L22 g21]dX21/ g22

d2L = Φ dX21/g22 , if Φ < 0, N. D.

= 0 + P1(P2. 2X1) -P2(-P1. 2X1 + P2. 2X2 )

= 4P1P2 X*1 - 2P2P2.X*2

= 4P2 (2/3. M) - 2P2(1/3.M)

= 2P2 M > 0

d2L < 0, N. D. (Max)

Note:

Maximise (i) U = x½y½ (ii) U = x.y

Let price x = p and price of y = q, and m = income

L = x½y½ + [m – px – qy]

Cobb-Douglas: if U = xayb and a+b=1 (a = ½ & b= ½ above): Then income expenditure share on x is a and on y is b

L = log x + log y + [m – px – qy]

Cobb-Douglas: if U = xayband a+b1 (a=1 & b=1 above) Then income expenditure share on x is a/a+b and on y is b/a+b

4. Integration

The demand and supply functions for a good are given by the equations and respectively. Determine the equilibrium price and quantity and calculate the consumer and producer surplus at equilibrium.

At equilibrium

So equilibrium

Thus equilibrium

Consumer Surplus

Difference between value to consumers and to the market…. Area above price line and under Demand curve

Producer Surplus

Difference between market value and total cost to producers… area below price line and above Supply curve

Total Surplus = CS + PS = 16 + 8 = 24