Example #10: problem
Setup: Consider the following two-country economy. The real
exchange rate is xed and equal to 1. Domestic consumption,
investment and taxes are given by
C = 110 + 0.4(Y − T ), I = 40, T = 25
Import and exports are given by
IM = 0.2Y, X = 0.2Y ∗
(asterisks denote foreign variables).
(a) Solve for domestic output in terms of G and Y ∗ .
(b) How does the government purchases multiplier compare to the
closed economy case?
(c) Suppose both countries are symmetric, just with asterisks
reversed. Solve for output in each country in terms of G, G∗ .
(d) How does the multiplier compare to the one you found in (b)?
(e) Suppose both countries have G = G∗ = 100. Calculate Y, Y ∗ .
Example #10: solution
• Part (a): The national income accounting identity is
Y = C + I + G + X − IM
• Plugging in what we know
Y = 110 + 0.4(Y − 25) + 40 + G + 0.2Y ∗ − 0.2Y
• Solving for Y gives
1
Y = 110 − 0.4(25) + 40 + G + 0.2Y ∗
1 − 0.4 + 0.2
1
= 140 + G + 0.2Y ∗
0.8
= 175 + 1.25G + 0.25Y ∗
Example #10: solution
• Part (b): In this open economy we have the multiplier
dY 1 1
= = = 1.25
dG 1 − 0.4 + 0.2 0.8
• If the economy was closed, we would have the multiplier
1 1
= = 1.67
1 − 0.4 0.6
• Hence the open economy multiplier is smaller. In the closed
economy there is no `leakage' to import demand.
Example #10: solution
• Part (c): We have for the domestic economy
Y = 175 + 1.25G + 0.25Y ∗ (1)
• Since the foreign country is symmetric, just reverse the asterisks
Y ∗ = 175 + 1.25G∗ + 0.25Y (2)
• We need to solve these two equations in two unknowns
Example #10: solution
• Part (c) cont: Plug equation (2) into equation (1)
∗
Y = 175 + 1.25G + 0.25 175 + 1.25G + 0.25Y
• Solve for Y
1
Y = 175 + 1.25G + 0.25 175 + 1.25G∗
1 − (0.25)(0.25)
1
= (1.25)(175) + 1.25G + (0.25)(1.25)G∗
1 − 0.0625
= 233.33 + 1.33G + 0.33G∗
• By symmetry, also have
Y ∗ = 233.33 + 1.33G∗ + 0.33G
Example #10: solution
• Part (d): Now the government purchases multiplier is
dY
= 1.33
dG
• This is larger than the multiplier (= 1.25) from part (b)
• In (b) we held Y ∗ xed. But here when we compute the multiplier
we are also taking into account the indirect eect of G on Y ∗ on
exports X back on to domestic Y
Example #10: solution
• Part (d) cont: Alternative approach. Start with
Y = C + I + G + X − IM
so
dY dC dX d(IM )
= +0+1+ −
dG dG dG dG
with
dC dY
= (0.4)
dG dG
dX dY ∗ dY
= (0.2) = (0.2)(0.25)
dG dG dG
d(IM ) dY
= (0.2)
dG dG
Example #10: solution
• Part (d) cont: Plugging all these in
dY dY dY dY
= (0.4) + 1 + (0.2)(0.25) − (0.2)
dG dG dG dG
• Solving for the multiplier
dY 1 1
= = = 1.33
dG 1 − 0.4 − (0.2)(0.25) + 0.2 0.75
• Same answer, but here we see clearly that the multiplier is larger
because of the change in Y ∗ on exports X back on to domestic Y .
Example #10: solution
• Part (e): If G = G∗ = 100 we simply have
Y = 233.33 + 1.33G + 0.33G∗
= 233.33 + 1.33(100) + 0.33(100)
= 400
• And because of the symmetry also
Y ∗ = 400
