ECO102 Lecture 06

ECO102 - Week 06.pdf

$T$ change indirect tax to make $A E_{0} = A E_{1}$
Test review
- $C = 90 + 0.75 (Y - 0.2 Y - 80)$
- $I = 10$
- $G = 20$
- $N X_{0} = 40 - 0.1 Y$
- $N X_{1} = 10 - 0.1 Y$
- $A E_{0} = 90 + 0.75 (Y - 0.2 Y - 80) + 10 + 20 + (40 - 0.1 Y)$
- $A E_{0} = 0.5 Y + 100.0$
  - $Y = 0.5 Y + 100$
  - $Y = 200.0$
- $A E_{1} = 90 + 0.75 (Y - 0.2 Y - 80) + 10 + 20 + (10 - 0.1 Y)$
- $A E_{1} = 0.5 Y + 70.0$
  - $Y = 0.5 Y + 70$
  - $Y = 140.0$
- Calculate $B$ and $n x$
  - $B = T + Y t - G$
  - $B = 80 + 0.2 Y - 20$
  - $B = 0.2 Y + 60$
  - $B = (0.2) (200) + 60 = 100.0 = b_{0}$
  - $n x_{0} = 40 - 0.1 (200) = 20.0$
- In ECO102 Tutorial 05 we did $G ↑$ to offset the shock
- Here we $T ↓$ to offset the shock.
- It's a $30 $$ decrease in AE.
- Since $Δ T$ is affected by $M P C$ as well as $S M$ . We need $T ↓ 40$
- $40 \cdot 0.75 = 30$
- Now $A E_{2} = A E_{0}$
- $Y_{2} = Y_{0}$
- $B = T + Y t - G$
- $B_{2} = 40 + 0.20 Y - 20$
- $B_{2} = 0.2 Y + 20$
- $B_{2} = (0.2) (200) + 20 = 60.0$
- $n x_{2} = 10 - 0.1 (200) = - 10.0$
- $\dots$
- Then graph it like normal.

8
- $C = 20 + 0.8 (Y - 10 - 0.2 Y)$
- $A E = 0.64 Y + 12.0 + \dots$
- $A E = 120 + 0.68 Y - 4 P$
- $A D_{0} = 300 - 10 P$
- $A D_{1} = 395 - 12.5 P$
- Vertical intercepts of $A D_{0}$ and $A D_{1}$ are 30.
- When tax rates change, it changes the slope, not the intercept.
9
- $A S = 10 P$
- $C = 10 + 0.8 (Y - 5 - 0.1 Y)$
- $I = 4$
- $G = 35$
- $N X = 5 - P - 0.1 Y$
- $A E = 10 + 0.8 (Y - 5 - 0.1 Y) + 4 + 35 + 5 - P - 0.1 Y$
- $A E = - P + 0.62 Y + 50.0$
- This is the same as the question above.
- $A E = 6 + 0.72 Y + 4 + 35 + 5 - P - 0.12 Y$
- $A E = - P + 0.6 Y + 50$
- Shortcut:
  - Make $b_{2} = 0$ .
  - If $A S = 10 P$
  - $Y = 10 P$
  - $P = 0.1 Y$
  - Sub in
- $A E = - 0.1 Y + 0.6 Y + 50$
- $A E = 0.5 Y + 50$
  - $Y = 0.5 Y + 50$
  - $Y = 100.0$
- $B = 5 + 0.1 Y - \hat{G} = 0$
  - $\hat{G} = 5 + 0.1 Y$
- Sub it in to Fake $A E$
- $A E = 6 + 0.72 Y + 4 + (5 + 0.1 Y) + 5 - (0.1 Y) - 0.12 Y$
- $A E = 0.6 Y + 20$
- $Y = 0.6 Y + 20$
  - $Y = 50.0$
- $\hat{G} = 5 + 0.1 (50) =↓ 10.0$
- $35 - 10 = 25 = G$
- Explain why $b_{0} = - 20 \to b_{2} = 0$ requires that $G ↓ 25$
  - This is because of spending multipliers.
  - The $25 ↓ G$ reduces $Y ↓ 50$ .
  - A $t = 0.1$ , then $Y t ↓ 5$ .
  - That's why $G ↓ 25$ is needed to offset the shock.
- What if we $T ↑$ to get $b_{2} = 0$
  - $T_{0} = 5$
  - $b_{0} = - 20$
  - When we increase the indirect taxes, we need to increase $T$ by $\frac{20}{0.72} = 27.7777777777778$ to get $b_{2} = 0$
  - #tk is this correct?
- Draw graphs for those curves above.
- $B_{0}$
- Solow growth model
  - $Y = A K^{α} L^{1 - α}$
  - Potential GDP as $Y^{⋆}$
  - #tk yt econ curtis
- LAS
  - Long-run aggregate supply
  - Crossing of $A D_{0}$ and $A S$ at $(Y_{0}, P_{0})$
  - Crossing of $A D_{1}$ and $A S$ at $(Y_{1}, P_{1})$
  - But we shoot over from $A \to B$ at $(Y_{1}, P_{0})$
  - Long-run aggregate supply is vertical at $Y^{⋆} = Y_{0}$
  - At $Y_{2} > Y^{⋆}$ , we have GDP above potential, which is not sustainable. Expansionary gap.
    - As $W ↑$ , the $A S$ curve shifts up, and we end up at $Y_{0}$ again, but with a higher price level.
    - $S M = 0$
- Solow model and LAS is on test 2 #tk
- Then we will look at impact of exchange rates.
A policy to balance the trade account
- $C = 13 + 0.8 (Y - 10 - 0.25 Y)$
- $I = 10$
- $G = 35$
- $N X = 5 - 0.5 P - 0.1 Y + e$
- $e = 5$
- $A S = 5 P$
- We just change the exchange rate to balance the trade account.
- If the exchange rate depreciates, then $e ↑$ , which increases $N X$ .
- $A E = 13 + 0.8 (Y - 10 - 0.25 Y) + 10 + 35 + (5 - 0.5 P - 0.1 Y + e)$
- $A E = - 0.5 P + 0.5 Y + e + 55.0$
- $A E = - 0.5 P + 0.5 Y + 60$
- Shortcut from before:
  - $A S = 5 P$
  - $Y = 5 P$
  - $P = \frac{Y}{5}$
- $A E = - 0.5 \frac{Y}{5} + 0.5 Y + 60$
- $A E = 0.4 Y + 60$
  - $Y = 0.4 Y + 60$
  - $Y = 100.0$
- $P_{0} = (0.2) (100) = 20.0$
- $N X_{0} = 5 - (0.5) (20) - 0.1 (100) - 5 = - 20.0$ ?
- #tk prof got $- 10$
- Goal is $N X_{2} = 0$
- $N X = 5 - 0.2 Y + e = 0$
- $0 = 5 - 0.2 Y + \hat{e}$
- $\hat{e} = 0.2 Y - 5$
- Sub into $A E$ to create a fake $A E$ curve
- $A E = 13 + 0.8 (Y - 10 - 0.25 Y) + 10 + 35 + (5 - 0.5 P - 0.1 Y + \hat{e})$
  - $A E = - 0.5 P + 0.5 Y + \hat{e} + 55.0$
  - $A E = - 0.5 \frac{Y}{5} + 0.5 Y + \hat{e} + 55.0$
  - $A E = 0.4 Y + \hat{e} + 55.0$
  - $A E = 0.4 Y + (0.2 Y - 5) + 55.0$
  - $A E = 0.6 Y + 50.0$
  - $Y = 0.6 Y + 50$
  - $Y = 125.0$
- $\hat{e} = (0.2) (125) - 5 = 20.0$ to balance the trade account.
- $A E = 13 + 0.8 (Y - 10 - 0.25 Y) + 10 + 35 + (5 - 0.5 P - 0.1 Y + 20)$
- $A E = - 0.5 P + 0.5 Y + 75.0$
- $Y = - 0.5 P + 0.5 Y + 75$
- $Y = 150.0 - P$
- Graph $N X$ with original, then $E_{0}$ then $\hat{E} = 20$
- As $P ↑$ , $N X ↓$
- As $e ↑$ , $N X ↑$
- $e$ or $G$ or $T$ as a policy tools have the same effect on $Y$ , but different effects on $N X$ and $B$ .
Eliminate consumer deficit.
- $A S = 10 P$
- $C = 35 + 0.8 (Y - 5 - 0.1 Y)$
- $I = 4$
- $G = 10$
- $N X = 5 - 0.12 Y - P$
- Canadians are likelier to be savers
- US are likelier to be borrowers
- In the early 2000s, canadians acted more like americans, and had a consumer deficit.
- #tk never been on exam before, but still good to know.
- $S = Y_{D} - C$
- $S = Y - T - Y t - C$
- $S = Y - 5 - 0.1 Y - (35 + 0.8 (Y - 5 - 0.1 Y))$
- $S = 0.18 Y - 36.0$
- $M P S = (1 - b) (1 - t)$
  - #tk flashcard of the above
- $M P S = (1 - 0.8) (1 - 0.1) = 0.18$
- $A S = 10 P$
- $Y = 10 P$
- $P = \frac{Y}{10}$
- $A E = 31 + 0.72 Y + 4 + 10 + 5 - 0.12 Y - 0.1 Y$
- $A E = 0.5 Y + 50$
  - $Y = 0.5 Y + 50$
  - $Y = 100.0$
- $S = 0.18 (100) - 36 = - 18.0$
- Net borrowing.
- To solve this. $\hat{S} = 0.18 Y - 36$
- $\hat{S} = 0$
- $0.18 Y - 36 = 0$
  - $Y = 200.0$
- $Y = 200$
- $A E = 0.5 (200) + 50 = 150.0$
- We need to increase $G$ by $50$ to get $A E = 200$ .
- $G_{0} = 10$ , means $G_{1} = 60$ to eliminate consumer deficit.
- $A E = 31 + 0.72 Y + 4 + 60 + 5 - 0.12 Y - 0.1 Y$
- $A E = 0.5 Y + 100$
  - $Y = 0.5 Y + 100$
  - $Y = 200.0$
- $A D_{1} = 250 - 2.5 P$
- Graph with $Y = X$
  - $A E_{0}$
  - $A E_{1}$
  - $A E_{2}$
- Graph with $A S$
  - $A D_{0}$
  - $A D_{1}$
- Graph with $S$
  - $S_{0}$
    - Negative
  - When $G ↑$ , it doesn't affect $S_{1}$
  - $S_{1} = S_{0}$
    - $0$
  - $S_{2} = S_{1} = S_{0}$
    - Positive
- Graph with $N X$
  - $N X_{0} = N X_{1}$
  - From Equilibrium $B \to C$
  - $P ↑ ⟹ N X ↓$
  - $N X_{2} < N X_{1}$
- Again, $P ↑ ⟹ N X ↓ ⟹ A E ↓$
- What if $T ↓$ to make $S_{2} = 0$
  - How much would they need to cut taxes by?
  - $50 ↑ G$ to get $S_{2} = 0$
  - Since taxes are multiplied.
  - $S ↑$
  - So $S_{0} \neq S_{1} \neq S_{2}$