money market problem set

1/ indeterminate.

• from money supply, increasing money supply will increase gdp since $M=YL(i)$ so $\uparrow M\Rightarrow \uparrow Y$. from goods market, $Z=c_0+c_1(Y-T)+\bar{I}+G$. so increasing $c_0$ will lead to an increase in $Z$, and $Y$ has to increase to match the new $Z$, so $\uparrow c_0\Rightarrow \uparrow Y$. thus, both changes cause $\uparrow Y$.
• but, increasing money supply will decrease interest rates based on negative relationship in $M=YL(i)$—since central bank buys bonds which increases demand which increases bond prices which means % between buy price and earning price for a bond is lower—so $\uparrow M\Rightarrow \downarrow i$. though, increasing $c_0$ will cause increase in interest rates also because of $M=YL(i)$, since $\uparrow c_0\Rightarrow \uparrow Y\Rightarrow \uparrow i$ based on the negative relationship via $L$.
• so, since the second part of the question isn’t completely given, indeterminate.

2/ false.

• atms allow for holding cash in a accessible place rather than needing to keep on hand, so overall demand for money for each person goes down, so $M_d$ goes down. for people to get rid of money for $M$ to reach $M_d$, they buy bonds which increases demand for bonds which decreases $i$, which by relationship $\downarrow M\Rightarrow \downarrow i$. so, investment part of it is incorrect.

3/ false.

• $T$ increases, so in $Z=c_0+c_1(Y-T)+\bar{I}+G$, $Y$ would have to decrease to keep equilibrium with $Z$ constant. this change in $Y$ will affect $M_d=YL(i)$, reducing $M_d$ and thus making an impact.

4/ true.

• contraction of gdp → $Y$ goes down, so $M_d=YL(i)$ keeping $M_d$ same requires $\downarrow i$. central bank buying bonds → more money but less interest → interest goes down, so right effect

5/

• $M_d=(Y-T)(0.3-i)$, $C=1+0.5(Y-T)$, $G=0.275$, $T=0.2Y$, $I=1-0.5i$
• citizens hold $25\%$ disposable income in cash
• banks hold $10\%$ of deposits on reserve with central bank

5a/

• given $i=5\%$
• $Y=Z$
• $=C+I+G$
• $=1+0.5(Y-T)+1-0.5i+0.275$
• $=1+0.5(Y-0.2Y)-0.5i+0.275$
• $Y-0.5(Y-0.2Y)=1+1-0.5i+0.275$
• $0.6Y=2.275-0.5i$
• $Y=\frac{2.25}{0.6}=3.75$

5b/

• given $i=5\%$
• $M_d=(Y-T)(0.3-i)$
• $=(Y-0.2Y)(0.3-i)$
• $=0.75$
• $M=M_d=0.75$

5c/

• given $M=0.052Y$
• $M=0.052Y=M_d=(Y-T)(0.3-i)$
• $0.052Y=(Y-0.2Y)(0.3-i)$
• $\frac{0.052Y}{0.8Y}=0.3-i$
• $0.065=0.3-i$
• $i=0.235=23.5\%$

5d/

• $Y=3.75$
• $=C+I+G$
• $=1+0.5(Y-T)+1-0.5i+G$
• $=1+0.5\times 0.8\times 3.75+1-0.5(0.235)+G$
• $=3.3825+G$
• $3.75-3.3825=G=0.3675$

6/

• $C=0.58Y$, Z = Y = \176$,$c_1 = 0.83$,$T = 0.31Y$,$I = 0.057Y - 1800i$,$G - T = 0.025Y$,$M_d = 0.6Y - 850i$,$M = $80$

6a/

• $G-T=0.025Y$
• $G-0.31Y=0.025Y$
• G = 0.025Y + 0.31Y = \58.96$

6b/

• $M_d=M=0.6Y-850i$
• \80 = 0.6($176) - 850i$
• $-25.6=-850i$
• $i=0.0301$

6c/

• $Y=c_0+c_1(Y-T)+I+G$
• \176 = c_0 + (0.83)($176 - 0.31($176)) + 0.057($176) - 1800(0.0301) + $58.96$
• \176 = c_0 + $115.60$
• c_0 = \60.4$