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explain the answer in more details on question a to e Consider a discrete-time v...
Jun 15, 2024
explain the answer in more details on question a to e
Solution
a
To solve for ptp_t in terms of mtm_t and Etpt+1E_t p_{t+1}, start with the given money demand equation: mtpt=cb(Etpt+1pt) m_t - p_t = c - b(E_t p_{t+1} - p_t) Rearrange to isolate ptp_t: mtc=ptb(Etpt+1pt) m_t - c = p_t - b(E_t p_{t+1} - p_t) mtc=pt(1+b)bEtpt+1 m_t - c = p_t(1 + b) - bE_t p_{t+1} pt(1+b)=mtc+bEtpt+1 p_t(1 + b) = m_t - c + bE_t p_{t+1} pt=mtc+bEtpt+11+b p_t = \frac{m_t - c + bE_t p_{t+1}}{1 + b}
b
Using the law of iterated projections, express Etpt+1E_t p_{t+1} in terms of Etmt+1E_t m_{t+1} and Etpt+2E_t p_{t+2}: Etpt+1=Etmt+1c+bEtpt+21+b E_t p_{t+1} = \frac{E_t m_{t+1} - c + bE_t p_{t+2}}{1 + b}
c
Iterate this process forward to express ptp_t in terms of mt,Etmt+1,Etmt+2,m_t, E_t m_{t+1}, E_t m_{t+2}, \ldots: pt=mtc1+b+b1+bEtpt+1 p_t = \frac{m_t - c}{1 + b} + \frac{b}{1 + b} E_t p_{t+1} Substitute Etpt+1E_t p_{t+1} iteratively: pt=mtc1+b+b1+b(Etmt+1c1+b+b1+bEtpt+2) p_t = \frac{m_t - c}{1 + b} + \frac{b}{1 + b} \left( \frac{E_t m_{t+1} - c}{1 + b} + \frac{b}{1 + b} E_t p_{t+2} \right) Continue this process: pt=i=0(b1+b)iEtmt+ic1+b p_t = \sum_{i=0}^{\infty} \left( \frac{b}{1 + b} \right)^i \frac{E_t m_{t+i} - c}{1 + b} Given the no-bubbles condition: limiEt[(b1+b)ipt+i]=0 \lim_{i \to \infty} E_t \left[ \left( \frac{b}{1 + b} \right)^i p_{t+i} \right] = 0 Thus: pt=i=0(b1+b)iEtmt+ic1+b p_t = \sum_{i=0}^{\infty} \left( \frac{b}{1 + b} \right)^i \frac{E_t m_{t+i} - c}{1 + b}
d
Intuitively, an increase in Etmt+iE_t m_{t+i} for any i > 0 raises ptp_t because higher expected future money supply increases the expected future price level, which in turn increases the current price level due to the forward-looking nature of price setting
e
Suppose expected money growth is constant, so Etmt+i=mt+giE_t m_{t+i} = m_t + g i. Solve for ptp_t in terms of mtm_t and gg: pt=i=0(b1+b)imt+gic1+b p_t = \sum_{i=0}^{\infty} \left( \frac{b}{1 + b} \right)^i \frac{m_t + g i - c}{1 + b} pt=mtc1+bi=0(b1+b)i+g1+bi=0i(b1+b)i p_t = \frac{m_t - c}{1 + b} \sum_{i=0}^{\infty} \left( \frac{b}{1 + b} \right)^i + \frac{g}{1 + b} \sum_{i=0}^{\infty} i \left( \frac{b}{1 + b} \right)^i Using the geometric series sum formula: i=0(b1+b)i=11b1+b=1+b \sum_{i=0}^{\infty} \left( \frac{b}{1 + b} \right)^i = \frac{1}{1 - \frac{b}{1 + b}} = 1 + b i=0i(b1+b)i=b1+b(1b1+b)2=b(1+b)2 \sum_{i=0}^{\infty} i \left( \frac{b}{1 + b} \right)^i = \frac{\frac{b}{1 + b}}{\left( 1 - \frac{b}{1 + b} \right)^2} = \frac{b}{(1 + b)^2} Thus: pt=mtc1+b(1+b)+g1+bb(1+b)2 p_t = \frac{m_t - c}{1 + b} (1 + b) + \frac{g}{1 + b} \frac{b}{(1 + b)^2} pt=mtc+gb(1+b)2 p_t = m_t - c + \frac{gb}{(1 + b)^2} An increase in gg raises ptp_t because it increases the expected future money supply, leading to higher current price levels
Answer
The final answers are as follows: - pt=mtc+bEtpt+11+bp_t = \frac{m_t - c + bE_t p_{t+1}}{1 + b} - Etpt+1=Etmt+1c+bEtpt+21+bE_t p_{t+1} = \frac{E_t m_{t+1} - c + bE_t p_{t+2}}{1 + b} - pt=i=0(b1+b)iEtmt+ic1+bp_t = \sum_{i=0}^{\infty} \left( \frac{b}{1 + b} \right)^i \frac{E_t m_{t+i} - c}{1 + b} - An increase in Etmt+iE_t m_{t+i} for any i > 0 raises ptp_t because it increases the expected future price level. - pt=mtc+gb(1+b)2p_t = m_t - c + \frac{gb}{(1 + b)^2}, and an increase in gg raises ptp_t.
Key Concept
The relationship between money supply, expected future money supply, and price levels.
Explanation
The price level ptp_t is influenced by the current money supply mtm_t and the expected future money supply Etmt+iE_t m_{t+i}. An increase in expected future money supply raises the current price level due to the forward-looking nature of price setting.
explain in details the question a
Solution
a
To solve for ptp_{t} in terms of mtm_{t} and Etpt+1E_{t} p_{t+1}, we start with the given money demand equation: mtpt=cb(Etpt+1pt) m_{t} - p_{t} = c - b \left( E_{t} p_{t+1} - p_{t} \right) First, we rearrange the terms to collect ptp_{t} on one side: mtpt=cbEtpt+1+bpt m_{t} - p_{t} = c - b E_{t} p_{t+1} + b p_{t} Next, we combine the ptp_{t} terms: mtc=pt(1+b)bEtpt+1 m_{t} - c = p_{t} (1 + b) - b E_{t} p_{t+1} Solving for ptp_{t}, we get: pt(1+b)=mtc+bEtpt+1 p_{t} (1 + b) = m_{t} - c + b E_{t} p_{t+1} Finally, we isolate ptp_{t}: pt=(b1+b)Etpt+1+(11+b)(mtc) p_{t} = \left( \frac{b}{1 + b} \right) E_{t} p_{t+1} + \left( \frac{1}{1 + b} \right) (m_{t} - c) Thus, the solution for ptp_{t} in terms of mtm_{t} and Etpt+1E_{t} p_{t+1} is: pt=(b1+b)Etpt+1+(11+b)(mtc) p_{t} = \left( \frac{b}{1 + b} \right) E_{t} p_{t+1} + \left( \frac{1}{1 + b} \right) (m_{t} - c)
Answer
The solution for ptp_{t} in terms of mtm_{t} and Etpt+1E_{t} p_{t+1} is: pt=(b1+b)Etpt+1+(11+b)(mtc) p_{t} = \left( \frac{b}{1 + b} \right) E_{t} p_{t+1} + \left( \frac{1}{1 + b} \right) (m_{t} - c)
Key Concept
Solving for the price level ptp_{t} in terms of the money stock mtm_{t} and the expected future price level Etpt+1E_{t} p_{t+1}.
Explanation
By rearranging the given money demand equation and isolating ptp_{t}, we can express the current price level as a function of the current money stock and the expected future price level.
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