## 经济代写|宏观经济学代写Macroeconomics代考|ECOS3007

2023年1月5日

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## 经济代写|宏观经济学代写Macroeconomics代考|Solving the Model with a Shifted Laplace Transform

The equations of Frisch’s model are the following ${ }^{30}$ :
\begin{aligned} & \dot{x}(t)=c-\lambda(r x(t)+s z(t)), \ & y(t)=m x(t)+\mu \dot{x}(t), \ & z(t)=\frac{1}{\varepsilon} \int_{t-\varepsilon}^t y(\tau) d \tau . \end{aligned}

As a preliminary, we can compute the equilibrium of the system, which will help us verify later on that our computations are right. After at least $\varepsilon$ time at the equilibrium, we know that the rate of change will be zero, and the past values will be the same as the equilibrium values. We have thus three equations to determine the three unknown equilibrium values:
\begin{aligned} & c=\lambda(r \bar{x}+s \bar{z}), \ & \bar{y}=m \bar{x}, \ & \bar{z}=\bar{y} . \end{aligned}
Solving for the equilibrium values, we get that
\begin{aligned} & \bar{x}=\frac{c}{\lambda(r+s m)}, \ & \bar{y}=\bar{z}=\frac{m c}{\lambda(r+s m)} . \end{aligned}
To find solutions satisfying this system, we reduce it to one equation by successive replacements: first, we replace $y_t$ in $z_t$ and we obtain $z_t=\frac{1}{\varepsilon} \int_{t-\varepsilon}^t m \cdot x(\tau)+\mu$. $\dot{x}(\tau) d \tau$. This can be simplified as $z_t=\frac{m}{\varepsilon} \int_{t-\varepsilon}^t x(\tau) d \tau+\frac{\mu}{\varepsilon}[x(t)-x(t-\varepsilon)]$. Putting this into (4.10), we obtain:
$$\dot{x}(t)=c-\lambda\left[r x(t)+s\left(\frac{m}{\varepsilon} \int_{t-\varepsilon}^t x(\tau) d \tau+\frac{\mu}{\varepsilon}[x(t)-x(t-\varepsilon)]\right)\right] .$$

## 经济代写|宏观经济学代写Macroeconomics代考|A Numerical Integration of Frisch’s System

The Laplace transform is useful because it gives us the sum of components used by Frisch in an elegant way and with some insights on the initial conditions along the way. Of course this is not the only way to solve the model, and one can also do a numerical integration to obtain a general solution without the infinite sum of components, which is inevitably truncated to obtain a finite sum; in this sense, there is an approximation in this method when we want to represent the final solution, but we will see that the numerical integration is not exempt from approximation as well when we manipulate the integral arising in $z(t)$. The system of discretized equations for $x, y$ and $z$ is:

\begin{aligned} x(t+h) & =x(t)+h(c-\lambda(r x(t)+s z(t))), \ y(t) & =m x(t)+\mu \frac{x(t+h)-x(t)}{h}, \ z(t) & =\frac{h}{\varepsilon}\left[\sum_{i=1}^{\frac{\varepsilon-h}{h}} y(t-i h)+\frac{y(t)+y(t-\varepsilon)}{2}\right], \end{aligned}
where the integral in $z$ has been approximated using a trapezoidal rule; this is a simultaneous system between $x(t+h), y(t)$ and $z(t)$, which we can reduce to one equation by substitutions. The important part in the last equation is $y(t)$, because the rest will be known, so that we can extract it and group what is known in a new variable $\Gamma$ :
\begin{aligned} z(t) & =\frac{h}{2 \varepsilon} y(t)+\frac{h}{\varepsilon}\left[\sum_{i=1}^{\frac{-h}{h}} y(t-i h)+\frac{y(t-\varepsilon)}{2}\right]=\frac{h}{2 \varepsilon} y(t)+\frac{h}{\varepsilon} \Gamma \ & =\frac{h}{2 \varepsilon}\left(m x(t)+\mu \frac{x(t+h)-x(t)}{h}\right)+\frac{h}{\varepsilon} \Gamma \ & =\frac{\mu}{2 \varepsilon} x(t+h)+\left(\frac{h m-\mu}{2 \varepsilon}\right) x(t)+\frac{h}{\varepsilon} \Gamma . \end{aligned}

# 宏观经济学代考

## 经济代写|宏观经济学代写Macroeconomics代考|Solving the Model with a Shifted Laplace Transform

Frisch 模型的方程式如下 ${ }^{30}$ :
$$\dot{x}(t)=c-\lambda(r x(t)+s z(t)), \quad y(t)=m x(t)+$$

$$c=\lambda(r \bar{x}+s \bar{z}), \quad \bar{y}=m \bar{x}, \bar{z}=\bar{y} .$$

$$\bar{x}=\frac{c}{\lambda(r+s m)}, \quad \bar{y}=\bar{z}=\frac{m c}{\lambda(r+s m)}$$

$$\dot{x}(t)=c-\lambda\left[r x(t)+s\left(\frac{m}{\varepsilon} \int_{t-\varepsilon}^t x(\tau) d \tau+\frac{\mu}{\varepsilon}[x(t)\right.\right.$$

## 经济代写|宏观经济学代写Macroeconomics代考|A Numerical Integration of Frisch’s System

$$x(t+h)=x(t)+h(c-\lambda(r x(t)+s z(t))), y(t)$$

$$z(t)=\frac{h}{2 \varepsilon} y(t)+\frac{h}{\varepsilon}\left[\sum_{i=1}^{\frac{-h}{h}} y(t-i h)+\frac{y(t-\varepsilon)}{2}\right]$$

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