# 经济代写|计量经济学代写Econometrics代考|Autoregressive models

Suppose the adjustment of the actual value of a variable $Y_t$ to its optimal (or desired) level (denoted by $Y_t^$ ) needs to be modelled. One way to do this is by using the partial adjustment model, which assumes that the change in actual $Y_t$ (that is, $Y_t-Y_{t-1}$ ) will be equal to a proportion of the optimal change $\left(Y_t^-Y_{t-1}\right)$, or:
$$Y_t-Y_{t-1}=\lambda\left(Y_t^*-Y_{t-1}\right)$$
where $\lambda$ is the adjustment coefficient, which takes values from 0 to 1 , and $1 / \lambda$ denotes the speed of adjustment.

Consider the two extreme cases: (a) if $\lambda=1$ then $Y_t=Y_t^$ and therefore the adjustment to the optimal level is instantaneous; while (b) if $\lambda=0$ then $Y_t=Y_{t-1}$, which means there is no adjustment of $Y_t$. Therefore, the closer $\lambda$ is to unity, the faster the adjustment will be. To understand this better, we can use a model from economic theory. Suppose $Y_t^$ is the desired level of inventories for a firm $i$, and that this depends on the level of sales of the firm $X_t$ :
$$Y_t^=\beta_1+\beta_2 X_t$$ Because there are ‘frictions’ in the market, there is bound to be a gap between the actual level of inventories and the desired one. Suppose also that only a part of the gap can be closed during each period. Then the equation that will determine the actual level of inventories will be given by: $$Y_t=Y_{t-1}+\lambda\left(Y_t^-Y_{t-1}\right)+u_t$$
That is, the actual level of inventories is equal to that at time $t-1$ plus an adjustment factor and a random component.

## 经济代写|计量经济学代写Econometrics代考|Computer example of the partial adjustment model

Consider the money-demand function:
$$M_t^=a Y_t^{b_1} R_t^{b_2} e_t^{u_t}$$ where the usual notation applies. Taking logarithms of this equation, we get: $$\ln M_t^=\ln a+b_1 \ln Y_t+b_2 \ln R_t+u_t$$

The partial adjustment hypothesis can be written as:
$$\frac{M_t}{M_{t-1}}=\left(\frac{M_t^}{M_{t-1}}\right)^\lambda$$ where, if we take logarithms, we get: $$\ln M_t-\ln M_{t-1}=\lambda\left(\ln M_t^-\ln M_{t-1}\right)$$
Substituting Equation (10.22) into Equation (10.24) we get:
\begin{aligned} \ln M_t-\ln M_{t-1} & =\lambda\left(\ln a+b_1 \ln Y_t+b_2 \ln R_t+u_t-\ln M_{t-1}\right) \ \ln M_t & =\lambda \ln a+\lambda b_1 \ln Y_t+\lambda b_2 \ln R_t+(1-\lambda) \ln M_{t-1}+\lambda u_t \end{aligned}
or:
$$\ln M_t=\gamma_1+\gamma_2 \ln Y_t+\gamma_3 \ln R_t+\gamma_4 \ln M_{t-1}+v_t$$
We shall use EViews to obtain OLS results for this model using data for the Italian economy (gross domestic product (GDP), the consumer price index (cpi) the M2 monetary aggregate $(M 2)$, plus the official discount interest rate $(R))$. The data are quarterly observations from $1975 \mathrm{q} 1$ to $1997 \mathrm{q} 4$. First we need to divide both $G D P$ and $M 2$ by the consumer price index in order to obtain real GDP and real money balances.

# 计量经济学代考

$$Y_t-Y_{t-1}=\lambda\left(Y_t^*-Y_{t-1}\right)$$

$$Y_t=\beta_1+\beta_2 X_t$$

$$Y_t=Y_{t-1}+\lambda\left(Y_t^{-} Y_{t-1}\right)+u_t$$

## 经济代写|计量经济学代写Econometrics代考|Computer example of the partial adjustment model

$$M_t^{=} a Y_t^{b_1} R_t^{b_2} e_t^{u_t}$$

$$\ln M_t=\ln a+b_1 \ln Y_t+b_2 \ln R_t+u_t$$

$$\backslash \operatorname{|rac}\left{M_{-} _t\left{\left{M_{-}{t-1}\right}=\backslash \text { left } \backslash \text { frac }\left{M_{-} t^{\wedge}\right}\left{M_{-}{t-1}\right} \backslash \text { right }\right)^{\wedge} \backslash\right. \text { lambda }$$

$$\ln M_t-\ln M_{t-1}=\lambda\left(\ln M_t^{-} \ln M_{t-1}\right)$$

$$\ln M_t-\ln M_{t-1}=\lambda\left(\ln a+b_1 \ln Y_t+b_2 \ln R_t+u_t-\ln M_{t-1}\right) \ln M_t=\lambda \ln a$$

$$\ln M_t=\gamma_1+\gamma_2 \ln Y_t+\gamma_3 \ln R_t+\gamma_4 \ln M_{t-1}+v_t$$

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