“计量”的意思是“以统方法做定研究”,“量”字为名词,构成动宾结构,这从其英文metric的含义亦可看出(与数学名词“度量空间metric space”情况类似),所以“量”字应读作“亮”(中国大陆《现代汉语辞典》2012年6月第6版“计量”条)。设若“计量”的“量”字读为“良”,则是两个动词词素的并列结构,含义略简。另如测智力的斯坦福一比奈智力量表(Stanford–Binet Intelligence Scale),按其内涵则应读“量”字为“良”,此亦可从英文scale的含义窥得。

问题 1.

Suppose that the classical regression model applies but that the true value of the constant is zero. In order to answer the following questions assume just one independent variable.

  1. Give the formulae for the two least squares slope estimators (the one with and the one without the constant).
  2. Calculate their variances.
  3. Compare the variance of the least squares slope estimator computed without a constant term with that of the estimator computed with an unnecessary constant term.

& y=\beta_1+\beta_2 x+\varepsilon \rightarrow \beta_2=\frac{\sum_i\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)}{\sum_i\left(x_i-\bar{x}\right)^2} \
& y=\tilde{\beta}_2 x+\varepsilon \rightarrow \tilde{\beta}_2=\frac{\sum_i x_i y_i}{\sum_i x_i^2}
\operatorname{Var}\left(\beta_2\right) & =\frac{\sigma^2}{\sum_i\left(x_i-\bar{x}\right)^2} \
\operatorname{Var}\left(\tilde{\beta}_2\right) & =\frac{\sigma^2}{\sum_i x_i^2}

  1. The ratio of these two variances is
    \frac{\operatorname{Var}\left(\tilde{\beta}_2\right)}{\operatorname{Var}\left(\beta_2\right)}= & \frac{\frac{\sigma^2}{\sum_i x_i^2}}{\frac{\sigma^2}{\sum_i\left(x_i-\bar{x}\right)^2}} \
    = & \frac{\sum_i\left(x_i-\bar{x}\right)^2}{\sum_i x_i^2} \
    & \sum\left(x_i-\bar{x}\right)^2=\sum\left(x_i^2-2 x_i \bar{x}+\bar{x}\right)=\sum x_i^2-2 n \bar{x} \bar{x}+n \bar{x}^2=\sum x_i^2-n \bar{x}^2 \
    = & \frac{\sum_i x_i^2-n \bar{x}^2}{\sum_i x_i^2} \
    = & 1-\frac{n \bar{x}^2}{\sum_i x_i^2} \leq 1
    It follows that fitting the constant term when it is unnecessary inflates the variance of the least squares estimator if the mean of the regressor is not zero.
问题 2.

Problem 2: (15 points)
Suppose that $y$ has the pdf $f(y \mid \mathbf{x})=\frac{1}{\boldsymbol{\beta}^{\prime} \mathbf{x}} e^{-y /\left(\boldsymbol{\beta}^{\prime} \mathbf{x}\right)}, y>0$. Then $E[y \mid \mathbf{x}]=$ $\boldsymbol{\beta}^{\prime} \mathbf{x}$ and $\operatorname{Var}[y \mid \mathbf{x}]=\left(\boldsymbol{\beta}^{\prime} \mathbf{x}\right)^2$. For this model, prove that the GLS and MLE estimators are the same, even though this distribution involves the same parameters in the conditional mean function and the disturbance variance.


问题 3.

Problem 3: (15 points)
The following model is estimated using a balanced panel of five firms over 20 years: $I_{i t}=\beta_1 F_{i t}+\beta_2 C_{i t}+\varepsilon_{i t}$, where the regressors are market value $(F)$ and capital $(C)$ and the dependent variable is investment $(I)$. Suppose that the true error structure of the model is $\varepsilon_{i t}=\alpha_i+\eta_{i t}$, where $\alpha$ is uncorrelated with the regressors.

  1. If the model is estimated as a fixed effects model, what will be the statistical properties, in terms of efficiency and consistency, of the estimates?
  2. The estimates for pooled OLS, fixed effects (using dummies) and random effects models are given in the table below. Use the statistics shown to decide whether the data support a fixed effects or random effects specification. Carefully explain your reasoning.
    Dependent Variable is Investment
    \hline Estimation & Constant & Market Value & Capital \
    \hline (a) OLS & $-48.030(-2.236)$ & $0.10509(9.236)$ & $0.30537(7.019)$ \
    (b) Fixed Effects & – & $0.10598(6.669)$ & $0.34666(14.348)$ \
    (c) Random Effects & $-61.575(-0.775)$ & $0.10549(6.859)$ & $0.34641(14.350)$ \

  1. If the individual effects are strictly uncorrelated with the regressors then a random effects model is the appropriate model. However, if a fixed effect model is estimated the estimates will be consistent but not efficient.
  2. Breush-Pagan LM test: Test statistic is 453.82 , the critical value from the chi-squared table is 3.84 , so the null hypothesis that random effects are not needed can be rejected.
    Hausman Test: Test statistic is 1.27 , the critical value from the chisquared table is 5.99 , so the null hypothesis of the random effects model cannot be rejected.
问题 4.

Problem 4: (15 points)
Consider the stochastic processes given below. For each process determine what the effects of first differencing the process, i.e. computing $y_t-y_{t-1}$, on autocorrelation are, e.g. reduction of the autocorrelation.

  1. $y_t=y_{t-1}+\varepsilon_t$, where $\varepsilon_t$ is normally distributed white noise.
  2. $y_t=\beta_0+\beta_1 t+\varepsilon_t$, where $\varepsilon_t$ is normally distributed white noise.
  3. $y_t=\boldsymbol{\beta}^{\prime} \mathbf{x}t+\varepsilon_t$, where $\varepsilon_t=\rho \varepsilon{t-1}+u_t$ and $u_t$ is normally distributed white noise.
    [Hint: Compare the autocorrelation of $\varepsilon_t$ and the autocorrelation of $\left.\left(\varepsilon_t-\varepsilon_{t-1}\right).\right]$

  1. $\Delta y_t=y_t-y_{t-1}=\varepsilon_t$, white noise, no more autocorrelation
  2. $\Delta y_t=y_t-y_{t-1}=\beta_1+\varepsilon_t-\varepsilon_{t-1}$. This is an MA(1) process with autocorrelation $\frac{\theta}{1+\theta^2}=\frac{1}{1+1}=\frac{1}{2}$.
  3. $\Delta y_t=y_t-y_{t-1}=\beta^{\prime}\left(x_t-x_{t-1}\right)+v_t$, where $v_t=\varepsilon_t-\varepsilon_{t-1}$.
    \operatorname{Var}\left(\varepsilon_t\right) & =\frac{\sigma_u^2}{1-\rho^2} \
    \operatorname{Var}\left(v_t\right) & =\operatorname{Var}\left(\varepsilon_t-\varepsilon_{t-1}\right)=\operatorname{Var}\left(\rho \varepsilon_{t-1}-\varepsilon_t+u_t\right) \
    & =\operatorname{Var}\left[(\rho-1) \varepsilon_{t-1}+u_t\right]=(\rho-1)^2 \frac{\sigma_u^2}{1-\rho^2}+\sigma_u^2=\frac{2 \sigma_u^2}{1+\rho} \
    \operatorname{Cov}\left[v_t, v_{t-1}\right] & =\operatorname{Cov}\left[\varepsilon_t-\varepsilon_{t-1}, \varepsilon_{t-1}-\varepsilon_{t-2}\right] \
    & =E\left[\varepsilon_t \varepsilon_{t-1}-\varepsilon_{t-1}^2-\varepsilon_t \varepsilon_{t-2}+\varepsilon_{t-1} \varepsilon_{t-2}\right] \
    & =\rho \frac{\sigma_u^2}{1-\rho^2}-\frac{\sigma_u^2}{1-\rho^2}-\rho^2 \frac{\sigma_u^2}{1-\rho^2}+\rho \frac{\sigma_u^2}{1-\rho^2}=\frac{\sigma_u^2\left(2 \rho-1-\rho^2\right)}{1-\rho^2} \
    & =\frac{\sigma_u^2(\rho-1)^2}{(\rho-1)(\rho+1)}=\frac{\sigma_u^2(\rho-1)}{\rho+1} \
    \frac{\operatorname{Cov}\left[v_t, v_{t-1}\right]}{\operatorname{Var}\left[v_t\right]} & =\frac{\rho-1}{2}
  4. Compare the two autocorrelations:
  5. $$
  6. \begin{aligned}
  7. |\rho| & >\left|\frac{\rho-1}{2}\right| \quad \quad \text { Assume } \rho>0 \text { and }|\rho|<1 \\ \rho & >-\frac{\rho-1}{2} \rightarrow \rho>\frac{1}{3}
  8. \end{aligned}
  9. $$
  10. If the original autocorrelation is greater than 1/3 (For economic data, this is likely to be fairly common.) the differenced process has a smaller autocorrelation.
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