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普通最小二乘估计的无偏性和一致性¶
本文证明了普通最小二乘估计的无偏性和一致性。 无偏性： \[ E\left[\beta^{O L S}\right]=\beta \]一致性 \[ \beta^{O L S}-\beta=\left(X^{\prime} X\right)^{-1} X^{\prime} u \stackrel{p}{\rightarrow} 0 \text { as } T \rightarrow \infty \] 无偏性¶为了证明普通最小二乘估计的无偏性，我们只需证明： \[ E\left[\beta^{O L S}\right]=\beta \]或 \[ E\left[\beta^{O L S}-\beta\right]=0 \] 将普通最小二乘估计量中的\(Y\)展开为\(X\beta+\mu\)：¶ \[ \begin{aligned} \beta^{O L S} & =\left(X^{\prime} X\right)^{-1} X^{\prime} Y \\ & =\left(X^{\prime} X\right)^{-1} X^{\prime}(X \beta+u) \\ & =\left(X^{\prime} X\right)^{-1} X^{\prime} X \beta+\left(X^{\prime} X\right)^{-1} X^{\prime} u \\ & =\beta+\left(X^{\prime} X\right)^{-1} X^{\prime} u \end{aligned} \] 再利用”误差的条件均值为零“这一假设，即\(E[u \mid X]=0\)¶ \[ \begin{aligned} E\left[\beta^{O L S} \mid X\right] & =E\left[\beta+\left(X^{\prime} X\right)^{-1} X^{\prime} u \mid X\right] \\ & =\beta+E\left[\left(X^{\prime} X\right)^{-1} X^{\prime} u \mid X\right] \\ & =\beta+\left(X^{\prime} X\right)^{-1} X^{\prime} E[u \mid X]\\ & =\beta \end{aligned} \] 最后根据迭代期望法则，证明无偏性¶ \[ E\left[\beta^{O L S}\right]=E\left[E\left[\beta^{O L S} \mid X\right]\right]=\beta \] 一致性¶为了证明一致性，我们只需证明： \[ \beta^{O L S}-\beta=\left(X^{\prime} X\right)^{-1} X^{\prime} u \stackrel{p}{\rightarrow} 0 ? \text { as } T \rightarrow \infty \] 在表达式中添加\(\frac{1}{T}\)，凑成平均的形式¶ \[ \beta^{O L S}-\beta=\left(\frac{1}{T} X^{\prime} X\right)^{-1} \frac{1}{T} X^{\prime} u \]注意 括号的部分需要求逆，因此括号里面乘以\(\frac{1}{T}\)等价于在括号外面乘以\(T\)。 将\(X^{\prime} X\)转换成“单个样本向量相乘得到矩阵，再求和”的形式¶ \[ \begin{aligned} X^{\prime} X&=\left[\begin{array}{cccc} 1 & 1 & \ldots & 1 \\ X_{11} & X_{21} & & X_{T 1} \\ X_{12} & X_{22} & & X_{T 2} \\ \ldots & \ldots & & \ldots \\ X_{1 K} & X_{2 K} & & X_{T K} \end{array}\right]\left[\begin{array}{ccccc} 1 & X_{11} & X_{12} & \ldots & X_{1 K} \\ 1 & X_{21} & X_{22} & \ldots & X_{2 K} \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 1 & X_{T 1} & X_{T 2} & \ldots & X_{T K} \end{array}\right] \\ & =\left[\begin{array}{ccccc} \mid & \mid & \mid & & \mid \\ X_1 & X_2 & X_3 & \ldots & X_T \\ \mid & \mid & \mid & & \mid \end{array}\right]\left[\begin{array}{ccc} - & X_1 & - \\ - & X_2 & - \\ - & X_3 & - \\ - & X_T & - \end{array}\right]\\ &=\sum_{t=1} ^T \underbrace{X_t} _{(K+1) \times 1} * \underbrace{X_t ^ {\prime}} _ {1 \times(K+1)} \\ \end{aligned} \] 将\(X^{\prime} \mu\)转换成“单个样本向量相乘得到向量，再求和”的形式¶ \[ \begin{aligned} X^{\prime} u & =\left[\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ X_{11} & X_{21} & & X_{T 1} \\ X_{12} & X_{22} & & X_{T 2} \\ \ldots & \cdots & & \ldots \\ X_{1 K} & X_{2 K} & & X_{T K} \end{array}\right]\left[\begin{array}{c} u_1 \\ u_2 \\ \ldots \\ u_T \end{array}\right] \\ & =\left[\begin{array}{ccccc} \mid & \mid & \mid & & \mid \\ X_1 & X_2 & X_3 & \ldots & X_T \\ \mid & \mid & \mid & & \mid \end{array}\right]\left[\begin{array}{c} u_1 \\ u_2 \\ \ldots \\ u_T \end{array}\right] \\ & =\sum_{t=1}^T \underbrace{X_t}_{(K+1) \times 1} * \underbrace{u_t}_{1 \times 1} \\ & \end{aligned} \] 当样本量足够大时，平均值趋近于期望值¶ \[ \begin{gathered} \beta^{O L S}-\beta=\left(\frac{1}{T} X^{\prime} X\right)^{-1} \frac{1}{T} X^{\prime} u \\ \frac{1}{T} X^{\prime} X=\frac{1}{T} \sum_{t=1}^T X_t X_t^{\prime} \stackrel{P}{\rightarrow} E\left[X_t X_t^{\prime}\right] \text { as } T \rightarrow \infty \\ \frac{1}{T} X^{\prime} u=\frac{1}{T} \sum_{t=1}^T X_t u_t \stackrel{P}{\rightarrow} E\left[X_t u_t\right] \text { as } T \rightarrow \infty \end{gathered} \] 再利用”误差的条件均值为零“这一假设，即\(E[u \mid X]=0\)¶ \[ \begin{aligned} E[u \mid X]&=0\\ \Rightarrow E\left[X_t u_t\right]&=E_{X} \left[E\left[X_t u_t \mid X \right] \right]\\ &=E_{X} \left[X_t E\left[u_t \mid X \right] \right]\\ &=E_{X} [\underbrace{X_t}_{(K+1) \times 1} \underbrace{0}_{1 \times 1}]\\ &=\underbrace{0}_{(K+1) \times 1} \end{aligned} \]因此 \[ \begin{aligned} \beta^{O L S}-\beta&= (\underbrace{\frac{1}{T} X^ {\prime} X} _ {E\left[X_t X_t ^ {\prime}\right]}) ^{-1} \underbrace{\frac{1}{T} X^ {\prime} u}_ {E\left[X_t u_t\right]}\\\ & \stackrel{p}{\rightarrow}\left(E\left[X_t X_t ^ {\prime}\right]\right) ^ {-1} E\left[X_t u_t\right] \\\ & \stackrel{p}{\rightarrow} \underbrace{0} _ {(K+1) \times 1} \end{aligned} \] 评论 |
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