如果你也在 怎样代写计量经济学Econometrics是将统计方法应用于经济数据,以赋予经济关系以经验内容。更确切地说,它是 “基于理论和观察的同步发展,通过适当的推理方法对实际经济现象进行定量分析”。 一本经济学入门教科书将计量经济学描述为允许经济学家 “从堆积如山的数据中筛选出简单的关系”。
计量经济学Econometrics的一个基本工具是多元线性回归模型。计量经济学理论使用统计理论和数理统计来评估和发展计量经济学方法。计量经济学家试图找到具有理想统计特性的估计器,包括无偏性、效率和一致性。应用计量经济学使用理论计量经济学和现实世界的数据来评估经济理论,开发计量经济学模型,分析经济历史和预测。
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经济代写|计量经济学代写Econometrics代考|The Linear Case
Let $X=\left(X_{1}, X_{2}\right)^{\mathrm{T}}$ be a bivariate random vector with distribution function
$$
F(x)=\int_{-\infty}^{x_{1}} \int_{-\infty}^{x_{2}} f\left(u_{1}, u_{2}\right) d u_{1} d u_{2}=\int_{\left(-\infty, x_{1}\right] \times\left(-\infty, x_{2}\right]} f(u) d u
$$
where $x=\left(x_{1}, x_{2}\right)^{\mathrm{T}}, u=\left(u_{1}, u_{2}\right)^{\mathrm{T}}$.
In this section I will derive the joint density of $Y=A X+b$, where $A$ is a (nonrandom) nonsingular $2 \times 2$ matrix and $b$ is a nonrandom $2 \times 1$ vector.
Recall from linear algebra (see Appendix I) that any square matrix $A$ can be decomposed into
$$
A=R^{-1} L \cdot D \cdot U
$$
where $R$ is a permutation matrix (possibly equal to the unit matrix $I$ ), $L$ is a lower-triangular matrix with diagonal elements all equal to $1, U$ is an uppertriangular matrix with diagonal elements all equal to 1 , and $D$ is a diagonal matrix. The transformation $Y=A X+b$ can therefore be conducted in five steps:
$$
\begin{aligned}
Z_{1} &=U X \
Z_{2} &=D Z_{1} \
Z_{3} &=L Z_{2} \
Z_{4} &=R^{-1} Z_{3} \
Y &=Z_{4}+b
\end{aligned}
$$
Therefore, I will consider the first four cases, $A=U, A=D, A=L$, and $A=R^{-1}$ for $b=0$ and then the case $A=I, b \neq 0$.
Let $Y=A X$ with $A$ an upper-triangular matrix:
$$
A=\left(\begin{array}{ll}
1 & a \
0 & 1
\end{array}\right) \text {. }
$$
Then
$$
Y=\left(\begin{array}{l}
Y_{1} \
Y_{2}
\end{array}\right)=\left(\begin{array}{c}
X_{1}+a X_{2} \
X_{2}
\end{array}\right)
$$
hence, the joint distribution function $H(y)$ of $Y$ is
$$
\begin{aligned}
H(y) &=P\left(Y_{1} \leq y_{1}, Y_{2} \leq y_{2}\right)=P\left(X_{1}+a X_{2} \leq y_{1}, X_{2} \leq y_{2}\right) \
&=E\left[I\left(X_{1} \leq y_{1}-a X_{2}\right) I\left(X_{2} \leq y_{2}\right)\right] \
&=E\left(E\left[I\left(X_{1} \leq y_{1}-a X_{2}\right) \mid X_{2}\right] I\left(X_{2} \leq y_{2}\right)\right) \
&=\int_{-\infty}^{y_{2}}\left(\int_{-\infty}^{y_{1}-a x_{2}} f_{1 \mid 2}\left(x_{1} \mid x_{2}\right) d x_{1}\right) f_{2}\left(x_{2}\right) d x_{2} \
&=\int_{-\infty}^{y_{2}}\left(\int_{-\infty}^{y_{1}-a x_{2}} f\left(x_{1}, x_{2}\right) d x_{1}\right) d x_{2}
\end{aligned}
$$
where $f_{1 \mid 2}\left(x_{1} \mid x_{2}\right)$ is the conditional density of $X_{1}$ given $X_{2}=x_{2}$ and $f_{2}\left(x_{2}\right)$ is the marginal density of $X_{2}$. If we take partial derivatives, it follows from (4.22) that for $Y=A X$ with $A$ given by (4.21),
$$
\begin{aligned}
h(y) &=\frac{\partial^{2} H(y)}{\partial y_{1} \partial y_{2}}=\frac{\partial}{\partial y_{2}} \int_{-\infty}^{y_{2}} f\left(y_{1}-a x_{2}, x_{2}\right) d x_{2} \
&=f\left(y_{1}-a y_{2}, y_{2}\right)=f\left(A^{-1} y\right) .
\end{aligned}
$$
经济代写|计量经济学代写Econometrics代考|The Nonlinear Case
If we denote $G(x)=A x+b, G^{-1}(y)=A^{-1}(y-b)$, then the result of Theorem $4.3$ reads $h(y)=f\left(G^{-1}(y)\right)\left|\operatorname{det}\left(\partial G^{-1}(y) / \partial y\right)\right|$. This suggests that Theorem $4.3$ can be generalized as follows:
Theorem 4.4: Let $X$ be $k$-variate, absolutely continuously distributed with joint density $f(x), x=\left(x_{1}, \ldots, x_{k}\right)^{\mathrm{T}}$, and let $Y=G(X)$, where $G(x)=$ $\left(g_{1}(x), \ldots, g_{k}(x)\right)^{\mathrm{T}}$ is a one-to-one mapping with inverse mapping $x=$ $G^{-1}(y)=\left(g_{1}^{}(y), \ldots, g_{k}^{}(y)\right)^{\mathrm{T}}$ whose components are differentiable in the components of $y=\left(y_{1}, \ldots, y_{k}\right)^{\mathrm{T}}$. Let $J(y)=\partial x / \partial y=\partial G^{-1}(y) / \partial y$, that is, $J(y)$ is the matrix with $i, j$ ‘s element $\partial g_{i}^{*}(y) / \partial y_{j}$, which is called the Jacobian. Then $Y$ is $k$-variate, absolutely continuously distributed with joint density $h(y)=f\left(G^{-1}(y)\right)|\operatorname{det}(J(y))|$ for $y$ in the set $G\left(\mathbb{R}^{k}\right)=\left{y \in \mathbb{R}^{k}: y=G(x)\right.$, $\left.f(x)>0, x \in \mathbb{R}^{k}\right}$ and $h(y)=0$ elsewhere.
This conjecture is indeed true. Its formal proof is given in Appendix 4.B.
An application of Theorem $4.4$ is the following problem. Consider the function
$$
\begin{aligned}
f(x) &=c \cdot \exp \left(-x^{2} / 2\right) & \text { if } x \geq 0, \
&=0 \quad \text { if } x<0 \end{aligned} $$ For which value of $c$ is this function a density? To solve this problem, consider the joint density $f\left(x_{1}, x_{2}\right)=c^{2} \exp \left[-\left(x_{1}^{2}+\right.\right.$ $\left.\left.x_{2}^{2}\right) / 2\right], x_{1} \geq 0, x_{2} \geq 0$, which is the joint distribution of $X=\left(X_{1}, X_{2}\right)^{\mathrm{T}}$, where $X_{1}$ and $X_{2}$ are independent random drawings from the distribution with density (4.26). Next, consider the transformation $Y=\left(Y_{1}, Y_{2}\right)^{\mathrm{T}}=G(X)$ defined by $$ \begin{aligned} &Y_{1}=\sqrt{X_{1}^{2}+X_{2}^{2}} \in(0, \infty) \\ &Y_{2}=\arctan \left(X_{1} / X_{2}\right) \in(0, \pi / 2) . \end{aligned} $$ The inverse $X=G^{-1}(Y)$ of this transformation is $$ \begin{aligned} &X_{1}=Y_{1} \sin \left(Y_{2}\right), \\ &X_{2}=Y_{1} \cos \left(Y_{2}\right) \end{aligned} $$ with Jacobian $$ J(Y)=\left(\begin{array}{ll} \partial X_{1} / \partial Y_{1} & \partial X_{1} / \partial Y_{2} \\ \partial X_{2} / \partial Y_{1} & \partial X_{2} / \partial Y_{2} \end{array}\right)=\left(\begin{array}{cc} \sin \left(Y_{2}\right) & Y_{1} \cos \left(Y_{2}\right) \\ \cos \left(Y_{2}\right) & -Y_{1} \sin \left(Y_{2}\right) \end{array}\right) . $$ Note that $\operatorname{det}[J(Y)]=-Y_{1}$. Consequently, the density $h(y)=h\left(y_{1}, y_{2}\right)=$ $f\left(G^{-1}(y)\right)|\operatorname{det}(J(y))|$ is $$ \begin{aligned} h\left(y_{1}, y_{2}\right) &=c^{2} y_{1} \exp \left(-y_{1}^{2} / 2\right) \quad \text { for } \quad y_{1}>0 \quad \text { and } \quad 0<y_{2}<\pi / 2, \
&=0 \text { elsewhere; }
\end{aligned}
$$
hence,
$$
\begin{aligned}
1 &=\int_{0}^{\infty} \int_{0}^{\pi / 2} c^{2} y_{1} \exp \left(-y_{1}^{2} / 2\right) d y_{2} d y_{1} \
&=c^{2}(\pi / 2) \int_{0}^{\infty} y_{1} \exp \left(-y_{1}^{2} / 2\right) d y_{1} \
&=c^{2} \pi / 2
\end{aligned}
$$
Thus, the answer is $c=\sqrt{2 / \pi}$ :
$$
\int_{0}^{\infty} \frac{\exp \left(-x^{2} / 2\right)}{\sqrt{\pi / 2}} d x=1 .
$$
计量经济学代写
经济代写|计量经济学代写ECONOMETRICS代考|THE LINEAR CASE
让X=(X1,X2)吨是具有分布函数的二元随机向量
F(X)=∫−∞X1∫−∞X2F(在1,在2)d在1d在2=∫(−∞,X1]×(−∞,X2]F(在)d在
在哪里X=(X1,X2)吨,在=(在1,在2)吨.
在本节中,我将推导出联合密度是=一种X+b, 在哪里一种是一个n这nr一种nd这米非单数2×2矩阵和b是非随机的2×1向量。
从线性代数中回忆s和和一种pp和nd一世X一世任何方阵一种可以分解为
一种=R−1大号⋅D⋅在
在哪里R是一个置换矩阵p这ss一世bl是和q在一种l吨这吨H和在n一世吨米一种吨r一世X$一世$,大号是一个对角元素都等于的下三角矩阵1,在是一个对角元素都等于 1 的上三角矩阵,并且D是对角矩阵。转型是=一种X+b因此可以分五个步骤进行:
从1=在X 从2=D从1 从3=大号从2 从4=R−1从3 是=从4+b
因此,我将考虑前四种情况,一种=在,一种=D,一种=大号, 和一种=R−1为了b=0然后是这种情况一种=一世,b≠0.
让是=一种X和一种上三角矩阵:
一种=(1一种 01).
然后
是=(是1 是2)=(X1+一种X2 X2)
因此,联合分布函数H(是)的是是
H(是)=磷(是1≤是1,是2≤是2)=磷(X1+一种X2≤是1,X2≤是2) =和[一世(X1≤是1−一种X2)一世(X2≤是2)] =和(和[一世(X1≤是1−一种X2)∣X2]一世(X2≤是2)) =∫−∞是2(∫−∞是1−一种X2F1∣2(X1∣X2)dX1)F2(X2)dX2 =∫−∞是2(∫−∞是1−一种X2F(X1,X2)dX1)dX2
在哪里F1∣2(X1∣X2)是条件密度X1给定X2=X2和F2(X2)是边际密度X2. 如果我们取偏导数,它遵循4.22那是为了是=一种X和一种由4.21,
H(是)=∂2H(是)∂是1∂是2=∂∂是2∫−∞是2F(是1−一种X2,X2)dX2 =F(是1−一种是2,是2)=F(一种−1是).
经济代写|计量经济学代写ECONOMETRICS代考|THE NONLINEAR CASE
如果我们表示G(X)=一种X+b,G−1(是)=一种−1(是−b), 那么定理的结果4.3读H(是)=F(G−1(是))|这(∂G−1(是)/∂是)|. 这表明定理4.3可以概括如下:
定理 4.4:让X是ķ-variate,绝对连续分布与联合密度F(X),X=(X1,…,Xķ)吨, 然后让是=G(X), 在哪里G(X)= (G1(X),…,Gķ(X))吨是具有逆映射的一对一映射X=$G^{-1}是=\left(g_{1}^{ }是, \ldots, g_{k}^{ }是\right)^{\mathrm{T}}在H这s和C这米p这n和n吨s一种r和d一世FF和r和n吨一世一种bl和一世n吨H和C这米p这n和n吨s这Fy=\左y_{1}, \ldots, y_{k}\righty_{1}, \ldots, y_{k}\right^{\mathrm{T}}.大号和吨Ĵ是=\partial x / \partial y=\partial G^{-1}是/ \部分y,吨H一种吨一世s,Ĵ是一世s吨H和米一种吨r一世X在一世吨H我, j‘s和l和米和n吨\partial g_{i}^{*}是/ \部分y_{j},在H一世CH一世sC一种ll和d吨H和Ĵ一种C这b一世一种n.吨H和n是一世sķ−在一种r一世一种吨和,一种bs这l在吨和l是C这n吨一世n在这在sl是d一世s吨r一世b在吨和d在一世吨Hj这一世n吨d和ns一世吨是H是=f\左G−1(是\ 右)| \ 操作员名称 {det}Ĵ(是)|F这r是一世n吨H和s和吨G\左\mathbb{R}^{k}\对\mathbb{R}^{k}\对=\left{y \in \mathbb{R}^{k}: y=GX\对。,\left.fX>0, x \in \mathbb{R}^{k}\right}一种ndH是=0$ 其他地方。
这个猜想确实是真的。其形式证明在附录 4.B 中给出。
定理的应用4.4是下面的问题。考虑函数
F(X)=C⋅经验(−X2/2) 如果 X≥0, =0 如果 X<0对于哪个值C这个函数是密度吗?为了解决这个问题,考虑联合密度F(X1,X2)=C2经验[−(X12+ X22)/2],X1≥0,X2≥0,这是联合分布X=(X1,X2)吨, 在哪里X1和X2是来自密度分布的独立随机图4.26. 接下来,考虑转换是=(是1,是2)吨=G(X)被定义为是1=X12+X22∈(0,∞)是2=反正切(X1/X2)∈(0,圆周率/2).逆X=G−1(是)这种转变是X1=是1罪(是2),X2=是1因(是2)与雅可比Ĵ(是)=(∂X1/∂是1∂X1/∂是2∂X2/∂是1∂X2/∂是2)=(罪(是2)是1因(是2)因(是2)−是1罪(是2)).注意这[Ĵ(是)]=−是1. 因此,密度H(是)=H(是1,是2)= F(G−1(是))|这(Ĵ(是))|是H(是1,是2)=C2是1经验(−是12/2) 为了 是1>0 和 0<是2<圆周率/2, =0 别处;
因此,
1=∫0∞∫0圆周率/2C2是1经验(−是12/2)d是2d是1 =C2(圆周率/2)∫0∞是1经验(−是12/2)d是1 =C2圆周率/2
因此,答案是C=2/圆周率:
∫0∞经验(−X2/2)圆周率/2dX=1.
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