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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 金融代写|金融实证代写Financial Empirical 代考|Empirical Coefficient of Determination

From $\mathbf{y}=\hat{\mathbf{x}}+\hat{\mathbf{u}}$ and $\overline{\hat{\mathbf{u}}}=0$ (i.e. $\overline{\mathbf{y}}=\overline{\hat{\mathbf{x}}}$ ) follows for empirical covariances or variances (for $\mathbf{x}, \mathbf{y} \in \mathbb{R}n$ defined as $s{\mathbf{x y}}=\frac{1}{n} \mathbf{x}^{\prime} \mathbf{y}-\overline{\mathbf{x}} \overline{\mathbf{y}}$ and $s_{\mathbf{x}}^2=s_{\mathbf{x x}}$. The bar stands for the arithmetical mean)
with “empirical coefficients of determination”
$$R_0^2=\frac{\hat{\mathbf{x}}^{\prime} \hat{\mathbf{x}}}{\mathbf{y}^{\prime} \mathbf{y}}, \quad R^2=\frac{s_{\hat{\mathbf{x}}}^2}{s_{\mathbf{y}}^2} \quad \text { und } \quad 0 \leq R^2 \leq R_0^2 \leq 1, \quad 0 \leq \frac{\hat{\mathbf{x}}^{\prime} \hat{\mathbf{u}}}{\mathbf{y}^{\prime} \mathbf{y}} \leq \frac{s_{\hat{\mathbf{x}} \hat{\mathbf{u}}}}{s_{\mathbf{y}}^2}<\frac{1}{2} .$$

1. For $R_0^2=1$ or $R^2=1$ is $\hat{\mathbf{u}}=0$ and $\hat{\mathbf{x}}^{\prime} \hat{\mathbf{u}}=\lambda_1 Q_1\left(\hat{x}_1\right)+\lambda_2 Q_2\left(\hat{x}_2\right)=0$. Therefore we have an interpolation of data with a smoothest solution (with $T_1 \hat{x}_1=0$, $\left.T_2 \hat{x}_2=0\right)$
2. For $R^2=0$ resp. $\hat{\mathbf{x}}=\overline{\hat{\mathbf{x}}} \mathbf{1}=\overline{\mathbf{y}} 1$ is $s_{\hat{\mathbf{u}}}^2=s_{\mathbf{y}}^2, s_{\hat{\mathbf{x}} \hat{u}}=0 \operatorname{resp} \cdot \lambda_1 Q_1\left(\hat{x}_1\right)+\lambda_2 Q_2\left(\hat{x}_2\right)=$ 0 . Consequently $\hat{x}_1(t)=\overline{\mathbf{y}}, \hat{x}_2(t)=0$ is a solution (with $T_1 \hat{x}_1=0, T_2 \hat{x}_2=0$, $\hat{\mathbf{x}}=\overline{\mathbf{y}} \mathbf{1}$ ), “consisting of no trend and no season”.

## 金融代写|金融实证代写Financial Empirical 代考|Properties of Monotonicity

To investigate the monotonicity of the functions
$$Q_1\left(\hat{x}_1 ; \lambda_1, \lambda_2\right), \quad Q_2\left(\hat{x}_2 ; \lambda_1, \lambda_2\right), \quad Q\left(\hat{\mathbf{x}}_1, \hat{\mathbf{x}}_2 ; \mathbf{y} ; \lambda_1, \lambda_2\right), \quad \text { and } \quad S\left(\hat{x}_1, \hat{x}_2 ; \mathbf{y} ; \lambda_1, \lambda_2\right)$$
we look at the partial derivatives in $\lambda_1, \lambda_2$. With the general rule of derivation $\frac{\mathrm{d} M^{-1}(x)}{\mathrm{d} x}=-M^{-1}(x) \frac{\mathrm{d} M(x)}{\mathrm{d} x} M^{-1}(x) \quad$ and especially $\quad \frac{\partial G}{\partial \lambda_i}=-\frac{1}{\lambda_i^2} G_i$ (because $G=\frac{1}{\lambda_1} G_1+\frac{1}{\lambda_2} G_2$ ) after Remark 5 holds
$$\frac{\partial A}{\partial \lambda_i}=-\left(I+A^* G\right)^{-1}\left(-A^* \frac{1}{\lambda_i^2} G_i\right)\left(I+A^* G\right)^{-1} A^*=\frac{1}{\lambda_i^2} A G_i A=\frac{1}{\lambda_i} W_i^{\prime} A, \quad i=1,2 .$$
The result is the “matrix of smoothness” of the weight function $\mathbf{w}_i(t)$.

• For the minimum $S\left(\hat{x}_1, \hat{x}_2 ; \mathbf{y}\right)$ holds
$$\frac{\partial S\left(\hat{x}_1, \hat{x}_2 ; \mathbf{y}\right)}{\partial \lambda_i}=\frac{1}{\lambda_t} \mathbf{y}^{\prime} W_i^{\prime} A \mathbf{y}=Q_i\left(\hat{x}_i\right) \geq 0, \quad i=1,2 .$$
Therefore $S\left(\hat{x}_1, \hat{x}_2 ; \mathbf{y} ; \lambda_1, \lambda_2\right)$ is monotonically not falling in both directions $\lambda_1$, $\lambda_2$. Furthermore it is convex (cf. Hebbel 2000):
• For $Q\left(\hat{\mathbf{x}}_1, \hat{\mathbf{x}}_2 ; \mathbf{y}\right)$ holds (because of symmetry)
$$\frac{\partial Q\left(\hat{\mathbf{x}}_1, \hat{\mathbf{x}}_2 ; \mathbf{y}\right)}{\partial \lambda_i}=\mathbf{y}^{\prime}\left(A \frac{\partial A}{\partial \lambda_i}+\frac{\partial A}{\partial \lambda_i} A\right) \mathbf{y}=\frac{2}{\lambda_i^2} \mathbf{y}^{\prime} W_i^{\prime} A A \mathbf{y}=\frac{2}{\lambda_i} \hat{\mathbf{x}}_i^{\prime} A \hat{\mathbf{u}} \geq 0, \quad i=1,2 .$$
• For $Q_i\left(\hat{x}_i\right)$ resp. $\lambda_i Q_i\left(\hat{x}_i\right)=\hat{\mathbf{x}}_i^{\prime} \hat{\mathbf{u}}=\frac{1}{\lambda_i} A G_i A$ hold (because of symmetry) with application of
$$\frac{\partial\left(A G_1 A\right)}{\partial \lambda_i}=\frac{1}{\lambda_i} W_i^{\prime} A G_1 A+\frac{1}{\lambda_i} A G_1 A W_i=2 \frac{\lambda_1}{\lambda_i} W_i^{\prime} A W_1 . \quad \frac{\partial\left(A G_2 A\right)}{\partial \lambda_i}=2 \frac{\lambda_2}{\lambda_i} W_i^{\prime} A W_2$$

the relationships
\begin{aligned} &\frac{\partial\left(\hat{\mathbf{x}}_1^{\prime} \hat{\mathbf{u}}\right)}{\partial \lambda_1}=-\frac{1}{\lambda_1} \hat{\mathbf{x}}_1^{\prime} \hat{\mathbf{u}}+\frac{2}{\lambda_1} \hat{\mathbf{x}}_1^{\prime} A \hat{\mathbf{x}}_1, \quad \frac{\partial\left(\hat{\mathbf{x}}_1^{\prime} \hat{\mathbf{u}}\right)}{\partial \lambda_2}=\frac{2}{\lambda_2} \hat{\mathbf{x}}_1^{\prime} A \hat{\mathbf{x}}_2 \geq 0, \ &\frac{\partial\left(\hat{\mathbf{x}}_2^{\prime} \hat{\mathbf{u}}\right)}{\partial \lambda_1}=\frac{2}{\lambda_1} \hat{\mathbf{x}}_2^{\prime} A \hat{\mathbf{x}}_1 \geq 0, \quad \frac{\partial\left(\hat{\mathbf{x}}_2^{\prime} \hat{\mathbf{u}}\right)}{\partial \lambda_2}=-\frac{1}{\lambda_2} \hat{\mathbf{x}}_2^{\prime} \hat{\mathbf{u}}+\frac{2}{\lambda_2} \hat{\mathbf{x}}_2^{\prime} A \hat{\mathbf{x}}_2 \end{aligned}
and analogously
\begin{aligned} &\frac{\partial Q_1\left(\hat{x}_1\right)}{\partial \lambda_1}=-\frac{2}{\lambda_1^2} \hat{\mathbf{x}}_1^{\prime} \hat{\mathbf{u}}+\frac{2}{\lambda_1^2} \hat{\mathbf{x}}_1^{\prime} A \hat{\mathbf{x}}_1 \leq 0, \quad \frac{\partial Q_1\left(\hat{x}_1\right)}{\partial \lambda_2}=\frac{2}{\lambda_1 \lambda_2} \hat{\mathbf{x}}_1^{\prime} A \hat{\mathbf{x}}_2 \geq 0, \ &\frac{\partial Q_2\left(\hat{x}_2\right)}{\partial \lambda_1}=\frac{2}{\lambda_1 \lambda_2} \hat{\mathbf{x}}_2^{\prime} A \hat{\mathbf{x}}_1 \geq 0, \quad \frac{\partial Q_2\left(\hat{x}_2\right)}{\partial \lambda_2}=-\frac{2}{\lambda_2^2} \hat{\mathbf{x}}_2^{\prime} \hat{\mathbf{u}}+\frac{2}{\lambda_2^2} \hat{\mathbf{x}}_2^{\prime} A \hat{\mathbf{x}}_2 \leq 0 . \end{aligned}

# 金融实证代考

## 金融代写|金融实证代写Financial Empirical 代考|Empirical Coefficient of Determination

$$R_0^2=\frac{\hat{\mathbf{x}}^{\prime} \hat{\mathbf{x}}}{\mathbf{y}^{\prime} \mathbf{y}}, \quad R^2=\frac{s_{\hat{\mathbf{x}}}^2}{s_{\mathbf{y}}^2} \quad \text { und } \quad 0 \leq R^2 \leq R_0^2 \leq 1, \quad 0 \leq \frac{\hat{\mathbf{x}}^{\prime} \hat{\mathbf{u}}}{\mathbf{y}^{\prime} \mathbf{y}} \leq \frac{s_{\hat{\mathbf{x}} \hat{\mathbf{u}}}}{s_{\mathbf{y}}^2}<\frac{1}{2} .$$

1. 为了 $R_0^2=1$ 或者 $R^2=1$ 是 $\hat{\mathbf{u}}=0$ 和 $\hat{\mathbf{x}}^{\prime} \hat{\mathbf{u}}=\lambda_1 Q_1\left(\hat{x}1\right)+\lambda_2 Q_2\left(\hat{x}_2\right)=0$. 因此，我们有一个具有最平滑解的数据揷值 (使用 $\left.T_1 \hat{x}_1=0, T_2 \hat{x}_2=0\right)$ 2. 为了 $R^2=0$ 分别 $\hat{\mathbf{x}}=\overline{\hat{\mathbf{x}}} \mathbf{1}=\overline{\mathbf{y}} 1$ 是 $s{\hat{\mathbf{u}}}^2=s_{\mathbf{y}}^2, s_{\hat{\mathbf{x}} \hat{\mathbf{u}}}=0 \operatorname{resp} \cdot \lambda_1 Q_1\left(\hat{x}_1\right)+\lambda_2 Q_2\left(\hat{x}_2\right)=0$. 最后 $\hat{x}_1(t)=\overline{\mathbf{y}}, \hat{x}_2(t)=0$ 是一个解决方案（与 $T_1 \hat{x}_1=0, T_2 \hat{x}_2=0, \hat{\mathbf{x}}=\overline{\mathbf{y}} \mathbf{1}$ ， “由无趋势和无季节组成”。

## 金融代写|金融实证代写Financial Empirical 代考|Properties of Monotonicity

$$Q_1\left(\hat{x}_1 ; \lambda_1, \lambda_2\right), \quad Q_2\left(\hat{x}_2 ; \lambda_1, \lambda_2\right), \quad Q\left(\hat{\mathbf{x}}_1, \hat{\mathbf{x}}_2 ; \mathbf{y} ; \lambda_1, \lambda_2\right), \quad \text { and } \quad S\left(\hat{x}_1, \hat{x}_2 ; \mathbf{y} ; \lambda_1, \lambda_2\right)$$
䖸们看一下偏导数 $\lambda_1, \lambda_2$. 用一般推导规则 $\frac{\mathrm{d} M^{-1}(x)}{\mathrm{d} x}=-M^{-1}(x) \frac{\mathrm{d} M(x)}{\mathrm{d} x} M^{-1}(x)$ 尤其是 $\frac{\partial G}{\partial \lambda i}=-\frac{1}{\lambda_i^2} G_i$ (因为 $G=\frac{1}{\lambda_1} G_1+\frac{1}{\lambda 2} G_2$ ) 在备注 5 成立后
$$\frac{\partial A}{\partial \lambda_i}=-\left(I+A^* G\right)^{-1}\left(-A^* \frac{1}{\lambda_i^2} G_i\right)\left(I+A^* G\right)^{-1} A^*=\frac{1}{\lambda_i^2} A G_i A=\frac{1}{\lambda_i} W_i^{\prime} A, \quad i=1,2 .$$

• 对于最低限度 $S\left(\hat{x}_1, \hat{x}_2 ; \mathbf{y}\right)$ 持有
$$\frac{\partial S\left(\hat{x}_1, \hat{x}_2 ; \mathbf{y}\right)}{\partial \lambda_i}=\frac{1}{\lambda_t} \mathbf{y}^{\prime} W_i^{\prime} A \mathbf{y}=Q_i\left(\hat{x}_i\right) \geq 0, \quad i=1,2 .$$
所以 $S\left(\hat{x}_1, \hat{x}_2 ; \mathbf{y} ; \lambda_1, \lambda_2\right)$ 不单调地在两个方向下降 $\lambda_1, \lambda_2$. 此外，它是凸的（参见 Hebbel 2000) :
• 为了 $Q\left(\hat{\mathbf{x}}_1, \hat{\mathbf{x}}_2 ; \mathbf{y}\right)$ 成立 (因为对称)
$$\frac{\partial Q\left(\hat{\mathbf{x}}_1, \hat{\mathbf{x}}_2 ; \mathbf{y}\right)}{\partial \lambda_i}=\mathbf{y}^{\prime}\left(A \frac{\partial A}{\partial \lambda_i}+\frac{\partial A}{\partial \lambda_i} A\right) \mathbf{y}=\frac{2}{\lambda_i^2} \mathbf{y}^{\prime} W_i^{\prime} A A \mathbf{y}=\frac{2}{\lambda_i} \hat{\mathbf{x}}_i^{\prime} A \hat{\mathbf{u}} \geq 0, \quad i=1,2 .$$
• 为了 $Q_i\left(\hat{x}_i\right)$ 分别 $\lambda_i Q_i\left(\hat{x}_i\right)=\hat{\mathbf{x}}_i^{\prime} \hat{\mathbf{u}}=\frac{1}{\lambda_i} A G_i A$ 保持 (因为对称) 与应用
$$\frac{\partial\left(A G_1 A\right)}{\partial \lambda_i}=\frac{1}{\lambda_i} W_i^{\prime} A G_1 A+\frac{1}{\lambda_i} A G_1 A W_i=2 \frac{\lambda_1}{\lambda_i} W_i^{\prime} A W_1 . \quad \frac{\partial\left(A G_2 A\right)}{\partial \lambda_i}=2 \frac{\lambda_2}{\lambda_i} W_i^{\prime} A W_2$$
关系
$$\frac{\partial\left(\hat{\mathbf{x}}_1^{\prime} \hat{\mathbf{u}}\right)}{\partial \lambda_1}=-\frac{1}{\lambda_1} \hat{\mathbf{x}}_1^{\prime} \hat{\mathbf{u}}+\frac{2}{\lambda_1} \hat{\mathbf{x}}_1^{\prime} A \hat{\mathbf{x}}_1, \quad \frac{\partial\left(\hat{\mathbf{x}}_1^{\prime} \hat{\mathbf{u}}\right)}{\partial \lambda_2}=\frac{2}{\lambda_2} \hat{\mathbf{x}}_1^{\prime} A \hat{\mathbf{x}}_2 \geq 0, \quad \frac{\partial\left(\hat{\mathbf{x}}_2^{\prime} \hat{\mathbf{u}}\right)}{\partial \lambda_1}=\frac{2}{\lambda_1} \hat{\mathbf{x}}_2^{\prime} A \hat{\mathbf{x}}_1 \geq 0, \quad \frac{\partial\left(\hat{\mathbf{x}}_2^{\prime} \hat{\mathbf{u}}\right)}{\partial \lambda_2}=-\frac{1}{\lambda_2} \hat{\mathbf{x}}_2^{\prime} \hat{\mathbf{u}}+\frac{2}{\lambda_2} \hat{\mathbf{x}}_2^{\prime} A \hat{\mathbf{x}}_2$$
和类似地
$$\frac{\partial Q_1\left(\hat{x}_1\right)}{\partial \lambda_1}=-\frac{2}{\lambda_1^2} \hat{\mathbf{x}}_1^{\prime} \hat{\mathbf{u}}+\frac{2}{\lambda_1^2} \hat{\mathbf{x}}_1^{\prime} A \hat{\mathbf{x}}_1 \leq 0, \quad \frac{\partial Q_1\left(\hat{x}_1\right)}{\partial \lambda_2}=\frac{2}{\lambda_1 \lambda_2} \hat{\mathbf{x}}_1^{\prime} A \hat{\mathbf{x}}_2 \geq 0, \quad \frac{\partial Q_2\left(\hat{x}_2\right)}{\partial \lambda_1}=\frac{2}{\lambda_1 \lambda_2} \hat{\mathbf{x}}_2^{\prime} A \hat{\mathbf{x}}_1 \geq 0, \quad \frac{\partial Q_2\left(\hat{x}_2\right)}{\partial \lambda_2}=-\frac{2}{\lambda_2^2} \hat{\mathbf{x}}_2^{\prime} \hat{\mathbf{u}}+\frac{2}{\lambda_2^2} \hat{\mathbf{x}}_2^{\prime} A \hat{\mathbf{x}}$$

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