### 统计代写|似然估计作业代写Probability and Estimation代考|Simple Linear Model

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等楖率论
• Advanced Mathematical Statistics 高等数理统计学
• (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

## 统计代写|似然估计作业代写Probability and Estimation代考|Estimation of the Intercept and Slope Parameters

First, we consider the LSE of the parameters. Using the model $(2.1)$ and the sample information from the normal distribution, we obtain the LSEs of $(\theta, \beta)^{\top}$ as
$$\left(\begin{array}{c} \tilde{\theta}{n} \ \tilde{\beta}{n} \end{array}\right)=\left(\begin{array}{c} \bar{y}-\tilde{\beta}{n} \bar{x} \ \frac{1}{Q}\left[x^{\top} \boldsymbol{Y}-\frac{1}{n}\left(\mathbf{1}{n}^{\top} \boldsymbol{x}\right)\left(\mathbf{1}_{n}^{\top} \boldsymbol{Y}\right)\right] \end{array}\right)$$

where
$$\bar{x}=\frac{1}{n} \mathbf{1}{n}^{\top} \boldsymbol{x}, \quad \bar{y}=\frac{1}{n} \mathbf{1}{n}^{\top} Y, \quad Q=\boldsymbol{x}^{\top} \boldsymbol{x}-\frac{1}{n}\left(\mathbf{1}{n}^{\top} \boldsymbol{x}\right)^{2} .$$ The exact distribution of $\left(\tilde{\theta}{n}, \tilde{\beta}{n}\right)^{\top}$ is a bivariate normal with mean $(\theta, \beta)^{\top}$ and covariance matrix $$\frac{\sigma^{2}}{n}\left(\begin{array}{cc} 1+\frac{n \bar{x}^{2}}{Q} & -\frac{n \bar{x}}{Q} \ \frac{n \bar{x}}{Q} & \frac{n}{Q} \end{array}\right)$$ An unbiased estimator of the variance $\sigma^{2}$ is given by $$s{n}^{2}=(n-2)^{-1}\left(\boldsymbol{Y}-\tilde{\theta}{n} \mathbf{1}{n}-\tilde{\beta}{n} \boldsymbol{x}\right)^{\top}\left(\boldsymbol{Y}-\tilde{\theta}{n} \mathbf{1}{n}-\tilde{\beta}{n} \boldsymbol{x}\right),$$
which is independent of $\left(\tilde{\theta}{n}, \tilde{\beta}{n}\right)$, and $(n-2) s_{n}^{2} / \sigma^{2}$ follows a central chi-square distribution with ( $n-2$ ) degrees of freedom (DF)

## 统计代写|似然估计作业代写Probability and Estimation代考|Test for Slope Parameter

Suppose that we want to test the null-hypothesis $\mathcal{H}{0}: \beta=\beta{o}$ vs. $\mathcal{H}{A}: \beta \neq \beta{o}$. Then, we use the likelihood ratio (LR) test statistic
$\mathcal{L}{n}^{(\sigma)}=\frac{\left(\tilde{\beta}{n}-\beta_{\mathrm{o}}\right)^{2} Q}{\sigma^{2}}, \quad$ if $\sigma^{2}$ is known
$\mathcal{L}{n}^{(s)}=\frac{\left(\tilde{\beta}{n}-\beta_{\mathrm{o}}\right)^{2} Q}{s_{n}^{2}}, \quad$ if $\sigma^{2}$ is unknown
where $\mathcal{L}{n}^{(\sigma)}$ follows a noncentral chi-square distribution with $1 \mathrm{DF}$ and noncentrality parameter $\Delta^{2} / 2$ and $\mathcal{L}{n}^{(s)}$ follows a noncentral $F$-distribution with $(1, m)$, where $m=n-2$ is DF and also the noncentral parameter is
$$\Delta^{2}=\frac{\left(\beta-\beta_{\mathrm{o}}\right)^{2} Q}{\sigma^{2}}$$
Under $\mathcal{H}{\mathrm{o}}, \mathcal{L}{n}^{(\sigma)}$ follows a central chi-square distribution and $\mathcal{L}{n}^{(s)}$ follows a central $F$-distribution. At the $\alpha$-level of significance, we obtain the critical value $\chi{1}^{2}(\alpha)$ or $F_{1, m}(\alpha)$ from the distribution and reject $\mathcal{H}{0}$ if $\mathcal{L}{n}^{(\sigma)}>\chi_{1}^{2}(\alpha)$ or $\mathcal{L}{n}^{(s)}>$ $F{1, n}(\alpha)$; otherwise, we accept $\mathcal{H}_{\mathrm{o}^{*}}$

## 统计代写|似然估计作业代写Probability and Estimation代考|PTE of the Intercept and Slope Parameters

This section deals with the problem of estimation of the intercept and slope parameters $(\theta, \beta)$ when it is suspected that the slope parameter $\beta$ may be $\beta_{o^{\circ}}$ From (2.30), we know that the LSE of $\theta$ is given by
$$\tilde{\theta}{n}=\bar{y}-\tilde{\beta}{n} \bar{x} .$$

If we know $\beta$ to be $\beta_{0}$ exactly, then the restricted least squares estimator (RLSE) of $\theta$ is given by
$$\hat{\theta}{n}=\bar{y}-\beta{\mathrm{o}} \bar{x} .$$
In practice, the prior information that $\beta=\beta_{\mathrm{o}}$ is uncertain. The doubt regarding this prior information can be removed using Fisher’s recipe of testing the null-hypothesis $\mathcal{H}{o}: \beta=\beta{o}$ against the alternative $\mathcal{H}{A}: \beta \neq \beta{o}$. As a result of this test, we choose $\tilde{\theta}{n}$ or $\hat{\theta}{n}$ based on the rejection or acceptance of $\mathcal{H}{\mathrm{a}}$. Accordingly, in case of the unknown variance, we write the estimator as $$\hat{\theta}{n}^{\mathrm{PT}}(\alpha)=\hat{\theta}{n} I\left(\mathcal{L}{n}^{(s)} \leq F_{1, m}(\alpha)\right)+\tilde{\theta}{n} I\left(\mathcal{L}{n}^{(s)}>F_{1, m}(\alpha)\right), \quad m=n-2$$
called the PTE, where $F_{1, n}(\alpha)$ is the $\alpha$-level upper critical value of a central $F$-distribution with $(1, m)$ DF and $I(A)$ is the indicator function of the set $A$. For more details on PTE, see Saleh (2006), Ahmed and Saleh (1988), Ahsanullah and Saleh (1972), Kibria and Saleh (2012) and, recently Saleh et al. (2014), among others. We can write PTE of $\theta$ as
$$\hat{\theta}{n}^{\mathrm{pT}}(\alpha)=\tilde{\theta}{n}+\left(\tilde{\beta}{n}-\beta{\mathrm{o}}\right) \bar{x} I\left(\mathcal{L}{n}^{(s)} \leq F{1, m}(\alpha)\right), \quad m=n-2$$
If $\alpha=1, \tilde{\theta}{n}$ is always chosen; and if $\alpha=0, \hat{\theta}{n}$ is chosen. Since $0<\alpha<1$, $\hat{\theta}{n}^{\mathrm{pT}}(\alpha)$ in repeated samples, this will result in a combination of $\tilde{\theta}{n}$ and $\hat{\theta}{n}$. Note that the PTE procedure leads to the choice of one of the two values, namely, either $\dot{\theta}{n}$ or $\hat{\theta}_{n}$. Also, the PTE procedure depends on the level of significance $\alpha$.

Clearly, $\hat{\beta}{n}$ is the unrestricted estimator of $\beta$, while $\beta{\mathrm{o}}$ is the restricted estimator. Thus, the PTE of $\beta$ is given by
$$\hat{\beta}{n}^{P \mathrm{~T}}(\alpha)=\tilde{\beta}{n}-\left(\tilde{\beta}{a}-\beta{\mathrm{o}}\right) I\left(\mathcal{L}{n}^{(s)} \leq F{1, m}(\alpha)\right), \quad m=n-2 .$$
Now, if $\alpha=1, \tilde{\beta}{n}$ is always chosen; and if $\alpha=0, \beta{\mathrm{o}}$ is always chosen.
Since our interest is to compare the LSE, RLSE, and PTE of $\theta$ and $\beta$ with respect to bias and the MSE, we obtain the expression of these quantities in the following theorem. First we consider the bias expressions of the estimators.

## 统计代写|似然估计作业代写Probability and Estimation代考|Estimation of the Intercept and Slope Parameters

(θ~n b~n)=(是¯−b~nX¯ 1问[X⊤是−1n(1n⊤X)(1n⊤是)])

X¯=1n1n⊤X,是¯=1n1n⊤是,问=X⊤X−1n(1n⊤X)2.的准确分布(θ~n,b~n)⊤是具有均值的双变量正态(θ,b)⊤和协方差矩阵σ2n(1+nX¯2问−nX¯问 nX¯问n问)方差的无偏估计σ2是（谁）给的sn2=(n−2)−1(是−θ~n1n−b~nX)⊤(是−θ~n1n−b~nX),

Δ2=(b−b这)2问σ2

## 统计代写|似然估计作业代写Probability and Estimation代考|PTE of the Intercept and Slope Parameters

θ~n=是¯−b~nX¯.

θ^n=是¯−b这X¯.

θ^np吨(一种)=θ~n+(b~n−b这)X¯一世(大号n(s)≤F1,米(一种)),米=n−2

b^n磷 吨(一种)=b~n−(b~一种−b这)一世(大号n(s)≤F1,米(一种)),米=n−2.

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