### 统计代写|似然估计作业代写Probability and Estimation代考|Bias and MSE Expressions

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• Statistical Inference 统计推断
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• 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|Bias and MSE Expressions

From (2.65), it is easy to see that the bias expression of $\hat{\theta}{n}^{\mathrm{RR}}(k)$ and $\hat{\beta}{n}^{\mathrm{RR}}(k)$, respectively, are given by
\begin{aligned} &b\left(\hat{\theta}{n}^{\mathrm{RR}}(k)\right)=-\frac{k}{Q+k} \beta \bar{x} \ &b\left(\hat{\beta}{n}^{\mathrm{RR}}(k)\right)=-\frac{k}{Q+k} \beta . \end{aligned}
Similarly, MSE expressions of the estimators are given by
\begin{aligned} \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{RR}}(k)\right) &=\frac{\sigma^{2}}{n}\left(1+\frac{n \bar{x}^{2} Q}{(Q+k)^{2}}\right)+\frac{k^{2} \beta^{2} \bar{x}^{2}}{(Q+k)^{2}} \ &=\frac{\sigma^{2}}{n}\left{\left(1+\frac{n \bar{x}^{2} Q}{(Q+k)^{2}}\right)+\frac{n \bar{x}^{2} k^{2} \beta^{2}}{(Q+k)^{2} \sigma^{2}}\right} \ &=\frac{\sigma^{2}}{n}\left{1+\frac{n \bar{x}^{2}}{Q(Q+k)^{2}}\left(Q^{2}+k^{2} \Delta^{2}\right)\right} \end{aligned} where $\Delta^{2}=\frac{Q \rho^{2}}{\sigma^{2}}$ and \begin{aligned} \operatorname{MSE}\left(\hat{\beta}{n}^{\mathrm{RR}}(k)\right) &=\frac{\sigma^{2} Q}{(Q+k)^{2}}+\frac{k^{2} \beta^{2}}{(Q+k)^{2}} \ &=\frac{\sigma^{2}}{Q(Q+k)^{2}}\left(Q^{2}+k^{2} \Delta^{2}\right) . \end{aligned}
Hence, the REff of these estimators are given by
\begin{aligned} \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{RR}}(k): \tilde{\theta}{n}\right) &=\left(1+\frac{n \bar{x}^{2}}{Q}\right)\left{1+\frac{n \bar{x}^{2}}{Q(Q+k)^{2}}\left(Q^{2}+k^{2} \Delta^{2}\right)\right}^{-1} \ \operatorname{REff}\left(\hat{\beta}{n}^{\mathrm{RR}}(k): \tilde{\beta}{n}\right) &=(Q+k)^{2}\left(Q^{2}+k \Delta^{2}\right)^{-1} \ &=\left(1+\frac{k}{Q}\right)\left{1+\frac{k \Delta^{2}}{Q^{2}}\right}^{-1} \end{aligned}

Note that the optimum value of $k$ is $Q \Delta^{-2}$. Hence,
\begin{aligned} \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{RR}}\left(Q \Delta^{-2}\right)\right) &=\frac{\sigma^{2}}{n}\left(1+\frac{n \bar{x}^{2} \Delta^{2}}{Q\left(1+\Delta^{2}\right)}\right) \ \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{RR}}\left(Q \Delta^{-2}\right): \tilde{\theta}{n}\right) &=\left(1+\frac{n \bar{x}^{2}}{Q}\right)\left{1+\frac{n \bar{x}^{2} \Delta^{2}}{Q\left(1+\Delta^{2}\right)}\right}^{-1} \ \operatorname{MSE}\left(\hat{\beta}{n}^{\mathrm{RR}}\left(Q \Delta^{-2}\right)\right) &=\frac{\sigma^{2}}{Q}\left(\frac{1+\Delta^{2}}{\Delta^{2}}\right)\left(\frac{Q}{1+Q}\right) \ \operatorname{REff}\left(\hat{\beta}{n}^{\mathrm{RR}}\left(Q \Delta^{-2}\right): \tilde{\beta}{n}\right) &=\left(\frac{1+\Delta^{2}}{\Delta^{2}}\right)\left(\frac{Q}{1+Q}\right) \end{aligned}

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

In this section, we consider the LASSO estimation of $(\theta, \beta)$ when it is suspected that $\beta$ may be 0 . For this case, the solution is given by
$$\left(\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda), \hat{\beta}{n}^{\mathrm{LASSO}}(\lambda)\right)^{\top}=\operatorname{argmin}{(\theta), \beta}}\left{\left|Y-\theta \mathbf{1}{n}-\beta x\right|^{2}+\sqrt{Q} \lambda \sigma|\beta|\right} .$$
Explicitly, we find
\begin{aligned} \hat{\theta}{n}^{\mathrm{LASSO}}(\lambda) &=\bar{y}-\hat{\beta}{n}^{\mathrm{LASSO}} \bar{x} \ \hat{\beta}{n}^{\mathrm{LASSO}} &=\operatorname{sgn}(\tilde{\beta})\left(|\tilde{\beta}|-\lambda \frac{\sigma}{\sqrt{Q}}\right)^{+} \ &=\frac{\sigma}{\sqrt{Q}} \operatorname{sgn}\left(Z{n}\right)\left(\left|Z_{n}\right|-\lambda\right)^{+}, \end{aligned}
where $Z_{n}=\sqrt{Q} \tilde{\beta}{n} / \sigma \sim \mathcal{N}(\Delta, 1)$ and $\Delta=\sqrt{Q} \beta / \sigma$. According to Donoho and Johnstone (1994), and results of Section 2.2.5, the bias and MSE expressions for $\hat{\beta}{n}^{\mathrm{LASSO}}(\lambda)$ are given by
\begin{aligned} b\left(\hat{\beta}{n}^{\mathrm{LASSO}}(\lambda)\right)=& \frac{\sigma}{\sqrt{Q}}{\lambda[\Phi(\lambda-\Delta)-\Phi(\lambda+\Delta)]\ &+\Delta[\Phi(-\lambda-\Delta)-\Phi(\lambda-\Delta)] \ &+[\phi(\lambda-\Delta)-\phi(\lambda+\Delta)]}, \ \operatorname{MSE}\left(\hat{\beta}{n}^{\mathrm{LASSO}}(\lambda)\right)=& \frac{\sigma^{2}}{Q} \rho_{\mathrm{ST}}(\lambda, \Delta), \end{aligned}
where
\begin{aligned} \rho_{\mathrm{ST}}(\lambda, \Delta)=& 1+\lambda^{2}+\left(1-\Delta^{2}-\lambda^{2}\right){\Phi(\lambda-\Delta)-\Phi(-\lambda-\Delta)} \ &-(\lambda-\Delta) \phi(\lambda+\Delta)-(\lambda+\Delta) \phi(\lambda-\Delta) \end{aligned}

## 统计代写|似然估计作业代写Probability and Estimation代考|Summary and Concluding Remarks

This chapter considers the location model and the simple linear regression model when errors of the models are normally distributed. We consider LSE, RLSE, PTE, SE and two penalty estimators, namely, the RRE and the LASSO estimator for the location parameter for the location model and the intercept and slope parameter for the simple linear regression model. We found that the RRE uniformly dominates LSE, PTE, SE, and LASSO. However, RLSE dominates all estimators near the null hypothesis. LASSO dominates LSE, PTE, and SE uniformly.

## 统计代写|似然估计作业代写Probability and Estimation代考|Bias and MSE Expressions

b(θ^nRR(ķ))=−ķ问+ķbX¯ b(b^nRR(ķ))=−ķ问+ķb.

\begin{aligned} \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{RR}}(k)\right) &=\frac{\sigma^{2}}{n }\left(1+\frac{n \bar{x}^{2} Q}{(Q+k)^{2}}\right)+\frac{k^{2} \beta^{2} \bar{x}^{2}}{(Q+k)^{2}} \ &=\frac{\sigma^{2}}{n}\left{\left(1+\frac{n \ bar{x}^{2} Q}{(Q+k)^{2}}\right)+\frac{n \bar{x}^{2} k^{2} \beta^{2}} {(Q+k)^{2} \sigma^{2}}\right} \ &=\frac{\sigma^{2}}{n}\left{1+\frac{n \bar{x} ^{2}}{Q(Q+k)^{2}}\left(Q^{2}+k^{2} \Delta^{2}\right)\right} \end{对齐}\begin{aligned} \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{RR}}(k)\right) &=\frac{\sigma^{2}}{n }\left(1+\frac{n \bar{x}^{2} Q}{(Q+k)^{2}}\right)+\frac{k^{2} \beta^{2} \bar{x}^{2}}{(Q+k)^{2}} \ &=\frac{\sigma^{2}}{n}\left{\left(1+\frac{n \ bar{x}^{2} Q}{(Q+k)^{2}}\right)+\frac{n \bar{x}^{2} k^{2} \beta^{2}} {(Q+k)^{2} \sigma^{2}}\right} \ &=\frac{\sigma^{2}}{n}\left{1+\frac{n \bar{x} ^{2}}{Q(Q+k)^{2}}\left(Q^{2}+k^{2} \Delta^{2}\right)\right} \end{对齐}在哪里Δ2=问ρ2σ2和MSE⁡(b^nRR(ķ))=σ2问(问+ķ)2+ķ2b2(问+ķ)2 =σ2问(问+ķ)2(问2+ķ2Δ2).

\begin{aligned} \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{RR}}(k): \tilde{\theta}{n}\right) &=\left (1+\frac{n \bar{x}^{2}}{Q}\right)\left{1+\frac{n \bar{x}^{2}}{Q(Q+k)^ {2}}\left(Q^{2}+k^{2} \Delta^{2}\right)\right}^{-1} \ \operatorname{REff}\left(\hat{\beta} {n}^{\mathrm{RR}}(k): \tilde{\beta}{n}\right) &=(Q+k)^{2}\left(Q^{2}+k \Delta ^{2}\right)^{-1} \ &=\left(1+\frac{k}{Q}\right)\left{1+\frac{k \Delta^{2}}{Q^ {2}}\right}^{-1} \end{对齐}\begin{aligned} \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{RR}}(k): \tilde{\theta}{n}\right) &=\left (1+\frac{n \bar{x}^{2}}{Q}\right)\left{1+\frac{n \bar{x}^{2}}{Q(Q+k)^ {2}}\left(Q^{2}+k^{2} \Delta^{2}\right)\right}^{-1} \ \operatorname{REff}\left(\hat{\beta} {n}^{\mathrm{RR}}(k): \tilde{\beta}{n}\right) &=(Q+k)^{2}\left(Q^{2}+k \Delta ^{2}\right)^{-1} \ &=\left(1+\frac{k}{Q}\right)\left{1+\frac{k \Delta^{2}}{Q^ {2}}\right}^{-1} \end{对齐}

\begin{aligned} \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{RR}}\left(Q \Delta^{-2}\right)\right) &=\ frac{\sigma^{2}}{n}\left(1+\frac{n \bar{x}^{2} \Delta^{2}}{Q\left(1+\Delta^{2} \right)}\right) \ \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{RR}}\left(Q \Delta^{-2}\right): \tilde {\theta}{n}\right) &=\left(1+\frac{n \bar{x}^{2}}{Q}\right)\left{1+\frac{n \bar{x }^{2} \Delta^{2}}{Q\left(1+\Delta^{2}\right)}\right}^{-1} \ \operatorname{MSE}\left(\hat{\ beta}{n}^{\mathrm{RR}}\left(Q \Delta^{-2}\right)\right) &=\frac{\sigma^{2}}{Q}\left(\frac {1+\Delta^{2}}{\Delta^{2}}\right)\left(\frac{Q}{1+Q}\right) \ \operatorname{REff}\left(\hat{\ beta}{n}^{\mathrm{RR}}\left(Q \Delta^{-2}\right): \tilde{\beta}{n}\right) &=\left(\frac{1+ \Delta^{2}}{\Delta^{2}}\right)\left(\frac{Q}{1+Q}\right) \end{aligned}\begin{aligned} \operatorname{MSE}\left(\hat{\theta}{n}^{\mathrm{RR}}\left(Q \Delta^{-2}\right)\right) &=\ frac{\sigma^{2}}{n}\left(1+\frac{n \bar{x}^{2} \Delta^{2}}{Q\left(1+\Delta^{2} \right)}\right) \ \operatorname{REff}\left(\hat{\theta}{n}^{\mathrm{RR}}\left(Q \Delta^{-2}\right): \tilde {\theta}{n}\right) &=\left(1+\frac{n \bar{x}^{2}}{Q}\right)\left{1+\frac{n \bar{x }^{2} \Delta^{2}}{Q\left(1+\Delta^{2}\right)}\right}^{-1} \ \operatorname{MSE}\left(\hat{\ beta}{n}^{\mathrm{RR}}\left(Q \Delta^{-2}\right)\right) &=\frac{\sigma^{2}}{Q}\left(\frac {1+\Delta^{2}}{\Delta^{2}}\right)\left(\frac{Q}{1+Q}\right) \ \operatorname{REff}\left(\hat{\ beta}{n}^{\mathrm{RR}}\left(Q \Delta^{-2}\right): \tilde{\beta}{n}\right) &=\left(\frac{1+ \Delta^{2}}{\Delta^{2}}\right)\left(\frac{Q}{1+Q}\right) \end{aligned}

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

\left(\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda), \hat{\beta}{n}^{\mathrm{LASSO}}(\lambda)\right)^ {\top}=\operatorname{argmin}{(\theta), \beta}}\left{\left|Y-\theta \mathbf{1}{n}-\beta x\right|^{2}+ \sqrt{Q} \lambda \sigma|\beta|\right} 。\left(\hat{\theta}{n}^{\mathrm{LASSO}}(\lambda), \hat{\beta}{n}^{\mathrm{LASSO}}(\lambda)\right)^ {\top}=\operatorname{argmin}{(\theta), \beta}}\left{\left|Y-\theta \mathbf{1}{n}-\beta x\right|^{2}+ \sqrt{Q} \lambda \sigma|\beta|\right} 。

θ^n大号一种小号小号这(λ)=是¯−b^n大号一种小号小号这X¯ b^n大号一种小号小号这=sgn⁡(b~)(|b~|−λσ问)+ =σ问sgn⁡(从n)(|从n|−λ)+,

\begin{对齐} b\left(\hat{\beta}{n}^{\mathrm{LASSO}}(\lambda)\right)=& \frac{\sigma}{\sqrt{Q}}{\ λ[\Phi(\lambda-\Delta)-\Phi(\lambda+\Delta)]\ &+\Delta[\Phi(-\lambda-\Delta)-\Phi(\lambda-\Delta)] \ & +[\phi(\lambda-\Delta)-\phi(\lambda+\Delta)]}, \ \operatorname{MSE}\left(\hat{\beta}{n}^{\mathrm{LASSO}}( \lambda)\right)=& \frac{\sigma^{2}}{Q} \rho_{\mathrm{ST}}(\lambda, \Delta), \end{aligned}\begin{对齐} b\left(\hat{\beta}{n}^{\mathrm{LASSO}}(\lambda)\right)=& \frac{\sigma}{\sqrt{Q}}{\ λ[\Phi(\lambda-\Delta)-\Phi(\lambda+\Delta)]\ &+\Delta[\Phi(-\lambda-\Delta)-\Phi(\lambda-\Delta)] \ & +[\phi(\lambda-\Delta)-\phi(\lambda+\Delta)]}, \ \operatorname{MSE}\left(\hat{\beta}{n}^{\mathrm{LASSO}}( \lambda)\right)=& \frac{\sigma^{2}}{Q} \rho_{\mathrm{ST}}(\lambda, \Delta), \end{aligned}

ρ小号吨(λ,Δ)=1+λ2+(1−Δ2−λ2)披(λ−Δ)−披(−λ−Δ) −(λ−Δ)φ(λ+Δ)−(λ+Δ)φ(λ−Δ)

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