# 统计代写|回归分析作业代写Regression Analysis代考|STA321

## 统计代写|回归分析作业代写Regression Analysis代考|Estimation and Practical

The parameter $\sigma^2$ is perhaps the most important parameter of a regression model because it measures prediction accuracy. As shown previously, another way to write the model is $Y=\beta_0+\beta_1 x+\varepsilon$, where $\varepsilon \sim \mathrm{N}\left(0, \sigma^2\right)$. Thus the prediction error terms are the $\varepsilon$ values, and these differ from zero with a variance of $\sigma^2$.

If the $\beta^{\prime}$ ‘s were known (as is true in simulations but not in reality), you could calculate errors $\varepsilon_i=\left{Y_i-\left(\beta_0+\beta_1 x_i\right)\right}$ and obtain an unbiased estimate of $\sigma^2$ as:
(An unbiased estimator of $\sigma^2$ ) $=\frac{1}{n} \sum_{i=1}^n \varepsilon_i^2$
This estimator is unbiased because each individual $\varepsilon_i^2$ is an unbiased estimator of $\sigma^2$, which you can see as follows:
$$\mathrm{E}\left(\varepsilon_i^2\right)=\mathrm{E}\left{\left(Y_i-\beta_0-\beta_1 x_i\right)^2\right}=\operatorname{Var}\left(Y_i \mid X=x_i\right)=\sigma^2 .$$
Self-study question: Using the fact that each $\varepsilon_i^2$ is an unbiased estimator of $\sigma^2$, along with the linearity and additivity properties of expectation, demonstrate mathematically that estimator $\frac{1}{n} \sum_{i=1}^n \varepsilon_i^2$ is an unbiased estimator of $\sigma^2$.

However, in practice, you cannot use this estimator because the $\beta$ ‘s are unknown; thus the $\varepsilon$ ‘s are unknown (or unobservable) as well. But you can use a similar estimator based on the residuals $e_i=\left{Y_i-\left(\hat{\beta}0+\hat{\beta}_1 x_i\right)\right}$, which are observable: (Another estimator of $\sigma^2$ ) $=\frac{1}{n} \sum{i=1}^n e_i^2$

## 统计代写|回归分析作业代写Regression Analysis代考|Standard Errors

The Gauss-Markov theorem states that the OLS estimator has minimum variance among linear unbiased estimators. What does “variance” of the OLS estimator refer to? Please look at Figure $3.1$ again: You can see that there is variability in the possible values of $\hat{\beta}_1$ ranging from $1.0$ to $2.0$. Variance of the estimator $\hat{\beta}_1$, denoted symbolically by $\operatorname{Var}\left(\hat{\beta}_1\right)$, refers to the variance of the distribution $p\left(\hat{\beta}_1\right)$ that is shown in Figure $3.1$.

If the assumptions of the Gauss-Markov model are true, then the following formula gives the exact variance of the OLS estimator $\hat{\beta}_1$.
Variance of the OLS estimator $\hat{\beta}_1$
$$\operatorname{Var}\left(\hat{\beta}_1\right)=\frac{\sigma^2}{(n-1) s_x^2}$$
In the formula for $\operatorname{Var}\left(\beta_1\right)$, note that $s_x^2=\sum\left(x_i-\hat{\mu}_x\right)^2 /(n-1)$ is the usual estimate of the variance of $X$. Note that the $\operatorname{Var}\left(\hat{\beta}_1\right)$ formula is conditional on the observed values of the $X$ data; this is apparent because $s_x^2$ is specifically a function of the observed $X$ data.

When coupled with unbiasedness of $\hat{\beta}_1$, smaller $\operatorname{Var}\left(\hat{\beta}_1\right)$ implies a more accurate estimate, i.e., an estimate that tends to be closer to $\beta_1$. Hence, we have the following interesting conclusions regarding the accuracy of the OLS estimate $\hat{\beta}_1$ :
The OLS estimate $\hat{\beta}_1$ of $\beta_1$ is more accurate when:

• $n$ is larger, and/or
• $s_x^2$ is larger, and/or
• $\sigma^2$ is smaller.
As mentioned above, the formula given for $\operatorname{Var}\left(\hat{\beta}_1\right)$ can be mathematically derived from the assumptions of the Gauss-Markov model. Violation of assumptions renders the formula incorrect. In particular, violation of the homoscedasticity assumption is the rationale for using heteroscedasticity-consistent standard errors, which are covered in Chapter 12.
Strangely enough, the mathematics needed to prove the variance formula is easier in the multiple regression model, so we will prove it later in Chapter 7. But for now, you should understand the assumptions that imply the result (e.g., the classical model) and the result itself (the formula for $\operatorname{Var}\left(\hat{\beta}_1\right)$ ) by using simulation: If you simulate many thousands of data sets from the same model, with the same sample size, and with the same $X$ data, then the sample variance estimate of the resulting thousands of $\hat{\beta}_1$ estimates will be (within simulation error) equal to $\sigma^2 /\left{(n-1) s_x^2\right}$. The simulation also clarifies the “conditional on observed values of the $X$ data” interpretation because the $X$ data are the same for every simulated data set.

# 回归分析代考

## 统计代写|回归分析作业代写回归分析代考|估计与实践

($\sigma^2$的无偏估计)$=\frac{1}{n} \sum_{i=1}^n \varepsilon_i^2$

$$\mathrm{E}\left(\varepsilon_i^2\right)=\mathrm{E}\left{\left(Y_i-\beta_0-\beta_1 x_i\right)^2\right}=\operatorname{Var}\left(Y_i \mid X=x_i\right)=\sigma^2 .$$

## 统计代写|回归分析作业代写回归分析代考|标准误差

OLS估计量的方差$\hat{\beta}_1$
$$\operatorname{Var}\left(\hat{\beta}_1\right)=\frac{\sigma^2}{(n-1) s_x^2}$$

$\beta_1$的OLS估计$\hat{\beta}_1$在以下条件下更准确:

• $n$较大，/或
• $s_x^2$较大，/或
• $\sigma^2$较小。如上所述，$\operatorname{Var}\left(\hat{\beta}_1\right)$给出的公式可以从高斯-马尔可夫模型的假设中数学推导出来。违反假设使公式不正确。特别是，违反同方差假设是使用异方差一致标准误差的基本原理，这将在第12章中介绍。奇怪的是，证明方差公式所需的数学在多元回归模型中更容易，所以我们将在后面的第7章中证明它。但是现在，您应该通过使用模拟来理解隐含结果(例如，经典模型)和结果本身($\operatorname{Var}\left(\hat{\beta}_1\right)$的公式)的假设:如果您模拟来自相同模型的数千个数据集，具有相同的样本量和相同的$X$数据，那么结果数千个$\hat{\beta}_1$估计的样本方差估计将(在模拟误差范围内)等于$\sigma^2 /\left{(n-1) s_x^2\right}$。模拟还阐明了“$X$数据的观测值有条件”解释，因为$X$数据对于每个模拟数据集都是相同的。

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