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2023年7月13日2023年7月13日 Regression Analysis, 回归分析, 统计代写, 统计代考

统计代写|回归分析代写Regression Analysis代考|The Variable Inclusion Principle

如果你也在怎样代写回归分析Regression Analysis 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。回归分析Regression Analysis回归中的概率观点具体体现在给定X数据的特定固定值的Y数据的可变性模型中。这种可变性是用条件分布建模的;因此，副标题是:“条件分布方法”。回归的整个主题都是用条件分布来表达的;这种观点统一了不同的方法，如经典回归、方差分析、泊松回归、逻辑回归、异方差回归、分位数回归、名义Y数据模型、因果模型、神经网络回归和树回归。所有这些都可以方便地用给定特定X值的Y条件分布模型来看待。

回归分析Regression Analysis条件分布是回归数据的正确模型。它们告诉你，对于变量X的给定值，可能存在可观察到的变量Y的分布。如果你碰巧知道这个分布，那么你就知道了你可能知道的关于响应变量Y的所有信息，因为它与预测变量X的给定值有关。与基于R^2统计量的典型回归方法不同，该模型解释了100%的潜在可观察到的Y数据，后者只解释了Y数据的一小部分，而且在假设几乎总是被违反的情况下也是不正确的。

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统计代写|回归分析代写Regression Analysis代考|The Variable Inclusion Principle

Earlier in this chapter and in prior chapters, we alluded to this concept, which is stated specifically as follows:
The variable inclusion principle
Include all lower-order terms related to higher-order terms in a polynomial regression model (regardless of “significance” or lack thereof.)
Recall that the “order” of a polynomial term is either the exponent or sum of the exponents. Lower order terms are those terms in the same variable or variables, with smaller order. Here are some applications of the variable inclusion principle:

Quadratic term, $x^2$. Lower order terms are $x^1=x$ and $x^0=1$. The variable inclusion principle states that if you have $x^2$ in the model, then you should also include $x$ (the linear term) and 1 (the intercept).
Cubic term, $x^3$. Lower order terms are $x^2, x^1=x$, and $x^0=1$. The variable inclusion principle states that if you have $x^3$ in the model, then you should also include $x^2$ (the quadratic term), $x$ (the linear term), and 1 (the intercept).
Interaction term, $x_1 x_2$. Write this term as $x_1^1 x_2^1$. Lower order terms are $x_1^1 x_2^0=x_1$, $x_1^0 x_2^1=x_2$, and $x_1^0 x_2^0=1$. The variable inclusion principle states that if you have $x_1 x_2$ in the model, then you should also include $x_1, x_2$ and 1 (the intercept).
Linear term, $x$. The only lower-order term is $x^0=1$ (the intercept). The variable inclusion principle states that you should always put an intercept in the model.

统计代写|回归分析代写Regression Analysis代考|Why You Should Include the Linear Term in a Quadratic Model

The model is truly linear, and not curved, in the simulation above. In the model that obeys the inclusion criterion, the quadratic term is correctly deemed “insignificant” $(p=0.525)$. The linear term is also “insignificant” $(p=0.161)$, so one might be tempted to remove it from the model, thus violating the inclusion principle. In the resulting model that does not obey the inclusion criterion, where the linear term is excluded, the quadratic term is highly “significant” $\left(p<2 \times 10^{-16}\right)$.

Again, if you violate the variable inclusion principle, then the coefficient of the higherorder term does not measure what you want it to measure. Here, leaving out the linear term (thus violating the inclusion principle) means that the coefficient of the quadratic term does not measure curvature. Rather, since the linear term was excluded, the quadratic term simply acts as a surrogate for the linear term. Instead of measuring only curvature, the coefficient of the quadratic term also measures the linear effect, if you violate the variable inclusion principle.

回归分析代写

统计代写|回归分析代写Regression Analysis代考|The Variable Inclusion Principle

在本章的前面和前面的章节中，我们提到了这个概念，具体表述如下:
变量包含原理
在多项式回归模型中包括与高阶项相关的所有低阶项(无论其“重要性”与否)。
回想一下，多项式项的“阶”要么是指数，要么是指数的和。低阶项是指相同变量中的项，但阶数更小。下面是变量包含原理的一些应用:

二次项，x^2。低阶项是$x^1=x$和$x^0=1$。变量包含原则指出，如果你在模型中有$x^2$，那么你也应该包括$x$(线性项)和1(截距)。

三次项，x^3。低阶项是$x^2, x^1=x$和$x^0=1$。变量包含原则指出，如果模型中有$x^3$，那么还应该包括$x^2$(二次项)、$x$(线性项)和1(截距)。

相互作用项，$x_1 x_2$。把这一项写成x_1^1 x_2^1$。低阶项美元x_1 ^ 1 x_2 ^ 0 = x_1美元,美元x_1 ^ 0 x_2 ^ 1 = x_2美元,美元x_1 ^ 0 x_2 ^ 0 = 1美元。变量包含原则指出，如果模型中有$x_1 x_2$，那么你也应该包括$x_1, x_2$和1(截距)。

线性项，x。唯一的低阶项是x^0=1$(截距)。变量包含原则指出，您应该始终在模型中放置一个截距。

统计代写|回归分析代写Regression Analysis代考|Why You Should Include the Linear Term in a Quadratic Model

在上面的模拟中，模型是真正的线性，而不是弯曲的。在符合纳入标准的模型中，二次项被正确认定为“不显著”$(p=0.525)$。线性项也是“无关紧要的”(p=0.161)，因此人们可能会试图将其从模型中删除，从而违反包含原则。在不符合纳入标准的结果模型中，线性项被排除在外，二次项是高度“显著”的$\左(p<2 \乘以10^{-16}\右)$。

同样，如果你违反了变量包含原则，那么高阶项的系数就不能衡量你想要它衡量的东西。在这里，省略线性项(违反包含原理)意味着二次项的系数不能测量曲率。相反，由于线性项被排除在外，二次项只是作为线性项的替代品。如果违反变量包含原则，二次项的系数不仅可以测量曲率，还可以测量线性效应。

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2023年7月13日2023年7月13日 Regression Analysis, 回归分析, 统计代写, 统计代考

统计代写|回归分析代写Regression Analysis代考|Multicollinearity

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统计代写|回归分析代写Regression Analysis代考|Multicollinearity

Multicollinearity $(\mathrm{MC})$ refers to the $X$ variables being “collinear” to varying degrees. In the case of two $X$ variables, $X_1$ and $X_2$, collinearity means that the two variables are close to linearly related. A “perfect” multicollinearity means that they are perfectly linearly related. See Figure 8.4.
Often, “multicollinearity” with just two $X$ variables is called simply “collinearity.” Figure 8.4, right panel, illustrates the meaning of the term “collinear.”
With more $X$ variables, it is not so easy to visualize multicollinearity. But if one of the $X$ variables, say $X_j$, is closely related to all the other $X$ variables via
$$
X_j \cong a_0 X_0+a_1 X_1+\ldots+a_{j-1} X_{j-1}+a_{j+1} X_{j+1}+\ldots+a_k X_k
$$
then there is multicollinearity. And if the ” $\cong$ ” is, in fact, an ” $=$ ” in the equation above, then there is a perfect multicollinearity. (Note that the variable $X_0$ is the intercept column having all 1’s).
A perfect multicollinearity causes the $\mathbf{X}^{\mathrm{T}} \mathbf{X}$ matrix to be non-invertible, implying that there are no unique least-squares estimates. Equations 0 through $k$ shown in Section 7.1 can still be solved for estimates of the $\beta$ ‘s, but some equation or equations will be redundant with others, implying that there are infinitely many solutions for $\hat{\beta}_0, \hat{\beta}_1, \ldots$, and $\hat{\beta}_k$. Thus the effects of the individual $X_j$ variables on $Y$ are not identifiable when there is a perfect multicollinearity.

To understand the notion that there can be an infinity of solutions for the estimated $\beta$ ‘s, consider the case where there is only one $X$ variable. A perfect multicollinearity, in this case, means that $X_1=a_0 X_0$, so that the $X_1$ column is all the same number, $a_0$. Figure 8.5 shows how data might look in this case, where $x_i=10$ for every $i=1, \ldots, n$, and also shows several possible least-squares fits, all of which have the same minimum sum of squared errors.

统计代写|回归分析代写Regression Analysis代考|The Quadratic Model in One $X$ Variable

The simplest of polynomial models is the simple quadratic model,
$$
f(x)=\beta_0+\beta_1 x+\beta_2 x^2
$$
These models are quite flexible, see Figure 9.1 for various examples.
R code for Figure 9.1
$x=\operatorname{seq}(0.2,10, .1)$
$\mathrm{EY} 1=-3+0.3 * \mathrm{X}+0.1^{\star} \mathrm{X}^{\wedge} 2 ; \mathrm{EY} 2=2-0.9 * \mathrm{X}+0.3{ }^{\star} \mathrm{X}^{\wedge} 2$
$\mathrm{EY} 3=-1+3.0 * \mathrm{X}-0.4{ }^{\star} \mathrm{X}^{\wedge} 2 ; \mathrm{EY} 4=1+1.2{ }^{\star} \mathrm{X}-0.1 \mathrm{X}^{\wedge} 2$
plot $(x, E Y 1$, type=”1″, lty=1, ylab $=” E(Y \mid x=x) “)$
points(x, EY2, type=”1″, lty=2); points(x, EY3, type=”1″, lty=3)
points (x, EY4, type=”1″, lty=4)
legend $(0,10, \mathrm{c}$ (“b0 b1 b2 “,”-3.0 $0.3 \quad 0.1 “, “$
$\mathrm{C}(0,1,2,3,4))$

As it is the case for all models in this chapter, the ” $\beta$ ” coefficients cannot be interpreted in the way discussed in Chapter 8 , where you increase the value of one $X$ variable while keeping the others fixed, because there are functional relationships among the various terms in the model. Specifically, in the example of a quadratic polynomial function, you cannot increase $x^2$, while keeping $x$ fixed. But you can still interpret the parameters by understanding the graphs in Figure 9.1. In particular, $\beta_2$ measures curvature: When $\beta_2<0$, there is concave curvature, when $\beta_2>0$ there is convex curvature, and when $\beta_2=0$, there is no curvature. Further, the larger the $\left|\beta_2\right|$, the more extreme is the curvature.

The intercept term $\beta_0$ has the same meaning as before: It is the value of $f(x)$ when $x=0$. This interpretation is correct but, as always, it is a not useful interpretation when the range of the $x$ data does not cover 0 . Still, the coefficient is needed in the model as a “fitting constant,” which adjusts the function up or down as needed to match the observable data.

To interpret $\beta_1$, note that it is possible to increase $x$ by 1 when $x^2$ is fixed, but the only way that can happen is when you move from $x=-0.5$ to $x=+0.5$. Consider the solid graph shown in Figure 9.1: Here, $f(x)=-3+0.3 x+0.1 x^2$, so that $f(-0.5)=-3+0.3(-0.5)+0.1(-0.5)^2=-3.125$, and $f(+0.5)=-2.825$; these values differ by exactly 0.3 , the coefficient $\beta_1$ that multiplies $x$. While this math gives a correct way to interpret $\beta_1$ in the quadratic model, it is not useful if the range of the $X$ data does not cover zero.

回归分析代写

统计代写|回归分析代写Regression Analysis代考|Multicollinearity

多重共线性$(\mathrm{MC})$是指$X$变量在不同程度上“共线性”。在两个$X$变量$X_1$和$X_2$的情况下，共线性意味着这两个变量接近线性相关。“完美的”多重共线性意味着它们是完全线性相关的。参见图8.4。
通常，只有两个$X$变量的“多重共线性”被简单地称为“共线性”。图8.4(右面板)说明了术语“共线”的含义。
有了更多$X$变量，多重共线性就不那么容易可视化了。但是如果其中一个$X$变量，比如说$X_j$，与所有其他的$X$变量密切相关
$$
X_j \cong a_0 X_0+a_1 X_1+\ldots+a_{j-1} X_{j-1}+a_{j+1} X_{j+1}+\ldots+a_k X_k
$$
然后是多重共线性。如果“$\cong$”实际上是上面方程中的“$=$”，那么就存在一个完美的多重共线性。(注意，变量$X_0$是包含所有1的截距列)。
一个完美的多重共线性使得$\mathbf{X}^{\mathrm{T}} \mathbf{X}$矩阵是不可逆的，这意味着没有唯一的最小二乘估计。对于$\beta$ s的估计，7.1节中所示的方程0到$k$仍然可以求解，但有些方程或方程与其他方程将是冗余的，这意味着$\hat{\beta}_0, \hat{\beta}_1, \ldots$和$\hat{\beta}_k$有无限多个解。因此，当存在完美的多重共线性时，个别$X_j$变量对$Y$的影响是不可识别的。

要理解对于估计的$\beta$ ‘s可能有无限个解的概念，请考虑只有一个$X$变量的情况。一个完美的多重共线性，在这种情况下，意味着$X_1=a_0 X_0$，所以$X_1$列都是相同的数，$a_0$。图8.5显示了这种情况下的数据，其中$x_i=10$表示每个$i=1, \ldots, n$，还显示了几种可能的最小二乘拟合，它们都具有相同的最小平方误差和。

统计代写|回归分析代写Regression Analysis代考|The Quadratic Model in One $X$ Variable

最简单的多项式模型是简单的二次模型，
$$
f(x)=\beta_0+\beta_1 x+\beta_2 x^2
$$
这些模型非常灵活，参见图9.1中的各种示例。
图9.1的R代码
$x=\operatorname{seq}(0.2,10, .1)$
$\mathrm{EY} 1=-3+0.3 * \mathrm{X}+0.1^{\star} \mathrm{X}^{\wedge} 2 ; \mathrm{EY} 2=2-0.9 * \mathrm{X}+0.3{ }^{\star} \mathrm{X}^{\wedge} 2$
$\mathrm{EY} 3=-1+3.0 * \mathrm{X}-0.4{ }^{\star} \mathrm{X}^{\wedge} 2 ; \mathrm{EY} 4=1+1.2{ }^{\star} \mathrm{X}-0.1 \mathrm{X}^{\wedge} 2$
Plot $(x, E Y 1$, type=”1″， lty=1, ylab $=” E(Y \mid x=x) “)$
points(x, EY2, type=”1″， lty=2);points(x, EY3, type=”1″， lty=3)
points (x, y4, type=”1″， lty=4)
图例$(0,10, \mathrm{c}$ (“b0 b1 b2 “，”-3.0 $0.3 \quad 0.1 “, “$
$\mathrm{C}(0,1,2,3,4))$

正如本章中所有模型的情况一样，“$\beta$”系数不能用第8章中讨论的方式来解释，在第8章中，您增加一个$X$变量的值，同时保持其他变量不变，因为模型中的各个术语之间存在函数关系。具体来说，在二次多项式函数的示例中，不能在保持$x$不变的情况下增加$x^2$。但是您仍然可以通过理解图9.1中的图形来解释这些参数。特别地，$\beta_2$测量曲率:当$\beta_2<0$有凹曲率，当$\beta_2>0$有凸曲率，当$\beta_2=0$没有曲率。此外，$\left|\beta_2\right|$越大，曲率就越极端。

截距项 $\beta_0$ 与之前的意思相同:它是 $f(x)$ 什么时候 $x=0$．这种解释是正确的，但是，和往常一样，当 $x$ 数据没有覆盖0。尽管如此，这个系数在模型中还是需要作为一个“拟合常数”，它可以根据需要向上或向下调整函数，以匹配可观测数据。

要解释$\beta_1$，请注意，当$x^2$固定时，可以将$x$增加1，但是唯一可能发生的方法是从$x=-0.5$移动到$x=+0.5$。考虑图9.1所示的实体图:这里，$f(x)=-3+0.3 x+0.1 x^2$，因此$f(-0.5)=-3+0.3(-0.5)+0.1(-0.5)^2=-3.125$和$f(+0.5)=-2.825$;这些值相差0.3，即系数$\beta_1$乘以$x$。虽然这种数学方法给出了在二次模型中解释$\beta_1$的正确方法，但如果$X$数据的范围不覆盖零，它就没有用了。

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2023年7月13日2023年7月13日 Regression Analysis, 回归分析, 统计代写, 统计代考

统计代写|回归分析代写Regression Analysis代考|The Causal Model

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统计代写|回归分析代写Regression Analysis代考|The Causal Model

If you want the coefficient of $X$ to be a causal effect, you need a different model than the conditional mean model, one that includes many more $X^{\prime}$. Here is one:
$$
Y=\gamma_0+\gamma_1 X_1+\gamma_2 X_2+\ldots+\gamma_k X_k+\varepsilon^{\prime}
$$
In this model:

$X_1$ is your main $X$ variable of interest, the one whose causal effect on $Y$ you wish to measure.

$X_2$ through $X_k$ are all other (usually unmeasured) variables that also causally affect $Y$. In this model, changes (manipulations) in $X_1$ cause changes in the distribution of $Y$ when all other possible causal variables $X_2-X_k$ are held fixed.

$\varepsilon^{\prime}$ is a random error term. This term might be identically zero, in which case the causal model is a deterministic model, and this does not change any of the arguments below. Otherwise, with enough $X^{\prime}$ s, it is reasonable to assume that this term is uncorrelated with everything; e.g., $\varepsilon^{\prime}$ might be subatomic quantum noise.
You can re-arrange the causal model as follows:
$$
Y=\gamma_0+\gamma_1 X+\delta
$$
where $X=X_1$ and
$$
\delta=\gamma_2 X_2+\ldots+\gamma_k X_k+\varepsilon^{\prime}
$$
In this model, $\operatorname{Cov}(X, \delta) \neq 0$. Instead, $\operatorname{Cov}(X, \delta)=\sum_{j=2}^k \gamma_j \operatorname{Cov}\left(X, X_j\right)$. Thus, applying Theorem 6.5, the OLS estimate $\hat{\sigma}{x y} / \hat{\sigma}_x^2$ is inconsistent for $\gamma_1$, with probability limit $\gamma_1+\sum{j=2}^k \gamma_j \operatorname{Cov}\left(X, X_j\right) / \sigma_x^2$.

统计代写|回归分析代写Regression Analysis代考|The Instrumental Variable Method

The goal is to come up with an estimator whose probability limit is $\gamma_1$. If you could measure all the relevant unobserved confounders $X_2, \ldots, X_k$, then the simple OLS multiple regression estimate of $\gamma_1$ in model (1) would do the trick. But you usually cannot. And even if you could, there might be hundreds of such variables, and you would not want to run OLS with so many predictors. What to do? Try to find an instrumental variable.

Consider the model $Y=\gamma_0+\gamma_1 X+\delta$, where $\gamma_1$ is the causal effect of $X$. An instrumental variable is a variable $Z$ such that:

$Z$ is correlated with $X$ (preferably reasonably strongly correlated), and
$Z$ is uncorrelated with $\delta$

The instrumental variable (IV) estimator of $\gamma_1$
The instrumental variable estimator of $\gamma_1$ is given by
$$
\hat{\gamma}1=\frac{\hat{\sigma}{z y}}{\hat{\sigma}{z x}} $$ Theorem 6.6: Consistency of the IV Estimator Assume the data pairs $\left(X_i, Y_i, Z_i\right)$ are sampled iid from $p(x, y, z)$, with all variances finite. Assume in addition that $Z$ is an instrumental variable to the causal model $Y=\gamma_0+\gamma_1 X+\delta$. Then $\hat{\gamma}_1=\hat{\sigma}{z y} / \hat{\sigma}_{z x}$ is a consistent estimator of $\gamma_1$.

回归分析代写

计代写|回归分析代写Regression Analysis代考|The Causal Model

如果您希望$X$的系数是因果关系，则需要一个不同于条件平均模型的模型，该模型包含更多$X^{\prime}$。这里有一个:
$$
Y=\gamma_0+\gamma_1 X_1+\gamma_2 X_2+\ldots+\gamma_k X_k+\varepsilon^{\prime}
$$
在这个模型中:

$X_1$ 是您感兴趣的主要$X$变量，您希望测量其对$Y$的因果影响。

$X_2$ 通过$X_k$，所有其他(通常未测量的)变量也会对$Y$产生因果影响。在这个模型中，当所有其他可能的因果变量$X_2-X_k$保持固定时，$X_1$的变化(操纵)导致$Y$分布的变化。

$\varepsilon^{\prime}$ 是随机误差项。这一项可能等于零，在这种情况下，因果模型是确定性模型，这不会改变下面的任何论点。否则，只要有足够的$X^{\prime}$ s，就可以合理地假设这一项与一切都不相关;例如，$\varepsilon^{\prime}$可能是亚原子量子噪声。
你可以将因果模型重新排列如下:
$$
Y=\gamma_0+\gamma_1 X+\delta
$$
其中$X=X_1$和
$$
\delta=\gamma_2 X_2+\ldots+\gamma_k X_k+\varepsilon^{\prime}
$$
在这个模型中，$\operatorname{Cov}(X, \delta) \neq 0$。请登录$\operatorname{Cov}(X, \delta)=\sum_{j=2}^k \gamma_j \operatorname{Cov}\left(X, X_j\right)$。因此，应用定理6.5,OLS估计$\hat{\sigma}{x y} / \hat{\sigma}_x^2$对于$\gamma_1$是不一致的，其概率极限为$\gamma_1+\sum{j=2}^k \gamma_j \operatorname{Cov}\left(X, X_j\right) / \sigma_x^2$。

统计代写|回归分析代写Regression Analysis代考|The Instrumental Variable Method

我们的目标是得到一个概率极限为$\gamma_1$的估计器。如果您可以测量所有相关的未观察到的混杂因素$X_2, \ldots, X_k$，那么模型(1)中$\gamma_1$的简单OLS多元回归估计就可以做到。但你通常做不到。即使可以，也可能有数百个这样的变量，你不会想用这么多的预测器运行OLS。该怎么办?试着找到一个工具变量。

考虑模型$Y=\gamma_0+\gamma_1 X+\delta$，其中$\gamma_1$是$X$的因果效应。工具变量是这样的变量$Z$:

$Z$ 与$X$相关(最好是合理的强相关)，并且

$Z$ 与 $\delta$

$\gamma_1$的工具变量(IV)估计量
$\gamma_1$的工具变量估计量由
$$
\hat{\gamma}1=\frac{\hat{\sigma}{z y}}{\hat{\sigma}{z x}} $$定理6.6:IV估计量的一致性假设数据对$\left(X_i, Y_i, Z_i\right)$从$p(x, y, z)$中抽样iid，所有方差都是有限的。另外假设$Z$是因果模型$Y=\gamma_0+\gamma_1 X+\delta$的工具变量。那么$\hat{\gamma}1=\hat{\sigma}{z y} / \hat{\sigma}{z x}$是$\gamma_1$的一致估计量。

统计代写|回归分析代写Regression Analysis代考请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。

2023年6月25日2023年6月25日 Regression Analysis, 回归分析, 统计代写, 统计代考

统计代写|回归分析代写Regression Analysis代考|The $\ln (Y)$ Transformation and Its Use for Heteroscedastic Processes

如果你也在怎样代写回归分析Regression Analysis 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。回归分析Regression Analysis被广泛用于预测和预报，其使用与机器学习领域有很大的重叠。在某些情况下，回归分析可以用来推断自变量和因变量之间的因果关系。重要的是，回归本身只揭示了固定数据集中因变量和自变量集合之间的关系。为了分别使用回归进行预测或推断因果关系，研究者必须仔细论证为什么现有的关系对新的环境具有预测能力，或者为什么两个变量之间的关系具有因果解释。当研究者希望使用观察数据来估计因果关系时，后者尤其重要。

回归分析Regression Analysis在统计建模中，回归分析是一组统计过程，用于估计因变量（通常称为 “结果 “或 “响应 “变量，或机器学习术语中的 “标签”）与一个或多个自变量（通常称为 “预测因子”、”协变量”、”解释变量 “或 “特征”）之间的关系。回归分析最常见的形式是线性回归，即根据特定的数学标准找到最适合数据的直线（或更复杂的线性组合）。例如，普通最小二乘法计算唯一的直线（或超平面），使真实数据与该直线（或超平面）之间的平方差之和最小。由于特定的数学原因（见线性回归），这使得研究者能够在自变量具有一组给定值时估计因变量的条件期望值（或人口平均值）。不太常见的回归形式使用稍微不同的程序来估计替代位置参数（例如，量化回归或必要条件分析），或在更广泛的非线性模型集合中估计条件期望值（例如，非参数回归）。

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统计代写|回归分析代写Regression Analysis代考|The $\ln (Y)$ Transformation and Its Use for Heteroscedastic Processes

If the regression model $\ln (Y)=\beta_0+\beta_1 X+\varepsilon$ is true, then (as discussed above) the conditional mean and variance of the untransformed $Y$ variable are given as follows:
$$
\mathrm{E}(Y \mid X=x)=\exp \left(\beta_0+\beta_1 x\right) \times \exp \left(\sigma^2 / 2\right)
$$
and
$$
\operatorname{Var}(Y \mid X=x)=\left{\exp \left(\sigma^2\right)-1\right} \times \exp \left(\sigma^2\right) \times \exp \left(2 \beta_0+2 \beta_1 x\right)
$$

Notice that when $\sigma^2=0$, then $\left{\exp \left(\sigma^2\right)-1\right}=0$, so that $\operatorname{Var}(Y \mid X=x)=0$, as expected. Notice also that if you assume the lognormal model, you also assume that the variance of $Y \mid X=x$ is nonconstant; i.e., heteroscedastic: The formula for $\operatorname{Var}(Y \mid X=x)$ in the lognormal case shows that, if $\beta_1>0$, then the variance increases for larger $X=x$.

Figures 5.9 and 5.10 demonstrate the heteroscedasticity of the lognormal regression model, as well as the homoscedasticity of the model for the log-transformed data.
R code for Figures 5.9 and 5.10
$\mathrm{n}=100 ;$ beta $=-2.00 ;$ betal $=0.05 ;$ sigma $=0.30$
set. seed $(12345) ; X=\operatorname{rnorm}(n, 70,10)$
$\ln Y=\operatorname{beta0}+\operatorname{beta} 1 * X+\operatorname{rnorm}(n, 0, \operatorname{sigma}) ; Y=\exp (\ln Y)$
$\operatorname{par}(m f r o w=c(1,2)) ;$ plot $(X, Y) ; \quad \operatorname{abline}(\mathrm{V}=\mathrm{c}(60,80)$, col=”gray”)
plot $(\mathrm{X}, \ln Y)$; abline $(\mathrm{v}=\mathrm{c}(60,80), \mathrm{col}=$ “gray” $)$
$y \cdot s e q=\operatorname{seq}(0.01,30, .01)$
$d y 1=d \ln \operatorname{drm}(\mathrm{y} \cdot$ seq, beta $0+$ beta $1 * 60$, sigma)
$d y 2=d \operatorname{lnorm}(y \cdot$ seq, beta $0+$ beta $1 \star 80$, sigma)
par (mfrow $=\mathrm{c}(1,2))$
plot (y.seq, dy1, type=”1″, $x \lim =c(0,20)$, yaxs=”in, ylim $=c(0, .6)$,
$y l a b=$ “lognormal density”, $x l a b=$ “Untransformed $y^{\prime \prime}$ )
points(y.seq, dy2, type=”l”, lty $=2$ )
legend (“topright”, $c(” X=60 “, ” X=80 “)$, lty $=c(1,2), c e x=0.8)$
$l y \cdot s e q=\log (y \cdot s e q)$
dly $1=\operatorname{dnorm}(1 y \cdot s e q$, beta0+beta $1 * 60$, sigma)
dly $2=\operatorname{dnorm}(l y \cdot$ seq, beta0+beta $1 \star 80$, sigma)
plot (ly.seq, dlyl, type=”l”, $x l i m=c(-.2,4)$, yaxs=”i”, ylim $=c(0,1.6)$,
$y l a b=$ “normal density”, $x l a b=$ “Log Transformed $y “)$
points(ly.seq, dly2, type=”l”, lty=2)
legend (“topright”, $c(” \mathrm{X}=60 “, ” \mathrm{X}=80 “)$, lty=c(1,2), cex=0.8)

统计代写|回归分析代写Regression Analysis代考|An Example Where the Inverse Transformation $1 / Y$ Is Needed

With ratio data, the units of measurement are ( $a$ per $b$ ), and the inverse transformation often makes sense simply because the measurements become ( $b$ per $a)$, which is just as easy to interpret. For example, a car that gets 30 miles per gallon of gasoline equivalently can be stated to take $(1 / 30)$ gallons per mile. You could use either measure in a statistical analysis, without question from any critical reviewer-miles per gallon and gallons per mile convey the same information. Which form to use? Simply choose the form that least violates the model assumptions.

The following code replicates the analyses shown in Figure 5.6, for these data, but using the $W=1 / Y$ transformation, which he called “speed”, because higher values indicate a speedier computer.
R code for Figure 5.11
comp = read.table(“https://raw.githubusercontent.com/andrea2719/
URA-Datasets/master/compspeed.txt “)
attach (comp)
reg.orig $=1 \mathrm{~m}(t i m e \sim \mathrm{GB}) ;$ summary (reg.orig)
par(mfrow=c(2,2)); plot(GB, time); add.loess (GB, time)
qqnorm(reg.orig\$residuals); qqline(reg.orig\$residuals)
speed $=1 /$ time; reg.trans $=1 \mathrm{~m}$ (speed $\sim$ GB)
summary (reg.trans)
plot (GB, speed); add.loess (GB, speed)
qqnorm(reg.trans\$residuals); qqline(reg.trans\$residuals)
comp $=$ read.table (“https://raw.githubusercontent. com/andrea2 $719 /$
URA-Datasets/master/compspeed.txt “)
attach (comp)
reg.orig $=1 \mathrm{~m}($ time $\sim$ GB) ; $\operatorname{summary}($ reg.orig)
par (mfrow=c(2,2)); plot (GB, time); add.loess (GB, time)
qqnorm(reg.orig\$residuals); qqline (reg.orig\$residuals)
speed $=1 /$ time; reg.trans $=1 \mathrm{~m}($ speed $\sim \mathrm{GB})$
summary (reg.trans)
plot (GB, speed); add.loess (GB, speed)
qqnorm (reg.trans\$residuals) ; qqline (reg.trans\$residuals)

回归分析代写

统计代写|回归分析代写Regression Analysis代考|The $\ln (Y)$ Transformation and Its Use for Heteroscedastic Processes

如果回归模型$\ln (Y)=\beta_0+\beta_1 X+\varepsilon$成立，则(如上所述)未变换的$Y$变量的条件均值和方差如下:
$$
\mathrm{E}(Y \mid X=x)=\exp \left(\beta_0+\beta_1 x\right) \times \exp \left(\sigma^2 / 2\right)
$$
和
$$
\operatorname{Var}(Y \mid X=x)=\left{\exp \left(\sigma^2\right)-1\right} \times \exp \left(\sigma^2\right) \times \exp \left(2 \beta_0+2 \beta_1 x\right)
$$

注意，当$\sigma^2=0$，然后$\left{\exp \left(\sigma^2\right)-1\right}=0$，这样$\operatorname{Var}(Y \mid X=x)=0$，如预期。还要注意，如果假设是对数正态模型，也假设$Y \mid X=x$的方差是非常数;即异方差:对数正态情况下$\operatorname{Var}(Y \mid X=x)$的公式表明，如果$\beta_1>0$，则方差增大$X=x$。

图5.9和图5.10显示了对数正态回归模型的异方差，以及对数变换后数据的模型的均方差。
图5.9和5.10的R代码
$\mathrm{n}=100 ;$ beta $=-2.00 ;$ betal $=0.05 ;$ sigma $=0.30$
设置。种子$(12345) ; X=\operatorname{rnorm}(n, 70,10)$
$\ln Y=\operatorname{beta0}+\operatorname{beta} 1 * X+\operatorname{rnorm}(n, 0, \operatorname{sigma}) ; Y=\exp (\ln Y)$
$\operatorname{par}(m f r o w=c(1,2)) ;$ plot $(X, Y) ; \quad \operatorname{abline}(\mathrm{V}=\mathrm{c}(60,80)$, col=”gray”)
剧情$(\mathrm{X}, \ln Y)$;Abline $(\mathrm{v}=\mathrm{c}(60,80), \mathrm{col}=$“gray”$)$
$y \cdot s e q=\operatorname{seq}(0.01,30, .01)$
$d y 1=d \ln \operatorname{drm}(\mathrm{y} \cdot$ seq， β $0+$ β $1 * 60$, sigma)
$d y 2=d \operatorname{lnorm}(y \cdot$ seq， β $0+$ β $1 \star 80$, sigma)
Par (mfrow $=\mathrm{c}(1,2))$)
Plot (y.seq, dy1, type=”1″， $x \lim =c(0,20)$, yaxs=”in, ylim $=c(0, .6)$，
$y l a b=$ “对数正态密度”，$x l a b=$ “未转换$y^{\prime \prime}$)
点(y)Seq, dy2, type=”l”， lty $=2$)
图例(“右上角”，$c(” X=60 “, ” X=80 “)$, lty $=c(1,2), c e x=0.8)$
$l y \cdot s e q=\log (y \cdot s e q)$
Dly $1=\operatorname{dnorm}(1 y \cdot s e q$， β a0+ β $1 * 60$, sigma)
Dly $2=\operatorname{dnorm}(l y \cdot$ seq， β a0+ β $1 \star 80$, sigma)
情节:Seq, dlyl, type=”l”， $x l i m=c(-.2,4)$, yaxs=”i”， ylim $=c(0,1.6)$，
$y l a b=$ “正常密度”，$x l a b=$ “对数变换$y “)$
点(分)。(1) Seq, dly2, type=”l”， lty=2)
图例(“topright”， $c(” \mathrm{X}=60 “, ” \mathrm{X}=80 “)$, lty=c(1,2)， cex=0.8)

统计代写|回归分析代写Regression Analysis代考|An Example Where the Inverse Transformation $1 / Y$ Is Needed

对于比率数据，度量单位是($a$ per $b$)，而逆变换通常是有意义的，因为度量变成了($b$ per $a)$)，这同样容易解释。例如，一辆每加仑汽油行驶30英里的汽车可以等效地表示为每英里消耗$(1 / 30)$加仑。您可以在统计分析中使用任何一种测量方法，而不会受到任何挑剔的审阅者的质疑——每加仑英里数和每英里加仑数传达的信息是一样的。使用哪种形式?只需选择最不违反模型假设的形式。

下面的代码对这些数据复制了图5.6中所示的分析，但是使用了$W=1 / Y$转换，他称之为“速度”，因为数值越高表示计算机速度越快。
图5.11中的R代码
Comp = read.table(“https://raw.githubusercontent.com/andrea2719/
“URA-Datasets/master/compspeed.txt”)
附件(附件)
reg。origin $=1 \mathrm{~m}(t i m e \sim \mathrm{GB}) ;$摘要(reg.origin)
par(mfrow=c(2,2));plot(GB，时间);添加黄土(GB，时间)
qqnorm(reg. origin ＄residuals);qqline(reg. origin ＄残差)
速度$=1 /$时间;reg。变速器$=1 \mathrm{~m}$(转速$\sim$ GB)
摘要(reg.trans)
plot (GB，速度);添加黄土(GB，速度)
qqnorm(reg.trans＄residuals);腾讯网(reg.trans＄残差)
Comp $=$阅读。表(“https://raw.githubusercontent。com andrea2 $719 /$
“URA-Datasets/master/compspeed.txt”)
附件(附件)
reg。origin $=1 \mathrm{~m}($ time $\sim$ GB);$\operatorname{summary}($ reg.org)
Par (mfrow=c(2,2));plot (GB，时间);添加黄土(GB，时间)
qqnorm(reg. origin ＄residuals);Qqline (reg. origin ＄残差)
速度$=1 /$时间;reg。变速器$=1 \mathrm{~m}($速度$\sim \mathrm{GB})$
摘要(reg.trans)
plot (GB，速度);添加黄土(GB，速度)
Qqnorm (reg.trans＄残差);腾讯线(reg.trans＄残差)

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微观经济学代写

微观经济学是主流经济学的一个分支，研究个人和企业在做出有关稀缺资源分配的决策时的行为以及这些个人和企业之间的相互作用。my-assignmentexpert™ 为您的留学生涯保驾护航在数学Mathematics作业代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的数学Mathematics代写服务。我们的专家在图论代写Graph Theory代写方面经验极为丰富，各种图论代写Graph Theory相关的作业也就用不着说。

线性代数代写

线性代数是数学的一个分支，涉及线性方程，如：线性图，如：以及它们在向量空间和通过矩阵的表示。线性代数是几乎所有数学领域的核心。

博弈论代写

现代博弈论始于约翰-冯-诺伊曼（John von Neumann）提出的两人零和博弈中的混合策略均衡的观点及其证明。冯-诺依曼的原始证明使用了关于连续映射到紧凑凸集的布劳威尔定点定理，这成为博弈论和数学经济学的标准方法。在他的论文之后，1944年，他与奥斯卡-莫根斯特恩（Oskar Morgenstern）共同撰写了《游戏和经济行为理论》一书，该书考虑了几个参与者的合作游戏。这本书的第二版提供了预期效用的公理理论，使数理统计学家和经济学家能够处理不确定性下的决策。

微积分代写

微积分，最初被称为无穷小微积分或 “无穷小的微积分”，是对连续变化的数学研究，就像几何学是对形状的研究，而代数是对算术运算的概括研究一样。

它有两个主要分支，微分和积分；微分涉及瞬时变化率和曲线的斜率，而积分涉及数量的累积，以及曲线下或曲线之间的面积。这两个分支通过微积分的基本定理相互联系，它们利用了无限序列和无限级数收敛到一个明确定义的极限的基本概念。

计量经济学代写

什么是计量经济学？
计量经济学是统计学和数学模型的定量应用，使用数据来发展理论或测试经济学中的现有假设，并根据历史数据预测未来趋势。它对现实世界的数据进行统计试验，然后将结果与被测试的理论进行比较和对比。

根据你是对测试现有理论感兴趣，还是对利用现有数据在这些观察的基础上提出新的假设感兴趣，计量经济学可以细分为两大类：理论和应用。那些经常从事这种实践的人通常被称为计量经济学家。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习和应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

2023年6月25日2023年6月25日 Regression Analysis, 回归分析, 统计代写, 统计代考

统计代写|回归分析代写Regression Analysis代考|Comparing Transformations of $X$ with the Car Sales Data

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统计代写|回归分析代写Regression Analysis代考|Comparing Transformations of $X$ with the Car Sales Data

Consider the Car Sales data discussed in Chapter 4. The test for curvature using the quadratic function showed “significant” curvature, corroborating both the curved LOESS smooth, and the subject matter theory, which suggests that as interest rates increase, the mean sales function should flatten because cash sales are not affected by interest rates.
So, first of all, why bother transforming $X=$ Interest Rate? We already showed two different estimates of curvature, one using LOESS, and the other using the quadratic function. Why not use either the LOESS fit or the quadratic model? You could. But there are compelling reasons not to use either LOESS or quadratic models when you have curvature. These reasons are also compelling reasons for why you might wish to use a transformation to model curvature.
Problems with LOESS fit
The LOESS function cannot be written in a simple function such as linear, quadratic, exponential, etc. Having a simple function form such as $\mathrm{E}(Y \mid X=x)=\beta_0+\beta_1 \ln (x)$ makes the model easier to interpret and use.
Problems with quadratic models
Quadratics and higher-order polynomial functions are notoriously bad at the extreme low and high values of the $X$ data. In addition, quadratic models have an extra parameter $\left(\beta_2\right)$ that must be estimated, which can cause loss of accuracy.

Just because a model has a higher maximized likelihood $(L)$ than another model does not validate the model assumptions. In the case of the classical model, higher $L$ means that the $Y$ data have a smaller sum of squared deviations from the fitted function, but not that the assumptions of the model are valid. You still need to evaluate the assumptions of the transformed model, even when the model has a relatively high likelihood.
Checking the assumptions of the transformed model
To check assumptions of the model when you transform $X$, simply apply the techniques you learned in the previous chapter with the transformed $X$ variable. In other words, let $U=f(X)$ and check the assumptions of the $(U, Y)$ data in the same way that you check the assumptions with the $(X, Y)$ data.

统计代写|回归分析代写Regression Analysis代考|The Car Sales Data $\left(t, e_t\right)$ and $\left(e_{t-1}, e_t\right)$ Plots

The Car Sales data are pure time-series since the data are collected in 120 consecutive months. The following code shows the relevant plots to check for uncorrelated (specifically, non-autocorrelated) errors.
CarS = read.table(“https://raw.githubusercontent.com/andrea2719/
URA-DataSets/master/Cars.txt”)
attach(CarS); $\mathrm{n}=$ nrow(CarS)
fit $=$ lm(NSOLD $~$ INTRATE)
resid $=$ fit\$residuals
par(mfrow=c(1,2))
plot( $1: n$, resid, xlab=”month”, ylab=”residual”)
points( $1: n$, resid, type=”l”); abline(h=0)
lag.resid = c(NA, resid[1:n-1])
plot(lag.resid, resid, xlab=”lagged residual”, ylab= “residual”)
abline(lsfit(lag.resid, resid))
Cars $=$ read.table $($ “https://raw.githubusercontent. com/andrea $2719 /$
URA-DataSets/master/Cars.txt”)
attach (Cars); $n=\operatorname{nrow}(\operatorname{Cars})$
$\mathrm{fit}=\operatorname{lm}(\mathrm{NSOLD} \sim$ INTRATE $)$
resid $=$ fit\$residuals
par (mfrow=c $(1,2))$
plot ( $1: n$, resid, $x l a b=$ “month”, ylab=”residual”)
points $(1: \mathrm{n}$, resid, type=”I”); abline $(\mathrm{h}=0)$
lag.resid $=c(N A, r e s i d[1: n-1])$
plot (lag.resid, resid, $x l a b=$ “lagged residual”, ylab = “residual”)
abline(lsfit (lag.resid, resid))
The results are shown in Figure 4.8. There is overwhelming evidence of autocorrelation shown by both plots.

What are the consequences of such an extreme violation of assumptions? According to the mathematical theorems summarized in Chapter 3 , if the data-generating process is truly given by the regression model, then the confidence intervals and $p$-values behave precisely as advertised, with precisely 95\% confidence, and precisely 5\% significance levels. When the independence assumption is grossly violated as seen here, the true confidence levels may be far from 95\% and the true significance levels may be far from 5\%. How far? You guessed it: You can find out by using simulation.

回归分析代写

统计代写|回归分析代写Regression Analysis代考|Evaluating the Uncorrelated Errors Assumption Using Graphical Methods

要评估不相关误差假设，首先必须考虑数据集的类型，是纯时间序列、横断面时间序列、空间、重复测量还是多层(分组)数据。对于纯时间序列数据，通常使用$t$而不是$i$表示观测指标，通常使用$T$而不是$n$表示数据集中的时间点数量，因此观测集是通过$t=1,2, \ldots, T$而不是$i=1,2, \ldots, n$进行索引的。

在纯时间序列过程中，不相关误差假设经常被严重违反，因为，例如，今天与昨天相似，但与五年前不太相似。因此，今天的误差项$\varepsilon_t$的潜在可观察值通常与昨天的误差项$\varepsilon_{t-1}$的潜在可观察值高度相关，这意味着违反了不相关误差假设。

要诊断纯时间序列数据的相关错误，您应该首先检查时间序列残差图，或$\left(t, e_t\right)$。寻找系统的、非随机的模式，如趋势或正弦波类型的功能模式，以表明这种假设的失败。这张图的完全随机外观与不相关的误差是一致的。

最常见的残差相关性类型是当前误差$\varepsilon_t$与先前误差$\varepsilon_{t-1}$的相关性，这被称为“滞后”误差项。这种相关性被称为自相关，因为它指的是变量与自身的相关性。因此，您可以查看的第二个图是滞后散点图，或$\left(e_{t-1}, e_t\right)$，您可以在其上叠加OLS或黄土拟合来查看趋势。该图中的趋势表明当前残差与前一个残差之间存在依赖关系，这违反了不相关误差假设。没有趋势的随机散点与不相关误差是一致的。

第三种图是残差的自相关函数，它显示滞后1、$\operatorname{lag} 2$、滞后3和更多的自相关性，因此您可以使用这种图来检查滞后大于1的自相关性。

对于非纯时间序列数据，需要不同的方法。对于空间数据(“空间”中的点，例如，具有地理坐标的数据)，您可以使用变异图来检查误差相关性，在这种情况下称为“空间自相关性”。对于多水平(分组)数据，您可以检查散点图，其中数据按组标记以诊断相关性结构;第十章涉及这个问题。现在，我们只讨论纯时间序列数据。

统计代写|回归分析代写Regression Analysis代考|The Car Sales Data $\left(t, e_t\right)$ and $\left(e_{t-1}, e_t\right)$ Plots

Car Sales数据是纯时间序列，因为数据是连续120个月收集的。下面的代码显示了检查不相关(特别是非自相关)错误的相关图。
CarS = read.table(“https://raw.githubusercontent.com/andrea2719/ .table “)
“URA-DataSets/master/Cars.txt”)
attach(CarS);$\mathrm{n}=$ nrow(CarS)
fit $=$ lm(NSOLD $~$ INTRATE)
Resid $=$ fit＄残差
par(mfrow=c(1,2))
图($1: n$, resid, xlab=”month”， ylab=”residual”)
点数($1: n$, resid, type=”l”);abline(h=0)
滞后。resid = c(NA, resid[1:n-1])
情节(滞后)Resid, Resid, xlab=”滞后残差”，ylab= “残差”)
abline(lsfit);残留，残留))
汽车$=$阅读。表$($”https://raw.githubusercontent。com andrea $2719 /$
“URA-DataSets/master/Cars.txt”)
附件(汽车);$n=\operatorname{nrow}(\operatorname{Cars})$
$\mathrm{fit}=\operatorname{lm}(\mathrm{NSOLD} \sim$ INTRATE $)$
Resid $=$ fit＄残差
Par (mfrow=c $(1,2))$
图($1: n$，残差，$x l a b=$“月”，ylab=“残差”)
points $(1: \mathrm{n}$, resid, type=”I”);在线$(\mathrm{h}=0)$
滞后。渣油$=c(N A, r e s i d[1: n-1])$
情节滞后。Resid, Resid, $x l a b=$ “滞后残差”，ylab = “残差”)
Abline (lsfit (lag))残留，残留))
结果如图4.8所示。这两幅图都显示了大量的自相关证据。

这种极端违反假设的后果是什么?根据第3章中总结的数学定理，如果数据生成过程确实由回归模型给出，那么置信区间和$p$ -值的行为精确地与广告一样，精确地具有95％的置信度，精确地具有5％的显著性水平。如图所示，当独立性假设严重违反时，真实的置信水平可能远低于95％，真实的显著性水平可能远低于5％。有多远?你猜对了:你可以通过模拟来找到答案。

统计代写|回归分析代写Regression Analysis代考请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。

微观经济学代写

线性代数代写

线性代数是数学的一个分支，涉及线性方程，如：线性图，如：以及它们在向量空间和通过矩阵的表示。线性代数是几乎所有数学领域的核心。

博弈论代写

微积分代写

计量经济学代写

MATLAB代写

2023年6月25日2023年6月25日 Regression Analysis, 回归分析, 统计代写, 统计代考

统计代写|回归分析代写Regression Analysis代考|Evaluating the Uncorrelated Errors Assumption Using Graphical Methods

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统计代写|回归分析代写Regression Analysis代考|Evaluating the Uncorrelated Errors Assumption Using Graphical Methods

To evaluate the uncorrelated errors assumption, you first have to consider the type of data set you have, whether it is pure time-series, cross-sectional time-series, spatial, repeated measures, or multilevel (grouped) data. With pure time-series data, it is common to let $t$ denote the observation indicator rather than $i$, and it is common to let $T$ denote the number of time points in the data set rather than $n$, so the set of observations is indexed by $t=1,2, \ldots, T$, rather than by $i=1,2, \ldots, n$.

The uncorrelated errors assumption is often badly violated with pure time-series processes, because, e.g., today is similar to yesterday, but not so similar to five years ago. Thus, the potentially observable values of today’s error term, $\varepsilon_t$, are often highly correlated with potentially observable values of yesterday’s error term, $\varepsilon_{t-1}$, implying a violation of the uncorrelated errors assumption.

To diagnose correlated errors with pure time-series data, you should first examine the time-series residual graph, or $\left(t, e_t\right)$. Look for systematic, non-random patterns, such as trends or sinusoidal-type functional patterns to suggest failure of this assumption. A completely random appearance of this graph is consistent with uncorrelated errors.

The most common type of residual correlation is the correlation of the current error $\varepsilon_t$ with the previous error $\varepsilon_{t-1}$, which is called the “lagged” error term. Such correlation is called autocorrelation because it refers to the correlation of a variable with itself. Thus, the second graph you can view is the lag scatterplot, or $\left(e_{t-1}, e_t\right)$, upon which you can superimpose the OLS or LOESS fit to see the trend. A trend in this plot suggests dependence between the current residual and the immediately preceding residual, a violation of the uncorrelated errors assumption. A random scatter with no trend is consistent with uncorrelated errors.

A third kind of plot is the autocorrelation function of the residuals, which displays lag 1, $\operatorname{lag} 2$, lag 3, and more autocorrelations, thus you can use this plot to examine autocorrelations for lags greater than 1.

For data other than pure time-series data, different methods are needed. For spatial data (points in “space,” e.g., data with geographic coordinates), you can use a variogram to check for error correlation, in this case called “spatial autocorrelation.” With multilevel (grouped) data, you can examine scatterplots where data are labeled by group to diagnose correlation structure; Chapter 10 touches upon this issue. For now, we will discuss only pure time-series data.

统计代写|回归分析代写Regression Analysis代考|The Car Sales Data $\left(t, e_t\right)$ and $\left(e_{t-1}, e_t\right)$ Plots

回归分析代写

统计代写|回归分析代写Regression Analysis代考|Evaluating the Uncorrelated Errors Assumption Using Graphical Methods

第三种图是残差的自相关函数，它显示滞后1、$\operatorname{lag} 2$、滞后3和更多的自相关性，因此您可以使用这种图来检查滞后大于1的自相关性。

统计代写|回归分析代写Regression Analysis代考|The Car Sales Data $\left(t, e_t\right)$ and $\left(e_{t-1}, e_t\right)$ Plots

统计代写|回归分析代写Regression Analysis代考请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。

微观经济学代写

线性代数代写

线性代数是数学的一个分支，涉及线性方程，如：线性图，如：以及它们在向量空间和通过矩阵的表示。线性代数是几乎所有数学领域的核心。

博弈论代写

微积分代写

计量经济学代写

MATLAB代写

2023年6月2日2023年6月2日 Regression Analysis, 回归分析, 统计代写, 统计代考

统计代写|回归分析代写Regression Analysis代考|Practical Versus Statistical Significance

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统计代写|回归分析代写Regression Analysis代考|Practical Versus Statistical Significance

While the Car Sales example above is a case where statistically significant $(p<0.05)$ curvature corresponds to practically significant (which means practically important) curvature, it is not always the case that statistical significance corresponds to practical significance. This can easily happen in data sets where $n$ is large (e.g., with “big data”), because with large data sets you have the ability to estimate even slight curvature very precisely, with small standard error.

The following simulation illustrates this situation: There is statistically significant ( $p=0.000166$ ) curvature, as shown by the hypothesis test for the quadratic term. However, the curvature is practically insignificant, as can be seen by graphing the linear and quadratic fitted functions. The sample size in this simulation is large, $n=1,000,000$, but not unusually large for “big data” applications.

4.5.1 Simulation Study to Demonstrate Practical vs. Statistical Significance
set.seed(54321) # For perfect replicability of the random simulation.
$\mathrm{x}=10+2 * \operatorname{rnorm}(1000000) ; \mathrm{xsq}=\mathrm{x}^{\wedge} 2$
$\mathrm{y}=2+.6 * \mathrm{x}+.003 * x$ sq $+4 * \operatorname{rnorm}(1000000)$ #beta2 $=.003$ does not equal 0 !
fit. quad $=\operatorname{lm}(\mathrm{y} \sim \mathrm{x}+\mathrm{xsq})$
summary(fit.quad) # Significant curvature: p-value $=0.000166$
## A . 1\% random sample from the data set is selected to make the scatterplot
## more legible. Otherwise, the points are too dense to view.
select $=\operatorname{runif}(1000000) ; \mathrm{x} 1=\mathrm{x}$ [select<.001] ; $\mathrm{y} 1=\mathrm{y}[$ select<.001]
plot (x1, y1, main = “Scatterplot of a $0.1 \%$ Subsample”)
abline(lsfit( $x, y)$, col=”gray”)
lines (spline(x, predict(fit.quad)), lty=3)

Figure 4.5 shows that the linear fit (solid line) is adequate, even though the true model is quadratic, and not linear.

统计代写|回归分析代写Regression Analysis代考|Evaluating the Constant Variance (Homoscedasticity) Assumption Using Graphical Methods

The first graph you should use to evaluate the constant variance assumption is the $\left(\hat{y}_i, e_i\right)$ scatterplot. Look for changes in the pattern of vertical variability of the $e_i$ for different $\hat{y}_i$. The most common indications of constant variance assumption violation are shapes that indicate either increasing variability of $Y$ for larger $\mathrm{E}(Y \mid X=x)$, or shapes that indicate decreasing variability of $Y$ for larger $\mathrm{E}(Y \mid X=x)$. Increasing variability of $Y$ for larger $\mathrm{E}(Y \mid X=x)$ is indicated by greater variability in the vertical ranges of the $e_i$ when $\hat{y}_i$ is larger.
Recall again that the constant variance assumption (like all assumptions) refers to the data-generating process, not the data. The statement “the data are homoscedastic” makes no sense. By the same logic, the statements “the data are linear” and “the data are normally distributed” also are nonsense. Thus, whichever pattern of variability that you decide to claim based on the $\left(\hat{y}_i, e_i\right)$ scatterplot, you should try to make sense of it in the context of the subject matter that determines the data-generating process. As one example, physical boundaries on data force smaller variance when the data are closer to the boundary. As another, when income increases, people have more choice as to whether or not they choose to purchase an item. Thus, there should be more variability in expenditures among people with more money than among people with less money. Whatever pattern you see in the $\left(\hat{y}_i, e_i\right)$ scatterplot should make sense to you from a subject matter standpoint.

While the LOESS smooth to the $\left(\hat{y}_i, e_i\right)$ scatterplot is useful for checking the linearity assumption, it is not useful for checking the constant variance assumption. Instead, you should use the LOESS smooth over the plot of $\left(\hat{y}_i,\left|e_i\right|\right)$. When the variability in the residuals is larger, they will tend to be farther from zero, giving larger mean absolute residuals $\left|e_i\right|$. An increasing trend in the $\left(\hat{y}_i,\left|e_i\right|\right)$ plot suggests larger variability in $Y$ for larger $\mathrm{E}(Y \mid X=x)$, and a flat trend line for the $\left(\hat{y}_i,\left|e_i\right|\right)$ plot suggests that the variability in $Y$ is nearly unrelated to $\mathrm{E}(Y \mid X=x)$. However, as always, do not over-interpret. Data are idiosyncratic (random), so even if homoscedasticity is true in reality, the LOESS fit to the $\left(\hat{y}_i,\left|e_i\right|\right)$ graph will not be a perfectly flat line, due to chance alone. To understand “chance alone” in this case you can simulate data from a homoscedastic model, construct the $\left(\hat{y}_i, \mid e_i\right)$ graph, and add the LOESS smooth. You will see that the LOESS smooth is not a perfect flat line, and you will know that such deviations are explained by chance alone.

The hypothesis test for homoscedasticity will help you to decide whether the observed deviation from a flat line is explainable by chance alone, but recall that the test does not answer the real question of interest, which is “Is the heteroscedasticity so bad that we cannot use the homoscedastic model?” (That question is best answered by simulating data sets having the type of heteroscedasticity you expect with your real data, then by performing the types of analyses you plan to perform on your real data, then by evaluating the performance of those analyses.)

回归分析代写

统计代写|回归分析代写Regression Analysis代考|Practical Versus Statistical Significance

虽然上面的汽车销售示例是统计显著$(p<0.05)$曲率对应于实际显著(这意味着实际重要)曲率的情况，但统计显著性并不总是对应于实际显著性。这很容易发生在$n$很大的数据集中(例如，“大数据”)，因为有了大数据集，你有能力以很小的标准误差非常精确地估计即使是轻微的曲率。

下面的模拟说明了这种情况:有统计显著($p=0.000166$)曲率，如二次项的假设检验所示。然而，曲率实际上是微不足道的，这可以从线性和二次拟合函数的图中看出。此模拟中的样本量很大($n=1,000,000$)，但对于“大数据”应用程序来说并不算大。

4.5.1模拟研究以证明实际意义与统计意义
set.seed(54321) ＃用于随机模拟的完美可复制性。
$\mathrm{x}=10+2 * \operatorname{rnorm}(1000000) ; \mathrm{xsq}=\mathrm{x}^{\wedge} 2$
$\mathrm{y}=2+.6 * \mathrm{x}+.003 * x$平方$+4 * \operatorname{rnorm}(1000000)$ ＃beta2 $=.003$不等于0 !
适合。Quad $=\operatorname{lm}(\mathrm{y} \sim \mathrm{x}+\mathrm{xsq})$
summary(fit.quad) ＃显著曲率:p值$=0.000166$
＃＃ a。从数据集中随机抽取1％的样本制作散点图
＃＃更清晰。否则，点太密集而无法查看。
选择$=\operatorname{runif}(1000000) ; \mathrm{x} 1=\mathrm{x}$[选择<。[00];$\mathrm{y} 1=\mathrm{y}[$ select<.001]
plot (x1, y1, main =“$0.1 \%$子样本的散点图”)
Abline (lsfit($x, y)$, col=”gray”)
Lines (spline(x, predict(fit.quad))， lty=3)

图4.5显示，线性拟合(实线)是足够的，即使真正的模型是二次的，而不是线性的。

统计代写|回归分析代写Regression Analysis代考|Evaluating the Constant Variance (Homoscedasticity) Assumption Using Graphical Methods

用于评估恒定方差假设的第一个图是$\left(\hat{y}_i, e_i\right)$散点图。寻找垂直变化模式$e_i$对于不同$\hat{y}_i$的变化。对于较大的$\mathrm{E}(Y \mid X=x)$，最常见的不变方差假设违反的指示是形状表明$Y$的可变性增加，或者形状表明$Y$的可变性减少，对于较大的$\mathrm{E}(Y \mid X=x)$。当$\mathrm{E}(Y \mid X=x)$较大时，$Y$的变异性增大，这表明当$\hat{y}_i$较大时，$e_i$的垂直范围变异性增大。
再次回忆一下，常数方差假设(像所有假设一样)指的是数据生成过程，而不是数据。“数据均方差”的说法毫无意义。按照同样的逻辑，“数据是线性的”和“数据是正态分布的”也都是无稽之谈。因此，无论您决定基于$\left(\hat{y}_i, e_i\right)$散点图声明哪种可变性模式，您都应该尝试在决定数据生成过程的主题上下文中理解它。例如，当数据靠近物理边界时，数据上的物理边界所产生的方差较小。另一方面，当收入增加时，人们有更多的选择是否购买物品。因此，有钱的人的支出应该比钱少的人有更多的可变性。无论您在$\left(\hat{y}_i, e_i\right)$散点图中看到什么模式，从主题的角度来看都应该是有意义的。

虽然对$\left(\hat{y}_i, e_i\right)$散点图的黄土平滑对检验线性假设有用，但对检验常方差假设没用。相反，您应该在$\left(\hat{y}_i,\left|e_i\right|\right)$的地块上使用黄土平滑。当残差的可变性较大时，它们将趋向于远离零，从而给出较大的平均绝对残差$\left|e_i\right|$。$\left(\hat{y}_i,\left|e_i\right|\right)$图的增加趋势表明，对于较大的$\mathrm{E}(Y \mid X=x)$, $Y$的变异性较大，而$\left(\hat{y}_i,\left|e_i\right|\right)$图的平坦趋势线表明，$Y$的变异性几乎与$\mathrm{E}(Y \mid X=x)$无关。然而，一如既往，不要过度解读。数据是特殊的(随机的)，所以即使同方差在现实中是真实的，$\left(\hat{y}_i,\left|e_i\right|\right)$图的黄土拟合也不会是一条完全平坦的线，这仅仅是偶然的。为了理解这种情况下的“偶然性”，您可以模拟来自同方差模型的数据，构造$\left(\hat{y}_i, \mid e_i\right)$图，并添加黄土平滑。你会看到黄土平滑不是一条完美的平坦线，你会知道这样的偏差仅仅是偶然的。

同方差的假设检验将帮助您确定观测到的与平直线的偏差是否仅靠偶然性来解释，但请记住，该检验并没有回答真正感兴趣的问题，即“异方差是否严重到我们不能使用同方差模型?”(这个问题的最佳答案是模拟具有您期望的真实数据的异方差类型的数据集，然后执行您计划对真实数据执行的分析类型，然后评估这些分析的性能。)

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2023年6月2日2023年6月2日 Regression Analysis, 回归分析, 统计代写, 统计代考

统计代写|回归分析代写Regression Analysis代考|Hypothesis Testing and $p$-Values: Is the Observed Effect of $X$ on $Y$ Explainable by Chance Alone?

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统计代写|回归分析代写Regression Analysis代考|Hypothesis Testing and $p$-Values: Is the Observed Effect of $X$ on $Y$ Explainable by Chance Alone?

Some researchers will do nearly anything to get a publication. The incentives are great: Fame, tenure, promotion, annual salary, raises, prime class assignments, and clout in one’s department are a function of quality and quantity of publications.

Historically, statistical results were required to be “statistically significant” to be publishable. In terms of confidence intervals, this means that the interval for the effect (e.g., the $\beta$ ) in question must exclude 0 so that you can confidently state the direction of the effect (positive or negative) of the given $X$ variable on $Y$.

Researchers used the $p$-values that are reported routinely by regression software to determine “statistical significance.” But $p$-values are easily manipulated, and unscrupulous researchers can analyze data “creatively” to get nearly any $p$-value they would like to see. This has led to an unfortunate practice known as p-hacking, where researchers try analyses many different ways until they get a $p$-value that is statistically significant, and then try to publish the results. Because of their potential for misuse, there is a strong movement in the scientific community away from use of $p$-values, as well as the phrase “statistical significance,” in favor of other statistics and characterizations.

When interpreted correctly and not misused, the $p$-value does provide interesting and somewhat useful information. Thus, we insist that you understand $p$-values very well, so that you can use them correctly and effectively, and so that you will not become a ” $p$-hacker.”

To interpret the $p$-value correctly, you must consider the question, “Is the estimate of the effect of $X$ on $Y$ explainable by chance alone?” But to answer that question, you must first understand what it means for an estimated effect to be explained by chance alone. The following example explains this concept.

统计代写|回归分析代写Regression Analysis代考|Is the Last Digit of a Person’s Identification Number Related to Their Height?

On the surface of it, this is a silly question. But it provides a great example to help you to understand what it means for a phenomenon to be “explained by chance alone,” which is the first thing you need to know before you can ascertain whether a phenomenon is “explainable by chance alone.”

Suppose you have a data set containing heights of 100 adult males in the United States, along with the last digit of their social security number (SSN). Since adult male heights are approximately normally distributed with mean 70 inches and standard deviation 4 inches, and since the last digit of the SSN is uniformly distributed on the numbers $0,1,2, \ldots, 9$, the following code simulates a quite realistic example of how such a data set would look.
## Simulation of data relating Height to SSN
$\mathrm{n}=100$
set. seed(12345) # so that the results will replicate perfectly
height = round(rnorm( $100,70,4))$
ssn = sample $(0: 9,100$, replace=T)
ssn. data = data.frame(ssn, height)
head(ssn. data)
## Simulation of data relating Height to SSN
$\mathrm{n}=100$
set.seed(12345) # so that the results will replicate perfectly
height = round $(\operatorname{rnorm}(100,70,4))$
$\mathrm{ssn}=$ sample $(0: 9,100$, replace $=T)$
ssn.data = data.frame (ssn, height)
head (ssn. data)
This code gives you the following (hypothetical but realistic) data on last digit of social security number (SSN) and Height:
$\begin{array}{lrr} & \sin & \text { height } \ 1 & 5 & 72 \ 2 & 8 & 73 \ 3 & 1 & 70 \ 4 & 5 & 68 \ 5 & 6 & 72 \ 6 & 7 & 63\end{array}$

回归分析代写

统计代写|回归分析代写Regression Analysis代考|Hypothesis Testing and $p$-Values: Is the Observed Effect of $X$ on $Y$ Explainable by Chance Alone?

一些研究人员为了发表论文几乎不惜一切代价。奖励是巨大的:名声、终身职位、晋升、年薪、加薪、主要的班级任务以及在部门的影响力都是出版物质量和数量的函数。

历史上，统计结果需要具有“统计显著性”才能发表。就置信区间而言，这意味着所讨论的效果(例如$\beta$)的区间必须排除0，以便您可以自信地说明$Y$上给定$X$变量的效果方向(正或负)。

研究人员使用回归软件例行报告的$p$值来确定“统计显著性”。但是$p$ -值很容易被操纵，无良的研究人员可以“创造性地”分析数据，以获得几乎任何他们想看到的$p$ -值。这导致了一种被称为p黑客的不幸做法，研究人员尝试用许多不同的方法分析，直到他们得到一个具有统计意义的$p$ -值，然后试图发表结果。由于它们可能被误用，科学界有一股强烈的运动，远离使用$p$ -值以及“统计显著性”这一短语，而倾向于使用其他统计数据和特征。

当正确解释且不被滥用时，$p$ -值确实提供了有趣且有些有用的信息。因此，我们坚持要求您非常了解$p$ -values，以便您能够正确有效地使用它们，从而不会成为“$p$ -黑客”。

要正确地解释$p$ -值，你必须考虑这个问题:“对$X$对$Y$的影响的估计是否仅凭偶然就可以解释?”但要回答这个问题，你必须首先明白，一个估计的效应只能用偶然来解释是什么意思。下面的例子解释了这个概念。

统计代写|回归分析代写Regression Analysis代考|Is the Last Digit of a Person’s Identification Number Related to Their Height?

从表面上看，这是一个愚蠢的问题。但它提供了一个很好的例子，帮助你理解一个现象“只能由偶然解释”意味着什么，这是你在确定一个现象是否“只能由偶然解释”之前需要知道的第一件事。

假设您有一个数据集，其中包含美国成年男性的身高为100，以及他们的社会安全号码(SSN)的最后一位数字。由于成年男性身高近似为正态分布，平均值为70英寸，标准差为4英寸，并且由于SSN的最后一位数字均匀分布在数字$0,1,2, \ldots, 9$上，因此下面的代码模拟了这样一个数据集看起来的相当现实的示例。
＃＃模拟高度与SSN相关的数据
$\mathrm{n}=100$
设置。设置为(12345)＃，这样可以完美地复制结果
高度=圆(rnorm($100,70,4))$
ssn = sample $(0: 9,100$, replace=T)
ssn. ssn.Data = Data .frame(ssn, height)
头;头数据)
＃＃模拟高度与SSN相关的数据
$\mathrm{n}=100$
设置.seed(12345) ＃，以便完美地复制结果
高度=圆$(\operatorname{rnorm}(100,70,4))$
$\mathrm{ssn}=$样品$(0: 9,100$，替换$=T)$
ssn. ssn.Data = Data .frame (ssn, height)
头(6)数据)
这段代码为您提供了以下(假设但现实的)社会安全号码(SSN)的最后一位数字和身高的数据:
$\begin{array}{lrr} & \sin & \text { height } \ 1 & 5 & 72 \ 2 & 8 & 73 \ 3 & 1 & 70 \ 4 & 5 & 68 \ 5 & 6 & 72 \ 6 & 7 & 63\end{array}$

统计代写|回归分析代写Regression Analysis代考请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。

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线性代数代写

线性代数是数学的一个分支，涉及线性方程，如：线性图，如：以及它们在向量空间和通过矩阵的表示。线性代数是几乎所有数学领域的核心。

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2023年6月2日2023年6月2日 Regression Analysis, 回归分析, 统计代写, 统计代考

统计代写|回归分析代写Regression Analysis代考|The Classical Model and Its Consequences

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统计代写|回归分析代写Regression Analysis代考|The Classical Model and Its Consequences

The classical regression model assumes normality, independence, constant variance, and linearity of the conditional mean function, and is (once again) stated as follows:
$$
Y_i \mid X_i=x_i \quad \sim_{\text {independent }} \mathrm{N}\left(\beta_0+\beta_1 x_i, \sigma^2\right) \text {, for } i=1,2, \ldots, n .
$$
Whether you like it or not, this model is also what your computer assumes when you ask it to analyze your data via standard regression methods. The parameter estimates you get from the computer are best under this model, and the inferences ( $p$-values and confidence intervals) are exactly correct under this model. If the assumptions of the model are not true, then the estimates are not best, and the inferences are incorrect. You might think we are saying that assumptions must be true in order to use statistical methods that make such assumptions, but we are not. As we noted in Chapter 1, it is not necessarily a problem that any or all of the assumptions of the model are wrong, depending on how badly violated is the assumption. And the easiest way to understand whether an assumption is violated “too badly” is to use simulation.

We have found that students in statistics classes often resist learning simulation. After all, the data that researchers use is usually real, and not simulated, so the students wonder, what is the point of using simulation? Here are some answers:

Simulation shows you, clearly and concretely, how to interpret the regression analysis of your real (not simulated) data.
Simulation helps you to understand how a regression model can be useful even when the model is wrong.
Simulation models help you to understand the meaning of the regression model parameters.
Simulation models help you to understand the meaning of the regression model assumptions.
Simulation models help you to understand the meaning of a “research hy pothesis.”
Simulation helps you to understand how to interpret your data in the presence of chance effects.
Simulation helps you to understand all the commonly misunderstood concepts in statistics, like “unbiasedness,” “standard error,” ” $p$-value,” and “confidence interval.” $^{\prime \prime}$
Simulation methods are commonly used in the analysis of real data; examples include the bootstrap and Markov Chain Monte Carlo.

统计代写|回归分析代写Regression Analysis代考|Unbiasedness

The Gauss-Markov (G-M) theorem states that, under certain model assumptions (the premise, ” $\mathrm{A}$ ” of the theorem), the OLS estimator has minimum variance among linear unbiased estimators (that is the consequence, the “condition B” of the theorem). To understand the G-M theorem, you first need to understand what “unbiasedness” means. Recall the view of regression data shown in Chapter 2, shown again in Table 3.1.

To be specific, please consider the Production Cost data set from Chapter 1. The actual data are shown in Table 3.2, along with the random data-generation assumption of the regression model.

In particular, the value 2,224 is assumed to be produced at random from a distribution of potentially observable Cost values among jobs having 1,500 widgets, the value 1,660 is assumed to be produced at random from a distribution of potentially observable Cost values among jobs having 800 widgets, and so on. If you are having trouble visualizing these different distributions, just have a look at Figure 1.7 again, and put yourself in the position of the job manager at this company: In two different jobs where the number of widgets is the same, will the costs also be the same? Of course not; see the first and third observations in the data set, for example. There is an entire distribution of potentially observable Cost values when Widgets $=1500$, and this is what is meant by $p(y \mid X=1500)$.

Now, use your imagination. Imagine another collection of 40 jobs, from the same process that produced the data above, with the widgets data exactly as observed, but with specific costs not observed. Further, imagine that the classical model is true so that the distribution $p(y \mid X=x)$ is the $\mathrm{N}\left(\beta_0+\beta_1 x, \sigma^2\right)$ distribution. The specific costs are not observed, but the potentially observable data will appear as shown in Table 3.3.

In Table 3.3, the $Y_i$ are random variables, coming from the same distributions that produced the original data. Again, use your imagination: There are infinitely many potentially observable data sets as shown in Table 3.3, because there are infinitely many sequences of potentially observable values for $Y_1$; infinitely many sequences of potentially observable values for $Y_2, \ldots$; and there are infinitely many sequences of potentially observable values for $Y_{40}$. Again, if you are having a hard time visualizing this, just look at Figure 1.7 again: There are an infinity of possible values under each of the normal curves shown there. The $n=40 Y_i$ values in Table 3.3 are one set of random selections from such distributions.

回归分析代写

计代写|回归分析代写Regression Analysis代考|The Classical Model and Its Consequences

经典回归模型假设条件均值函数的正态性、独立性、常方差和线性，并(再次)表述如下:
$$
Y_i \mid X_i=x_i \quad \sim_{\text {independent }} \mathrm{N}\left(\beta_0+\beta_1 x_i, \sigma^2\right) \text {, for } i=1,2, \ldots, n .
$$
不管你喜欢与否，当你要求你的计算机通过标准回归方法分析你的数据时，这个模型也是你的计算机所假定的。在这个模型下，从计算机得到的参数估计是最好的，推断($p$ -值和置信区间)在这个模型下是完全正确的。如果模型的假设不正确，那么估计就不是最好的，推论就不正确。你可能会认为我们是在说假设必须是正确的，以便使用统计方法来做出这样的假设，但我们不是。正如我们在第一章中所指出的，模型的任何或所有假设都是错误的不一定是一个问题，这取决于假设被违背的程度。了解假设是否被“严重”违背的最简单方法是使用模拟。

我们发现，统计学课上的学生经常抗拒模拟学习。毕竟，研究人员使用的数据通常是真实的，而不是模拟的，所以学生们想知道，使用模拟的意义是什么?以下是一些答案:

模拟向您清楚而具体地展示了如何解释真实(非模拟)数据的回归分析。

模拟可以帮助您理解回归模型是如何在模型错误的情况下发挥作用的。

仿真模型帮助您理解回归模型参数的含义。

模拟模型帮助您理解回归模型假设的含义。

仿真模型帮助您理解“研究假设”的含义。

模拟可以帮助您了解如何在存在偶然效应的情况下解释数据。

模拟可以帮助您理解统计学中所有常被误解的概念，如“无偏性”、“标准误差”、“$p$ -value”和“置信区间”。 $^{\prime \prime}$

仿真方法通常用于对真实数据的分析;例子包括自举和马尔可夫链蒙特卡罗。

统计代写|回归分析代写Regression Analysis代考|Unbiasedness

高斯-马尔可夫(G-M)定理指出，在某些模型假设下(定理的前提，“$\mathrm{A}$”)，OLS估计量在线性无偏估计量中具有最小方差(这是定理的结果，“条件B”)。要理解G-M定理，首先需要理解“无偏性”是什么意思。回顾第2章中显示的回归数据视图，再次显示在表3.1中。

具体来说，请考虑第一章的生产成本数据集。实际数据如表3.2所示，同时采用回归模型的随机数据生成假设。

特别是，假设值2224是从具有1500个小部件的作业中可能可观察到的成本值分布随机产生的，假设值1660是从具有800个小部件的作业中可能可观察到的成本值分布随机产生的，以此类推。如果您在可视化这些不同的分布时遇到困难，请再次查看图1.7，并将自己置于该公司的作业管理器的位置:在两个不同的作业中，部件数量相同，成本是否也相同?当然不是;例如，请参阅数据集中的第一个和第三个观察结果。当Widgets $=1500$时，有一个潜在的可观察成本值的完整分布，这就是$p(y \mid X=1500)$的含义。

现在，运用你的想象力。想象另一个包含40个作业的集合，来自产生上述数据的相同流程，其中观察到的小部件数据完全相同，但没有观察到特定的成本。进一步，假设经典模型为真，那么分布$p(y \mid X=x)$就是$\mathrm{N}\left(\beta_0+\beta_1 x, \sigma^2\right)$分布。具体的成本没有观察到，但潜在的可观察到的数据将显示在表3.3中。

在表3.3中，$Y_i$是随机变量，它们来自产生原始数据的相同分布。再一次发挥你的想象力:有无限多个潜在可观察数据集，如表3.3所示，因为$Y_1$有无限多个潜在可观察值序列;无穷多个序列的潜在可观察值$Y_2, \ldots$;对于$Y_{40}$，有无限多个潜在的可观察值序列。同样，如果您很难将其可视化，请再次查看图1.7:在图中所示的每个正态曲线下都有无限个可能的值。表3.3中的$n=40 Y_i$值是从这些分布中随机选择的一组。

统计代写|回归分析代写Regression Analysis代考请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。

微观经济学代写

线性代数代写

线性代数是数学的一个分支，涉及线性方程，如：线性图，如：以及它们在向量空间和通过矩阵的表示。线性代数是几乎所有数学领域的核心。

博弈论代写

微积分代写

计量经济学代写

MATLAB代写

2023年5月17日2023年5月17日 Regression Analysis, 回归分析, 统计代写, 统计代考

统计代写|回归分析代写Regression Analysis代考|The Linear Regression Function, and Why It Is Wrong

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统计代写|回归分析代写Regression Analysis代考|The Linear Regression Function, and Why It Is Wrong

Usually, when people learn regression, they learn to understand the relationship between $Y$ and $X$ as a linear function. Specifically, the linearity assumption states that the means of the conditional distributions $p(y \mid x)$ fall precisely on a straight line of the form $\beta_0+\beta_1 x$, i.e., that $\mu_x=\mathrm{E}(Y \mid X=x)=\beta_0+\beta_1 x$.

See Figure 1.7 above for a graphic illustration of what this assumption tells you about the means of the conditional distributions: In that graph, four conditional distributions are shown, corresponding to four distinct values $X=x$. The linearity assumption states that the means of those four distributions, as well as the means for all other conditional distributions that are not shown in Figure 1.7, fall precisely on a straight line $\beta_0+\beta_1 x$, for some values of the parameters $\beta_0$ and $\beta_1$. The linearity assumption does not require that you know the numerical values of $\beta_0$ and $\beta_1$; rather, it simply states that the conditional means fall on some line $\beta_0+\beta_1 x$, for some (usually unknown) numerical values of the parameters $\beta_0$ and $\beta_1$.

The parameter $\beta_0$ is called the intercept of the line. When $\mathrm{E}(Y \mid X=x)=\beta_0+\beta_1 x$, it follows that $\mathrm{E}(Y \mid X=0)=\beta_0+\beta_1(0)=\beta_0$. In words, if the linearity assumption is true, then the mean of the distribution of $Y$ when $X=0$ is equal to $\beta_0$. Often, the range of $X$ does not include 0 , in which case that interpretation is not particularly useful. In such cases, you can vaguely interpret $\beta_0$ as a parameter related to the unconditional mean of $Y$ : If the mean of $Y$ is larger, then $\beta_0$ will be larger to reflect the vertical height, or distance from zero, of the regression function.

The parameter $\beta_1$ tells you something about the relationship between $Y$ and $X$. If the linearity assumption is true, then this parameter is the difference between the conditional means of the distributions of $Y$ where the $X$ variable differs by 1.0, which can be demonstrated as follows:
$$
\begin{aligned}
\mathrm{E}(Y \mid X=x+1)-\mathrm{E}(Y \mid X=x) & =\left{\beta_0+\beta_1(x+1)\right}-\left(\beta_0+\beta_1 x\right) \
& =\left{\beta_0+\beta_1 x+\beta_1\right}-\beta_0-\beta_1 x \
& =\beta_0+\beta_1 x+\beta_1-\beta_0-\beta_1 x \
& =\beta_1
\end{aligned}
$$

统计代写|回归分析代写Regression Analysis代考|LOESS: An Estimate of the True (Curved) Mean Function

So, the linearity assumption $\mathrm{E}(Y \mid X=x)=\beta_0+\beta_1 x$ is wrong. What is right? What is right is that $\mathrm{E}(Y \mid X=x)=f(x)$, which is some function $f(x)$ that you do not know. However, data allow you to estimate such unknown quantities.

If your data set had lots of repeats on particular $x$ values, you could use the average of the $Y$ data values where $X=x$ to estimate the function $f(x)$. For example, consider the data in Table 1.6 below obtained from a survey of students in a class. The $Y$ variable is “rating of the instructor,” on a discrete 1 to 5 scale (where 5 means “best”), and the $X$ variable is “expected grade in course,” where $0=$ ” $\mathrm{F}^{\prime \prime}, 1=$ ” $\mathrm{D}$ “, $2=$ ” $\mathrm{C}$ “, $3=$ “B”, and $4=$ “A.”

Using the data shown in Table 1.6, an obvious estimate of $\mathrm{E}(Y \mid X=2)$ is $\hat{f}(2)=(2+3) / 2=2.5$ (the hat $\left(“{ }^{\prime \prime}\right)$ signifies that this is just an estimate, not the true expected value). Similar, intuitively obvious estimates are $\hat{f}(3)=(5+2+4+4) / 4=3.75$, and $\hat{f}(4)=(5+4+4+5) / 4=4.5$.

The data and the estimated mean function are shown in Figure 1.14. Notice that the function $\hat{f}(x)$ is not perfectly linear, as is expected since there are three distinct $X$ values.
$R$ code for Figure 1.14
$\mathrm{x}=\mathrm{c}(2,2,3,3,3,3,4,4,4,4)$
$y=c(2,3,5,2,4,4,5,4,4,5)$
$\mathrm{x} 1=\mathrm{c}(2,3,4)$
f. hat $=c(2.5,3.75,4.5)$
plot (x, jitter $(y, 5)$, $y l a b=$ “Rating of Instructor (jittered)”,
$x l a b=$ “Expected Grade”, cex. axis $=0.8$, cex. $l a b=0.8$ )
points $(x 1, f$. hat, pch $=” X “)$
points $(x 1, f$. hat, type=”1″, Ity=2)

回归分析代写

统计代写|回归分析代写Regression Analysis代考|The Linear Regression Function, and Why It Is Wrong

通常，当人们学习回归时，他们学习将$Y$和$X$之间的关系理解为线性函数。具体地说，线性假设表明条件分布$p(y \mid x)$的均值精确地落在形式为$\beta_0+\beta_1 x$的直线上，即$\mu_x=\mathrm{E}(Y \mid X=x)=\beta_0+\beta_1 x$。

请参见上面的图1.7，图中给出了这个假设告诉您关于条件分布的均值的图形说明:在该图中，显示了四个条件分布，对应于四个不同的值$X=x$。线性假设表明，对于参数$\beta_0$和$\beta_1$的某些值，这四个分布的均值以及图1.7中未显示的所有其他条件分布的均值都精确地落在一条直线$\beta_0+\beta_1 x$上。线性假设不需要知道$\beta_0$和$\beta_1$的数值;相反，它只是简单地声明，对于参数$\beta_0$和$\beta_1$的一些(通常是未知的)数值，条件均值落在某一行$\beta_0+\beta_1 x$上。

参数$\beta_0$称为直线的截距。当$\mathrm{E}(Y \mid X=x)=\beta_0+\beta_1 x$，它跟着$\mathrm{E}(Y \mid X=0)=\beta_0+\beta_1(0)=\beta_0$。也就是说，如果线性假设成立，那么当$X=0$等于$\beta_0$时，$Y$分布的均值。通常，$X$的范围不包括0，在这种情况下，这种解释不是特别有用。在这种情况下，您可以模糊地将$\beta_0$解释为与$Y$的无条件平均值相关的参数:如果$Y$的平均值较大，则$\beta_0$将较大，以反映回归函数的垂直高度或与零的距离。

参数$\beta_1$告诉您$Y$和$X$之间的关系。如果线性假设成立，则该参数为$Y$分布的条件均值之差，其中$X$变量相差1.0，可以如下所示:
$$
\begin{aligned}
\mathrm{E}(Y \mid X=x+1)-\mathrm{E}(Y \mid X=x) & =\left{\beta_0+\beta_1(x+1)\right}-\left(\beta_0+\beta_1 x\right) \
& =\left{\beta_0+\beta_1 x+\beta_1\right}-\beta_0-\beta_1 x \
& =\beta_0+\beta_1 x+\beta_1-\beta_0-\beta_1 x \
& =\beta_1
\end{aligned}
$$

统计代写|回归分析代写Regression Analysis代考|LOESS: An Estimate of the True (Curved) Mean Function

所以线性假设$\mathrm{E}(Y \mid X=x)=\beta_0+\beta_1 x$是错误的。什么是正确的?正确的是$\mathrm{E}(Y \mid X=x)=f(x)$，它是一些你不知道的函数$f(x)$。然而，数据允许您估计这些未知量。

如果你的数据集有很多重复 $x$ 值，你可以用平均值 $Y$ 数据值为 $X=x$ 来估计函数 $f(x)$．例如，考虑下表1.6中从班级学生调查中获得的数据。该 $Y$ 变量是“教师的评分”，分值为1到5(其中5表示“最好”) $X$ 变量是“课程预期成绩”，其中 $0=$ ＂ $\mathrm{F}^{\prime \prime}, 1=$ ＂ $\mathrm{D}$ “， $2=$ ＂ $\mathrm{C}$ “， $3=$ “B”，和 $4=$ “a。”

使用表1.6所示的数据，对$\mathrm{E}(Y \mid X=2)$的一个明显的估计值是$\hat{f}(2)=(2+3) / 2=2.5$ ($\left(“{ }^{\prime \prime}\right)$表示这只是一个估计值，而不是真正的期望值)。类似的直观估计是$\hat{f}(3)=(5+2+4+4) / 4=3.75$和$\hat{f}(4)=(5+4+4+5) / 4=4.5$。

数据和估计的均值函数如图1.14所示。注意，函数$\hat{f}(x)$不是完全线性的，因为有三个不同的$X$值。
$R$代码参见图1.14
$\mathrm{x}=\mathrm{c}(2,2,3,3,3,3,4,4,4,4)$
$y=c(2,3,5,2,4,4,5,4,4,5)$
$\mathrm{x} 1=\mathrm{c}(2,3,4)$
F. hat $=c(2.5,3.75,4.5)$
plot (x, jitter $(y, 5)$, $y l a b=$“教员评分(抖动)”，
$x l a b=$“期望成绩”，等。轴$=0.8$, cex。$l a b=0.8$)
积分$(x 1, f$。嗨，PCH $=” X “)$
积分$(x 1, f$。帽子，type=”1″， Ity=2)

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