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For a simple linear regression model \(Y i=\alpha \beta x_{i}+\epsilon_{i}, \quad i=1,2 \ldots \ldots n\) where the \(\epsilon_{i}\) 's are independent and normally distributed with zero means and equal variances \(a \sim\), show that, \(Y\) and $$ 2 ?=\frac{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right) Y_{i}}{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}} $$ have zero covariance.

Short Answer

Expert verified
The calculation shows that the covariance of \(Y\) and \(\hat{\beta}\) is zero, which means changes in the value of \(Y\) doesn't correspond to changes in the value of \(\hat{\beta}\), and vice versa.

Step by step solution

01

Understand the regression model

The equation \(Y i=\alpha \beta x_{i}+\epsilon_{i}, \quad i=1,2 \ldots \ldots n\) describes a simple linear regression model, where \(Y_i\) is the dependent variable, \(x_i\) is the independent variable, \(\alpha\) is the intercept of the regression line, \(\beta\) is the slope of the regression line, and \(\epsilon_i\) represent the errors - deviations of the observed \(Y_i\) from the line described by \(\alpha + \beta x_{i}\).
02

Analyze the given equation

The given formula for \(\hat{\beta}\) is \(\hat{\beta} =\frac{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right) Y_{i}}{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}}\). This is the OLS estimator for the slope coefficient, \(\beta\), in a simple linear regression model. This equation calculates the slope of the line that minimizes the sum of the squared residuals.
03

Calculate Covariance

First write out what covariance is for these two variables: \(Cov(Y, \hat{\beta}) = E[(Y - E[Y])(\hat{\beta} - E[\hat{\beta}])] = E[Y\hat{\beta}] - E[Y]E[\hat{\beta}]\). Substituting for the expected values of \(Y\) and \(\hat{\beta}\), and aligning that with the original model, we see that \(E[\epsilon] = 0\) and therefore covariance becomes zero: \(Cov(Y, \hat{\beta}) = 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Covariance
Covariance is a statistical concept that measures how much two random variables change together. It indicates whether an increase in one variable corresponds to an increase or decrease in another. When two variables have a positive covariance, it means they tend to increase or decrease together. On the other hand, a negative covariance indicates that as one variable increases, the other tends to decrease.

Covariance is calculated using the formula: \[Cov(X, Y) = E[(X - E[X])(Y - E[Y])]\] where \(E[X]\) and \(E[Y]\) are the expected values (means) of \(X\) and \(Y\).

In the context of a simple linear regression model, we often examine the covariance between the dependent variable \(Y\) and the estimator of the slope \(\hat{\beta}\). If this covariance is zero, it suggests that there is no linear association between \(Y\) and \(\hat{\beta}\) in terms of their deviations from their means. This can happen if, for instance, the errors \(\epsilon_i\) in the regression model are appropriately accounted for.

Since the covariance focuses on variability in a non-standardized way, care should be taken to interpret its magnitude, which depends on the units of the variables involved.
Ordinary Least Squares
Ordinary Least Squares (OLS) is a method for estimating the parameters of a linear regression model. It aims to find the optimal values for the intercept and slope parameters that minimize the sum of squared differences between the observed values and the values predicted by the linear model.

In mathematical terms, the OLS estimator for the slope \(\beta\) in a simple linear regression is given by: \[\hat{\beta} = \frac{\sum_{i=1}^{n}(x_{i}-\bar{x})Y_{i}}{\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}}\] Here, \(\bar{x}\) is the mean of the independent variable \(x_i\), and \(Y_i\) are the dependent variable values. The formula essentially weighs the products of deviations in \(x_i\) and \(Y_i\) according to how much \(x_i\) deviates from its mean.

The OLS method is popular because it provides the best linear unbiased estimators (BLUE) when certain conditions, such as normally distributed errors with constant variance, are met. In practice, OLS is often implemented using various software tools that help compute these estimators quickly and accurately. However, understanding the underlying mathematical principles, like the derivation of \(\hat{\beta}\), helps students grasp why OLS is reliable and widely used.
Independent Variables
In regression analysis, independent variables, also known as predictors or explanatory variables, form the basis upon which we predict the response variable. In a simple linear regression, the model can be written as \(Y = \alpha + \beta x_i + \epsilon_i\), where \(x_i\) is the independent variable.

Independent variables are crucial as they explain the variability in the dependent variable \(Y\). Their choice and accuracy significantly impact the effectiveness and interpretability of the regression model. It is essential to ensure that these variables do not have a perfect linear relationship with each other, a phenomenon known as multicollinearity, which can distort the regression outcomes.

When selecting independent variables, researchers often consider:
  • Relevance: The variable should be theoretically or empirically associated with the dependent variable.
  • Availability: Reliable and accurate data should be accessible.
  • Linearity: The relationship between the independent variables and the dependent variable should be linear or transformable to linear.
By thoughtfully selecting independent variables, researchers can develop robust models that offer accurate predictions and meaningful insights.

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Most popular questions from this chapter

The following data are a result of an investigation as to the effect of reaction temperature \(x\) on percent conversion of a chemical process \(y .\) [See Myers and Montgomery (2002).] Fit a simple linear regression, and use a lack-of-fit test to determine if the model is adequate. Discuss. $$ \begin{array}{ccc} & \text { Temperature } & \text { Conversion } \\ \text { Observation } & \left({ }^{\circ} \mathbf{C}\right), x & \frac{\circ}{o}, y \\ \hline 1 & 200 & 43 \\ 2 & 250 & 78 \\ 3 & 200 & 69 \\ 4 & 250 & 73 \\ 5 & 189.65 & 48 \\ 6 & 260.35 & 78 \\ 7 & 225 & 65 \\ 8 & 225 & 74 \\ 9 & 225 & 76 \\ 10 & 225 & 79 \\ 11 & 225 & 83 \\ 12 & 225 & 81 \end{array} $$

A study was made by a retail merchant to determine the relation between weekly advertising expenditures and sales. The following data were recorded: $$ \begin{array}{cc} \text { Advertising Costs (\$) } & \text { Sales (\$) } \\ \hline 40 & 385 \\ 20 & 400 \\ 25 & 395 \\ 20 & 365 \\ 30 & 475 \\ 50 & 440 \\ 40 & 490 \\ 20 & 420 \\ 50 & 560 \\ 40 & 525 \\ 25 & 480 \\ 50 & 510 \end{array} $$ (a) Plot a scatter diagram. (b) Find the equation of the regression line to predict weekly sales from advertising expenditures. (c) Estimate the weekly sales when advertising costs are \(\$ 35 .\) (d) Plot the residuals versus advertising costs. Comment.

The Statistics Consulting Center at Virginia Polytechnic Institute and State University analyzed data on normal woodchucks for the Department of Veterinary Medicine. The variables of interest were bodyweight in grams and heart weight in grams. It was also of interest to develop a linear regression equation in order to determine if there is a significant linear relaionship between heart weight and total body weight. Use heart weight as the independent variable and body weight as the dependent variable and fit a simple linear regression using the following data. In addition, test he hypothesis \(H Q: \quad 3=0\) versus \(H_{1}: 0 \neq 0\). Draw conclusions. $$ \begin{array}{cc} \text { Body Weight } & \text { Heart Weight } \\ \text { grains) } & \text { (grams) } \\ \hline 4050 & 11.2 \\ 2465 & 12.4 \\ 3120 & 10.5 \\ 5700 & 13.2 \\ 2595 & 9.8 \\ 3640 & 11.0 \\ 2050 & 10.8 \\ 4235 & 10.4 \\ 2935 & 12.2 \\ 4975 & 11.2 \\ 3690 & 10.8 \\ 2800 & 14.2 \\ 2775 & 12.2 \\ 2170 & 10.0 \\ 2370 & 12.3 \\ 2055 & 12.5 \\ 2025 & 11.8 \\ 2645 & 16.0 \\ 2675 & 13.8 \end{array} $$

Observations on the yield of a chemical reaction taken at various temperatures were recorded as follows: $$ \begin{array}{cccc} \underline{\mathrm{S}}\left({ }^{\circ} \mathrm{C}\right) & y(\%) & \boldsymbol{x}\left({ }^{\circ} \mathrm{C}\right) & y(\%) \\ \hline 150 & 75.4 & 150 & 77.7 \\ 150 & 81.2 & 200 & 84.4 \\ 200 & 85.5 & 200 & 85.7 \\ 250 & 89.0 & 250 & 89.4 \\ 250 & 90.5 & 300 & 94.8 \\ 300 & 96.7 & 300 & 95.3 \end{array} $$ (a) Plot the data. (b) Does it appear from the plot as if the relationship is linear? (c) Fit a simple linear regression and test for lack of fitt. (d) Draw conclusions based on your result in (c).

A study conducted at. VPI\&SU to determine if certain static arm-strength measures have an influence on the "dynamic lift" characteristics of an individual. Twenty-five individuals were subjected to strength tests and then were asked to perform a weight-lifting test in which weight was dynamically lifted overhead. The data are given here. (a) Estimate \(\alpha\) and 0 for the linear regression curve \(\mu_{Y \mid x}=a+0 x\) (b) Find a point estimate of \(\mu_{Y \mid 30}\). (c) Plot the residuals versus the \(X\) s (arm strength). Comment. $$ \begin{array}{c|c|c} & \text { Arm } & \text { Dynamic } \\ \text { Individual } & \text { Strength, } \mathrm{x} & \text { Lift, } y \\ \hline 1 & 17.3 & 71.7 \\ 2 & 19.3 & 48.3 \\ 3 & 19.5 & 88.3 \\ 4 & 19.7 & 75.0 \\ 5 & 22.9 & 91.7 \\ 6 & 23.1 & 100.0 \\ 7 & 26.4 & 73.3 \\ 8 & 26.8 & 65.0 \\ 9 & 27.6 & 75.0 \\ 10 & 28.1 & 88.3 \\ 11 & 28.2 & 68.3 \\ 12 & 28.7 & 96.7 \end{array} $$ $$ \begin{array}{c|c|c} & \text { Arm } & \text { Dynamic } \\ \text { Individual } & \text { Strength, } \boldsymbol{x} & \text { Lift, } \boldsymbol{y} \\ \hline 13 & 29.0 & 76.7 \\ 14 & 29.6 & 78.3 \\ 15 & 29.9 & 60.0 \\ 16 & 29.9 & 71.7 \\ \mathbf{1 7} & \mathbf{3 0 . 3} & 85.0 \\ 18 & 31.3 & 85.0 \\ \mathbf{1 9} & 36.0 & 88.3 \\ 20 & 39.5 & 100.0 \\ 21 & 40.4 & 100.0 \\ 22 & 44.3 & 100.0 \\ 23 & 44.6 & 91.7 \\ 24 & 50.4 & 100.0 \\ 25 & 55.9 & 71.7 \end{array} $$

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