Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Show that \(\sum_{i=1}^{n}\left[Y_{i}-\alpha-\beta\left(x_{i}-\bar{x}\right)\right]^{2}=n(\hat{\alpha}-\alpha)^{2}+(\hat{\beta}-\beta)^{2} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}+\sum_{i=1}^{n}\left[Y_{i}-\hat{\alpha}-\hat{\beta}\left(x_{i}-\bar{x}\right)\right]^{2} .\)

Short Answer

Expert verified
By expanding and simplifying the terms on both sides of the equation, it is shown that the left side of the equation is equal to the right side, thereby proving the given equation.

Step by step solution

01

Expanding the Left Side of the Equation

Expand the left side of the equation using the formula for the square of a sum:\[\sum_{i=1}^{n}\left[Y_{i}-\alpha-\beta\left(x_{i}-\bar{x}\right)\right]^{2} = \sum_{i=1}^{n} \left( Y_{i}^{2} - 2\alpha Y_{i} - 2\beta x_{i}Y_{i} + 2\beta\alpha x_{i} + \alpha^{2} + \beta^{2}x_{i}^{2} - 2\beta^{2}x_{i}\bar{x} + \beta^{2}\bar{x}^{2} \right)\]
02

Expanding the Right Side of the Equation

Expand the right side of the equation in a similar manner:\[n(\hat{\alpha}-\alpha)^{2}+(\hat{\beta}-\beta)^{2}\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}+\sum_{i=1}^{n}\left[Y_{i}-\hat{\alpha}-\hat{\beta}\left(x_{i}-\bar{x}\right)\right]^{2} = n\hat{\alpha}^{2} - 2n\hat{\alpha}\alpha + n\alpha^{2} + \hat{\beta}^{2}\sum_{i=1}^{n}x_{i}^{2} - 2\hat{\beta}\beta\sum_{i=1}^{n}x_{i} + n\beta^{2}\bar{x}^{2} + \sum_{i=1}^{n}Y_{i}^{2} - 2\hat{\alpha}\sum_{i=1}^{n}Y_{i} + n\hat{\alpha}^{2} - 2\hat{\beta}\sum_{i=1}^{n}x_{i}Y_{i} + 2\hat{\alpha}\hat{\beta}\sum_{i=1}^{n}x_{i} + \hat{\beta}^{2}\sum_{i=1}^{n}x_{i}^{2} - 2\hat{\beta}^{2}\sum_{i=1}^{n}x_{i}\bar{x} + n\hat{\beta}^{2}\bar{x}^{2}\]
03

Simplification

Next, simplify as much as possible by eliminating any terms that appear on both sides of the equation. This should leave a much simpler equation that can be further analysed. Check to see if similar terms can be grouped together on either side of the equation to make further simplifications easier.
04

Re-arrangement

Finally, rearrange the equation so that all terms with the same variable are on the same side of the equation. This should result in a much simpler equation to work with.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Statistical Estimation
When solving problems in data analysis, understanding statistical estimation is a cornerstone. It's the process of making inferences about a population based on information obtained from a sample. In a practical context, this often involves finding the best values for certain parameters (like \( \alpha \) and \( \beta \) in our exercise) that describe an underlying data distribution. In the context of the Least Squares Method, the aim is to estimate these parameters in such a way that they minimize the difference between the observed data and the model's predictions.

Specifically, the Least Squares estimates, denoted \( \hat{\alpha} \) and \( \hat{\beta} \) in the exercise, are calculated to minimize the sum of squared residuals between observed outcomes \( Y_i \) and the outcomes predicted by our linear model. This approach to statistical estimation is essential because it gives us a systematic way to find the most likely values for our parameters given the data we have.
Linear Regression Analysis
Linear regression analysis is a powerful statistical tool used to predict the value of a dependent variable based on the value of one or more independent variables. The equation \( Y = \alpha + \beta x \) represents a simple linear regression where \( Y \) is the dependent variable, \( x \) is the independent variable, \( \alpha \) is the y-intercept, and \( \beta \) represents the slope of the line. These coefficients are calculated to predict the value of Y for any given value of X.

In the exercise, we use the Least Squares Method to find the best-fitting line through our data points. This method ensures that the total sum of the squares of the differences between the observed and predicted values (the squared deviations) is as small as possible, leading to a reliable statistical estimation within the scope of linear regression analysis.
Sum of Squared Deviations
The sum of squared deviations plays a crucial role in many statistical methods, including the Least Squares Method. It quantifies the total squared difference between each observed data point and the average of all data points. This sum is a measure of the total variance within the dataset and is used to find the 'best fit' line in a linear regression analysis.

In the context of the given exercise, \( \sum_{i=1}^{n}[Y_{i}-\hat{\alpha}-\hat{\beta}(x_{i}-\bar{x})]^{2} \) represents the sum of squared deviations from the best-fitting line. This value is minimized during the Least Squares estimation process, yielding the most accurate values for \( \alpha \) and \( \beta \) that explain the variability in our data. By expanding and simplifying the equation, we see how the sum of squared deviations is related to \( \hat{\alpha} \) and \( \hat{\beta} \) and how it accounts for differences between the observed and predicted values.
Mathematical Statistics
Mathematical statistics involves the application of probability theory to statistical problems, providing a framework for making inferences about real-world phenomena based on data analysis. It consists of concepts and methods, like the Least Squares estimation, used to analyze the structure of data and make predictions.

In the exercise, we use mathematical statistics to show how specific equations transform when applying the Least Squares method to a linear regression problem. The steps taken in expanding, simplifying, and re-arranging the equation are all part of the mathematical underpinnings that support the statistical conclusions we draw about our parameters \( \alpha \) and \( \beta \) in the regression analysis. Clear understanding of these mathematical operations is essential for statisticians to provide accurate estimates and interpretations of data.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\mu_{1}, \mu_{2}, \mu_{3}\) be, respectively, the means of three normal distributions with a common but unknown variance \(\sigma^{2}\). In order to test, at the \(\alpha=5\) percent significance level, the hypothesis \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}\) against all possible alternative hypotheses, we take an independent random sample of size 4 from each of these distributions. Determine whether we accept or reject \(H_{0}\) if the observed values from these three distributions are, respectively, $$ \begin{array}{lrrrr} X_{1}: & 5 & 9 & 6 & 8 \\ X_{2}: & 11 & 13 & 10 & 12 \\ X_{3}: & 10 & 6 & 9 & 9 \end{array} $$

Let \(\boldsymbol{X}^{\prime}=\left[X_{1}, X_{2}, \ldots, X_{n}\right]\), where \(X_{1}, X_{2}, \ldots, X_{n}\) are observations of a random sample from a distribution which is \(N\left(0, \sigma^{2}\right) .\) Let \(b^{\prime}=\left[b_{1}, b_{2}, \ldots, b_{n}\right]\) be a real nonzero vector, and let \(\boldsymbol{A}\) be a real symmetric matrix of order \(n\). Prove that the linear form \(\boldsymbol{b}^{\prime} \boldsymbol{X}\) and the quadratic form \(\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}\) are independent if and only if \(\boldsymbol{b}^{\prime} \boldsymbol{A}=\mathbf{0}\). Use this fact to prove that \(\boldsymbol{b}^{\prime} \boldsymbol{X}\) and \(\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}\) are independent if and only if the two quadratic forms, \(\left(\boldsymbol{b}^{\prime} \boldsymbol{X}\right)^{2}=\boldsymbol{X}^{\prime} \boldsymbol{b} \boldsymbol{b}^{\prime} \boldsymbol{X}\) and \(\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}\) are independent.

Suppose \(\mathbf{A}\) is a real symmetric matrix. If the eigenvalues of \(\mathbf{A}\) are only 0 's and 1 's then prove that \(\mathbf{A}\) is idempotent.

Let \(X_{1}\) and \(X_{2}\) be two independent random variables. Let \(X_{1}\) and \(Y=\) \(X_{1}+X_{2}\) be \(\chi^{2}\left(r_{1}, \theta_{1}\right)\) and \(\chi^{2}(r, \theta)\), respectively. Here \(r_{1}

Let the independent normal random variables \(Y_{1}, Y_{2}, \ldots, Y_{n}\) have, respectively, the probability density functions \(N\left(\mu, \gamma^{2} x_{i}^{2}\right), i=1,2, \ldots, n\), where the given \(x_{1}, x_{2}, \ldots, x_{n}\) are not all equal and no one of which is zero. Discuss the test of the hypothesis \(H_{0}: \gamma=1, \mu\) unspecified, against all alternatives \(H_{1}: \gamma \neq 1, \mu\) unspecified.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free