A. Rewrite the data in Example 1 with new x-coordinates in...

a. Rewrite the data in Example 1 with new $x$-coordinates in mean deviation form. Let $X$ be the associated design matrix. Why are the columns of $X$ orthogonal? b. Write the normal equations for the data in part (a), and solve them to find the least-squares line, $y=\beta_{0}+\beta_{1} x^{*}$ where $x^{*}=x-5.5 .$

a. Rewrite the data in Example 1 with new $x$-coordinates in mean deviation form. Let $X$ be the associated design matrix. Why are the columns of $X$ orthogonal?
b. Write the normal equations for the data in part (a), and solve them to find the least-squares line, $y=\beta_{0}+\beta_{1} x^{*}$ where $x^{*}=x-5.5 .$

Key Concepts

Design Matrix

The design matrix is a structured arrangement of the predictor variable values along with a column of ones representing the intercept term. This matrix is used in linear regression to set up the system of equations necessary for estimating the model parameters by linking the observed data with the regression coefficients.

Mean Deviation Form

Mean deviation form involves recentering the predictor variables by subtracting their mean from each observation, thus creating a new variable with a mean of zero. This transformation simplifies computations and interpretation in regression analysis, as it can reduce multicollinearity and makes the intercept term represent the overall mean response.

Orthogonal Columns

In the context of a design matrix for regression, columns are orthogonal when their inner product (or sum of products) is zero. When the predictor variable is centered (mean deviation form), its column becomes uncorrelated with the constant (intercept) column, which simplifies the estimation process because the normal equations decouple, allowing independent calculation of regression coefficients.

Normal Equations

Normal equations are derived by setting the derivative of the sum of squared residuals with respect to each parameter equal to zero. These equations form a system that, when solved, provide the least squares estimates of the model parameters. The simplicity of the normal equations can be enhanced through the orthogonality of the design matrix columns.

Least Squares Estimation

Least squares estimation is a method used to determine the best-fitting line through the data by minimizing the sum of the squared differences between the observed responses and the values predicted by the linear model. This approach provides estimates for the intercept and slope that make the overall prediction error as small as possible.

Transcript

00:01 Okay, so question is write x coordinate of example 1 in mean deviation form where the data of example 1 is given here.

00:08 And first is why columns are orthogonal of this obtained matrix and b write normal equation and find the least square feet line for this where x star is x minus 5 .5.

00:24 Okay, so to find the mean deviation form, we need to find the mean of the given x values.

00:30 So we have x here, 2, 5, 7, 8.

00:36 So summation of x will be 2 plus 5 plus 7 plus 8 and mean will be summation of x over n which is 4.

00:49 So adding this we get 10 plus 12.

00:53 So that is 22 by 4 and that is equals to 5 .5.

01:01 Okay, so we found our mean.

01:04 So therefore, we can write a new data as 2 minus 5 .5.

01:09 So we'll write this as x star, which is 2 minus 5 .5, that is minus 3 .5.

01:17 Then we have 5 minus 5 .5.

01:20 So minus 0 .5.

01:21 7 minus 5 .5 is 1 .5, right? and then 8 minus 5 .5 is 2 .5 so we got our extra now a normal equation will be so we know normal equation is the form is given as y equals to b0 plus beta 1 extra so y is same as previous 1 2 3 3 then beta not and beta 1 are unknown parameter so we can write it as vector form and here for beta not we'll have 1 1 1 and 1 right and for beta 1 coefficient is x star which is minus 3 .5 minus 0 .5 1 .5 and 2 .5 now we need to show why the columns are orthogonal so this is our x this is our and this is a beta okay so we need to show why column of x or orthogonal so orthogonal we obtain by doing inner product right so we'll do a inner product of column one and column two of x so x column one times x column two right and we need to do inner product okay so that is equals to 1 times minus 3 .5, 1 times minus 0 .5, 1 .5 and 2 .5.

02:59 So that gives us minus 3 .5 plus minus 0 .5 plus 1 .5 plus 2 .5 and that is equals to 0.

03:11 So indeed by solving we know it is orthogonal since we are getting 0, right? a vector is orthogonal when its inner product is 0.

03:24 So why it is happening? so let's take the variables and then we can understand...

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