{"id":226461,"date":"2025-06-05T09:42:08","date_gmt":"2025-06-05T09:42:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=226461"},"modified":"2025-06-05T09:42:10","modified_gmt":"2025-06-05T09:42:10","slug":"x2-n-ssxx-ssxx-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/05\/x2-n-ssxx-ssxx-2\/","title":{"rendered":"x^2 + n SSxx SSxx 2"},"content":{"rendered":"\n

x^2 + n SSxx SSxx 2. Show that (X’X)^-1 = in simple linear regression 3. 1 SSxx SSxx<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To show that (X\u2032X)\u22121=[1n+x\u02c92SSXX\u2212x\u02c9SSXX\u2212x\u02c9SSXX1SSXX](\\mathbf{X}’\\mathbf{X})^{-1} = \\begin{bmatrix} \\frac{1}{n} + \\frac{\\bar{x}^2}{SS_{XX}} & -\\frac{\\bar{x}}{SS_{XX}} \\\\ -\\frac{\\bar{x}}{SS_{XX}} & \\frac{1}{SS_{XX}} \\end{bmatrix}<\/p>\n\n\n\n

in simple linear regression, let\u2019s walk through the derivation.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Simple Linear Regression Design Matrix<\/h3>\n\n\n\n

In simple linear regression, we model: yi=\u03b20+\u03b21xi+\u03b5iy_i = \\beta_0 + \\beta_1 x_i + \\varepsilon_i<\/p>\n\n\n\n

The design matrix X\\mathbf{X} is: X=[1x11x2\u22ee\u22ee1xn],size n\u00d72\\mathbf{X} = \\begin{bmatrix} 1 & x_1 \\\\ 1 & x_2 \\\\ \\vdots & \\vdots \\\\ 1 & x_n \\\\ \\end{bmatrix}, \\quad \\text{size } n \\times 2<\/p>\n\n\n\n

Then, X\u2032X\\mathbf{X}’\\mathbf{X} is: X\u2032X=[n\u2211xi\u2211xi\u2211xi2]=[nnx\u02c9nx\u02c9\u2211xi2]\\mathbf{X}’\\mathbf{X} = \\begin{bmatrix} n & \\sum x_i \\\\ \\sum x_i & \\sum x_i^2 \\\\ \\end{bmatrix} = \\begin{bmatrix} n & n\\bar{x} \\\\ n\\bar{x} & \\sum x_i^2 \\\\ \\end{bmatrix}<\/p>\n\n\n\n

\n\n\n\n

Step 2: Define SSXXSS_{XX}<\/h3>\n\n\n\n

SSXX=\u2211(xi\u2212x\u02c9)2=\u2211xi2\u2212nx\u02c92SS_{XX} = \\sum (x_i – \\bar{x})^2 = \\sum x_i^2 – n\\bar{x}^2<\/p>\n\n\n\n

\n\n\n\n

Step 3: Invert X\u2032X\\mathbf{X}’\\mathbf{X}<\/h3>\n\n\n\n

We want to find: (X\u2032X)\u22121=[nnx\u02c9nx\u02c9\u2211xi2]\u22121(\\mathbf{X}’\\mathbf{X})^{-1} = \\left[ \\begin{array}{cc} n & n\\bar{x} \\\\ n\\bar{x} & \\sum x_i^2 \\\\ \\end{array} \\right]^{-1}<\/p>\n\n\n\n

Use the inverse of a 2\u00d72 matrix:<\/p>\n\n\n\n

If A=[abcd]A = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}, then A\u22121=1ad\u2212bc[d\u2212b\u2212ca]A^{-1} = \\frac{1}{ad – bc} \\begin{bmatrix} d & -b \\\\ -c & a \\end{bmatrix}<\/p>\n\n\n\n

Apply this to X\u2032X\\mathbf{X}’\\mathbf{X}:<\/p>\n\n\n\n

    \n
  • a=na = n<\/li>\n\n\n\n
  • b=nx\u02c9b = n\\bar{x}<\/li>\n\n\n\n
  • d=\u2211xi2d = \\sum x_i^2<\/li>\n<\/ul>\n\n\n\n

    Then: det\u2061(X\u2032X)=n\u2211xi2\u2212n2x\u02c92=n(\u2211xi2\u2212nx\u02c92)=nSSXX\\det(\\mathbf{X}’\\mathbf{X}) = n \\sum x_i^2 – n^2 \\bar{x}^2 = n (\\sum x_i^2 – n\\bar{x}^2) = n SS_{XX}<\/p>\n\n\n\n

    Thus, (X\u2032X)\u22121=1nSSXX[\u2211xi2\u2212nx\u02c9\u2212nx\u02c9n](\\mathbf{X}’\\mathbf{X})^{-1} = \\frac{1}{n SS_{XX}} \\begin{bmatrix} \\sum x_i^2 & -n\\bar{x} \\\\ -n\\bar{x} & n \\end{bmatrix}<\/p>\n\n\n\n

    Now simplify each entry:<\/p>\n\n\n\n

    Top-left: \u2211xi2nSSXX=nx\u02c92+SSXXnSSXX=x\u02c92SSXX+1n\\frac{\\sum x_i^2}{n SS_{XX}} = \\frac{n\\bar{x}^2 + SS_{XX}}{n SS_{XX}} = \\frac{\\bar{x}^2}{SS_{XX}} + \\frac{1}{n}<\/p>\n\n\n\n

    Top-right and bottom-left: \u2212nx\u02c9nSSXX=\u2212x\u02c9SSXX\\frac{-n\\bar{x}}{n SS_{XX}} = -\\frac{\\bar{x}}{SS_{XX}}<\/p>\n\n\n\n

    Bottom-right: nnSSXX=1SSXX\\frac{n}{n SS_{XX}} = \\frac{1}{SS_{XX}}<\/p>\n\n\n\n

    \n\n\n\n

    Final Result:<\/h3>\n\n\n\n

    (X\u2032X)\u22121=[1n+x\u02c92SSXX\u2212x\u02c9SSXX\u2212x\u02c9SSXX1SSXX](\\mathbf{X}’\\mathbf{X})^{-1} = \\begin{bmatrix} \\frac{1}{n} + \\frac{\\bar{x}^2}{SS_{XX}} & -\\frac{\\bar{x}}{SS_{XX}} \\\\ -\\frac{\\bar{x}}{SS_{XX}} & \\frac{1}{SS_{XX}} \\end{bmatrix}<\/p>\n\n\n\n

    \n\n\n\n

    Explanation (Summary)<\/h3>\n\n\n\n

    This derivation uses the structure of the design matrix in simple linear regression and the formula for the inverse of a 2\u00d72 matrix. We rewrite the sums using sample mean x\u02c9\\bar{x} and the corrected sum of squares SSXXSS_{XX} to express the inverse in terms familiar in regression analysis. This matrix plays a key role in computing the variance-covariance matrix of the estimated coefficients and thus is fundamental in statistical inference.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    x^2 + n SSxx SSxx 2. Show that (X’X)^-1 = in simple linear regression 3. 1 SSxx SSxx The Correct Answer and Explanation is: To show that (X\u2032X)\u22121=[1n+x\u02c92SSXX\u2212x\u02c9SSXX\u2212x\u02c9SSXX1SSXX](\\mathbf{X}’\\mathbf{X})^{-1} = \\begin{bmatrix} \\frac{1}{n} + \\frac{\\bar{x}^2}{SS_{XX}} & -\\frac{\\bar{x}}{SS_{XX}} \\\\ -\\frac{\\bar{x}}{SS_{XX}} & \\frac{1}{SS_{XX}} \\end{bmatrix} in simple linear regression, let\u2019s walk through the derivation. Step 1: Simple Linear Regression Design […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-226461","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226461","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=226461"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226461\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=226461"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=226461"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=226461"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}