STAT3613 Tutorial 04

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Nov 24, 2024

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT2313/3613 Marketing Engineering Tutorial 4 Review of Factor Analysis (a) Notation p - number of variables in the data set. m - number of selected common factors. F = { F 1 , F 2 , . . . , F m } | - unobservable and random factors. X = { X 1 , X 2 , . . . , X p } | - continuous random variables. μ = { μ 1 , μ 2 , . . . , μ p } | - mean of X 1 , . . . , X p . Σ - p × p sample covariance matrix for X . s ij = s ji - sample covariance between X i and X j . Γ - p × p correlation matrix for X . r ij = r ji - covariance between X i and X j . L = l 11 l 12 . . . l 1 m l 21 l 22 . . . l 2 m . . . . . . . . . . . . l p 1 l p 2 . . . l pm - matrix of factor loadings. l ij - factor loadings (covariance) of the i th variable and the j th factor. h 2 i - communality (portion of variance explained by the m factors) of X i . = { 1 , 2 , . . . , p } | - additional sources of variation/errors/specific factors for the variables X 1 , . . . , X p . Ψ = ψ 1 0 . . . 0 0 ψ 2 . . . 0 . . . . . . . . . . . . 0 0 . . . ψ p - matrix of uniqueness or specific variance. ψ i - uniqueness or specific variance of the specific factor i for X i . (b) Objective The basic idea is that it may be possible to describe a set of p variables X 1 , X 2 , . . . , X p in terms of a smaller number of unobservable random quantities called factors F 1 , F 2 , . . . , F m , and hence illustrate the relationship between these variables by approximating the covariance or correlation matrix. 1
(c) Factor analysis model (i) Model form: (In matrix, X - μ = LF + ) X 1 - μ 1 = l 11 F 1 + l 12 F 2 + . . . + l 1 m F m + 1 X 2 - μ 2 = l 21 F 1 + l 22 F 2 + . . . + l 2 m F m + 2 . . . X p - μ p = l p 1 F 1 + l p 2 F 2 + . . . + l pm F m + p (ii) Assumptions 1. X is linearly dependent upon F and . 2. E ( F ) = 0 E ( F j ) = 0 for j = 1 , . . . , m 3. Cov( F ) = I var( F j ) = 1 for j = 1 , . . . , m 4. E ( ) = 0 E ( i ) = 0 for i = 1 , . . . , p 5. Cov( ) = Ψ = ψ 1 0 . . . 0 0 ψ 2 . . . 0 . . . . . . . . . . . . 0 0 . . . ψ p var( i ) = ψ i for i = 1 , . . . , p 6. F and are independent Cov( F , ) = 0 (iii) Properties 1. Cov( X ) = Σ = Cov( LF + ) = L Cov( F ) L | + Cov( ) = LL | + Ψ where the ( i, i ) th element is var( X i ) = l 2 i 1 + l 2 i 2 + . . . + l 2 im + ψ i = h 2 i + ψ i and the ( i, j ) th element is Cov( X i , X k ) = l i 1 l k 1 + l i 2 l k 2 + . . . + l im l km , for i 6 = j . 2. Cov( X , F ) = L where the ( i, j ) th element is Cov( X i , F j ) = l ij . Note: L is not unique: Let T be any m × m (non-random) orthogonal matrix such that T T | = T | T = I Define a new loading L * and factor F * L * = LT , F * = T | F We have Σ * = L * ( L * ) | + Ψ = LT T | L | + Ψ = LL | + Ψ = Σ E ( F * ) = T | E ( F ) = 0 var( F * ) = T | var( F ) T = T | IT = I 2
Communality and specific variance do not change l 2 i 1 + . . . + l 2 im = ( l * i 1 ) 2 + . . . + ( l * im ) 2 ψ i = ψ * i σ ii = σ 2 ii Example 1. The covariance matrix for three standardized random variables Z 1 , Z 2 and Z 3 is given as Γ = 1 . 0 0 . 63 0 . 45 0 . 63 1 . 0 0 . 35 0 . 45 0 . 35 1 . 0 . (a) Show that the covariance matrix can be generated by the one factor model: Z 1 = 0 . 9 F 1 + 1 , Z 2 = 0 . 7 F 1 + 2 , Z 3 = 0 . 5 F 1 + 3 , where var( F 1 ) = 1 , Cov( i , F 1 ) = 0 ( i = 1 , 2 , 3) and Ψ = 0 . 19 0 0 0 0 . 51 0 0 0 0 . 75 . That is, write Γ in the form Γ = LL | + Ψ . (b) Calculate communalities h 2 i , i = 1 , 2 , 3 , and interpret these quantities. (c) Calculate Corr( Z i , F 1 ) , i = 1 , 2 , 3 . Which variable might carry the greatest weight in "naming" the common factor? Why? Solution (a) The one factor model: ( p = 3 , m = 1) Z 1 Z 2 Z 3 = l 11 l 21 l 31 F 1 + 1 2 3 Z = LF + Hence, var( Z ) = L var( F 1 ) L | + var( ) + L Cov( F 1 , ) + Cov( F 1 , ) L | = LL | + Ψ 3
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