TAMS39 Lecture 10 Principal Component Analysis Factor Analysis

Transcription

1 TAMS39 Lecture 10 Principal Component Analysis Factor Analysis Martin Singull Department of Mathematics Mathematical Statistics Linköping University, Sweden

2 Content - Lecture Principal component analysis (PCA) Factor analysis (FA) TAMS39 Lecture 10 1/36

3 Principal component analysis (PCA) A principal component analysis (PCA) is concerned with explaining the variance-covariance structure of a set of variables through a few linear combinations of these variables. Its general objectives are data reduction, and interpretation. Although p components are required to reproduce the total system variability, often much of this variability can be accounted for by a small number k of the principle components. TAMS39 Lecture 10 2/36

4 Let x = (x 1,..., x p ) be a random vector with the covariance Σ. Algebraically, the principal components are particular linear combinations of the p random variables. Geometrically, these linear combinations represent the selection of a new coordinate system obtained by rotating the original system with x 1,..., x p as the coordinate axes. TAMS39 Lecture 10 3/36

5 Consider the linear combinations y 1 = a 1x,. y p = a px, with the variances and covariances var(y i ) = a iσa i, cov(y i, y k ) = a iσa k, i = 1,...p, i, k = 1,...p. The principal components are those uncorrelated linear combinations y 1,..., y p whose variance above are as large as possible. TAMS39 Lecture 10 4/36

6 First principal component = linear combination y 1 = a 1 x that maximize var(a 1 x) subject to a 1 a 1 = 1.. ith principal component = linear combination y i = a i x that maximize var(a i x) subject to a i a i = 1 and cov(a i x, a kx) = 0 for k < i, i = 1,..., p. TAMS39 Lecture 10 5/36

7 Let λ 1,..., λ p > 0 be the eigevalue of the matrix Σ and let H = (h 1,..., h p ) be an m m orthogonal matrix such that H ΣH = diag(λ 1,..., λ p ) = Λ, so that h i is a eigenvector of Σ corresponding to the eigenvalue λ i. We now have that the covariance between any linear combination a x and a linear combination based on a eigenvector h ix is given by cov(a x, h ix) = a Σh i = λ i a h i. Hence, cov(a x, h ix) = 0 is the same as a and h i to be orthogonal. TAMS39 Lecture 10 6/36

8 Theorem For k = 1,..., p λ k = max a a=1, a h i =0, i=1,...,k 1 a Σa = h kσh k. TAMS39 Lecture 10 7/36

9 TAMS39 Lecture 10 8/36

10 Measures of total variation Note that in transforming to principal components the measure trσ and Σ of total variations are unchanged, for p trσ = trh ΣH = trλ = λ i, Σ = H ΣH = Λ = i=1 p λ i. Note also that k i=1 λ i is the variance of the first k principal components. In principal component analysis the hope is that for some small k, this variance is close to trσ, i.e., the first k principle components explain most of the variation in x, and the remaining p k principal components contribute little. i=1 TAMS39 Lecture 10 9/36

11 Sample principle component analysis Assume that x 1,..., x n are iid N p (µ, Σ), i.e., X = (x 1,..., x n ) N p,n (µ1, Σ, I ). The MLE of Σ is n 1 n S where S is the sample covariance matrix given by S = 1 n 1 X ( I n 1 n (1 n1 n ) 1 1 ) n X. The MLE of the λ i s, the ordered (assumed distinct) eigenvalues of Σ, are n 1 n ˆλ i, where ˆλ i s are the ordered eigenvalues of S. The ˆλ i s are distinct with probability one, since n > p. TAMS39 Lecture 10 10/36

12 The sample principle components are defined as First sample principal component = linear combination ŷ 1 = a 1 x that maximize the sample variance a 1 Sa 1 subject to a 1 a 1 = 1.. ith smaple principal component = linear combination ŷ i = a i x that maximize the sample variance a i Sa i subject to a i a i = 1 and a i Sa k = 0 for k < i, i = 1,..., p. TAMS39 Lecture 10 11/36

13 We now have a similar theorem as above. Theorem For k = 1,..., p ˆλ k = max a Sa = a a=1, a ĥ ĥ ksĥk, i =0, i=1,...,k 1 where ĥk is a eigenvector of the MLE n 1 n S of Σ corresponding to the eigenvalue ˆλ k. TAMS39 Lecture 10 12/36

14 Asymptotic distributions Assume that x 1,..., x n are iid N p (µ, Σ). Assume also that the eigenvalues of Σ are distinct and positive, so that 0 < λ p <... < λ 1. The sampling distribution of the MLEs ˆλ i and ĥi are difficult to derive and beyond the scope of this course. We shall simply summarize some results. TAMS39 Lecture 10 13/36

15 Asymptotic mean and variance Tractable expressions for the exact moments of the eigenvalues of S are unknown, but asymptotic expressions for some of these have been found by Lawley (1956). Lawley has shown that if λ i is a distinct eigenvalue of Σ the mean and variance of ˆλ i can be expand for large n as E(ˆλ i ) = λ i + λ i n p j=1,j i λ j λ i λ j + O(n 2 ) and var(ˆλ i ) = 2λ2 i n 1 1 n p j=1,j i ( λj ) 2 + O(n 3 ). λ i λ j TAMS39 Lecture 10 14/36

16 Asymptotic distribution, cont. Let and then we have E i = λ i λ = (λ 1,..., λ p ) p k=1,k i λ k (λ k λ i ) 2 h kh k, 1. n(ˆλ λ) N p (0, 2Λ 2 ), where Λ = diag(λ 1,..., λ p ), i.e., n(ˆλi λ i ) 2λi N(0, 1) for i = 1,..., p, 2. n(ĥi h i ) N p (0, E i ), 3. ˆλ i and ĥi is independently distributed for i = 1,..., p. TAMS39 Lecture 10 15/36

17 Result 1 implies that, for large n, the ˆλ i are independently distributed. Using result 1 one can also construct confidence interval for the λ i s. Result 2 implies that the elements of each ĥi are correlated, and the correlation depends to a large extent on the separation of the eigenvalues λ 1,..., λ p (which are unknown) and the sample size n. TAMS39 Lecture 10 16/36

18 H : λ k+1 = λ k+2 =... = λ p = λ Suppose we want to test the hypothesis that the last p k eigenvalues of the covariance matrix Σ are equal to λ. That is, we want to test H : λ k+1 = λ k+2 =... = λ p = λ vs. A H, where λ is unknown. This is the so-called isotropy test for the eigenvalues. Typically one conducts a series of isotropy tests starting with k = p 2 and increasing k until the null hypothesis is accepted. TAMS39 Lecture 10 17/36

19 PCA LRT The LRT (based in normality) for the hypothesis H is based in the statistic Q = p ˆλ j=k+1 j ( 1 p ˆλ ) p k, p k j=k+1 j where ˆλ 1 ˆλ 2... ˆλ p are the eigenvalues of the sample covariance matrix S based on f = n 1 degrees of freedom. It has been shown by Lawley (1956) that Q = ( f k 1 6 ( 2(p k) p k χ 2 ( 1 2 (p k)(p k + 1) 1 ). )) ln Q Hence, H is rejected if Q > χ 2 1 α ( 1 2 (p k)(p k + 1) 1). TAMS39 Lecture 10 18/36

20 Example PCA The weekly rates of return for five stocks (JP Morgan, Citibank, Wells faro, Royal Dutch Shell, and ExxonMobil) listed in the NY Stock Exchange were determined for the period january 2004 through December The weekly rates of return redefined as current week closing price - previous week closing price, previous week closing price adjusted for stock splits and dividends. The observations in 103 successive weeks appear to be independently distributed, but the rates of return across stocks are correlated, because as on expects, stocks tend to move together in response to general economics conditions. TAMS39 Lecture 10 19/36

21 Let x 1,..., x 5 denote observed weekly rates of return for the stocks given above. Then x = (0.0011, , , , ) R = , 1 and where R is the sample correlation matrix R = D 1/2 D S = diag(s 11,..., s 55 ). S SD 1/2 S, with We note that R is the sample covariance matrix for the standardize observations z i = x i x i sii, i = 1,..., 5. TAMS39 Lecture 10 20/36

22 The eigenvalues and corresponding normalized eigenvectors of R are ˆλ 1 = 2.437, ĥ 1 = (0.469, 0.532, 0.465, 0.387, 0.361), ˆλ 2 = 1.407, ĥ 2 = ( 0.368, 0.236, 0.315, 0.585, 0.606), ˆλ 3 = 0.501, ĥ 3 = ( 0.604, 0.136, 0.772, 0.093, 0.109), ˆλ 4 = 0.400, ĥ 4 = (0.363, 0.629, 0.289, 0.381, 0.493), ˆλ 5 = 0.255, ĥ 5 = (0.384, 0.496, 0.071, 0.595, 0.498). TAMS39 Lecture 10 21/36

23 Using the standardize variables, we obtain the first two sample principle components: ŷ 1 = ĥ 1z = 0.469z z z z z 5, ŷ 2 = ĥ 2z = 0.368z z z z z 5, and these components, which account for ˆλ 1 + ˆλ 2 p = = 0.77, i.e., 77% of the total (standardize) sample variance, have interesting interpretations. TAMS39 Lecture 10 22/36

24 The first component is a roughly equally weighted sum, or index, of the five stocks. This component might be called a general stock-market component, or, simply, a market component. The second component represent a contrast between banking stocks and the oil stocks. It might be called an industry component. Thus, we see that most of the variation in these stock return is due to market activity and uncorrelated industry activity. TAMS39 Lecture 10 23/36

25 Factor analysis Closely related to PCA is factor analysis. Factor analysis is a statistical method used to study the dimensionality of a set of variables. In factor analysis, latent variables represent unobserved constructs and are referred to as factors or dimensions. The essential purpose of factor analysis is to describe, if possible, the covariance relationships among many variables in terms of a few underlying, but unobservable, random quantities called factors. TAMS39 Lecture 10 24/36

26 Factor analysis Example Suppose we wish to judge the abilities of high school students entering university. We may give them a test of 50 questions. These 50 questions, however, may fall into a few categories, such as reading comprehension, mathematics, and arts. Here we only have three factors. TAMS39 Lecture 10 25/36

27 The score of any randomly selected high school student on the ith question, denoted by y i, can be modeled in the form y i = µ i + λ i1 f 1 + λ i2 f 2 + λ i3 f 3 + ε i, i = 1,..., 50, where µ i is the mean for y i. Without loss of generality we can assume that f j iid N(0, 1) for j = 1, 2, 3, and independently distributed of the errors ε i iid N(0, ψ i ) for i = 1,..., 50. TAMS39 Lecture 10 26/36

28 Factor analysis Model In matrix notation, the model with k factors and p characteristics of a subject can be written as y = µ + Λf + ε, where y = (y 1,..., y p ), µ = (µ 1,..., µ p ), and Ψ = diag(ψ 1,..., ψ p ). ε = (ε 1,..., ε p ) N p (0, Ψ), f = (f 1,..., f k ) N k (0, I k ), λ λ 1k Λ =.. : p k, λ p1... λ pk TAMS39 Lecture 10 27/36

29 Factor analysis Covariance matrix Since, cov(f ) = I k, cov(ε) = Ψ and cov(f, ε) = 0, it follows that the covariance of y is given by cov(y) = Λ cov(f )Λ + cov(ε) = ΛΛ + Ψ Σ. Note that the value of Σ is unchanged if Λ is post-multiplied by any k k orthogonal matrix. Hence, there is no unique choice of Λ. TAMS39 Lecture 10 28/36

30 Factor analysis Number of factors The number of parameters that need to be estimated is pk for Λ and p for the diagonal elements of Ψ, totalling to p(k + 1). The number of quantities available for estimating is p(p + 1)/2 elements in the sample covariance matrix S. Thus in principle the number of factors that can be selected to represent the data should be less than or equal to (p 1)/2. We have noted that the value of Σ is unchanged if Λ is postmultiplied by any k k orthogonal matrix. Thus, the effective number of parameters is not p(k + 1), but p(k + 1) k(k 1)/2. Hence, p(k + 1) k(k 1) 2 p(p + 1) 2 k 1 2 ((2p + 1) 8p + 1). TAMS39 Lecture 10 29/36

31 Factor analysis Uniqueness It should be mentioned, however, that if k satisfies the inequality, it does not necessary imply that a solution exist, let alone uniqueness. The inequality above has been arrived by requiring that the number of unknowns should be less than or equal to the number of equations available. To get a unique solution, we not only require that k satisfies the inequality but also that the k k matrix Λ Ψ 1 Λ is a diagonal matrix with diagonal elements that are ordered from largest to smallest (see Lawley and Maxwell, 1970). TAMS39 Lecture 10 30/36

32 Factor analysis MLE The estimate cannot be obtained explicitly, and iterative methods have to be used. To obtain the unique ML solution, it follows that when factor analysis is carried out on the correlation matrix R = D 1/2 S SD 1/2 S, where D S = diag(s 11,..., s pp ), we need to solve the equations k ˆλ 2 ij + ˆψ i = 1, i = 1,..., p, j=1 ( ) ( Ψ 1/2 R Ψ 1/2 Λ) Ψ 1/2 = Ψ 1/2 Λ D, where D is the diagonal matrix D = I + D and D is the diagonal matrix D = Λ Ψ 1 Λ (uniqueness condition). TAMS39 Lecture 10 31/36

33 Factor analysis Choosing the number of factors In factor analysis, we seek a diagonal matrix, Ψ with positive diagonal elements such that Σ Ψ is a positive semidefinite matrix of rank k. It can be shown that such a k will always be larger than the number of eigenvalues of the population correlation matrix that are greater than one (see Guttman, 1954). Since the population correlation matrix can be estimated by the sample correlation matrix, R, a rule of thumb often used in statistical packages chooses k to be the number of eigenvalues of R greater than 1. This choice of k can be used as an initial guess for the number of factors. TAMS39 Lecture 10 32/36

34 With this choice of k, we may test its adequacy by testing the hypothesis H : Σ = ΛΛ + Ψ vs. A : Σ ΛΛ + Ψ, where Λ is a p k matrix. The MLE of Σ under the alternative is S and under H with k factors is given by Σ k = Λ k Λ k + Ψ, where Λ k and Ψ are the MLEs for k factors. TAMS39 Lecture 10 33/36

35 Factor analysis LRT One can show that an asymptotic test statistic, based on the LRT, is given by ( ) S 2 ln χ 2 (g), Σ where g = 1 2 ((p k)2 (p + k)). However, Bartlett (1954) suggested replacing 2 in the above expression by n (2p + 4k + 11)/6 to get a better approximation. This factor is known as Bartlett s correction. The hypothesis H is rejected if ( n ) 2p + 4k + 11 ln 6 ( ) S > χ 2 1 α(g). Σ TAMS39 Lecture 10 34/36

36 Example Factor analysis The stock-price dat above is analyzed assuming an k = 2 factor model and using the ML method. Maximum likelihood Estimated factor loadings Specific variances Variable ˆλ1 ˆλ2 ˆψi (= 1 ĥ2 i ) 1. JP Morgan Citibank Wells Fargo Royal Dutch Shell Texaco TAMS39 Lecture 10 35/36

37 TAMS39 Lecture 10 36/36

38 Linköping University - Research that makes a difference