Covariance and Correlation Matrix

Transcription

1 Covariance and Correlation Matrix Given sample {x n } N 1, where x Rd, x n = x 1n x 2n. x dn sample mean x = 1 N N n=1 x n, and entries of sample mean are x i = 1 N N n=1 x in sample covariance matrix is a d d matrix Z with entries Z ij = N 1 1 N n=1 (x in x i )(x jn x j ) sample correlation matrix is a d d matrix C with P N n=1 (x in x i )(x jn x j ) entries C ij = N 1 1 σ xi σ xj, where σ xi and σ xj are the sample standard deviations p. 168

2 Covariance and Correlation Matrix Example Given sample: [ ] [ 1.2, 0.9 C = [ ] Z =, [ [ ], [ ] ] , x = = ] [ [ ] ] Observe, if sample is z-normalized (x new ij standard deviation 1) then C equals Z. See cov(), cor(), scale() in R. = x ij x i σ xi, mean 0, p. 169

3 Principal Component Analysis with NN Principal Component Analysis (PCA) is a technique for dimensionality reduction lossy data compression feature extraction data visualization Idea: orthogonal projection of the data onto a lower dimensional linear space, such that the variance of the projected data is maximized. u 1 p. 170

4 Maximize Variance of Projected Data Given data {x n } N 1 where x n has dimensionality d. Goal: project data onto a space having dimensionality m < d while maximizing the variance of the projected data. Let us consider the projection onto one-dimensional space (m = 1). Define direction of this space using a d-dimensional vector u 1. Mean of the projected data is u T 1 x, where x is sample mean N x = 1 N n=1 x n p. 171

5 Maximize Variance of Projected Data (cont.) Variance of projected data is given by 1 N N n=1 ( u T 1 x n u T 1 x) 2 = u T 1 Su 1 where S is the data covariance matrix defined by S = 1 N N (x n x)(x n x) T n=1 Goal: maximize the projected variance u T 1 Su 1 with respect to u 1. Prevent u 1 growing to infinity, use constrain u T 1 u 1 = 1, gives optimization problem: maximize u T 1 Su 1 subject to u T 1 u 1 = 1 p. 172

6 Maximize Variance of Projected Data (cont.) Lagrangian form (one Lagrange multiplier λ 1 ): L(u 1,λ 1 ) = u T 1 Su 1 λ 1 (u T 1 u 1 1) Set derivative with respect to u 1 to zero, L(u 1,λ 1 ) u 1 = 0 gives Su 1 = λ 1 u 1 last term says that u 1 must be an eigenvector of S. Finally by left-multiplying by u T 1 and making use of ut 1 u 1 = 1 one can see that the variance is given by u T 1 Su 1 = λ 1. Observe, that variance is maximized when u 1 equals to the eigenvector having largest eigenvalue λ 1. p. 173

7 Second Principal Component Second eigenvector u 2 should also be of unit length and orthogonal to u 1 (after projection uncorrelated to u T 1 x). maximize u T 2 Su 2 subject to u T 2 u 2 = 1, u T 2 u 1 = 0 Lagrangian form (two Lagrange multipliers λ 1,λ 2 ): L(u 2,λ 1,λ 2 ) = u T 2 Su 2 λ 2 (u T 2 u 2 1) λ 1 (u T 2 u 1 0) This gives solution u T 2 Su 2 = λ 2 which implies that u 2 should be eigenvector of S with second largest eigenvalue λ 2. Other dimensions are given by the eigenvectors with decreasing eigenvalues. p. 174

8 PCA Example First and second eigenvector Projection on first eigenvector data.xy[,2] data.xy[,1] Projection on both orthogonal eigenvectors data.x.eig[,2] cbind(data.x.eig.1, rep(0, N))[,2] cbind(data.x.eig.1, rep(0, N))[,1] data.x.eig[,1] p. 175

9 Proportion of Variance In image and speech processing problems the inputs are usually highly correlated. If dimensions are highly correlated, then there will be small number of eigenvectors with large eigenvalues (m d). As a result, a large reduction in dimensionality can be attained. Proportion of variance explained, digit class 1 (USPS database) λ 1 + λ λ m λ 1 + λ λ m λ d Proportion of variance Eigenvectors p. 176

10 PCA Second Example 256 Segment image in = 1024 image pieces of size : x 1,x 2,...,x 1024 R 64 Determine mean: x = 1024 n=1 x i Determine covariance matrix S and the m eigenvectors u 1,u 2,...,u m having the largest corresponding eigenvalues λ 1,λ 2,...,λ m Create eigenvector matrix U, where u 1,u 2,...,u m are column vectors Project image pieces x i into subspace as follows: z T i = U T (x T i x T ) p. 177

11 PCA Second Example (cont.) Reconstruct image pieces by back-projecting it to the eti = UzTi + x. Note, mean is added original space as x (substracted step before) because data is not normalized 1.0 Original image 0.8 Proportion of variance Proportion of variance explained in image Eigenvectors Reconstructed with 16 eigenvectors Reconstructed with 32 eigenvectors Reconstructed with 48 eigenvectors Reconstructed with 64 eigenvectors p. 178

12 PCA with a Neural Network V w 1 w 2 w d V = w T x = d w j x j j=1 x 1 x 2 x d Apply Hebbian learning rule w i = ηv x i, such that after some update steps weight vector w should point in direction of maximum variance. p. 179

13 PCA with a Neural Network (cont.) Suppose that there is a stable equilibrium point for w such that the average weight change is zero 0 = w i = V x i = V j w j x j x i = j C ij w j = Cw. Angle brackets indicates an average over the input distribution P(x) and C denotes the correlation matrix with C ij x i x j, or C xx T Note, C is symmetric (C ij = C ji ) and positive semi-definite which implies that its eigenvalues are positive or zero and eigenvectors can be taken as orthogonal. p. 180

14 PCA with a Neural Network (cont.) At our hypothetical equilibrium point, w is an eigenvector of C with eigenvalue 0 Never stable, because C has some pos. eigenvalues and some corresponding eigenvector would grow exponentially constrain the growth of w, e.g. renormalization ( w = 1) after each update step more elegant idea: adding a weight decay proportional to V 2 to Hebbian learning rule (Oja s Rule) w i = ηv (x i V w i ) p. 181

15 PCA with a Neural Network Example Oja s Rule (blue vector), largest eigenvector (red vector) data.xy[,2] data.xy[,1] p. 182

16 Some insights into Oja s Rule Oja s rule converges to a weight vector w with following properties: w = 1 (unit length), eigenvector direction: w lies in a maximal eigenvector direction of C, variance maximization: w lies in a direction that maximizes V 2 Oja s learning rule is still limited, because we can construct only the first principal component of the z-normalized data. p. 183

17 Construct the first m principal components Single-layer network with the i-th output V i given by V i = j w ijx j = w T i x, w i is the weight vector for the i-th output Oja s m-unit learning rule Sanger s learning rule w ij = ηv i (x j w ij = ηv i (x j d k=1 i k=1 V k w kj ) V k w kj ) Both rules reduce to Oja s 1-unit rule for the m = 1 and i = 1 case p. 184

18 Oja s and Sanger s Rule In both cases the w i vectors converge to orthogonal unit vectors Weight vectors become in Sanger s rule exactly the first m principal components, in order w i = ±c i, where c i is normalized eigenvector of the correlation matrix C belonging to the i-th largest eigenvalue λ i Oja s m-unit rule converges to the m weight vectors that span the same subspace as the first m eigenvectors, but do not find the eigenvector directions themselves p. 185

19 Linear Auto-Associative Network x 1 reconstructed features x d m extracted features z 1 z m reconstruction extraction x 1 original features Network is training to perform identity mapping Idea: bottleneck units represents significant features in the input data Train network by minimizing the sum-of-square error 1 N ( ) d 2 n=1 k=1 y k (x (n) ) x (n) 2 k ) p. 186 x d

20 Linear Auto-Associative Network (cont.) Equivalent to the Oja s/sanger s update rule, this type of learning can be considered as unsupervised learning, since no independent target data is provided Error function has a unique global minimum when hidden units have linear activations functions At this minimum the network performs a projection onto the m-dimensional sub-space which is spanned by the first m principal components of the data Note, however, that these vectors need not to be orthogonal or normalized p. 187

21 Non-Linear Auto-Associative Network x1 reconstructed features xd linear non-linear linear z1 zm m extracted features non-linear x1 original features xd p. 188