Probabilistic & Unsupervised Learning. Latent Variable Models

Transcription

1 Probabilistic & Unsupervised Learning Latent Variable Models Maneesh Sahani Gatsby Computational Neuroscience Unit, and MSc ML/CSML, Dept Computer Science University College London Term 1, Autumn 2018

2 Exponential family models Simple, single-stage generative models. Easy, often closed-form expressions for learning and model comparison.... but limited in expressiveness.

3 Exponential family models Simple, single-stage generative models. Easy, often closed-form expressions for learning and model comparison.... but limited in expressiveness. What about distributions like these?

4 Exponential family models Simple, single-stage generative models. Easy, often closed-form expressions for learning and model comparison.... but limited in expressiveness. What about distributions like these? In each case, data may be generated by combining and transforming latent exponential family variates.

5 Latent variable models Explain correlations in x by assuming dependence on latent variables z (e.g. objects, illumination, pose) (e.g. object parts, surfaces) z P[θ z] x z P[θ x] (e.g. edges) p(x, z; θ x, θ z) = p(x z; θ x)p(z; θ z) p(x; θ x, θ z) = dz p(x z; θ x)p(z; θ z) (retinal image, i.e. pixels)

6 Latent variable models Describe structured distributions. Correlations in high-dimensional x may be captured by fewer parameters. Capture an underlying generative process. z may describe causes of x. help to separate signal from noise. Combine exponential family distributions into richer, more flexible forms. P(z), P(x z) and even P(x, z) may be in the exponential family P(x) rarely is. (Exception: Linear Gaussian models).

7 Latent variables and Gaussians Gaussian correlation can be composed from latent components and uncorrelated noise. x N ( [ ]) 3 2 0, 2 3

8 Latent variables and Gaussians Gaussian correlation can be composed from latent components and uncorrelated noise. x N ( [ ]) 3 2 0, 2 3 z N (0, 1) x N ( [ ] [ ]) z, 1 0 1

9 Probabilistic Principal Components Analysis (PPCA) If the uncorrelated noise is assumed to be isotropic, this model is called PPCA.

10 Probabilistic Principal Components Analysis (PPCA) If the uncorrelated noise is assumed to be isotropic, this model is called PPCA. Data: D = X = {x 1, x 2,..., x N}; x i R D Latents: Z = {z 1, z 2,..., z N}; z i R K K Linear generative model: x d = Λ dk z k + ɛ d k=1 z k are independent N (0, 1) Gaussian factors ɛ d are independent N (0, ψ) Gaussian noise K <D z 1 z 2 z K x 1 x 2 x D

11 Probabilistic Principal Components Analysis (PPCA) If the uncorrelated noise is assumed to be isotropic, this model is called PPCA. Data: D = X = {x 1, x 2,..., x N}; x i R D Latents: Z = {z 1, z 2,..., z N}; z i R K K Linear generative model: x d = Λ dk z k + ɛ d k=1 z k are independent N (0, 1) Gaussian factors ɛ d are independent N (0, ψ) Gaussian noise K <D z 1 z 2 z K x 1 x 2 x D Model for observations x is a correlated Gaussian: p(z) = N (0, I) p(x z) = N (Λz, ψi) where Λ is a D K matrix.

12 Probabilistic Principal Components Analysis (PPCA) If the uncorrelated noise is assumed to be isotropic, this model is called PPCA. Data: D = X = {x 1, x 2,..., x N}; x i R D Latents: Z = {z 1, z 2,..., z N}; z i R K K Linear generative model: x d = Λ dk z k + ɛ d k=1 z k are independent N (0, 1) Gaussian factors ɛ d are independent N (0, ψ) Gaussian noise K <D z 1 z 2 z K x 1 x 2 x D Model for observations x is a correlated Gaussian: p(z) = N (0, I) p(x z) = N (Λz, ψi) p(x) = p(z)p(x z)dz where Λ is a D K matrix.

13 Probabilistic Principal Components Analysis (PPCA) If the uncorrelated noise is assumed to be isotropic, this model is called PPCA. Data: D = X = {x 1, x 2,..., x N}; x i R D Latents: Z = {z 1, z 2,..., z N}; z i R K K Linear generative model: x d = Λ dk z k + ɛ d k=1 z k are independent N (0, 1) Gaussian factors ɛ d are independent N (0, ψ) Gaussian noise K <D z 1 z 2 z K x 1 x 2 x D Model for observations x is a correlated Gaussian: p(z) = N (0, I) Note: E x [f (x)] = E z [ Ex z [f (x)] ] V x [x] = E z [V [x z]] + V z [E [x z]] p(x z) = N (Λz, ψi) ] ) p(x) = p(z)p(x z)dz = N (E z [Λz], E z [Λzz T Λ T + ψi where Λ is a D K matrix.

14 Probabilistic Principal Components Analysis (PPCA) If the uncorrelated noise is assumed to be isotropic, this model is called PPCA. Data: D = X = {x 1, x 2,..., x N}; x i R D Latents: Z = {z 1, z 2,..., z N}; z i R K K Linear generative model: x d = Λ dk z k + ɛ d k=1 z k are independent N (0, 1) Gaussian factors ɛ d are independent N (0, ψ) Gaussian noise K <D z 1 z 2 z K x 1 x 2 x D Model for observations x is a correlated Gaussian: p(z) = N (0, I) Note: E x [f (x)] = E z [ Ex z [f (x)] ] V x [x] = E z [V [x z]] + V z [E [x z]] p(x z) = N (Λz, ψi) ] ) ( ) p(x) = p(z)p(x z)dz = N (E z [Λz], E z [Λzz T Λ T + ψi = N 0, ΛΛ T + ψi where Λ is a D K matrix.

15 Multivariate Gaussians and latent variables Two models: p(x) = N (0, Σ) p(z) = N (0, I) p(x z) = N (Λz, ψi) ( ) p(x) = N 0, ΛΛ T + ψi Descriptive density model: correlations are captured by off-diagonal elements of Σ. Σ has D(D+1) free parameters. 2 Only constrained to be positive definite. Simple ML estimate. Interpretable causal model: correlations captured by common influence of latent variable. ΛΛ T + ψi has DK + 1 free parameters. For K < D covariance structure is constrained ( blurry pancake ) ML estimation is more complex.

16 PPCA likelihood The marginal distribution on x gives us the PPCA likelihood: log p(x Λ, ψ) = N 2π(ΛΛ 2 log T + ψi) 1 [ 2 Tr (ΛΛ T + ψi) ] 1 xx T n }{{} NS To find the ML values of (Λ, ψ) we could optimise numerically (gradient ascent / Newton s method), or we could use a different iterative algorithm called EM which we ll introduce soon. In fact, however, ML for PPCA is more straightforward in principle, as we will see by first considering the limit ψ 0. [Note: We may also add a constant mean µ to the output, so as to model data that are not distributed around 0. In this case, the ML estimate µ = 1 N n xn and we can define S = 1 N n (x µ)(x µ)t in the likelihood above.]

17 The ψ 0 limit As ψ 0, the latent model can only capture K dimensions of variance.

18 The ψ 0 limit As ψ 0, the latent model can only capture K dimensions of variance.

19 The ψ 0 limit As ψ 0, the latent model can only capture K dimensions of variance.

20 The ψ 0 limit As ψ 0, the latent model can only capture K dimensions of variance.

21 The ψ 0 limit As ψ 0, the latent model can only capture K dimensions of variance. In a Gaussian model, the ML parameters will find the K -dimensional space of most variance.

22 Principal Components Analysis This leads us to an (old) algorithm called Principal Components Analysis (PCA). Assume data D = {x i} have zero mean (if not, subtract it). Find direction of greatest variance λ (1). 5 λ (1) = argmax v =1 (x T nv) 2 n x x x Find direction orthogonal to λ (1) with greatest variance λ (2). Find direction orthogonal to {λ (1), λ (2),..., λ (n 1) } with greatest variance λ (n). Terminate when remaining variance drops below a threshold.

23 Eigendecomposition of a covariance matrix The eigendecomposition of a covariance matrix makes finding the PCs easy.

24 Eigendecomposition of a covariance matrix The eigendecomposition of a covariance matrix makes finding the PCs easy. Recall that u is an eigenvector, with scalar eigenvalue ω, of a matrix S if Su = ωu u can have any norm, but we will define it to be unity (i.e., u T u = 1).

25 Eigendecomposition of a covariance matrix The eigendecomposition of a covariance matrix makes finding the PCs easy. Recall that u is an eigenvector, with scalar eigenvalue ω, of a matrix S if Su = ωu u can have any norm, but we will define it to be unity (i.e., u T u = 1). For a covariance matrix S = xx T (which is D D, symmetric, positive semi-definite): In general there are D eigenvector-eigenvalue pairs (u (i), ω (i) ), except if two or more eigenvectors share the same eigenvalue (in which case the eigenvectors are degenerate any linear combination is also an eigenvector).

26 Eigendecomposition of a covariance matrix The eigendecomposition of a covariance matrix makes finding the PCs easy. Recall that u is an eigenvector, with scalar eigenvalue ω, of a matrix S if Su = ωu u can have any norm, but we will define it to be unity (i.e., u T u = 1). For a covariance matrix S = xx T (which is D D, symmetric, positive semi-definite): In general there are D eigenvector-eigenvalue pairs (u (i), ω (i) ), except if two or more eigenvectors share the same eigenvalue (in which case the eigenvectors are degenerate any linear combination is also an eigenvector). The D eigenvectors are orthogonal (or orthogonalisable, if ω (i) = ω (j) ). Thus, they form an orthonormal basis. i u (i) u (i) T = I.

27 Eigendecomposition of a covariance matrix The eigendecomposition of a covariance matrix makes finding the PCs easy. Recall that u is an eigenvector, with scalar eigenvalue ω, of a matrix S if Su = ωu u can have any norm, but we will define it to be unity (i.e., u T u = 1). For a covariance matrix S = xx T (which is D D, symmetric, positive semi-definite): In general there are D eigenvector-eigenvalue pairs (u (i), ω (i) ), except if two or more eigenvectors share the same eigenvalue (in which case the eigenvectors are degenerate any linear combination is also an eigenvector). The D eigenvectors are orthogonal (or orthogonalisable, if ω (i) = ω (j) ). Thus, they form an orthonormal basis. i u (i) u (i) T = I. Any vector v can be written as ( ) T v = u (i) u (i) v = (u T (i) v)u (i) = v (i) u (i) i i i

28 Eigendecomposition of a covariance matrix The eigendecomposition of a covariance matrix makes finding the PCs easy. Recall that u is an eigenvector, with scalar eigenvalue ω, of a matrix S if Su = ωu u can have any norm, but we will define it to be unity (i.e., u T u = 1). For a covariance matrix S = xx T (which is D D, symmetric, positive semi-definite): In general there are D eigenvector-eigenvalue pairs (u (i), ω (i) ), except if two or more eigenvectors share the same eigenvalue (in which case the eigenvectors are degenerate any linear combination is also an eigenvector). The D eigenvectors are orthogonal (or orthogonalisable, if ω (i) = ω (j) ). Thus, they form an orthonormal basis. i u (i) u (i) T = I. Any vector v can be written as ( ) T v = u (i) u (i) v = (u T (i) v)u (i) = i i i v (i) u (i) The original matrix S can be written: S = i ω (i) u (i) u (i) T = UWU T where U = [u (1), u (2),..., u (D) ] collects the eigenvectors and W = diag [ (ω (1), ω (2),..., ω (D) ) ].

29 PCA and eigenvectors The variance in direction u (i) is (x T u (i) ) 2 = u T (i) xx T u (i) = u T (i) Su (i) = u T (i) ω (i) u (i) = ω (i)

30 PCA and eigenvectors The variance in direction u (i) is (x T u (i) ) 2 = u T (i) xx T u (i) = u T (i) Su (i) = u T (i) ω (i) u (i) = ω (i) The variance in an arbitrary direction v is ( )) 2 (x T v) 2 = (x T v (i) u (i) = i ij v (i) u (i) T Su (j) v (j) = ij v (i) ω (j) v (j) u (i) T u (j) = i v 2 (i)ω (i)

31 PCA and eigenvectors The variance in direction u (i) is (x T u (i) ) 2 = u T (i) xx T u (i) = u T (i) Su (i) = u T (i) ω (i) u (i) = ω (i) The variance in an arbitrary direction v is ( )) 2 (x T v) 2 = (x T v (i) u (i) = i ij v (i) u (i) T Su (j) v (j) = ij v (i) ω (j) v (j) u (i) T u (j) = i v 2 (i)ω (i) If v T v = 1, then i v 2 (i) = 1 and so argmax v =1 (x T v) 2 = u (max) The direction of greatest variance is the eigenvector the largest eigenvalue.

32 PCA and eigenvectors The variance in direction u (i) is (x T u (i) ) 2 = u T (i) xx T u (i) = u T (i) Su (i) = u T (i) ω (i) u (i) = ω (i) The variance in an arbitrary direction v is ( )) 2 (x T v) 2 = (x T v (i) u (i) = i ij v (i) u (i) T Su (j) v (j) = ij v (i) ω (j) v (j) u (i) T u (j) = i v 2 (i)ω (i) If v T v = 1, then i v 2 (i) = 1 and so argmax v =1 (x T v) 2 = u (max) The direction of greatest variance is the eigenvector the largest eigenvalue. In general, the PCs are exactly the eigenvectors of the empirical covariance matrix, ordered by decreasing eigenvalue.

33 PCA and eigenvectors The variance in direction u (i) is (x T u (i) ) 2 = u T (i) xx T u (i) = u T (i) Su (i) = u T (i) ω (i) u (i) = ω (i) The variance in an arbitrary direction v is ( )) 2 (x T v) 2 = (x T v (i) u (i) = i ij v (i) u (i) T Su (j) v (j) = ij v (i) ω (j) v (j) u (i) T u (j) = i v 2 (i)ω (i) If v T v = 1, then i v 2 (i) = 1 and so argmax v =1 (x T v) 2 = u (max) The direction of greatest variance is the eigenvector the largest eigenvalue. In general, the PCs are exactly the eigenvectors of the empirical covariance matrix, ordered by decreasing eigenvalue. 100 The eigenspectrum shows how the variance 80 is distributed across dimensions; can identify transitions that might separate signal from noise, or the number of PCs that capture a predetermined fraction of variance eigenvalue (variance) eigenvalue number fractional variance remaining eigenvalue number

34 PCA subspace The K principle components define the K -dimensional subspace of greatest variance x x 2 x 1 Each data point x n is associated with a projection ˆx n into the principle subspace. K ˆx n = x n(k) λ (k) k=1 This can be used for lossy compression, denoising, recognition,...

35 Example of PCA: Eigenfaces vismod.media.mit.edu/vismod/demos/facerec/basic.html

36 Example of PCA: Genetic variation within Europe Novembre et al. (2008) Nature 456:98-101

37 Example of PCA: Genetic variation within Europe Novembre et al. (2008) Nature 456:98-101

38 Example of PCA: Genetic variation within Europe Novembre et al. (2008) Nature 456:98-101

39 Equivalent definitions of PCA Find K directions of greatest variance in data. Find K -dimensional orthogonal projection that preserves greatest variance. Find K -dimensional vectors z i and matrix Λ so that ˆx i = Λz i is as close as possible (in squared distance) to x i.... (many others)

40 Another view of PCA: Mutual information Problem: Given x, find z = Ax with columns of A unit vectors, s.t. I(z; x) is maximised (assuming that P(x) is Gaussian). I(z; x) = H(z) + H(x) H(z, x) = H(z)

41 Another view of PCA: Mutual information Problem: Given x, find z = Ax with columns of A unit vectors, s.t. I(z; x) is maximised (assuming that P(x) is Gaussian). I(z; x) = H(z) + H(x) H(z, x) = H(z) So we want to maximise the entropy of z.

42 Another view of PCA: Mutual information Problem: Given x, find z = Ax with columns of A unit vectors, s.t. I(z; x) is maximised (assuming that P(x) is Gaussian). I(z; x) = H(z) + H(x) H(z, x) = H(z) So we want to maximise the entropy of z. What is the entropy of a Gaussian? H(z) = dz p(z) ln p(z) = 1 2 ln Σ + D (1 + ln 2π) 2

43 Another view of PCA: Mutual information Problem: Given x, find z = Ax with columns of A unit vectors, s.t. I(z; x) is maximised (assuming that P(x) is Gaussian). I(z; x) = H(z) + H(x) H(z, x) = H(z) So we want to maximise the entropy of z. What is the entropy of a Gaussian? H(z) = dz p(z) ln p(z) = 1 2 ln Σ + D (1 + ln 2π) 2 Therefore we want the distribution of z to have largest volume (i.e. det of covariance matrix). Σ z = AΣ xa T = AUW xu T A T

44 Another view of PCA: Mutual information Problem: Given x, find z = Ax with columns of A unit vectors, s.t. I(z; x) is maximised (assuming that P(x) is Gaussian). I(z; x) = H(z) + H(x) H(z, x) = H(z) So we want to maximise the entropy of z. What is the entropy of a Gaussian? H(z) = dz p(z) ln p(z) = 1 2 ln Σ + D (1 + ln 2π) 2 Therefore we want the distribution of z to have largest volume (i.e. det of covariance matrix). Σ z = AΣ xa T = AUW xu T A T So, A should be aligned with the columns of U which are associated with the largest eigenvalues (variances).

45 Another view of PCA: Mutual information Problem: Given x, find z = Ax with columns of A unit vectors, s.t. I(z; x) is maximised (assuming that P(x) is Gaussian). I(z; x) = H(z) + H(x) H(z, x) = H(z) So we want to maximise the entropy of z. What is the entropy of a Gaussian? H(z) = dz p(z) ln p(z) = 1 2 ln Σ + D (1 + ln 2π) 2 Therefore we want the distribution of z to have largest volume (i.e. det of covariance matrix). Σ z = AΣ xa T = AUW xu T A T So, A should be aligned with the columns of U which are associated with the largest eigenvalues (variances). Projection to the principal component subspace preserves the most information about the (Gaussian) data.

46 Linear autoencoders: From supervised learning to PCA output units hidden units ˆx 1 ˆx 2 ˆx D Q P z 1 z K decoder generation encoder recognition input units x 1 x 2 x D Learning: argmin ˆx x 2 ˆx = Qz z = Px P,Q At the optimum, P and Q perform the projection and reconstruction steps of PCA. (Baldi & Hornik 1989).

47 ML learning for PPCA l = N 2 log 2πC N 2 Tr [ C 1 S ] l Λ = N 2 where C = ΛΛ T + ψi ( log C Λ Λ Tr [ C 1 S ] ) = N ( C 1 Λ + C 1 SC 1 Λ ) So at the stationary points we have SC 1 Λ = Λ. This implies either: Λ = 0, which turns out to be a minimum. C = S ΛΛ T = S ψi. Now rank(λλ T ) K rank(s ψi) K S has D K eigenvalues = ψ and Λ aligns with space of remaining eigenvectors. or, taking the SVD: Λ = ULV T : S(ULV T VLU T + ψi) 1 ULV T = ULV T VL 1 S(UL 2 U T + ψi) 1 U = U U(L 2 + ψi) = (UL 2 U T + ψi)u (UL 2 U T + ψi) 1 U = U(L 2 + ψi) 1 SU(L 2 + ψi) 1 = U (L 2 + ψi) SU = U (L 2 + ψi) }{{} diagonal columns of U are eigenvectors of S with eigenvalues given by l 2 i + ψ. Thus, Λ = ULV T spans a space defined by K eigenvectors of S; and the lengths of the column vectors of L are given by the eigenvalues ψ (V selects an arbitrary basis in the latent space). Remains to show (we won t, but it s intuitively reasonable) that the global ML solution is attained when Λ aligns with the K leading eigenvectors.

48 PPCA latents In PCA the noise is orthogonal to the subspace, and we can project x n ˆx n trivially. In PPCA, the noise is more sensible (equal in all directions). But what is the projection?

49 PPCA latents In PCA the noise is orthogonal to the subspace, and we can project x n ˆx n trivially. In PPCA, the noise is more sensible (equal in all directions). But what is the projection? Find the expected value z n = E [z n x n] and then take ˆx n = Λz n.

50 PPCA latents In PCA the noise is orthogonal to the subspace, and we can project x n ˆx n trivially. In PPCA, the noise is more sensible (equal in all directions). But what is the projection? Find the expected value z n = E [z n x n] and then take ˆx n = Λz n. Tactic: write p(z n, x n θ), consider x n to be fixed. What is this as a function of z n? p(z n, x n) = p(z n)p(x n z n)

51 PPCA latents In PCA the noise is orthogonal to the subspace, and we can project x n ˆx n trivially. In PPCA, the noise is more sensible (equal in all directions). But what is the projection? Find the expected value z n = E [z n x n] and then take ˆx n = Λz n. Tactic: write p(z n, x n θ), consider x n to be fixed. What is this as a function of z n? p(z n, x n) = p(z n)p(x n z n) = (2π) K 2 exp{ 1 2 zt nz n} 2πΨ 1 2 exp{ 1 2 (xn Λzn)T Ψ 1 (x n Λz n)}

52 PPCA latents In PCA the noise is orthogonal to the subspace, and we can project x n ˆx n trivially. In PPCA, the noise is more sensible (equal in all directions). But what is the projection? Find the expected value z n = E [z n x n] and then take ˆx n = Λz n. Tactic: write p(z n, x n θ), consider x n to be fixed. What is this as a function of z n? p(z n, x n) = p(z n)p(x n z n) = (2π) K 2 exp{ 1 2 zt nz n} 2πΨ 1 2 exp{ 1 2 (xn Λzn)T Ψ 1 (x n Λz n)} = c exp{ 1 2 [zt nz n + (x n Λz n) T Ψ 1 (x n Λz n)]}

53 PPCA latents In PCA the noise is orthogonal to the subspace, and we can project x n ˆx n trivially. In PPCA, the noise is more sensible (equal in all directions). But what is the projection? Find the expected value z n = E [z n x n] and then take ˆx n = Λz n. Tactic: write p(z n, x n θ), consider x n to be fixed. What is this as a function of z n? p(z n, x n) = p(z n)p(x n z n) = (2π) K 2 exp{ 1 2 zt nz n} 2πΨ 1 2 exp{ 1 2 (xn Λzn)T Ψ 1 (x n Λz n)} = c exp{ 1 2 [zt nz n + (x n Λz n) T Ψ 1 (x n Λz n)]} = c exp{ 1 2 [zt n(i + Λ T Ψ 1 Λ)z n 2z T nλ T Ψ 1 x n]}

54 PPCA latents In PCA the noise is orthogonal to the subspace, and we can project x n ˆx n trivially. In PPCA, the noise is more sensible (equal in all directions). But what is the projection? Find the expected value z n = E [z n x n] and then take ˆx n = Λz n. Tactic: write p(z n, x n θ), consider x n to be fixed. What is this as a function of z n? p(z n, x n) = p(z n)p(x n z n) = (2π) K 2 exp{ 1 2 zt nz n} 2πΨ 1 2 exp{ 1 2 (xn Λzn)T Ψ 1 (x n Λz n)} = c exp{ 1 2 [zt nz n + (x n Λz n) T Ψ 1 (x n Λz n)]} = c exp{ 1 2 [zt n(i + Λ T Ψ 1 Λ)z n 2z T nλ T Ψ 1 x n]} = c exp{ 1 2 [zt nσ 1 z n 2z T nσ 1 µ + µ T Σ 1 µ]}

55 PPCA latents In PCA the noise is orthogonal to the subspace, and we can project x n ˆx n trivially. In PPCA, the noise is more sensible (equal in all directions). But what is the projection? Find the expected value z n = E [z n x n] and then take ˆx n = Λz n. Tactic: write p(z n, x n θ), consider x n to be fixed. What is this as a function of z n? p(z n, x n) = p(z n)p(x n z n) = (2π) K 2 exp{ 1 2 zt nz n} 2πΨ 1 2 exp{ 1 2 (xn Λzn)T Ψ 1 (x n Λz n)} = c exp{ 1 2 [zt nz n + (x n Λz n) T Ψ 1 (x n Λz n)]} = c exp{ 1 2 [zt n(i + Λ T Ψ 1 Λ)z n 2z T nλ T Ψ 1 x n]} = c exp{ 1 2 [zt nσ 1 z n 2z T nσ 1 µ + µ T Σ 1 µ]} So Σ = (I + Λ T Ψ 1 Λ) 1 = I βλ and µ = ΣΛ T Ψ 1 x n = βx n. Where β = ΣΛ T Ψ 1.

56 PPCA latents In PCA the noise is orthogonal to the subspace, and we can project x n ˆx n trivially. In PPCA, the noise is more sensible (equal in all directions). But what is the projection? Find the expected value z n = E [z n x n] and then take ˆx n = Λz n. Tactic: write p(z n, x n θ), consider x n to be fixed. What is this as a function of z n? p(z n, x n) = p(z n)p(x n z n) = (2π) K 2 exp{ 1 2 zt nz n} 2πΨ 1 2 exp{ 1 2 (xn Λzn)T Ψ 1 (x n Λz n)} = c exp{ 1 2 [zt nz n + (x n Λz n) T Ψ 1 (x n Λz n)]} = c exp{ 1 2 [zt n(i + Λ T Ψ 1 Λ)z n 2z T nλ T Ψ 1 x n]} = c exp{ 1 2 [zt nσ 1 z n 2z T nσ 1 µ + µ T Σ 1 µ]} So Σ = (I + Λ T Ψ 1 Λ) 1 = I βλ and µ = ΣΛ T Ψ 1 x n = βx n. Where β = ΣΛ T Ψ 1. Thus, ˆx n = Λ(I + Λ T Ψ 1 Λ) 1 Λ T Ψ 1 x n = x n Ψ(ΛΛ T + Ψ) 1 x n

57 PPCA latents In PCA the noise is orthogonal to the subspace, and we can project x n ˆx n trivially. In PPCA, the noise is more sensible (equal in all directions). But what is the projection? Find the expected value z n = E [z n x n] and then take ˆx n = Λz n. Tactic: write p(z n, x n θ), consider x n to be fixed. What is this as a function of z n? p(z n, x n) = p(z n)p(x n z n) = (2π) K 2 exp{ 1 2 zt nz n} 2πΨ 1 2 exp{ 1 2 (xn Λzn)T Ψ 1 (x n Λz n)} = c exp{ 1 2 [zt nz n + (x n Λz n) T Ψ 1 (x n Λz n)]} = c exp{ 1 2 [zt n(i + Λ T Ψ 1 Λ)z n 2z T nλ T Ψ 1 x n]} = c exp{ 1 2 [zt nσ 1 z n 2z T nσ 1 µ + µ T Σ 1 µ]} So Σ = (I + Λ T Ψ 1 Λ) 1 = I βλ and µ = ΣΛ T Ψ 1 x n = βx n. Where β = ΣΛ T Ψ 1. Thus, ˆx n = Λ(I + Λ T Ψ 1 Λ) 1 Λ T Ψ 1 x n = x n Ψ(ΛΛ T + Ψ) 1 x n This is not the same projection. PPCA takes into account noise in the principal subspace.

58 PPCA latents In PCA the noise is orthogonal to the subspace, and we can project x n ˆx n trivially. In PPCA, the noise is more sensible (equal in all directions). But what is the projection? Find the expected value z n = E [z n x n] and then take ˆx n = Λz n. Tactic: write p(z n, x n θ), consider x n to be fixed. What is this as a function of z n? p(z n, x n) = p(z n)p(x n z n) = (2π) K 2 exp{ 1 2 zt nz n} 2πΨ 1 2 exp{ 1 2 (xn Λzn)T Ψ 1 (x n Λz n)} = c exp{ 1 2 [zt nz n + (x n Λz n) T Ψ 1 (x n Λz n)]} = c exp{ 1 2 [zt n(i + Λ T Ψ 1 Λ)z n 2z T nλ T Ψ 1 x n]} = c exp{ 1 2 [zt nσ 1 z n 2z T nσ 1 µ + µ T Σ 1 µ]} So Σ = (I + Λ T Ψ 1 Λ) 1 = I βλ and µ = ΣΛ T Ψ 1 x n = βx n. Where β = ΣΛ T Ψ 1. Thus, ˆx n = Λ(I + Λ T Ψ 1 Λ) 1 Λ T Ψ 1 x n = x n Ψ(ΛΛ T + Ψ) 1 x n This is not the same projection. PPCA takes into account noise in the principal subspace. As ψ 0, the PPCA estimate the PCA value.

59 PPCA latents principal subspace

60 PPCA latents PCA projection principal subspace

61 PPCA latents PPCA noise PPCA latent prior principal subspace

62 PPCA latents PPCA posterior PPCA latent prior PPCA noise PPCA projection principal subspace

63 Factor Analysis If dimensions are not equivalent, equal variance assumption is inappropriate. Data: D = X = {x 1, x 2,..., x N}; x i R D Latents: Z = {z 1, z 2,..., z N}; z i R K K Linear generative model: x d = Λ dk z k + ɛ d k=1 z k are independent N (0, 1) Gaussian factors ɛ d are independent N (0, Ψ dd) Gaussian noise K <D Model for observations x is still a correlated Gaussian: p(z) = N (0, I) p(x z) = N (Λz, Ψ) p(x) = p(z)p(x z)dz = N ( ) 0, ΛΛ T + Ψ where Λ is a D K, and Ψ is D D and diagonal. z 1 z 2 z K x 1 x 2 x D Dimensionality Reduction: Finds a low-dimensional projection of high dimensional data that captures the correlation structure of the data.

64 Factor Analysis (cont.) z 1 z 2 z K x 1 x 2 x D ML learning finds Λ ( common factors ) and Ψ ( unique factors or uniquenesses ) given data parameters (corrected for symmetries): DK + D K (K 1) 2 If number of parameters > D(D+1) 2 model is not identifiable (even after accounting for rotational degeneracy discussed later) no closed form solution for ML params: N (0, ΛΛ T + Ψ)

65 Factor Analysis projections Our analysis for PPCA still applies: ˆx n = Λ(I + Λ T Ψ 1 Λ) 1 Λ T Ψ 1 x n = x n Ψ(ΛΛ T + Ψ) 1 x n but now Ψ is diagonal but not spherical. Note, though, that Λ is generally different from that found by PPCA.

66 Factor Analysis projections Our analysis for PPCA still applies: ˆx n = Λ(I + Λ T Ψ 1 Λ) 1 Λ T Ψ 1 x n = x n Ψ(ΛΛ T + Ψ) 1 x n but now Ψ is diagonal but not spherical. Note, though, that Λ is generally different from that found by PPCA. And Λ is not unique: the latent space may be transformed by an arbitrary orthogonal transform U (U T U = UU T = I) without changing the likelihood.

67 Factor Analysis projections Our analysis for PPCA still applies: ˆx n = Λ(I + Λ T Ψ 1 Λ) 1 Λ T Ψ 1 x n = x n Ψ(ΛΛ T + Ψ) 1 x n but now Ψ is diagonal but not spherical. Note, though, that Λ is generally different from that found by PPCA. And Λ is not unique: the latent space may be transformed by an arbitrary orthogonal transform U (U T U = UU T = I) without changing the likelihood. z = Uz and Λ = ΛU T Λ z = ΛU T Uz = Λz

68 Factor Analysis projections Our analysis for PPCA still applies: ˆx n = Λ(I + Λ T Ψ 1 Λ) 1 Λ T Ψ 1 x n = x n Ψ(ΛΛ T + Ψ) 1 x n but now Ψ is diagonal but not spherical. Note, though, that Λ is generally different from that found by PPCA. And Λ is not unique: the latent space may be transformed by an arbitrary orthogonal transform U (U T U = UU T = I) without changing the likelihood. z = Uz and Λ = ΛU T Λ z = ΛU T Uz = Λz l = 1 2π(ΛΛ 2 log T + Ψ) xt (ΛΛ T + Ψ) 1 x

69 Factor Analysis projections Our analysis for PPCA still applies: ˆx n = Λ(I + Λ T Ψ 1 Λ) 1 Λ T Ψ 1 x n = x n Ψ(ΛΛ T + Ψ) 1 x n but now Ψ is diagonal but not spherical. Note, though, that Λ is generally different from that found by PPCA. And Λ is not unique: the latent space may be transformed by an arbitrary orthogonal transform U (U T U = UU T = I) without changing the likelihood. z = Uz and Λ = ΛU T Λ z = ΛU T Uz = Λz l = 1 2π(ΛU 2 log T UΛ T + Ψ) xt (ΛU T UΛ T + Ψ) 1 x

70 Factor Analysis projections Our analysis for PPCA still applies: ˆx n = Λ(I + Λ T Ψ 1 Λ) 1 Λ T Ψ 1 x n = x n Ψ(ΛΛ T + Ψ) 1 x n but now Ψ is diagonal but not spherical. Note, though, that Λ is generally different from that found by PPCA. And Λ is not unique: the latent space may be transformed by an arbitrary orthogonal transform U (U T U = UU T = I) without changing the likelihood. z = Uz and Λ = ΛU T Λ z = ΛU T Uz = Λz l = 1 2π(ΛU 2 log T UΛ T + Ψ) xt (ΛU T UΛ T + Ψ) 1 x = 1 2 log 2π( Λ ΛT + Ψ) xt ( Λ Λ T + Ψ) 1 x

71 Gradient methods for learning FA Optimise negative log-likelihood: l = 1 2 log 2π(ΛΛT + Ψ) xt (ΛΛ T + Ψ) 1 x w.r.t. Λ and Ψ (need matrix calculus) subject to constraints. No spectral short-cut exists. Likelihood can have more than one (local) optimum, making it difficult to find the global value. For some data ( Heywood cases ) likelihood may grow unboundedly by taking one or more Ψ dd 0. Can eliminate by assuming a prior on Ψ with zero density at Ψ dd = 0, but results sensitive to precise choice of prior. Expectation maximisation (next week) provides an alternative approach to maximisation, but doesn t solve these issues.

72 FA vs PCA PCA and PPCA are rotationally invariant; FA is not If x Ux for unitary U, then λ PCA (i) Uλ PCA (i) FA is measurement scale invariant; PCA and PPCA are not If x Sx for diagonal S, then λ FA (i) Sλ FA (i) FA and PPCA define a probabilistic model; PCA does not [Note: it may be tempting to try to eliminate the scale-dependence of (P)PCA by pre-processing data to equalise total variance on each axis. But P(PCA) assume equal noise variance. Total variance has contributions from both ΛΛ T and noise, so this approach does not exactly solve the problem.]

73 Canonical Correlations Analysis Data vector pairs: D = {(u 1, v 1), (u 2, v 2)... } in spaces U and V. Classic CCA Find unit vectors υ 1 U, φ 1 V such that the correlation of u T i υ 1 and v T i φ 1 is maximised. As with PCA, repeat in orthogonal subspaces. Probabilistic CCA Generative model with latent z i R K : z N (0, I) u N (Υz, Ψ u) Ψ u 0 v N (Φz, Ψ v) Ψ v 0 Block diagonal noise.

74 Limitations of Gaussian, FA and PCA models Gaussian, FA and PCA models are easy to understand and use in practice. They are a convenient way to find interesting directions in very high dimensional data sets, eg as preprocessing However, they make strong assumptions about the distribution of the data: only the mean and variance of the data are taken into account. The class of densities which can be modelled is too restrictive. 1 x i x i1 By using mixtures of simple distributions, such as Gaussians, we can expand the class of densities greatly.

75 Mixture Distributions 1 x i x i1 A mixture distribution has a single discrete latent variable: s iid i Discrete[π] x i s i P si [θ si ] Mixtures arise naturally when observations from different sources have been collated. They can also be used to approximate arbitrary distributions.

76 The Mixture Likelihood The mixture model is s iid i Discrete[π] x i s i P si [θ si ]

77 The Mixture Likelihood The mixture model is s iid i Discrete[π] x i s i P si [θ si ] Under the discrete distribution P(s i = m) = π m; π m 0, k π m = 1 m=1

78 The Mixture Likelihood The mixture model is s iid i Discrete[π] x i s i P si [θ si ] Under the discrete distribution P(s i = m) = π m; π m 0, k π m = 1 m=1 Thus, the probability (density) at a single data point x i is k P(x i) = P(x i s i = m)p(s i = m) m=1

79 The Mixture Likelihood The mixture model is s iid i Discrete[π] x i s i P si [θ si ] Under the discrete distribution P(s i = m) = π m; π m 0, k π m = 1 m=1 Thus, the probability (density) at a single data point x i is P(x i) = = k P(x i s i = m)p(s i = m) m=1 k π mp m(x i; θ m) m=1

80 The Mixture Likelihood The mixture model is s iid i Discrete[π] x i s i P si [θ si ] Under the discrete distribution P(s i = m) = π m; π m 0, k π m = 1 m=1 Thus, the probability (density) at a single data point x i is P(x i) = = k P(x i s i = m)p(s i = m) m=1 k π mp m(x i; θ m) m=1 The mixture distribution (density) is a convex combination (or weighted average) of the component distributions (densities).

81 Approximation with a Mixture of Gaussians (MoG) The component densities may be viewed as elements of a basis which can be combined to approximate arbitrary distributions. Here are examples where non-gaussian densities are modelled (aproximated) as a mixture of Gaussians. The red curves show the (weighted) Gaussians, and the blue curve the resulting density. Uniform Triangle Heavy tails Given enough mixture components we can model (almost) any density (as accurately as desired), but still only need to work with the well-known Gaussian form.

82 Clustering with a MoG

83 Clustering with a MoG

84 Clustering with a MoG In clustering applications, the latent variable s i represents the (unknown) identity of the cluster to which the ith observation belongs. Thus, the latent distribution gives the prior probability of a data point coming from each cluster. P(s i = m π) = π m Data from the mth cluster are distributed according to the mth component: P(x i s i = m) = P m(x i) Once we observe a data point, the posterior probability distribution for the cluster it belongs to is P(s i = m x i) = Pm(xi)πm m Pm(xi)πm This is often called the responsibility of the mth cluster for the ith data point.

85 The MoG likelihood Each component of a MoG is a Gaussian, with mean µ m and covariance matrix Σ m. Thus, the probability density evaluated at a set of n iid observations, D = {x 1... x n} (i.e. the likelihood) is p(d {µ m}, {Σ m}, π) = The log of the likelihood is = n i=1 m=1 n i=1 m=1 k π m N (x i µ m, Σ m) k π m 1 2πΣm e 1 2 (x i µ m) T Σ 1 m (x i µ m) log p(d {µ m}, {Σ m}, π) = n log i=1 k m=1 π m 1 2πΣm e 1 2 (x i µ m) T Σ 1 m (x i µ m) Note that the logarithm fails to simplify the component density terms. A mixture distribution does not lie in the exponential family. Direct optimisation is not easy.

86 Maximum Likelihood for a Mixture Model The log likelihood is:

87 Maximum Likelihood for a Mixture Model The log likelihood is: L = n k log π mp m(x i; θ m) i=1 m=1

88 Maximum Likelihood for a Mixture Model The log likelihood is: L = n k log π mp m(x i; θ m) i=1 m=1 Its partial derivative wrt θ m is L θ m = n i=1 π m k m=1 πmpm(xi; θm) P m(x i; θ m) θ m

89 Maximum Likelihood for a Mixture Model The log likelihood is: L = n k log π mp m(x i; θ m) i=1 m=1 Its partial derivative wrt θ m is L θ m = n i=1 π m P m(x i; θ m) k πmpm(xi; θm) θ m m=1 or, using P/ θ = P log P/ θ,

90 Maximum Likelihood for a Mixture Model The log likelihood is: L = n k log π mp m(x i; θ m) i=1 m=1 Its partial derivative wrt θ m is L θ m = n i=1 π m P m(x i; θ m) k πmpm(xi; θm) θ m m=1 or, using P/ θ = P log P/ θ, = n π mp m(x i; θ m) k m=1 πmpm(xi; θm) }{{} i=1 log P m(x i; θ m) θ m

91 Maximum Likelihood for a Mixture Model The log likelihood is: L = n k log π mp m(x i; θ m) i=1 m=1 Its partial derivative wrt θ m is L θ m = n i=1 π m P m(x i; θ m) k πmpm(xi; θm) θ m m=1 or, using P/ θ = P log P/ θ, = = n π mp m(x i; θ m) k m=1 πmpm(xi; θm) }{{} n i=1 i=1 log P m(x i; θ m) θ m r im log P m(x i; θ m) θ m

92 Maximum Likelihood for a Mixture Model The log likelihood is: L = n k log π mp m(x i; θ m) i=1 m=1 Its partial derivative wrt θ m is L θ m = n i=1 π m P m(x i; θ m) k πmpm(xi; θm) θ m m=1 or, using P/ θ = P log P/ θ, = = n π mp m(x i; θ m) k m=1 πmpm(xi; θm) }{{} n i=1 i=1 log P m(x i; θ m) θ m r im log P m(x i; θ m) θ m And its partial derivative wrt π m is L π m = n P m(x i; θ m) k m=1 πmpm(xi; θm) i=1

93 Maximum Likelihood for a Mixture Model The log likelihood is: L = n k log π mp m(x i; θ m) i=1 m=1 Its partial derivative wrt θ m is L θ m = n i=1 π m P m(x i; θ m) k πmpm(xi; θm) θ m m=1 or, using P/ θ = P log P/ θ, = = n π mp m(x i; θ m) k m=1 πmpm(xi; θm) }{{} n i=1 i=1 log P m(x i; θ m) θ m r im log P m(x i; θ m) θ m And its partial derivative wrt π m is L π m = n P m(x i; θ m) n k = m=1 πmpm(xi; θm) i=1 i=1 r im π m

94 MoG Derivatives For a MoG, with θ m = {µ m, Σ m} we get L µ m = n i=1 L = 1 Σ 1 m 2 r imσ 1 m (x i µ m) n i=1 r im ( Σ m (x i µ m)(x i µ m) T ) These equations can be used (along with Lagrangian derivatives wrt π m that enforce normalisation) for gradient based learning; e.g., taking small steps in the direction of the gradient (or using conjugate gradients).

95 The K-means Algorithm The K-means algorithm is a limiting case of the mixture of Gaussians (c.f. PCA and Factor Analysis).

96 The K-means Algorithm The K-means algorithm is a limiting case of the mixture of Gaussians (c.f. PCA and Factor Analysis). Take π m = 1/k and Σ m = σ 2 I, with σ 2 0.

97 The K-means Algorithm The K-means algorithm is a limiting case of the mixture of Gaussians (c.f. PCA and Factor Analysis). Take π m = 1/k and Σ m = σ 2 I, with σ 2 0. Then the responsibilities become binary r im δ(m, argmin x i µ l 2 ) l with 1 for the component with the closest mean and 0 for all other components. We can then solve directly for the means by setting the gradient to 0.

98 The K-means Algorithm The K-means algorithm is a limiting case of the mixture of Gaussians (c.f. PCA and Factor Analysis). Take π m = 1/k and Σ m = σ 2 I, with σ 2 0. Then the responsibilities become binary r im δ(m, argmin x i µ l 2 ) l with 1 for the component with the closest mean and 0 for all other components. We can then solve directly for the means by setting the gradient to 0. The k-means algorithm iterates these two steps: ( ) assign each point to its closest mean set r im = δ(m, argmin x i µ l 2 ) l ( ) update the means to the average of their assigned points set µ m = i r imx i i r im

99 The K-means Algorithm The K-means algorithm is a limiting case of the mixture of Gaussians (c.f. PCA and Factor Analysis). Take π m = 1/k and Σ m = σ 2 I, with σ 2 0. Then the responsibilities become binary r im δ(m, argmin x i µ l 2 ) l with 1 for the component with the closest mean and 0 for all other components. We can then solve directly for the means by setting the gradient to 0. The k-means algorithm iterates these two steps: ( assign each point to its closest mean set r im = δ(m, argmin update the means to the average of their assigned points ) x i µ l 2 ) l ( set µ m = i r imx i This usually converges within a few iterations, although the fixed point depends on the initial values chosen for µ m. The algorithm has no learning rate, but the assumptions are quite limiting. i r im )

100 A preview of the EM algorithm We wrote the k-means algorithm in terms of binary responsibilities. Suppose, instead, we used the fractional responsibilities from the full (non-limiting) MoG, but still neglected the dependence of the responsibilities on the parameters. We could then solve for both µ m and Σ m. The EM algorithm for MoGs iterates these two steps: Evaluate the responsibilities for each point given the current parameters. Optimise the parameters assuming the responsibilities stay fixed: µ m = i r imx i i r im(x i µ m)(x i µ m) T and Σ m = i r im i r im Although this appears ad hoc, we will see (later) that it is a special case of a general algorithm, and is actually guaranteed to increase the likelihood at each iteration.

101 Issues There are several problems with these algorithms: slow convergence for the gradient based method gradient based method may develop invalid covariance matrices local minima; the end configuration may depend on the starting state how do you adjust k? Using the likelihood alone is no good. singularities; components with a single data point will have their covariance going to zero and the likelihood will tend to infinity. We will attempt to address many of these as the course goes on.