Model selection in penalized Gaussian graphical models

Transcription

1 University of Groningen ernst January 2014

2 Penalized likelihood generates a PATH of solutions Consider an experiment: Γ genes measured across T time points. Assume n iid samples y (1),..., y (n), where y (i) = (y (i) 1,..., y (i) ΓT ). Assume Y (i) N(0, K 1 ), then Likelihood: l(k y) n {log( K ) tr(sk)}. 2 AIM: Optimization of penalized likelihood: subject to K 0; ˆK λ := argmax K {l(k y)} K 1 1/λ... for λ in some range!!; some factorial colouring F.

3 Between proximity and truth True process for data Y : Y Q. A statistical model is a collection of measures: M i = {P θ θ Θ i }... and typically we consider several: M = {M 1,..., M k }. What is the best model? Proximity : the model that is closest to the truth: min i k KL(Pˆθi ; Q). Truth : the model that is most likely to be the truth: max i k P(M i Y ).

4 Truth With flat prior on Θ M and M, model probability for M M is: p(m y) p(y M)/p(y) e n l(θ) θ Θ M e l(ˆθ) Θ M e 1 2 (θ ˆθ) t n 2 θ 2 l(ˆθ)(θ ˆθ) θ e l(ˆθ) (2π) p/2 n p/2 2 θ l(ˆθ) 2 where p = dim(θ M ) and l(ˆθ) = l(ˆθ)/n. Schwarz (1978) ignored terms not depending on n: p(m y) e l(ˆθ) n p/2, for large n. 1/2,

5 BIC: Bayesian information criterion By applying the log and ignoring constant terms c: BIC(M) = 2l(ˆθ) + p log(n). Minimizing BIC corresponds to maximizing posterior model probability. Define model weights W (M), W (M) = e BIC(M)/2, which rescaled correspond to posterior model probabilities, p(m i y) = W (M i ) M M W (M).

6 BIC for Gaussian graphical models The BIC for an estimated Gaussian graphical model K: BIC( K) = n( log K + Tr(S K)) + log(n), p K where p K = {unique non-zeroes in K}, which is less than p(p 1)/2.

7 BIC for Gaussian graphical models The BIC for an estimated Gaussian graphical model K: BIC( K) = n( log K + Tr(S K)) + p K log(n), where p K = {unique non-zeroes in K}, which is less than p(p 1)/2. But... BIC is an asymptotic approximation. What is n?? For penalized estimate K are number of parameters not smaller??

8 ebic: extended Bayesian information criterion The BIC for an estimated Gaussian graphical model K: ebic γ ( K) = n( log K + Tr(S K)) + log(n) + log(p), p K 4γp K where p K = {unique non-zeroes in K} p = number of nodes γ = tuning paramter(0 γ 1) If γ = 0 ordinary BIC γ = 1 additional sparsity γ = 0.5 good trade-off (Foygel & Drton, 2010)

9 Proximity Let the true density of the data Y be: Y q = dq. The Kullback-Leibler (KL) divergence between fitted and true model KL(ˆθ i ) = q(y) log q(y)dy q(y) log p(y; ˆθ i )dy

10 Proximity Let the true density of the data Y be: Y q = dq. The Kullback-Leibler (KL) divergence between fitted and true model KL(ˆθ i ) = q(y) log q(y)dy q(y) log p(y; ˆθ i )dy = C E q (l(ˆθ i )

11 Proximity Let the true density of the data Y be: Y q = dq. The Kullback-Leibler (KL) divergence between fitted and true model KL(ˆθ i ) = q(y) log q(y)dy q(y) log p(y; ˆθ i )dy = C E q (l(ˆθ i ) C l(ˆθ i ) + p i where p(.; θ) is density ( associated ) with P θ. pi = Trace Ji 1 K i dim(θ i ), where J i and K i are Fisher informations using model M i : ( 2 log f (Y, J i = E ˆθ ) ( ) i ) log(f (Y, ˆθi ) g θ θ t, K i = V g. θ

12 AIC: Akaike s information criterion By applying the log and ignoring constant terms c: AIC(M) = 2l(ˆθ) + 2p. Minimizing AIC corresponds to maximizing posterior model probability. Define Akaike weights W (M), W (M) = e AIC(M)/2, which rescaled correspond to probability weights that add up to one, p(m i ) = W (M i ) M M W (M).

13 AIC for Gaussian graphical models The AIC for an estimated Gaussian graphical model K: AIC( K) = n( log K + Tr(S K)) +, 2p K where p K = {unique non-zeroes in K}, which is less than p(p 1)/2.

14 AIC for Gaussian graphical models The AIC for an estimated Gaussian graphical model K: AIC( K) = n( log K + Tr(S K)) + 2p K, where p K = {unique non-zeroes in K}, which is less than p(p 1)/2. But... AIC is an asymptotic approximation. What is n?? For penalized estimate K are number of parameters not smaller?? In the next slides, we propose three alternatives.

15 1. Exact AIC KL(K 0 K) = 1 2 {Tr( KΣ 0 ) log KΣ 0 p} Scaling this by 2 and ignoring a constant 2KL(K 0 K) = {log K Tr( KS)} {Tr( K(Σ 0 S))}. We can write this as 2KL(K 0 K) = 2l( K) {Tr( K(Σ 0 S))}. Definition (Degrees of freedom in Gaussian graphical model) Let Y N(0, K 1 0 ) and K an estimate of K 0 : df K = 1 2 {Tr( K(Σ 0 S))}.

16 Approximate Exact AIC We obviously don t know Σ 0, but we can estimate it: for some tuning paramter π. Σ 0 = πs (1 π)diag{σ 2 11,..., σ 2 pp}, Definition (Approximate Exact AIC) Let Y N(0, K 1 0 ) and K an estimate of K 0 : AIC( K) = 2l( K) + 2 df, where df = 1 2 {Tr( K( Σ 0 S))}.

17 2. Exact GIC Problem of AIC: ˆK λ is not a MLE. GIC for M-estimator ˆK derived by Konishi & Kitagawa (1996): n GIC = 2 l k ( ˆK; x k ) + 2tr{R 1 Q}, (1) k=1 where R and Q are square matrices of order p 2 given by R = 1 n {Dψ(x k, K)} n K= ˆK, k=1 Q = 1 n ψ(x k, K)Dl k (K) n K= ˆK. k=1 Problem: Bias term in (1) requires inversion of d 2 d 2 matrix R.

18 Approximate GIC (Abbruzzo, Vujacic, Wit) We derived an explicit estimator of KL that avoids matrix inversion: where ĜIC(λ) = 2l( ˆK λ ) + 2 df GIC, df GIC = nk=1 c(s k I λ ) c( ˆK λ (S k I λ ) ˆK λ ) 2n where c() is the vectorize operator, S k = x k x T k, I λ = 1 * ( ˆK λ!= 0), an indicator matrix. c(s I λ) c( ˆK λ (S I λ ) ˆK λ ), 2

19 Approximate GIC: Simulations df_kl df_gic df_aic KL GIC AIC λ λ Bias term (left) and KL divergence (right) estimates for n = 100, d = 50.

20 3. Exact Cross-validation (CV) Ignoring an additive constant, the KL divergence is: KL(K 0 K) = 1 2 log K + E Σ0 1 2 {Tr( KXX t )} The idea of cross-validation is to replace the final expectation by KL(K 0 K) 1 2n n {log K (k) + Tr( K (k) x k xk)} t k=1 Clearly, this would require recalculating the estimate K (k) many times.

21 Approximate Cross-validation (Vujacic, Abbruzzo, Wit) We write the KL divergence as follows, KL(K 0 ˆK λ ) = 1 n l( ˆK λ ) + bias, where l(k) = n{log K tr(ks)}/2 and bias = tr( ˆK λ (K 1 0 S))/2. Definition LOOCV-inspired estimate of Kullback-Leibler divergence KLCV (λ) = 1 ni=1 n l( ˆK 1 c[( ˆK λ S i ) I λ ] ( λ )+ ˆK λ ˆK λ )c[(s S i ) I λ ], n(n 1) where c() is the vectorize operator, S k = x k x T k, I λ = 1 * ( ˆK λ!= 0), an indicator matrix.

22 Proof LOOCV = 1 n f (S i, ˆK ( i) ) 2n i=1 = 1 2 f (S, ˆK) 1 n [f (S i, ˆK ( i) ) f (S i, ˆK)] 2n i=1 1 n l( ˆK) 1 n [ df (Si, ˆK) ] c( ˆK ( i) ˆK). 2n dω i=1 Matrix differential calculus: df (S i, ˆK)/dΩ = c( ˆK 1 S i ). The term c( ˆK ( i) ˆK) is obtained via the Taylor expansion 0 df (S, ˆK) dω + d 2 f (S, ˆK) dω 2 c( ˆK ( i) ˆK) + d 2 f (S, ˆK) dωds c(s( i) S). Inserting: df (S, ˆK)/dΩ = c( ˆK 1 S), d 2 f (S, ˆK)/dΩdS = I p 2, d 2 f (S, ˆK)/dΩ 2 = ˆK 1 ˆK 1 and consequently c( ˆK ( i) ˆK) = ( ˆK ˆK)c(S ( i) S).

23 KLCV simulation results d=100 KL ORACLE KLCV AIC GACV n= (0.37) (0.45) (0.28) (19.94) n= (0.34) (0.39) (0.41) (2.77) n= (0.27) (0.33) (0.81) (1.40) n= (0.19) (0.23) (0.48) (0.52) n= (0.07) (0.08) (0.08) (0.07 )

24 Conclusions BIC aims to find true model AIC aims to come closest to the truth BIC gives sparser model than AIC (typically) AIC/BIC have asymptotic issues. What are number of parameters for penalized inference? Improved versions are available!