Steins method for normal approximation of linear statistics of - PowerPoint PPT Presentation

Stein’s method for normal approximation of linear statistics of beta-ensembles Gaultier Lambert (joint work with Michel Ledoux and Christian Webb) UZH University of Z¨ urich Stein Method 1 Normal approximation for Gibbs measures 2 β -ensembles 3

Stein Method

Stein Method When is a random variable Gaussian? Lemma (Stein ’72) Let X be a real-valued random variable. X is N (0 , σ 2 ) if and only if for all f ∈ C ∞ b ( R ), � � Xf ( X ) − σ 2 f ′ ( X ) E = 0 .

Stein Method When is a random variable Gaussian? Lemma (Stein ’72) Let X be a real-valued random variable. X is N (0 , σ 2 ) if and only if for all f ∈ C ∞ b ( R ), � � Xf ( X ) − σ 2 f ′ ( X ) E = 0 . Let ( X n ) n ∈ N be a sequence of random variables. If for all f ∈ C ∞ b ( R ), � � X n f ( X n ) − f ′ ( X n ) n →∞ E lim = 0 , then X n ⇒ N (0 , 1).

Stein Method Proof: Let Z be N (0 , 1). It suffices to show that for any ϕ ∈ C ∞ ( R ), 0 ≤ ϕ ≤ 1, n →∞ E ϕ ( X n ) = E ϕ ( Z ) . lim

Stein Method Proof: Let Z be N (0 , 1). It suffices to show that for any ϕ ∈ C ∞ ( R ), 0 ≤ ϕ ≤ 1, n →∞ E ϕ ( X n ) = E ϕ ( Z ) . lim Consider the equation: f ′ ( x ) = xf ( x ) − ϕ ( x ) + E ϕ ( Z ) . The solution is given by ˆ x e − t 2 / 2 � � f ( x ) = − e x 2 / 2 ϕ ( t ) − E ϕ ( Z ) dt −∞ ˆ ∞ e − t 2 / 2 � � = e x 2 / 2 ϕ ( t ) − E ϕ ( Z ) dt x

Stein Method Proof: Let Z be N (0 , 1). It suffices to show that for any ϕ ∈ C ∞ ( R ), 0 ≤ ϕ ≤ 1, n →∞ E ϕ ( X n ) = E ϕ ( Z ) . lim Consider the equation: f ′ ( x ) = xf ( x ) − ϕ ( x ) + E ϕ ( Z ) . The solution is given by ˆ x e − t 2 / 2 � � f ( x ) = − e x 2 / 2 ϕ ( t ) − E ϕ ( Z ) dt −∞ ˆ ∞ e − t 2 / 2 � � = e x 2 / 2 ϕ ( t ) − E ϕ ( Z ) dt x Thus, we have � � X n f ( X n ) − f ′ ( X n ) E ϕ ( X n ) = E ϕ ( Z ) + E √ and, since f is smooth and bounded by 2 π � ϕ � L ∞ , the last term converges to zero as n → ∞ .

Stein Method Probability metrics If X is a complete separable metric space, the space of Borel probability measures on X equipped with the topology of weak convergence is completely metrizable. An example of a metric is given by � � d ( X , Y ) = sup E ϕ ( X ) − E ϕ ( Y ) ϕ ∈ F where F is a sufficiently big class of functions ϕ : X → R .

Stein Method Probability metrics If X is a complete separable metric space, the space of Borel probability measures on X equipped with the topology of weak convergence is completely metrizable. An example of a metric is given by � � d ( X , Y ) = sup E ϕ ( X ) − E ϕ ( Y ) ϕ ∈ F where F is a sufficiently big class of functions ϕ : X → R . The total variation distance : � � d TV ( X , Y ) = sup E ϕ ( X ) − E ϕ ( Y ) . � ϕ � L ∞ ≤ 1 / 2 The Kolmogorov distance (for real-valued random variables): � � � � d K ( X , Y ) = sup � P [ X ≤ x ] − P [ Y ≤ x ] � . x ∈ R

Stein Method Kantorovich or Wasserstein metrics For any q ≥ 1, define the Wasserstein q distance between two random variables: ( x , y ) E [ | x − y | q ] 1 / q . W q ( X , Y ) = inf where the infimum is taken over all couplings or random vectors ( x , y ) such that d d x = X and y = Y .

Stein Method Kantorovich or Wasserstein metrics For any q ≥ 1, define the Wasserstein q distance between two random variables: ( x , y ) E [ | x − y | q ] 1 / q . W q ( X , Y ) = inf where the infimum is taken over all couplings or random vectors ( x , y ) such that d d x = X and y = Y . It turns out that � � W 1 ( X , Y ) = sup E ϕ ( X ) − E ϕ ( Y ) . � ϕ � Lip ≤ 1

Stein Method Kantorovich or Wasserstein metrics For any q ≥ 1, define the Wasserstein q distance between two random variables: ( x , y ) E [ | x − y | q ] 1 / q . W q ( X , Y ) = inf where the infimum is taken over all couplings or random vectors ( x , y ) such that d d x = X and y = Y . It turns out that � � W 1 ( X , Y ) = sup E ϕ ( X ) − E ϕ ( Y ) . � ϕ � Lip ≤ 1 The bounded-Lipschitz distance : � � d BL ( X , Y ) = sup E ϕ ( X ) − E ϕ ( Y ) . � ϕ � Lip ≤ 1 � ϕ � L ∞ ≤ 1

Stein Method Berry-Essen theorem Let ( X j ) j ∈ N be i.i.d. real-valued random variables with mean zero and variance 1 and define S N = X 1 + · · · + X N √ . N Theorem (Barbour, Hall, ’84) There exists C > 0 such that � ≤ C E | X 1 | 3 � � � E [ S N ≤ x ] − E [ Z ≤ x ] √ d K ( S N , Z ) = sup . N x ∈ R

Stein Method Berry-Essen theorem Let ( X j ) j ∈ N be i.i.d. real-valued random variables with mean zero and variance 1 and define S N = X 1 + · · · + X N √ . N Theorem (Barbour, Hall, ’84) There exists C > 0 such that � ≤ C E | X 1 | 3 � � � E [ S N ≤ x ] − E [ Z ≤ x ] √ d K ( S N , Z ) = sup . N x ∈ R We will now show the easier claim: d BL ( S N , Z ) ≤ C E | X 1 | 3 √ . N

Stein Method Regularity of the solution of Stein’s equation Stein’s Lemma The solution of Stein’s equation satisfy √ � f � L ∞ ≤ 2 π � ϕ � L ∞ , � f ′ � L ∞ ≤ min {� ϕ � Lip , 2 � ϕ � L ∞ } � f ′ � Lip ≤ 4 � ϕ � L ∞ + 2 � ϕ � Lip

Stein Method Proof: For all t ∈ [0 , 1], let N = tX 1 + X 2 + · · · + X N S t √ . N Let ϕ ∈ L ∞ ( R ), Lipschitz–continuous, and f be the corresponding solution of Stein’s equation. By Taylor’s theorem, we have � � √ X 1 f ( S 1 E [ S N f ( S N )] = N ) N E � � �� √ √ f ( S 0 N ) + f ′ ( S T = N ) X 1 / N E X 1 N � � � � √ X 1 f ( S 0 X 2 1 f ′ ( S T = N E N ) + E N ) where T is uniform in [0 , 1] and independent of ( X j ) j ∈ N .

Stein Method Proof: For all t ∈ [0 , 1], let N = tX 1 + X 2 + · · · + X N S t √ . N Let ϕ ∈ L ∞ ( R ), Lipschitz–continuous, and f be the corresponding solution of Stein’s equation. By Taylor’s theorem, we have � � √ X 1 f ( S 1 E [ S N f ( S N )] = N ) N E � � �� √ √ f ( S 0 N ) + f ′ ( S T = N ) X 1 / N E X 1 N � � � � √ X 1 f ( S 0 X 2 1 f ′ ( S T = N E N ) + E N ) where T is uniform in [0 , 1] and independent of ( X j ) j ∈ N . By independence, � � � � � � X 1 f ( S 0 X 2 1 f ′ ( S 0 f ′ ( S 0 E N ) = 0 and E N ) = E N ) , so that � �� � � � X 2 f ′ ( S T N ) − f ′ ( S 0 f ′ ( S 0 E [ S N f ( S N )] = E N ) + E N ) . 1

Stein Method Proof: Thus, we have � �� � � � � � S N f ( S N ) − f ′ ( S N ) X 2 f ′ ( S T N ) − f ′ ( S 0 f ′ ( S 0 N ) − f ′ ( S 1 E = E N ) + E N ) 1 Since f ′ is Lipschitz–continuous and S T N − S 0 N = TX 1 N , we obtain √ � | X 1 | 3 � � � ≤ 3 � f ′ � Lip S N f ( S N ) − f ′ ( S N ) √ . E E 2 N

Stein Method Proof: Thus, we have � �� � � � � � S N f ( S N ) − f ′ ( S N ) X 2 f ′ ( S T N ) − f ′ ( S 0 f ′ ( S 0 N ) − f ′ ( S 1 E = E N ) + E N ) 1 Since f ′ is Lipschitz–continuous and S T N − S 0 N = TX 1 N , we obtain √ � | X 1 | 3 � � � ≤ 3 � f ′ � Lip S N f ( S N ) − f ′ ( S N ) √ . E E 2 N By Stein’s equation: f ′ ( x ) = xf ( x ) − ϕ ( x ) + E ϕ ( Z ) , we conclude that, if � ϕ � Lip ≤ 1 and � ϕ � L ∞ ≤ 1, then � | X 1 | 3 � 9 E [ ϕ ( S N ) − ϕ ( Z )] ≤ √ . E N so that � | X 1 | 3 � d BL ( S N , Z ) ≤ 9 E √ . N

Stein Method Theorem (Chatterjee, Meckes ’08) Let ( X t ) t ≥ 0 be a family of random variables in L 2 ( P ) with the same law. Assume that X 0 is F –measurable and that there exists constants Λ , σ > 0 such that � X t − X 0 � � � lim = − Λ X 0 + R t → 0 E � F t � ( X t − X 0 ) 2 � � � = 2Λ σ 2 + S lim t → 0 E � F t � | X t − X 0 | 2+ α � = o ( t ) , α > 0 . E t → 0 Then W 1 ( X 0 , σ Z ) ≤ E | R | + E | S | . Λ

Stein Method Theorem (Chatterjee, Meckes ’08) Let ( X t ) t ≥ 0 be a family of random variables in L 2 ( P ) with the same law. Assume that X 0 is F –measurable and that there exists constants Λ , σ > 0 such that � X t − X 0 � � � lim = − Λ X 0 + R t → 0 E � F t � ( X t − X 0 ) 2 � � � = 2Λ σ 2 + S lim t → 0 E � F t � | X t − X 0 | 2+ α � = o ( t ) , α > 0 . E t → 0 Then W 1 ( X 0 , σ Z ) ≤ E | R | + E | S | . Λ Applications in random matrix theory: Chatterjee ’09 D¨ obler–Stolz ’11 ’14 Fulman ’12 Webb ’16 Joyner–Smilansky ’15 ’17

Stein Method Proof: Let ϕ be a Lipschitz continuous and consider the solution of the equation: h ′′ ( x ) − xh ′ ( x ) = ϕ ( x ) − E ϕ ( Z ) . (1) Since X t and X 0 have the same law, for any t > 0, E [ h ( X t ) − h ( X )] = 0 . By Taylor’s theorem, we have � � � E | X t − X 0 | 2+ α � ( X t − X 0 ) h ′ ( X 0 ) + ( X t − X 0 ) 2 h ′′ ( X 0 ) = O . E 2 t → 0 By assumption ( σ 2 = 1), we also have � ( X t − X 0 ) � h ′ ( X 0 ) + ( X t − X 0 ) 2 h ′′ ( X 0 ) lim = 0 t → 0 E 2 t t � � � � − X 0 h ′ ( X 0 ) + σ 2 h ′′ ( X 0 ) Rh ′ ( X 0 ) + Sh ′′ ( X 0 ) / 2 = Λ E + E . Using equation (1), since � h ′ � L ∞ , � h ′′ � L ∞ ≤ � ϕ � Lip , we obtain E ϕ ( X 0 ) − E ϕ ( Z ) ≤ E | R | + E | S | . Λ