Conditional Sampling for Max Stable Random Fields

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Conditional Sampling for Max Stable Random Fields Yizao Wang Department of Statistics, the University of Michigan April 30th, 0 th GSPC at Duke University, Durham, North Carolina Joint work with Stilian A. Stoev

An Illustrating Example Given the model and some observations, how to do prediction? 0.7 1 1 0 1 0. 3.3 9.7 1.97 0.9 7. 1 0 1 Figure: A sample from the de Haan Pereira Model (de Haan and Pereira 00): a stationary (moving maxima) max stable random field. Parameters: ρ = 0., β 1 = 1., β = 0.7.

An Illustrating Example Four conditional samplings from the de Haan Pereira Model. 1 0 1 0. 3.3 0.7 9.7 1.97 0.9 7. 1 1 0 1 1 0 1 0.7 0. 3.3 9.7 1.97 0.9 7. 1 0 1 1 0 1 0.7 0. 3.3 9.7 1 1 0 1 0.7 0. 3.3 9.7 1.97 0.9.97.97 0.9 7. 1 0 1 0. 3.3 0.7 9.7 1 0.9 7. 7. 1 1 0 1 1 0 1 1 0 1 Difficulties: Analytical formula often impossible. Naïve Monte Carlo method does not apply.

Conditional Sampling for Max Stable Random Fields Our contribution: Obtained explicit formula of regular conditional probability for max linear models (including a large class of max stable random fields). Developed efficient software (R package maxlinear) for large scale conditional samplings. Potential applications in prediction for extremal phenomena, e.g., environmental and financial problems.

Max Linear Models Formula: p X i = a i,j Z j, 1 i n, denoted by X = A Z, j=1 Z j f Zj independent continuous, nonnegative random variables A = (a i,j ) 1 i n,1 j p, a i,j 0.

Max Linear Models Formula: X i = p a i,j Z j, 1 i n, denoted by X = A Z, j=1 Z j f Zj independent continuous, nonnegative random variables A = (a i,j ) 1 i n,1 j p, a i,j 0. Z 1,..., Z p independent α Fréchet: P(Z > t) = exp{ σ α t α } {X i } 1 i n is an α Fréchet process approximation of arbitrary max stable process (random field) with p sufficiently large. Modeling spatial extremes arising in meteorology, geology, and environmental applications.

Conditional Sampling for Max Linear Models Consider the model (A known) 0.7 1 X = A Z. Observations: X = x. What is the conditional distribution of Z, given X = x? Prediction on Y = B Z, given X = x. 1 0 1 1 0 1 0. 3.3 9.7 1.97 0.9 7.

Conditional Sampling for Max Linear Models Consider the model (A known) 0.7 1 X = A Z. Observations: X = x. What is the conditional distribution of Z, given X = x? Prediction on Y = B Z, given X = x. 1 0 1 1 0 1 0. 3.3 9.7 1.97 0.9 7. Remarks: Theoretical issue: P(Z E X = x) is not well defined. Rigorous treatment: ν(x, E) : R n + B R p [0, 1], regular conditional + probability.

Conditional Sampling for Max Linear Models Consider the model (A known) 0.7 1 X = A Z. Observations: X = x. What is the conditional distribution of Z, given X = x? Prediction on Y = B Z, given X = x. 1 0 1 1 0 1 0. 3.3 9.7 1.97 0.9 7. Remarks: Theoretical issue: P(Z E X = x) is not well defined. Rigorous treatment: ν(x, E) : R n + B R p [0, 1], regular conditional + probability. Computational issue: dim(a) = n p, dim(b) = n B p, n small, n B, p large.

A Toy Example Consider X = A Z with ( 1 1 1 A = 1 0 0 ), i.e. { X1 = Z 1 Z Z 3 X = Z 1 We have (in)equality constraints: Z 1 min(x 1, X ) =: ẑ 1 Z X 1 =: ẑ Z 3 X 1 =: ẑ 3.

A Toy Example Consider X = A Z with ( 1 1 1 A = 1 0 0 ), i.e. { X1 = Z 1 Z Z 3 X = Z 1 We have (in)equality constraints: Z 1 min(x 1, X ) =: ẑ 1 Z X 1 =: ẑ Z 3 X 1 =: ẑ 3. Two cases: (i) (red) If 0 < X 1 = X = a, then ẑ 1 = ẑ = ẑ 3 = a and Z 1 = ẑ 1, Z ẑ, Z 3 ẑ 3. (ii) (blue) If 0 < a = X < X 1 = b, then ẑ 1 = a, ẑ = ẑ 3 = b and Z 1 = ẑ 1, Z = ẑ, Z 3 ẑ 3 or Z 1 = ẑ 1, Z ẑ, Z 3 = ẑ 3.

A Toy Example Consider X = A Z with ( 1 1 1 A = 1 0 0 ), i.e. { X1 = Z 1 Z Z 3 X = Z 1 We have (in)equality constraints: Z 1 min(x 1, X ) =: ẑ 1 Z X 1 =: ẑ Z 3 X 1 =: ẑ 3. Two cases: (i) (red) If 0 < X 1 = X = a, then ẑ 1 = ẑ = ẑ 3 = a and Z 1 = ẑ 1, Z ẑ, Z 3 ẑ 3. (ii) (blue) If 0 < a = X < X 1 = b, then ẑ 1 = a, ẑ = ẑ 3 = b and Z 1 = ẑ 1, Z = ẑ, Z 3 ẑ 3 or Z 1 = ẑ 1, Z ẑ, Z 3 = ẑ 3. When Z j = ẑ j, we say Z j hits ẑ j different hitting scenarios.

Intuition of Conditional Distribution of Z X = x Define ẑ j := min 1 i n x i /a i,j and C(A, x) := {z R p + : x = A z} Need a distribution on C(A, x), for each x.

Intuition of Conditional Distribution of Z X = x Define ẑ j := min 1 i n x i /a i,j and C(A, x) := {z R p + : x = A z} Need a distribution on C(A, x), for each x. Partition of C(A, x), according to equality constraints: (blue) C(A, x) = C {1,} (A, x) C {1,3} (A, x) C {1,,3} (A, x) with C {1,} (A, x) = {z 1 = ẑ 1, z = ẑ, z 3 < ẑ 3 } C {1,3} (A, x) = {z 1 < ẑ 1, z = ẑ, z 3 = ẑ 3 } C {1,,3} (A, x) = {z 1 = ẑ 1, z = ẑ, z 3 = ẑ 3 }

Intuition of Conditional Distribution of Z X = x Define ẑ j := min 1 i n x i /a i,j and C(A, x) := {z R p + : x = A z} Need a distribution on C(A, x), for each x. Partition of C(A, x), according to equality constraints: (blue) C(A, x) = C {1,} (A, x) C {1,3} (A, x) C {1,,3} (A, x) with C {1,} (A, x) = {z 1 = ẑ 1, z = ẑ, z 3 < ẑ 3 } C {1,3} (A, x) = {z 1 < ẑ 1, z = ẑ, z 3 = ẑ 3 } C {1,,3} (A, x) = {z 1 = ẑ 1, z = ẑ, z 3 = ẑ 3 } Hitting scenarios: J {1,..., p} : C J (A, x). Conditional distribution: mixture of distributions, indexed by hitting scenarios: J (A, x) = {J : C J (A, x) }.

Intuition of Conditional Distribution of Z X = x C {1,} (A, x) = {z 1 = ẑ 1, z = ẑ, z 3 < ẑ 3 } C {1,3} (A, x) = {z 1 < ẑ 1, z = ẑ, z 3 = ẑ 3 } C {1,,3} (A, x) = {z 1 = ẑ 1, z = ẑ, z 3 = ẑ 3 } Define ν J on all the hitting scenarios J: ν J (x, E) := δẑj (π j (E)) P{Z j π j (E) Z j < ẑ j }, j J j J c

Intuition of Conditional Distribution of Z X = x C {1,} (A, x) = {z 1 = ẑ 1, z = ẑ, z 3 < ẑ 3 } C {1,3} (A, x) = {z 1 < ẑ 1, z = ẑ, z 3 = ẑ 3 } C {1,,3} (A, x) = {z 1 = ẑ 1, z = ẑ, z 3 = ẑ 3 } Define ν J on all the hitting scenarios J: ν J (x, E) := δẑj (π j (E)) P{Z j π j (E) Z j < ẑ j }, j J j J c It suffices to concentrate on relevant hitting scenarios J r (A, x) = {J J (A, x) : J = r} with r = min{ J : J J (A, x)}. Clearly, C {1,,3} is negligible compared to C {1,} and C {1,3} (C {1,,3} has lower dimension).

A Toy Example Consider X = A Z with ( 1 1 1 A = 1 0 0 In this case, ) { X1 = Z, i.e. 1 Z Z 3 X = Z 1 ẑ 1 = min(x 1, X ), ẑ = ẑ 3 = X 1. Two cases: (i) (red) If X 1 = X = a with 0 < a, then, ẑ 1 = ẑ = ẑ 3 = a and Z 1 = ẑ 1, Z ẑ, Z 3 ẑ 3. J (A, X) = {{1}, {1, }, {1, 3}, {1,, 3}} and J r (A, X) = {{1}}. (ii) (blue) If 0 < a = X < X 1 = b, then ẑ 1 = a, ẑ = ẑ 3 = b and Z 1 = ẑ 1, Z = ẑ, Z 3 ẑ 3 or Z 1 = ẑ 1, Z ẑ, Z 3 = ẑ 3. J (A, X) = {{1, }, {1, 3}, {1,, 3}} and J r (A, X) = {{1, }, {1, 3}}. Different X different hitting distributions different hitting scenarios.

Conditional Distribution for Max Linear Models Theorem (W and Stoev 0). The regular conditional probability ν(x, E) of Z w.r.t. X equals: E R R p where ν(x, E) = J J r (A,x) p J (A, x) w J := j J +, p J (A, x)ν J (x, E), P X a.a. x A (R p +) ẑ j f Zj (ẑ j ) j J c F Zj (ẑ j ), J J r (A,x) Proof is involved with the definition of regular conditional probability. p J = 1.

Conditional Distribution for Max Linear Models Theorem (W and Stoev 0). The regular conditional probability ν(x, E) of Z w.r.t. X equals: E R R p where ν(x, E) = J J r (A,x) p J (A, x) w J := j J +, p J (A, x)ν J (x, E), P X a.a. x A (R p +) ẑ j f Zj (ẑ j ) j J c F Zj (ẑ j ), J J r (A,x) Proof is involved with the definition of regular conditional probability. Algorithm I for conditional sampling Z X = x: (1) compute ẑ j, J (A, x), r(j (A, x)) and p J (A, x), and () sample Z ν(x, ). p J = 1.

Conditional Distribution for Max Linear Models Theorem (W and Stoev 0). The regular conditional probability ν(x, E) of Z w.r.t. X equals: E R R p where ν(x, E) = J J r (A,x) p J (A, x) w J := j J +, p J (A, x)ν J (x, E), P X a.a. x A (R p +) ẑ j f Zj (ẑ j ) j J c F Zj (ẑ j ), J J r (A,x) Proof is involved with the definition of regular conditional probability. Algorithm I for conditional sampling Z X = x: (1) compute ẑ j, J (A, x), r(j (A, x)) and p J (A, x), and () sample Z ν(x, ). p J = 1. Not the end of the story! We haven t discussed identification of J (A, x), which is closely related to the NP hard set covering problem.

Set Covering Problem Let H = (h i,j ) n p with h i,j {0, 1}. Write [m] {1,,, m}, m N. The column j [p] covers the row i [n], if h i,j = 1. The goal is to cover all rows with least columns. This is equivalent to solving min δ j, subject to h i,j δ j 1, i [n]. (1) δ j {0,1} j [p] An example: j [p] H = j [p] 1 1 0 1 0 1 0 1 1. Minimum cost coverings are columns {1, }, {1, 3}, and {, 3}.

Identification of J r (A, x) and Set Covering Problem J J r (A, x) 1 1 a solution of (1) with h i,j = 1 {ai,j ẑ j =x i } and δ j = 1 {j J}.

Identification of J r (A, x) and Set Covering Problem J J r (A, x) 1 1 a solution of (1) with h i,j = 1 {ai,j ẑ j =x i } and δ j = 1 {j J}. A toy example: consider X = A Z with ( ) { 1 1 1 X1 = Z A =, i.e. 1 Z Z 3. 1 0 0 X = Z 1 Then, (blue) 0 < a = X < X 1 = b H = ( 0 1 1 1 0 0 The minimum cost coverings are columns {1, } and {1, 3}. ).

Identification of J r (A, x) and Set Covering Problem J J r (A, x) 1 1 a solution of (1) with h i,j = 1 {ai,j ẑ j =x i } and δ j = 1 {j J}. A toy example: consider X = A Z with ( ) { 1 1 1 X1 = Z A =, i.e. 1 Z Z 3. 1 0 0 X = Z 1 Then, (blue) 0 < a = X < X 1 = b H = ( 0 1 1 1 0 0 The minimum cost coverings are columns {1, } and {1, 3}. ). Write J r (H) = J r (A, x), with H referred to as the hitting matrix.

Identification of J r (A, x) and Set Covering Problem J J r (A, x) 1 1 a solution of (1) with h i,j = 1 {ai,j ẑ j =x i } and δ j = 1 {j J}. A toy example: consider X = A Z with ( ) { 1 1 1 X1 = Z A =, i.e. 1 Z Z 3. 1 0 0 X = Z 1 Then, (blue) 0 < a = X < X 1 = b H = ( 0 1 1 1 0 0 The minimum cost coverings are columns {1, } and {1, 3}. ). Write J r (H) = J r (A, x), with H referred to as the hitting matrix. Set covering problem is NP hard.

Simple Cases of Set Covering Problem H 1 = Two types of H: 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 It takes a while to solve, H = 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 0 1 0 0 0 0 1 1 0 1 0 0 0 1 0 1 1 0 0 0 r(j (H 1 )) = and J r (H 1 ) =. Clearly, columns 1 and are dominating. Therefore, r(j (H )) = and J (H ) = {{1, }}..

Simple Cases of Set Covering Problem H 1 = Two types of H: 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 It takes a while to solve, H = 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 0 1 0 0 0 0 1 1 0 1 0 0 0 1 0 1 1 0 0 0 r(j (H 1 )) = and J r (H 1 ) =. Clearly, columns 1 and are dominating. Therefore, r(j (H )) = and J (H ) = {{1, }}.. Lemma (W and Stoev) W.p.1, H has nice structure. example

Factorization of Regular Conditional Probability Formula Theorem (W and Stoev) With probability one, ν(x, E) = r ν (s) (X, E) with {Z j } j J (s) X = x ν (s) (x, ). s=1 Blocking structure: r s=1 J(s) = {1,..., p} and r s=1 Is = {1,..., n}.

Factorization of Regular Conditional Probability Formula Theorem (W and Stoev) With probability one, ν(x, E) = r ν (s) (X, E) with {Z j } j J (s) X = x ν (s) (x, ). s=1 Blocking structure: r s=1 J(s) = {1,..., p} and r s=1 Is = {1,..., n}. Conditional independence: restricted on X = x, X i p a i,j Z j = j=1 j J (s) a i,j Z j, i I s.

Factorization of Regular Conditional Probability Formula Theorem (W and Stoev) With probability one, ν(x, E) = r ν (s) (X, E) with {Z j } j J (s) X = x ν (s) (x, ). s=1 Blocking structure: r s=1 J(s) = {1,..., p} and r s=1 Is = {1,..., n}. Conditional independence: restricted on X = x, X i p a i,j Z j = j=1 j J (s) a i,j Z j, i I s. Algorithm II {Z j } j J (1) ν (1) (x, ) {Z j } j J (r) ν (r) (x, ) Z d = ν(x, ) = J J r p J ν J (x, )

Computational Efficiency Time (in secs) of identification of the blocking structure for Algorithm II: p \ n 1 0 00 0.03 (0.0) 0.13 (0.03) 0. (0.0) 1. (0.09) 000 0.11 (0.0) 0.0 (0.0) 1.00 (0.0).9 (0.33)

Computational Efficiency Time (in secs) of identification of the blocking structure for Algorithm II: p \ n 1 0 00 0.03 (0.0) 0.13 (0.03) 0. (0.0) 1. (0.09) 000 0.11 (0.0) 0.0 (0.0) 1.00 (0.0).9 (0.33) Comparison of two formulas: J J r (A,x) p J ν J (x, ) = ν(x, ) = Example Suppose, given X = x, r s=1 w (s) j j J (s) ν (s) j (x, ) } {{ } ν (s) (x, ) r = and J (s) =, s []. Then, to use ν(x, ) requires memory for J r (A, x) = weights, while to apply {ν (s) (x, )} s [] requires only.

Applications Simulations based on the de Haan Pereira model (de Haan and Pereira 00). Computational tools for prediction: real data analysis on maxima rainfall data at Bourgogne, France.

De Haan Pereira Model A stationary max stable random field model: De Haan and Pereira (00): with φ(t 1, t ) := X t = e R φ(t u)m α (du), t = (t 1, t ) R. β 1 β { π 1 ρ exp 1 [ β (1 ρ ) 1 t1 ρβ 1 β t 1 t + βt ] }. Consistent estimators known for ρ, β 1, β.

De Haan Pereira Model A stationary max stable random field model: De Haan and Pereira (00): with φ(t 1, t ) := X t = e R φ(t u)m α (du), t = (t 1, t ) R. β 1 β { π 1 ρ exp 1 [ β (1 ρ ) 1 t1 ρβ 1 β t 1 t + βt ] }. Consistent estimators known for ρ, β 1, β. A discretized version: X t = h /α φ(t u j1j )Z j1j, q j 1,j q 1 with u j1j = ((j 1 + 1/)h, (j + 1/)h) and Z j1j s i.i.d. 1 Fréchet.

1 Simulations 1 1 0 1 1 1 0 1 1 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 Figure: conditional samplings from de Haan Pereira model. Parameters: ρ = 0, β 1 = 1, β = 1.

Simulations 1 0 1 0. 3.3 0.7 9.7 1.97 0.9 7. 1 1 0 1 1 0. 3.3 0.7 9.7.97 0.9 7. 1 0 1 0 1 1 0 1 Figure: 9% quantile of the conditional marginal deviation. Parameters: ρ = 0., β 1 = 1., β = 0.7.

Review Obtained explicit formula of regular conditional probability for max linear models (including a large class of max stable random fields). Investigated the conditional independence structure. Developed efficient software (R package maxlinear) for large scale conditional samplings. Potential applications in prediction for extremal phenomena, e.g., environmental and financial problems.

Review Obtained explicit formula of regular conditional probability for max linear models (including a large class of max stable random fields). Investigated the conditional independence structure. Developed efficient software (R package maxlinear) for large scale conditional samplings. Potential applications in prediction for extremal phenomena, e.g., environmental and financial problems. Thank you. Website: http: //www.stat.lsa.umich.edu/~yizwang/software/maxlinear/

Auxiliary Results

Regular Conditional Probability The regular conditional probability ν of Z given σ(x), is a function such that ν : R n B R p [0, 1], + (i) ν(x, ) is a probability measure, for all x R n, (ii) The function ν(, E) is measurable, for all Borel sets E B R p. (iii) For all E R R p and D R R n, (P X ( ) := P(X )): P(Z E, X D) = ν(x, E)P X (dx). () We will first guess a formula for ν and then prove (). D

A Heuristic Proof Consider a neighbor of C J (A, x) (of P measure 0) C J (A, x) = { z R p + : z j = ẑ j, j J, z k < ẑ k, k J c} C ɛ J (A, x) := { z R p + : z j [ẑ j (1 ɛ), ẑ j (1 + ɛ)], j J, z k < ẑ k (1 ɛ), k J c} for small enough ɛ > 0 and let C ɛ (A, x) := J J (A,x) C J ɛ (A, x). The sets A (C ɛ (A, x)) shrink to the point x, as ɛ 0. Proposition (W and Stoev 0). For all x A (R p +), we have, as ɛ 0, Remarks: P(Z E Z C ɛ (A, x)) ν(x, E), E R R p +. Proved by Taylor expansion. The choice of CJ ɛ (A, x) is delicate.

Nice Structure of the Hitting Matrix H Blocks of matrix H: Write i 1 j i if h i1,j = h i,j = 1. Define an equivalence relation on [n]: i 1 i, if i 1 = ĩ0 j 1 ĩ 1 j jm ĩm = i. (3) r blocks: the equivalence relation (3) induces [n] = r s=1 Is. Further, J (s) := J (s) := {j [p] : h i,j = 1 for all i I s } {j [p] : h i,j = 1 for some i I s } Example: H = Two blocks: 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 0 1 0 0 0 0 1 1 0 1 0 0 0 1 0 1 1 0 0 0 1. I 1 = {1,, 3}, J (1) = {}, J (1) = {,, 7},. I = {,,, 7}, J () = {1}, J () = {1,, 3, }. Lemma (W and Stoev 0). W.p.1, the hitting matrix H of max linear model X = A Z has nice structure: (i) r = r(j (H)) = r(j (A, x)) and (ii) J (s) is nonempty s [r].

When We Have Bad H? Another Toy Example Consider Different hitting matrices: X = A Z with A = 1 0 1 1 1 0 1 Z 1 > Z, Z > Z 3 H = Z 1 < Z < Z 1, Z > Z 3 H = Z 1 = Z, Z > Z 3 H =. 1 0 0 1 0 0 0 1 1 1 0 0 0 1 0 0 1 1 1 0 0 1 1 0 0 1 1

Factorization of Regular Conditional Probability Formula Theorem (W and Stoev 0). With probability one, we have (i) for all J [p], J J r (A, A Z) if and only if J can be written as J = {j 1,..., j r } with j s J (s), s [r], (ii) For the regular conditional probability ν(x, E), ν(x, E) = r ν (s) (X, E) with ν (s) (X, E) = s=1 where for all j J (s), w (s) j = ẑ j f Zj (ẑ j ) ν (s) j (x, E) = δ πj (E)(ẑ j ) k J (s) \{j} k J (s) \{j} F Zk (ẑ k ) (s) j J (s) w j ν (s) (s) j J (s) w j j (X, E), P(Z k π k (E) Z k < ẑ k ).

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