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Detecting mixtures in multivariate extremes S.H.A. Tendijck - - PowerPoint PPT Presentation

Detecting mixtures in multivariate extremes S.H.A. Tendijck Lancaster University January 31, 2020 Motivating application 2 / 16 Motivating application 2 / 16 Two types of waves Swell versus wind waves: 3 / 16 Contents 1 Crash course in


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Detecting mixtures in multivariate extremes

S.H.A. Tendijck

Lancaster University January 31, 2020

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Motivating application

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Motivating application

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Two types of waves

Swell versus wind waves:

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Contents

1 Crash course in underlying theory 2 My model

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Overview

1 Crash course in underlying theory 2 My model

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Univariate extremes

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5 10 x 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

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Univariate extremes

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5 10 x 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

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Univariate extremes

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5 10 x 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

5 10

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Univariate extremes

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5 10 x 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

5 10

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Multivariate extremes

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Multivariate extremes

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Multivariate extremes

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Multivariate extremes

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Conditional extremes

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Conditional extremes

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Conditional extremes

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Conditional extremes

Heffernan-Tawn model: Y = αX + Z for (X,Y ) on standard margins and Z some residual distribution, independent of X.

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Conditional extremes

Heffernan-Tawn model: Y ∣(X > u) = αX + Z for (X,Y ) on standard margins and Z some residual distribution, independent of X.

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Conditional extremes

Heffernan-Tawn model: Y ∣(X > u) = αX + X βZ for (X,Y ) on standard margins and Z some residual distribution, independent of X.

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Conditional extremes

Heffernan-Tawn model: Y ∣(X > u) = αX + X βZ for (X,Y ) on standard margins and Z some residual distribution, independent of X. Pros:

  • It can capture both asymptotic dependence (α = 1) and asymptotic

independence (α < 1);

  • Many bivariate distributions follow this structure asymptotically;
  • Extends well to multivariate distributions.

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Conditional extremes

Heffernan-Tawn model: Y ∣(X > u) = αX + X βZ for (X,Y ) on standard margins and Z some residual distribution, independent of X. Pros:

  • It can capture both asymptotic dependence (α = 1) and asymptotic

independence (α < 1);

  • Many bivariate distributions follow this structure asymptotically;
  • Extends well to multivariate distributions.

Cons:

  • It doesn’t capture mixture structures;
  • Data needs to be on standard margins;
  • Inconsistent in modelling X∣Y and Y ∣X when both are large.

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Mixtures in extremes

2 4 6 8 10 12 14

XL

  • 2

2 4 6 8 10

YL

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Mixtures in extremes

2 4 6 8 10 12 14

XL

  • 2

2 4 6 8 10

YL

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Mixtures in extremes

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XL

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2 4 6 8 10

YL

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Mixtures in extremes

The Heffernan-Tawn model extends to Y ∣(X > u) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ α1X + X β1Z1 with probability p; α2X + X β2Z2 with probability 1 − p.

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Mixtures in extremes

The Heffernan-Tawn model extends to Y ∣(X > u) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ α1X + X β1Z1 with probability p; α2X + X β2Z2 with probability 1 − p. What do we want:

  • Fit the model;
  • Estimate the number of mixture components;
  • Estimate the mixture probabilities.

Methods:

1 Quantile-Regression model; 2 Fitting a Heffernan-Tawn mixture model directly.

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Quantile Regression

How do we estimate the 90% conditional quantile of Y given X?

2 4 6 8 10 12

X

  • 2

2 4 6 8 10 12

Y

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Quantile Regression

How do we estimate the 90% conditional quantile of Y given X?

2 4 6 8 10 12

X

  • 2

2 4 6 8 10 12

Y

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Quantile Regression

How do we estimate the 90% conditional quantile of Y given X?

2 4 6 8 10 12

X

  • 2

2 4 6 8 10 12

Y

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Quantile Regression

How do we estimate the 90% conditional quantile of Y given X?

2 4 6 8 10 12

X

  • 2

2 4 6 8 10 12

Y

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Quantile Regression

How do we estimate the 90% conditional quantile of Y given X?

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X

  • 2

2 4 6 8 10 12

Y

Minimise the L1 distance to the line, while keeping 90% below.

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Overview

1 Crash course in underlying theory 2 My model

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My model

We assume the Heffernan-Tawn model holds, i.e., Y ∣(X > u) = αX + X βZ. Our quantile regression model is given by qτ(x) = αx + xβz.

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My model

We assume the Heffernan-Tawn model holds, i.e., Y ∣(X > u) = αX + X βZ. Our quantile regression model is given by qτ(x) = c + αx + xβz.

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My model

We assume the Heffernan-Tawn model holds, i.e., Y ∣(X > u) = αX + X βZ. Our quantile regression model is given by qτ(x) = c + αx + xβz. For stability, we fit simultaneously for τ = 0.05,0.15,...,0.95. We get 13 estimated parameters: (ˆ α, ˆ β, ˆ c, ˆ z1,..., ˆ z10).

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My model

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X

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5 10 15

Y Logistic Model

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My model

We assume a mixture HT model holds, i.e., Y ∣(X > u) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ α1X + X β1Z1 with probability 1 − p, α2X + X β2Z2 with probability p. where α1 > α2.

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My model

We assume a mixture HT model holds, i.e., Y ∣(X > u) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ α1X + X β1Z1 with probability 1 − p, α2X + X β2Z2 with probability p. where α1 > α2. Our quantile regression model is given by qτ(x) ∼ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ c1 + α1x + xβ1z if τ > p, c2 + α2x + xβ2z if τ < p,

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My model

We assume a mixture HT model holds, i.e., Y ∣(X > u) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ α1X + X β1Z1 with probability 1 − p, α2X + X β2Z2 with probability p. where α1 > α2. Our quantile regression model is given by qτ(x) ∼ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ c1 + α1x + xβ1z if τ > p, c2 + α2x + xβ2z if τ < p, For stability, we fit simultaneously for τ = 0.05,0.15,...,0.95. We get 17 estimated parameters: (ˆ p, ˆ α1, ˆ α2, ˆ β1, ˆ β2, ˆ c1, ˆ c2, ˆ z1, ..., ˆ z10).

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My model

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X

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5 10 15

Y Asymmetric Logistic Model

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Estimating the number of components

2 3 4 5 6 7 8 9 10 11 X

  • 6
  • 4
  • 2

2 4 6 8 10 Y Best fit with 1 mixture(s)

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Estimating the number of components

2 3 4 5 6 7 8 9 10 11 X

  • 6
  • 4
  • 2

2 4 6 8 10 Y Best fit with 1 mixture(s) 2 3 4 5 6 7 8 9 10 11 X

  • 6
  • 4
  • 2

2 4 6 8 10 12 Y Best fit with 2 mixture(s)

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Estimating the number of components

2 3 4 5 6 7 8 9 10 11 X

  • 6
  • 4
  • 2

2 4 6 8 10 Y Best fit with 1 mixture(s) 2 3 4 5 6 7 8 9 10 11 X

  • 6
  • 4
  • 2

2 4 6 8 10 12 Y Best fit with 2 mixture(s) 2 3 4 5 6 7 8 9 10 11 X

  • 6
  • 4
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2 4 6 8 10 12 Y Best fit with 3 mixture(s)

Question: How can we compare?

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Estimating the number of components

2 3 4 5 6 7 8 9 10 11 X

  • 6
  • 4
  • 2

2 4 6 8 10 Y Best fit with 1 mixture(s) 2 3 4 5 6 7 8 9 10 11 X

  • 6
  • 4
  • 2

2 4 6 8 10 12 Y Best fit with 2 mixture(s) 2 3 4 5 6 7 8 9 10 11 X

  • 6
  • 4
  • 2

2 4 6 8 10 12 Y Best fit with 3 mixture(s)

Question: How can we compare? Method: 10-fold cross-validation.

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Estimating the number of components

1 2 3 4 5 6 7 8 9

Number of components

1.188 1.19 1.192 1.194 1.196 1.198 1.2 1.202 1.204 1.206

Cross-Validation Statistics

104

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Estimating the number of components

1.185 1.19 1.195 1.2 1.205 1.21

CV statistics density

104 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 9 8 7 6 5 4 3 2 1

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Assessing the model fit

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X

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5 10 15 20

Y

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Assessing the model fit

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X

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5 10 15 20

Y

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Assessing the model fit

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X

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5 10 15 20

Y

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Assessing the model fit

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5 10 15 20

Y

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Assessing the model fit

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Y

Method ˆ p 95% confidence interval Simulation Quantile Regression

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Assessing the model fit

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X

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5 10 15 20

Y

Method ˆ p 95% confidence interval Simulation 1.11 ⋅ 1e − 5 (1.00, 1.21) ⋅ 1e − 5 Quantile Regression 0.90 ⋅ 1e − 5 ??

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Problems

Is this method already perfect?

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Problems

Is this method already perfect? No, there are just a couple of minor issues:

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Problems

Is this method already perfect? No, there are just a couple of minor issues:

1 Cross-Validation statistics are not necessarily convex;

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Problems

Is this method already perfect? No, there are just a couple of minor issues:

1 Cross-Validation statistics are not necessarily convex; 2 Not trivial how to fit this framework into a Bayesian setting;

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Problems

Is this method already perfect? No, there are just a couple of minor issues:

1 Cross-Validation statistics are not necessarily convex; 2 Not trivial how to fit this framework into a Bayesian setting; 3 Not trivial how to quantify uncertainty;

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Problems

Is this method already perfect? No, there are just a couple of minor issues:

1 Cross-Validation statistics are not necessarily convex; 2 Not trivial how to fit this framework into a Bayesian setting; 3 Not trivial how to quantify uncertainty; 4 Overfitting is not really penalised;

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Problems

Is this method already perfect? No, there are just a couple of minor issues:

1 Cross-Validation statistics are not necessarily convex; 2 Not trivial how to fit this framework into a Bayesian setting; 3 Not trivial how to quantify uncertainty; 4 Overfitting is not really penalised; 5 It is not clear how to efficiently extend it to d-dimensions;

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Problems

Is this method already perfect? No, there are just a couple of minor issues:

1 Cross-Validation statistics are not necessarily convex; 2 Not trivial how to fit this framework into a Bayesian setting; 3 Not trivial how to quantify uncertainty; 4 Overfitting is not really penalised; 5 It is not clear how to efficiently extend it to d-dimensions; 6 It is not very quick;

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Problems

Is this method already perfect? No, there are just a couple of minor issues:

1 Cross-Validation statistics are not necessarily convex; 2 Not trivial how to fit this framework into a Bayesian setting; 3 Not trivial how to quantify uncertainty; 4 Overfitting is not really penalised; 5 It is not clear how to efficiently extend it to d-dimensions; 6 It is not very quick; 7 It is only suited for picking up mixtures that fan out;

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Problems

Is this method already perfect? No, there are just a couple of minor issues:

1 Cross-Validation statistics are not necessarily convex; 2 Not trivial how to fit this framework into a Bayesian setting; 3 Not trivial how to quantify uncertainty; 4 Overfitting is not really penalised; 5 It is not clear how to efficiently extend it to d-dimensions; 6 It is not very quick; 7 It is only suited for picking up mixtures that fan out; 8 Standard tests for testing number of mixture components based on

the likelihood is not (directly) applicable;

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Problems

Is this method already perfect? No, there are just a couple of minor issues:

1 Cross-Validation statistics are not necessarily convex; 2 Not trivial how to fit this framework into a Bayesian setting; 3 Not trivial how to quantify uncertainty; 4 Overfitting is not really penalised; 5 It is not clear how to efficiently extend it to d-dimensions; 6 It is not very quick; 7 It is only suited for picking up mixtures that fan out; 8 Standard tests for testing number of mixture components based on

the likelihood is not (directly) applicable;

9 Fitting a Heffernan-Tawn mixture model using Dirichlet processes

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Detecting mixtures in multivariate extremes

S.H.A. Tendijck

Lancaster University January 31, 2020

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Estimating the change point

How to estimate the change point using only 10 quantile levels? For example, assume that it is optimal to split up the quantile levels according to: {0.05,0.15,...,0.45},{0.55,0.65,...,0.95} We have then estimates (α1,β1,c1) for the first set, and (α2,β2,c2) for the second set. Define Q = {qz ∶ qz(x) = c1 + α1x + xβ1z, or qz(x) = c2 + α2x + xβ2z, z ∈ R}. Pick p = 0.5;

1 Choose the q ∈ Q such that q represents best the pth quantile of Y

given X;

2 Find out to which of the models the fitted regression model belongs; 3 Next iteration: estimate the change point in [0.45,0.5] or

[0.5,0.55].

4 Repeat

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Estimating the change point

0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 quantile 20 40 60 80 100 120 140 Change point estimate | 1 change point Meth1 Meth2

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