Approximation of Heavy-tailed distributions via infinite dimensional phase type distributions

Transcription

1 1 / 36 Approximation of Heavy-tailed distributions via infinite dimensional phase type distributions Leonardo Rojas-Nandayapa The University of Queensland ANZAPW March, Barossa Valley, SA, Australia.

2 Summary of Mogens talk... 2 / 36

3 3 / 36 Phase type Distributions of first passage times associated to a Markov Jump Process with a finite number of states. Facts 1. Dense class in the nonnegative distributions. 2. Mathematically tractable. 3. However, it is a subclass of light tailed distributions.

4 4 / 36 Key problem For most applications, a phase type approximation will suffice. However, for certain important problems this fail. A partial solution Consider Markov Jump processes with infinite number of states. This class contains heavy tailed distribution but also many other degenerate cases.

5 5 / 36 A more refined approach: Infinite mixtures of PH Consider X a random variable with phase type distribution, and let S be a nonnegative discrete random variable. Then the distribution of the random variable Y := S X is an infinite mixture of phase type distributions. Such a class is mathematically tractable and obviously dense in the nonnegative distributions.

6 6 / 36 Let F, G and H be the cdf of Y, S and X respectively. Then F (y) = 0 G(y/s)dH(s), y 0. The right hand side is called the Mellin Stieltjes transform. If G is phase type, the F is absolutely continuous with density f (y) = 0 g(y/s)dh(s), y 0.

7 Some interesting questions related to this class of distributions? 7 / 36

8 8 / What are the possible tail behaviors in this class? 2. How to approximate a theoretic heavy tailed distribution? 3. Given a sample y 1,..., y n. How to estimate the parameters of a distribution in this class?

9 Heavy Tails 9 / 36

10 10 / 36 Recall that... Heavy tailed distributions A nonnegative random variable X has a heavy tailed distribution iff E [e θx ] =, θ > 0. Equivalently, lim sup x Otherwise we say X is light tailed. P(X > x) e θx =, θ > 0.

11 11 / 36 Characterization via Convolutions Let X 1, X 2 be nonnegative iid rv with unbounded support, 1. It holds that lim inf x P(X 1 + X 2 > x) P(X 1 > x) Moreover, X 1 is heavy tailed iff lim inf x P(X 1 + X 2 > x) P(X 1 > x) = Furthermore, X is subexponential iff P(X 1 + X 2 > x) lim = 2. x P(X 1 > x)

12 12 / 36 Heavy tailed random variables arise as Exponential transformations of some light tailed random variables (normal lognormal, exponential Pareto). As weak limits of normalized maxima (also normalized sums). As reciprocals of certain random variables. Random sums of light tailed distributions. Products of light tailed distributions.

13 Problem 1: What are the possible tail behaviours in this class? 13 / 36

14 14 / 36 Theorem Let X be a phase type distribution and S be a nonnegative random variable. Define Y := S X Y is heavy tailed iff S has unbounded support.

15 15 / 36 Example Let X exp(λ) and S any nonnegative discrete distribution supported over a countable set {s 1, s 2,... }. P(S X > y) k=1 lim y e θy = lim e λy/s k P(S = s k ) y e θy =. (choose s k large enough and such that λ/s k < θ).

16 16 / 36 Theorem (generalization) Let S H 1 and X H 2, and let Y := S X. 1. If there exist θ > 0 and ξ(x) a nonnegative function such that lim sup e θx( ) H 1 (x/ξ(x)) + H 2 (ξ(x)) = 0, (1) x then Y is light tailed. 2. If there exists ξ(x) a nonnegative function such that for all θ > 0 it holds that lim sup x e θx H 1 (x/ξ(x)) H 2 (ξ(x)) =, (2) then Y is heavy tailed.

17 17 / 36 Example Let S Weibull(λ, p) and Y Weibull(β, q). Then Y := S X is light tailed iff 1 p + 1 q 1. Otherwise it is heavy tailed.

18 What are the possible maximum domains of attraction? 18 / 36

19 19 / 36 Frechét domain of attraction Breiman s Lemma If S is a α regularly varying random variable and there exists δ > 0 such that E [X α+δ ] <, then Y := S X has an α regularly varying distribution. Moreover P(Y > x) E [X]P(S > x).

20 20 / 36 Breiman s Lemma If L 1/S is a α regularly varying function then Y := S X has an α regularly varying distribution. Moreover P(Y > x) E [X]P(S > x).

21 Gumbel domain of attraction and Subexponentiality Let V (x) = ( 1) η 1 L (η 1) 1/S (x).1 Theorem If V ( ) is a von Mises function, then F MDA(Λ). Moreover, if lim inf x then F is subexponential. V (tx)v (x) V > 1, t > 1, (tx)v (x) 1 η is the largest dimension among the Jordan blocks associated to the largest eigenvalue of the sub intensity matrix 21 / 36

22 Problem 2: How to approximate theoretical heavy tailed distributions? 22 / 36

23 23 / 36 An approach for approximating theoretic distributions Take a sequence of Erlang rvs X n Erlang(n, n). Consider increasing sequences S n := {s n,0, s n,1, s n,2,... }, n N such that s n,0 = 0, s n,i as i and n := sup{s n,i+1 s n,i : i N} 0, as n 0 For each n, define the cdf s F n (x) = F(s n,i+1 ), x [s n,i, s n,i+1 ).

24 24 / 36 Construct a sequences of random variables S n F n, X n Erlang(n, n). Since X n 1, then Slutsky s theorem implies that S n X n X F. Note: It can be modified and will also work for light tailed distributions.

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26 26 / 36 Ruin problem Definition Let F and G be two distribution functions with non-negative support. The relative error from F to G from 0 to s is defined by sup x [0,s] F(x) G(x). F(x)

27 27 / 36 Theorem Fix u > 0 and let F and G be such that F I (x) G I (x) sup ɛ(u) x [0,u] F I (x) where ɛ : R + R + is some non-decreasing function. Let F I be the integrated tail of F and let G and approximation of F. Further assume that F dominates G stochastically. Then ψ F (u) ψ G (u) ɛ(u)e (Number of down-crossings; A), where A := {Ruin happens in the last down-crossing}.

28 Problem 3: Statistical inference? 28 / 36

29 Key Idea See the first passages times as incomplete data set from the evolution of the Markov Jump Process. Employ the EM algorithm for estimating the parameters. 29 / 36

30 30 / 36 Density of an infinite mixture of PH f A (y) = π i (θ)αe T y/s i t/s i, x 0. i=1 Complete Data Likelihood l c (θ, α, T ) = L i log π i (θ) + i=1 i=1 k l T kl s i Z i k + i=1 k=1 i=1 k=1 p Bk i log α k + p A i k log ( tk s i ) i=1 k l i=1 k=1 ( ) Nkl i log Tkl s i p t k s i Z i k.

31 EM estimators Assume (θ, α, T ) is the current set of parameters in the EM algorithm. The one step EM-estimators are thus given by θ = argmax Θ (α U i t) log π i (Θ), i=1 diag (α)u t α =, N T kl = ( diag (R) 1 T R) kl, k l, t = α U diag (t) diag (R) 1, where R = R(θ, α, T ) = U i = U i (θ, α, T ) = N i=1 j=1 N j=1 π i (θ) s i ( π i (θ) J yj /s i ; α, T, t ) s i f A (y j ; θ, α, T ), exp ( T y j s i ) f A (y j ; θ, α, T ), 31 / 36

32 32 / 36 In the above J i (y) := y 0 e T i (y u) t i αe T i u du. Proposition Let c = max{t kk : k = 1,..., p} and set K := c 1 T + I. Define D(s) := D(s; α, T ) = 1 c s K r tαk s r r=0 Then J(y; θ, α, T ) = s=0 (cy) s+1 e cy D(s). (s + 1)!

33 33 / 36 An alternative model Let r (0, 1). f B (y) = r αe T y t + (1 r) i=1 π i (θ) λq i y q 1 e λ i y, Γ(q 1)

34 34 / 36 Proposition The EM estimators take the form r = rαut N α = diag (α)ut αut T kl = ( diag (R) 1 T R ) kl t = αu diag (t) diag (R) 1 θ = argmax Θ w i log π i (Θ) λ = i=1 i=1 i=1 / v i w i. s i

35 35 / 36 where R = R(α, T, λ, θ) = U = U(α, T, λ, θ) = N j=1 N j=1 w i = w i (α, T, λ, θ) = π i (θ) v i = v i (α, T, λ, θ) = π i (θ) J ( y j ; α, T, t ) f B (y j ; r, α, T, θ, λ). exp ( T y j ) f B (y j ; r, α, T, θ, λ) N j=1 N j=1 g(y j ; q, λ/s i ) f B (y j ; r, α, T, θ, λ) y j g(y j ; q, λ/s i ) f B (y j ; r, α, T, θ, λ)

36 36 / 36 Advantages Fast. Recovers the parameters of simulated data. Works reasonable well with real data. Converges to the MLE estimators most of the times. Provides a better approximation for the tails. Disadvantages There are some (minor) numerical issues yet to be resolved. MLE estimators are not appropriate for estimating tail parameters. EVT methods could be implemented. Sensitive with respect to the tail parameters. Kullback-Leibler does not work properly. Though we now have the first simpler approach.