Reliable Inference in Conditions of Extreme Events. Adriana Cornea

Transcription

1 Reliable Inference in Conditions of Extreme Events by Adriana Cornea University of Exeter Business School Department of Economics ExISta Early Career Event October 17, 2012

2 Outline of the talk Extreme events & fat-tails Bootstrap Two papers joint with K. M. Abadir (Imperial College): Approximating moments by nonlinear transformations, with an application to bootstrapping for fat-tails Bootstrapping with fat-tailed asymmetry

3 Black Monday: Extreme event Black Monday, 19 Oct. 1987: Fall of more than 20% Average daily % change, Jan. 1 to Oct. 19, 1987: µ = 0.055% Standard deviation, Jan. 1 to Oct. 19, 1987: σ = 0.95%

4 Assume normal distribution N(µ, σ 2 ) normal density f(x) µ = 0, σ 2 = 0.2 µ = 0, σ 2 = 1.0 µ = 0, σ 2 = 5.0 µ = 2, σ 2 = x Normal densities

5 µ = 0.055%, σ = 0.95% Pr(x < 22) How small is ? (Stock and Watson (2007)) The world population is about 6 billion. Probability of winning a lottery among all living people is The universe has existed for 15 billion years, or about sec. Probability of choosing a particular second at random from all seconds since the beginning of time is There are approximately molecules of gas in the first kilometer above the earth s surface. Probability of choosing one is It is extremely unlikely that the return distribution is normal. Tails of the return distribution: much fatter (heavier) than those of the normal distribution.

6 Normal distribution N(0, 1) is extremely light-tailed Pr( X > x) 1 2πxe x 2 /2 (1) Many economic & financial variables are fat-tailed power laws Pr( X > x) C x α (2) α: tail index (fat-tailedness measure) E X p exists for p < α E X p does not exist for p α Empirics: economics losses from earthquakes (α (0.6, 1.5)), income (α (1.5, 3)), wealth (α 1.5), returns on many stocks (α (2, 4) infinite fourth moment)

7 Light vs. fat tails Normal N(25,1) Asymmetric α stable, α=1.2 Lévy, α = 0.5 density f(x) x Tails of asymmetric α-stable distributions with α = 1.2 < 2 are fatter than those of the normal distribution (for which α = 2). Tails of Lévy distribution are fatter than those of the asymmetric α-stable distribution with α = 1.2.

8 Bootstrap Pulling yourself by the bootstraps. Illustration in the Dec. 04 issue of Significance

9 More serious bootstrap Suppose we have an i.i.d. sample: x = (x 1,, x n ) from an unknown distribution F (θ) Compute ˆθ and a statistic of interest t(n, ˆθ) for testing H 1 : θ = θ 0 vs. H 1 : θ θ 0 A simple bootstrap draws randomly with replacement from the empirical distribution function of x say B times Bootstrap sample x j = (x n 1, x 1, x 5, x 10,, x 5 ), j = 1,, B For each x j compute ˆθ j and t j (n, ˆθ j) Reject H 0 if 1 ( N B j=1 I t j (n, ˆθ j) < t(n, ˆθ) ) is smaller than 0.05 or 0.10 Or build a 90% confidence interval for θ with limits ˆθ (ˆθ 0.95 ˆθ), ˆθ (ˆθ 0.05 ˆθ)

10 Different bootstraps Naive or nonparametric bootstrap: i.i.d. Wild bootstrap: heteroskedasticity Block bootstrap, subsampling: autocorrelation M out of n bootstrap, subsampling: non-smooth statistics (max(x)), x has an infinite variance distribution Rich literature: Efron (1979), Godfrey (2009), Shao & Tu (1995), Politis, Romano & Wolf (1999), Good (1994), Efron & Tibshirani (1993) and many more

11 Bootstrap and fat-tails Let s take θ = E(x) and assume x F unknown and var(x) = We want to build a 90% CI about E(x) Naive bootstrap not valid if var(x) = : Athreya (1989), Knight (1989) Previous work: m out of n bootstrap, subsampling: Politis, Romano & Wolf (1999) parametric bootstrap: Cornea & Davidson (2011) Nothing works if tails are fat and asymmetric

12 An idea Take any transformation of x, x = g(y) such that var(y) < For simplicity we take x = exp(y) Suppose x has a Pareto distribution, F x (u) = 1 u α, 1 < α < 2 Then, F y (w) = 1 e αw E(x) = α/(α 1), E(y) = 1/α, hence E(x) = 1/(1 E(y)) Upper/lower limits of 90% bootstrap CI for E(x) are x n ( x 0.95 x n ), x n ( x 0.05 x n ) ( ) ( ) x 0.95 := 1/ 1 ȳ(b+1)(0.95), x 0.05 := 1/ 1 ȳ(b+1)0.05

13 Some simulations to illustrate Naive bootstrap M out of n bootstrap α = α = α = Table 1: Bootstrap coverage probabilities for E(x) without transformation; B = 399; 10,000 replications

14 Naive bootstrap α = α = α = Table 2: Naive bootstrap coverage probabilities for E(x) using the transformation; B = 399; 10,000 replications

15 Relaxing the assumptions In reality we do not know F x and F y and the link between E(x) and E(y) We can use power series expansions of exp(y) Raw expansion Centered expansion x = k j=0 y j j! + R k (3) k x = e E y e y E y = e E y (y E y) j j! j=0 + R c k (4) Higher-order terms create the problems when x has a fat-tailed distribution Bounding R k by a low power term will (hopefully) solve the problem

16 A crazy expansion Let 1 i 2 and (y E y) /m ζ + 2πi y, where i y Z, m N, and ζ ( π, π] Then we have the expansion x = e E y e 2πmiy (exp ( )) ζ im e E y e ( ) 2πmiy im ξ i k + ϱ x,k ξk := k j=0 ζj /(i j j!) iy is random, but m is deterministic and to be chosen later Binomial expansion gives x = e E y e 2πmiy Re(ξ im k ) + Rc x,k

17 This expansion allows us to conclude that Rx,k c = O ( p ζ k+1 ) And find an accurate bound for Rθ,k c, θ = E(x) R c θ,k [ e E y E e ( )] 2πmiy ξ im ζ k+1 k H (k + 1)! ξ k Denote ψ = ζ k+1 (k+1)! ξ k Where 1 2e m sin 1 ψ cos (m log (1 ψ )) + e 2m sin 1 ψ H( ψ ) := 1 + e m sin 1 ψ 1 + e mπ [ ) 0, 1 e π/m [ for ψ 1 e π/m, 1 ] (1, ) respectively

18 Application: transformation-based bootstrap Letting y := log(x) and z := e E y e 2πmiy Re(ξ im k ), we have x = z + Rx,k c and the remainder has the bound Bc x,k By applying the triangle inequality twice, z B c x,k x z + Bc x,k which can be used to build conservative CIs for E(x) To do this, consider an i.i.d. sample x 1,, x n and compute x n := 1 n n x j, j=1 z + n := 1 n n z j, B c x := 1 n,k n j=1 n j=1 B c x j,k By the triangle inequality, t 1,n x n t 2,n, t 1,n := z + n B c x n,k, t 2,n := z + n +B c x n,k

19 We can bootstrap t 1,n and t 2,n instead of x n for appropriate choice of k and m If Bx,k c is too large then the CI is too conservative (we don t want that). If k or m, then Bx,k c vanishes and z coincides with x and we are back to the original invalid bootstrap. Thus k, m have to be finite and their value chosen depending on the thickness of the tail of x, α.

20 In practice, first estimate α (using for instance Hill (1975) method) Then, for extreme quantile (99%) take k = 1 and an estimate of m is given by the integer part of exp ( n 1/ α α 2)

21 n = 100 n = 1000 m m α = α = α = Table 3: Transformation-based bootstrap coverage probabilities for E(x), k = 1, B = 399, 10,000 replications; data from Pareto(α)

22 For lower quantiles take k = 2, an estimate of m is given by the integer part of exp ( n 1/2 1.16α n 1 α 2) exp (40.36n 1/2 sign. level 1)

23 n = 100 n = 1000 m m α = α = α = Table 4: Transformation-based bootstrap coverage probabilities for E(x), k = 2, B = 399, 10,000 replications; data from Pareto(α)