Ústav teorie informace a automatizace RESEARCH REPORT. Andrea Karlová: Stable distributions: On densities. No October 2011

Transcription

1 Akademie věd České republiky Ústav teorie informace a automatizace Academy of Sciences of the Czech Republic Institute of Information Theory and Automation RESEARCH REPORT Andrea Karlová: Stable distributions: On densities No. 237 October 211 ÚTIA AV ČR, P. O. Box 18, Prague, Czech Republic Telex: atom c, Fax: (+42) (2) utia@utia.cas.cz

2 This report constitutes an unrefereed manuscript which is intended to be submitted for publication. Any opinions and conclusions expressed in this report are those of the author(s) and do not necessarily represent the views of the Institute. 2

3 Stable Distributions: On Densities. Andrea Karlová September 26, 211 Contents 1 Introduction 2 2 Basic Properties 2 3 Integral Representation 5 4 Representations via Expansions 7 5 Incomplete Gammma Function 9 References 12 1

4 1 Introduction William Feller stated in 1] that densitites of stable distributions are probably not known. V.M. Zolotarev spent a lot of effort on deriving aproximative formulas via series expansions and integral representations for stable densities. In this report we investigate derivation of stable densities and stable distribution function. We further investigate relation between stable distribution functions and incomplete gamma functions. We define stable distribution using its characteristic exponent. We use parametrization discussed in Karlová?]. Consider probability distribution F with its characteristic function ˆF. We will call F stable distribution iff its characteristic function is of the following form: where ˆF (k) = e ψ(k), with c >, p 1, p 2 1 and p 1 + p 2 = 1. 2 Basic Properties ψ(k) = c k e i(p1 p2) 2 sgn k, (1.1) Let us start with expressing stable density via inverse Fourier transformation: f(x;, p 1 ) = 1 e ikx c k e i(p 1 p 2 ) 2 sgn k dk. (2.1) We denote the integrand as g(k, p 1 ). Next, we show simple, yet very useful, property of stable density: For x : f(x;, p 1 ) = 1 = 1 = f( x;, p 2 ). f(x;, p 1 ) = f( x,, p 2 ). (2.2) ] e ikx ck e i(p 1 p 2 ) 2 + e ikx ck e i(p 1 p 2 ) 2 dk = ] e ik( x) ck e i(p 2 p 1 ) 2 + e ik( x) ck e i(p 2 p 1 ) 2 dk = We conclude that g(k, x; p 1 )+g(k, x; p 1 ) = g(k, x; p 2 )+g(k, x; p 2 ). Therefore without loss of generality we can consider x and formulate consequent relation: f(x;, p 1 ) = 1 Re e ikx ck e i(p 1 p 2 ) 2 dk = (2.3) = 1 Re e ikx ck e i(p 1 p 2 ) 2 dk. 2

5 Let us compute real and imaginary part of the integrand g(k, x; p 1 ). Assume k > : g(k, x; p 1 ) = e ikx ck e i(p 1 p 2 ) 2 = Thus we conclude, that: f(x,, p 1 ) = 1 = e ck cos(p 1 p 2) 2 ] coskx ck sin(p 1 p 2 )/2]] ie ck cos(p 1 p 2) 2 ] sinkx ck sin(p 1 p 2 )/2]]. e ck cos(p 1 p 2) 2 ] coskx ck sin(p 1 p 2 )/2]]dk. (2.4) From previous, we can compute f(,, 1/2) by direct integration. We use substitution t = ck and obtain: f(,, 1/2) = c1/ t 1 1 e t dt = 1 c1/ Γ + 1 ]. By assuming p 1 = p 2, the integrand became substantially simplified. Direct computation of f(;, p 1 ) in case of p 1 p 2 is, however, considerably more complicated. We make use of following lemma. Lemma 2.1. For < θ < 2 and 1 < < 2: e k cos θ cosk sin θ]dk = Γ ] cos θ, e k cos θ sink sin θ]dk = Γ ] sin(θ + ). Preliminarily, we start with notation. Having two contours L 1 : a 1, b 1 ] C, L 2 : a 2, b 2 ] C with L 1 (b 1 ) = L 2 (a 2 ), we denote the oriented sum of contours as L 1 L 2. The orientation of the contour is denoted by and so by L 2 we denote contour with oposite orientation to contour L 2. If the countour L is closed, we denote the its interior as intl. Next, we prove the lemma. Proof. Consider a complex valued function g(z) = e z e iθ dz, for z C. For < r 1 < r 2 < we define a contour L as: L = L 1 L 2 L 3 L 4, where (2.5) L 1 (t) = t for t r 1, r 2 ]; L 2 (t) = r 2 e it for t, θ]; L 3 (t) = te iθ for t r 1, r 2 ]; L 4 (t) = r 1 e it for t, θ]. Next, we integrate g(z) along contour L: r2 θ e z e iθ dz = e t e iθ dt + i r 2 e it r 2 ei(t+θ) dt L r 1 r2 r 1 e iθ e t dt i θ r 1 e it r 1 ei(t+θ) dt. 3

6 Because contour L is closed and integrand g(z) is regular everywhere in interior of contour L, we can apply Cauchy Theorem and obtain: e z e iθ dz =. L Next, we estimate values of integrals over arcs L 2 and L 4. Let us start with arc L 2. We make use of relation exp z = exp (Re z). θ θ ir 2 e it r 2 ei(t+θ) dt = r 2 ( sin t + i cos t)e r 2 cos(t+θ)+i sin(t+θ)] dt Because... r 2 θ e r 2 cos(t+θ) dt = r 2 θ r 2 e r 2 cos(t+θ) dt. r 2 (+1)θ θ (+1)θ e r 2 cos(y) dy = r 2 (+1)θ e r 2 y dy < 1 e r 2 (+1)θ r 1 2 θ e r 2 sin(y) dy, for r 2. We proved that integrals along arcs L 2, L 4 converges to, which implies following equity: e t e iθ dt = e iθ 1 ] e t dt = e iθ Γ + 1. (2.6) Finally, we compare real and imaginary part of later equity: e t cos 1 ] θ cost sin θ]dt = Γ + 1 cos θ, e t cos 1 ] θ sin t sin θ]dt = Γ + 1 sin(θ + ), where we used relation sin θ = sin(θ + ). Q.E.D Returning to problem of computation f(,, p 1 ) for any admissible p 1, we use Lemma 1.1 with θ = (p 1 p 2 ) 2 and substitute t = ck. We obtain: f(;, p 1 ) = 1 c1/ Γ + 1 ] cos(p 1 p 2 )/2]. (2.7) The previous discussion served us as a motivation for investigating analytic continuation of integrand g(k,, p 1 ) in complex plane. We also realized the importance of choice of suitable contours for evaluation of the integral. Analytic extension and choice of contour are closely contected. 4

7 Let us conclude this section with a useful property of stable distribution function. Using (2.2), we obtain: F (x;, p 1 ) = x f(y;, p 1 )dy = = 1 F ( x;, p 2 ), and thus F (x;, p 1 ) + F ( x;, p 2 ) = 1. 3 Integral Representation x f(y;, p 2 )dy = A density of stable distribution in the form of integral representation is: for x > f(x;, β) = x1/( 1) ( 1) /2 β/ e x/( 1) U(ϕ;β) U(ϕ; β)dϕ. (3.1) Using the following relation: ] e x/( 1) U(ϕ;β) = x 1 x1/( 1) e x /( 1) U(ϕ;β) U(ϕ; β), (3.2) we have equality: f(x;, β) = x1/( 1) ( 1) /2 β/ e x/( 1) U(ϕ;β) U(ϕ; β)dϕ = 1 /2 β/ Using the symmetry f(x;, β) = f( x;, β), we have for x < : For x = : f(x;, β) = x 1/( 1) ( 1) /2 β/ ] e x/( 1) U(ϕ;β) dϕ. x (3.3) e x /( 1) U(ϕ; β) U(ϕ; β)dϕ. (3.4) f(;, β) = 1 Γ(1 + 1/) cos(β/). (3.5) Let us verify that f(x;, β)dx = 1: f(x;, β)dx = = f(x;, β)dx + f(x;, β)dx + f(x;, β)dx = f(x;, β)dx. f(x;, β)dx + f( x;, β)dx = (3.6) 5

8 Using (3.2) and Fubini theorem: f(x;, β)dx = ( 1 /2 /2 β/ ] ) e x/( 1) U(ϕ;β) dϕ dx = 1 x /2 β/ ] dϕ e x/( 1) U(ϕ;β) dx = x = 1 dϕ = β and f(x;, β)dx = β/ ( 1 /2 β/ x ] ) e x/( 1) U(ϕ; β) dϕ dx = 1 The distribution function of stable distribution is: for x > where x and so f(y;, β)dy = For x < F (x;, β) = F (x;, β) = x = 1 x ( 1 /2 β/ x /2 β/ f(y;, β)dy = 1 x f(y;, β)dy /2 β/ ] ) e y/( 1) U(ϕ;β) dϕ dy = (3.7) y ] dϕ e y/( 1) U(ϕ;β) dy = 1 x y F (x;, β) = 1 1 f(y;, β)dy = x /2 β/ /2 β/ e x/( 1) U(ϕ;β) dϕ. (3.8) f( y;, β)dy = x f(y;, β)dy = dϕ = 1 2 β. e x/( 1) U(ϕ;β) dϕ = 1 /2 e x /( 1) U(ϕ; β) dϕ = 1 F ( x;, β). (3.9) β/ We simply deduce that F (x;, β) + F ( x;, β) = 1. For x = : F (;, β) = 1 2 β (3.1) 6

9 4 Representations via Expansions In this section we derive aproximative formulas for stable densities and stable distribution functions. Starting with relation (2.3), the integrand can be continously extended into complex plain and we denote it as g(z; p 1 ). We see that complex valued function g(z; p 1 ) is holomorphic. For < r 1 < r 2 < we define contour L as: L L = L 1 L 2 L 3 L 4, where (4.1) L 1 (t) = t for t r 1, r 2 ]; L 2 (t) = r 2 e it for t, (p 1 p 2 ) 2 ]; L 3 (t) = te i(p1 p2) 2 for t r1, r 2 ]; L 4 (t) = r 1 e it for t, (p 1 p 2 ) 2 ]. Let us integrate g(z; p 1 ) along contour L: e izx z e i(p 1 p 2 ) 2 dz = + i i r2 exp{ itx t e i(p1 p2) 2 }dt+ r 1 (p1 p 2) 2 r2 r 1 (p1 p 2) 2 r 2 exp{ ir 2 e it x r 2 e it i(p1 p2) 2 }e it dt e i(p1 p2) 2 exp{ ite i(p 1 p 2) 2 x t }dt r 1 exp{ ir 1 e it x r 1 e it i(p1 p2) 2 }e it dt. Because contour L is closed and integrand g(z, p 1 ) is holomorphic, we can apply Cauchy Theorem which gives us: e izx z e i(p 1 p 2 ) 2 dz =. L For r 1, r 2, the integrals along arcs L 2, L 4 converges to and we get following equity: e itx t e i(p 1 p 2 ) 2 dt = e i(p1 p2) 2 e ite i(p 1 p 2 ) 2 x t dt. (4.2) Using above equity, we can express stable density in (2.3) as: f(x;, p 1 ) = 1 Re e i(p1 p2) 2 e ite i(p 1 p 2 ) 2 x ct dt. (4.3) Let us substitute t = y 1/ : f(x;, p 1 ) = 1 Re e i(p1 p2) 2 e iy 1/ e i(p 1 p 2 ) 2 x e cy y 1 dy (4.4) 7

10 and expand exp{ iy 1/ e i(p1 p2) 2 x} into series: e iy1/ e i(p 1 p 2 ) 2 x iy 1/ e i(p1 p2) 2 x] k = k! k= y k/ e ipk x k =. k! k= We use linearity of integral and rewrite (4.4) as: f(x;, p 1 ) = 1 Re = 1 = 1 k= k=1 k= x k e ipk+i(p1 p2) 2 k! y k+1 1 e cy dy = x k cos( p 2 k + (p 1 p 2 ) k+1 2 )Γ( + 1) = (k + 1)!c (k+1)/ sin(p 2 k)γ(k/ + 1) k!c k/ x k 1. Using the relation we can express later sum via gamma functions as: f(x;, p 1 ) = k=1 Γ(k/ + 1) x k 1. (4.5) Γ(k + 1)Γ(p 2 k)γ(1 p 2 k) ck/ To determine expansion for stable distribution function, we can can integrate previous series: F (x;, p 1 ) F (;, p 1 ) = and so: = F (x;, p 1 ) = p 2 + k= x k=1 f(y;, p 1 ) = 1 k=1 Γ(k/ + 1) Γ(k + 1)Γ(p 2 k)γ(1 p 2 k) k=1 sin(p 2 k)γ(k/ + 1) k!kc k/ x k = x k kc k/ p 2 Γ(k/ + 1) x k Γ(k + 1)B(1 + p 2 k, 1 p 2 k) c = k/ Γ(k/ + 1) x k = p 2 Γ(k + 1)B(1 + p 2 k, 1 p 2 k) c. k/ Let us summerize the results into the lemma: Lemma 4.1. For stable density f(x;, p 1 ) and x > the following holds: f(x;, p 1 ) = 1 k=1 sin(p 2 k)γ(k/ + 1) k!c k/ x k 1. (4.6) Let us note that first part of the lemma is summerized Feller XVII.7 Lemma 1, p.583 in 1]. The method how the result is derived is different from the one presented. 8

11 5 Incomplete Gammma Function Recall that Gamma function, a natural extension of factorial, is defined by integral: Γ(a) = x a 1 e x dx, (5.1) where a C, Ra >. Having space of test functions S, ϕ(x) = e x S for x R + and so we can represent gamma function as: Γ(a) = x 1 1 {x>}, e x. (5.2) A very natural idea is investigation of integrals: where we denote Γ(a) = x y a 1 e y dy + γ(a, x) = Γ(a, x) = x x x y a 1 e y dy, (5.3) y a 1 e y dy, (5.4) y a 1 e y dy, (5.5) and call functions Γ(a, x), γ(a, x) incomplete gamma functions. Expansion for incomplete gamma function: a 1 Γ(a, x) = (a 1)!e x k= x k, k >. (5.6) k! Lemma 5.1. Random variable X with Poisson distribution with parameter a > has probability distribution in terms of incomplete Gamma function as follows: P {X x} = Γ(a + 1, x) Γ(a + 1). (5.7) Lemma 5.2. Random variable X with standart Gaussian distribution has a probability distribution in terms of incomplete Gamma function as follows: Γ( 1 2, x2 2 ) 2 P {X x} = for x, 1 Γ( 1 2, x 2 (5.8) 2 ) 2 for x >. Proof. Starting from Gaussian probability distribution function we use substitution to get expression via incomplete gamma function. Assume first x > : P {X x} = 1 x e y2 1 2 dy = 1 x e y2 2 dy 9

12 Incomplete Gamma Function Γ(1/2,x) Γ(1/2,x 2 ) Γ(1/2,x 2 /2) Γ(1/2,x 2 /2) Γ(1/2,x) Γ(1/2,x 2 ) Γ(1/2,x 2 /2) Γ(1/2,x 2 /2) Figure 1: Incomplete Gamma Function Substitute z = y2 dz 2, then dy = 2z and: For x P {X x} = x 2 /2 P {X x} = 1 x z 1 2 e z dz = 1 Γ( 1 2, x 2 2 ) 2. e y2 1 2 dy = x e y2 2 Γ( 1 dy = 2, x 2 2 ) 2. Q.E.D Lemma 5.3. Random variable X with Lévy distribution has a probability dis- 1

13 tribution function in terms of incomplete Gamma function as follows: Proof. P {X x} = Γ( 1 2, 1 2x ). (5.9) P {X x} = 1 x Substitute z = 1 dz 2y, then dy = 2z 2 P {X x} = 1 1/2x and: y 3 2 e 1 2y dy (2z) 3 2 2z 2 e z dz = 1 1/2x z 1 2 e z dz = Γ( 1 2, 1 2x ). Definition 5.1. Class of Compound Incomplete Gamma Functions: Q.E.D Γ(, f(x)) P {X x} = g(x) Γ() (5.1) Let us derive characteristic function: d Γ(, f(x)) (5.11) dx 11

14 References 1] Feller W. (1971). An Introduction to Probability Theory and Its Applications, Vol. 2, 2nd Edn., Wiley. 2] Orey S. (1968). On Continuity Properties of Infinitely Divisible Distribution Functions. Annals of Mathematical Statistics, Vol. 39, No 3, pp ] Zolotarev V.M. (1986). One-dimensional stable distributions. Amer. Math. Soc., Providence, RI. 12