Asymptotic Expansions of the Stable Distribution

Karina Arias-Calluari[1],∗ Fernando Alonso-Marroquin [1,2], and Michael Harre [1]

[1]School of Civil Engineering, University of Sydney, Sydney NSW 2006 and[2]Computational Physics IfB, ETH Zurich, CH-8093, Zurich, Switzerland

(Dated: December 7, 2017)

The stable distribution is the attractor of distributions which hold power laws with infinite variance. This function doesnot have an explicit expression and no uniform solution has been proposed yet. This paper presents a uniform analyticalapproximation for the stable distribution based on matching power series expansions. For this solution, the trans-stable functionis defined as an auxiliary function to evaluate the stable distribution. The trans-stable function removes the numerical issuesof the calculations of the stable function and it presents the same asymptotic behaviour as the stable distribution function. Toobtain the uniform analytical approximation two types of approximations are developed. The first one is called “inner solution”and it is an asymptotic expansion around x = 0. The second refers to “outer solution” and it is a series expansion for x→∞.Then, the uniform solution is proposed as a result of an asymptotic matching between inner and outer solutions. Finally, theresults of analytical approximation were compared to the numerical results of the stable distribution function, making thisuniform solution valid to be applied as an analytical approximation.

I. INTRODUCTION

A wide range of natural and social phenomena ex-hibit a power-law in the probability distribution of largeevents. These tails are characterized by the asymptoticrelation f(x) ∼ 1/x1+α, where x is the size of the events.

For 0 < α ≤ 1, the distribution has an indefinite meanvalue. Examples that fall in this range are the affectedareas of bushfires [1], wildfires and rockfalls [2], intensityand duration of rains [3], neuron firing rates and neuralavalanches [4, 5], call holding times and packed inter-arrival data of the internet [6], war intensities [7] andpersonal identity losses on internet [8].

For 1 < α ≤ 2, the distribution has a defined meanvalue but still exihibits an infinite variance. A differentrange of examples have been found, such as landslideareas triggered by earthquakes [7, 9], heavy rains andsnowmelts [9], a probability distribution of monthly riverflows [10], daily returns in currency exchange rates [11],the effects of networks in price returns [12], daily returnsof Dow Jones index [11] and benefits and returns of topHollywood movies [13].

Section 35 of the book by Gnedenko and Kolmogorov[14] shows that the normal distribution is an “attractor”of distributions with finite variances. On the other hand,the attractor of distribution holding power laws with in-finite variances corresponds to the more general “stablelaw”.

The fundamental concept of attractors is formulated asfollows: If a normalized sum of a set of independent, iden-tically distributed random variables {X1, X2, X3....XN}satisfies:

limN→∞

1

σN

(N∑i=1

Xi − µN

)= X. (1)

Then X belongs to the stable law. The coefficients µN

∗ School of Civil Engineering; kari0293@uni.sydney.edu.au

and σN represent the centering and normalizing valuesrespectively [14].

The Gnedenko-Kolmogorov theorem is a generalizationof the classical central limit theorem which states thatnormalized sum of independent random variables withfinite variance in Eq. (1) converges to a variable that isnormally distributed [14, 15]. This is the case of distribu-tions with power-law tails with finite variance α ≥ 2. Thenormalized coefficient is σN =

√N and the centering co-

efficient is µN = NE[X], where N represents the lengthof the sum and E[X] refers to expected value [16, 17]. Onthe other hand, for independent random variables holdingpower laws with infinite variance1 0 < α < 2, Uchaikinand Zorotalev [17, 18] show that X in Eq. (1) follows asymmetric stable law if the normalization coefficient isσN = N1/α and the centering coefficient is µN = 0 forα ≤ 1 or µN = NE[X] for α > 1.

The stable distribution function is given by an integralthat has analytical solution only for few cases —normaland Cauchy distributions— in the remaining cases, itdoes not have a closed-form expression. The numericalsolution of the stable distribution has numerical oscilla-tions specifically in the tails. For some cases it displaysapparent discontinuity in logarithmic scale plots becauseof negative values obtained from the numerical solution[20]. Because of that, it can be said that the numericalsolution of the stable distribution is not reliable, becausethe probability density function must be always positive.

Analytical expressions in terms of power series werepresented by different authors. Feller [19], Elliot [21] andZolotarev [18] used power series to obtain converging al-gorithms of the stable distribution function in two ranges,the first one for α < 1 and the second for α > 1 forsymmetric distributions. However, some of the proposedseries do not converge to the stable function, and some

1Note: Infinite variance is observed for 0 < α < 2. This character-istic occurs for 0 < α ≤ 1, as a consequence of not having a well-defined expected value E [X]. For 1 < α < 2, the integral in thevariance definition diverges [17–19].

2

of them are only applicable for extreme values x→ 0 orx → ∞. Mantegna [22] presented a quite similar solu-tion to Elliot [21] but his algorithm is only valid whenx→∞ and 0.75 < α ≤ 1.95. Nolan [23] presented his al-gorithms considering asymmetric distributions for largeevents x → ∞ focusing on tail behaviour only. Thus,the stable distribution function does not have an explicitexpression [24, 25] and no uniform solution of the stabledistribution has been proposed until now [18, 19, 23].

Due to the absence of an explicit expresion, numericalsolutions were developed to evaluate the stable distri-bution function by using numerical recursive quadraturemethods [26–28]. Nolan [26, 29] develops a numerical so-lution for the estimation of stable parameters through amaximun likelihood method for each data set of x. How-ever, Nolan’s method converges only for α > 0.4 andthe convergence to the stable distribution function seemsto be not accurate enough. Despite this fact, Nolan’smethod constitutes an important method that is still be-ing used [27].

Apart from the numerical issues in the evaluation ofthe stable distribution, some authors have pointed outits infinite variance as a drawback [30–32]. To avoid theinfinite variance of the stable distribution function, sev-eral truncations are proposed. The truncations make thevariance finite, consequently the distribution function ofthe sum of independent random variables converges tothe normal distribution due to the central limit theo-rem for a large N value. Nevertheless, a time series inthe real phenomena can exhibit infinite variance, one ofthese cases is the variance of price fluctuations on stockmarkets, which increases when the time frame is enlarged[33, 34].

The aim of this paper is to formulate a uniform an-alytical approximation for the stable distribution func-tion based on a series expansion. To achieve this aimwe propose several regularizations of the inner and outerseries expansions to ensure convergence. This will be animportant tool to get the most accurate approximationreducing numerical errors (oscillations) when the stablefunction is evaluated.

This paper is divided in two parts. The first partintroduces the stable distribution and the trans-stablefunction. They are defined by Fourier transformations insections II and III respectively. The trans-stable func-tion is shown to be identical to the stable distributionfor α < 1 and it has the same asymptotic behaviourfor α > 1 for large events. The second part refers tosection IV and it deals with the closest form— analyti-cal approximations— of the stable distribution. For thispurpose, two types of approximations are developed. Thefirst approximation refers to the inner expansion thatconverges asymptotically to the stable distribution asx → 0. The second approximation refers to the outerexpansion that converges asymptotically as x→∞. Forthe outer expansion two cases are presented, one is ob-tained from the stable function in subsection IV B andthe second one from trans-stable function in subsection

IV C. Finally, the uniform solution in section V is pro-posed as a result of matching the inner and the outer so-lution. The analytical equation of uniform solution pro-posed in this paper gives an approximated solution of thestable distribution function over the range −∞ < x <∞.

II. STABLE DISTRIBUTION FUNCTION

The stable distribution is presented in Section 34 of theGnedenko-Kolmogorov book [14] as the Fourier transfor-mation:

f(x;α, β, σ, µ) =1

2π

∞∫−∞

ϕ(t;α, β, σ, µ)eixtdt, (2)

where ϕ(t) is defined as the characteristic function,

ϕ(t;α, β, σ, µ) = e(itµ−|σt|α(1−iβsgn(t)Φ)). (3)

The four parameters involved are: the stability param-eter α ∈ 〈0 2], the skewness parameter β ∈ [−1 +1] , thescale parameter σ ∈ (0 +∞) , and the location parameterµ ∈ (−∞+∞). The parameter α constitutes the char-acteristic exponent of the asymptotic power-law in thetails and it determines whether the mean value and thevariance exist. The stable distribution with 0 < α ≤ 1does not have a mean value and it has a define varianceonly for α = 2 [35].

The function sgn (t) represents the sign of t and thefunction Φ is defined as:

Φ =

{tan

(πα2

)α 6= 1,

− 2π log |t| α = 1.

(4)

The stable distribution is the family of all attractorsof normalized sums of independent and identically dis-tributed random variables. The most well-known stabledistribution functions are the Cauchy distribution withα = 1 and the normal distribution function with α = 2.Both functions have β = 0, which means they are sym-metric distributions about their mean [18].

In this paper we will focus on symmetric distributions(β = 0). For this case the stable distribution can benormalized as follows:

f(x;α, β = 0, σ, µ) = Re

{S

(x− µσ

, α

)}, (5)

where the general distribution function is given by thefollowing equation:

S(x;α) =1

π

∞∫0

e−tα

eixtdt. (6)

The real part of this function corresponds to the normal-ized stable distribution,

s(x;α) = Re(S(x;α)).

3

Consequently,

s(x;α) =1

π

∞∫0

e−tα

cos(tx)dt. (7)

III. TRANS-STABLE FUNCTION

Zolotarev (1986) used the term “trans-stable” to referto a power series expansion that converges to the stabledistribution for 0 < α < 1 only [18]. In this paper, trans-stable is the function which one of its solutions originatesZolotarev series when the series expansions are appliedaround x→∞. First we define the complex trans-stablefunction in the range of 0 < α < 2. For α < 1, the sta-ble distribution and the trans-stable function are identi-cal. For α > 1, the trans-stable function and the stabledistribution present the same asymptotic behaviour forx → ∞. Consequently, our trans-stable function can beused to find a numerical aproximation of the stable dis-tribution function for α > 1 for large events.

First, the complex trans-stable function is defined asan integral over the path C in the complex plane:

GC (x;α) =1

π

∫C

I (x, z;α) dz, (8)

where

I (x, z;α) = e−zαeixz. (9)

The relation of this function to the stable S(x;α) andthe trans-stable T (x;α) functions is obtained by choosinga particular path C in the complex plane. Then, thestable distribution and trans-stable function are given byEq. (10) and (11) respectively:

S(x;α) = G[0,∞)(x;α) =1

π

∞∫0

e−tα

eixtdt, (10)

T (x;α) = G[0,i∞)(x;α) =1

π

i∞∫0

e−tα

eixtdt. (11)

First it will be shown that for 0 < α ≤ 1, bothstable S(x;α) and trans-stable T (x;α) functions areidentical. For 1 < α < 2 it will be demonstrated thatboth functions exhibit the same asymptotic behaviourwhen x→∞.

This demonstration is based on the evaluation of thecomplex trans-stable integral Eq. (8) using polar repre-sentation for α ≤ 1 and rectangular representation forα > 1 on the complex integrand. The demonstrationsare presented in the following subsections.

Im( )z

Re( )z

3C

2C

1C

4C

i

i

0Figure 1. Contour integration for Eq. (8)

1. For 0 < α ≤ 1

Here we will show that for 0 < α ≤ 1 the stable andtrans-stable functions are identical. This demonstrationwill be done by considering the closed contour shownin Figure 1. Since the complex function in Eq. (9) isanalytical over the complex plane, the integral over theclosed contour Eq. (8) is zero,

∮I (x, z;α) dz = 0. (12)

Let us take the contour in Figure 1 that can be dividedinto four straight paths so that:

4∑1

∫Ci

I (x, z;α) dz = 0. (13)

Now, we will take the limit when τ → ∞ in Figure 1.Using Eq. (10,11,13) the following equation is obtained:

S(x;α)−T (x;α) = −limτ→∞

3∑i=2

∫Ci

I (x, z;α) dz. (14)

To evaluate the right hand side in Eq. (14) it is con-venient to use the polar representation of the complexnumber z = reiθ and express Eq. (9) in polar coordinates:

I (x, z;α) = eg(x,r,θ;α)+ih(x,r,θ;α), (15)

g(x, r, θ;α) = −rα cos (θα)− rx sin θ, (16)

h(x, r, θ;α) = −rα sin (θα) + rx cos θ. (17)

4

It will be adopted the nomenclature of signal theory,where the polar notation separates the effects of instan-taneous amplitude |I| = eg and its instantaneous phaseh of a complex function [36]. Consequently, g(x, r, θ;α)represents the attenuation factor and h(x, r, θ;α) repre-sents the oscillation factor.

Now let us notice that limr→∞g(x, r, θ;α) = −∞ for0 < α ≤ 1 at any value of x. This statement is based onthe fact that cos(θα) ≥ 0 in the first quadrant for α ≤ 1.Consequently, limr→∞I (x, z;α) = 0 so that the integralof the right side of the Eq. (14) vanishes at τ → ∞,therefore:

S(x;α) = T (x;α) if 0 < α ≤ 1. (18)

So, the Eq. (18) will allow to use the trans-stable functionT (x;α) instead of the stable distribution function S(x;α)for 0 < α ≤ 1 in the numerical integration. This is withthe aim to remove numerical oscillation, specifically inthe tails. It is noticeable that the integration of the trans-stable function T (x;α) in Eq. (11)is performed over theimaginary axis. Applying the following change of variablet→ −it (formally done by defining u = −it so that du =−idt and later replacing the dummy variable u by t insidethe integral), the trans-stable function is converted intoa Laplace transformation. Consequently, the integrationis performed over the real axis. The Fourier and Laplacerepresentations for T (x;α) are shown in Eq. (19),

T (x;α) =1

π

i∞∫0

e−tα

eixtdt =1

π

∞∫0

e−(it)αe−xtidt. (19)

Figure 2 compares the Fourier representation of thestable distribution function S(x;α) and the Laplace rep-resentation of trans-stable function T (x;α). The inte-gration is performed using a recursive adaptive Simpsonquadrature method [37]. It is evident that the Laplacerepresentation removes the oscillations of the Fourier rep-resentation of the stable distribution for α < 1.

It is important to add that the stable distribution func-tion and trans-stable function hold the same value fortheir Fourier and Laplace transform representations. Thedifference between each transform representation is theaxis in which each function is integrated. The expressionsare shown in Table I.

Stable distributionS(x;α)

Trans-stable functionT (x;α)

Fouriertransform

1π

∞∫0

e−tα

eixtdt 1π

i∞∫0

e−tα

eixtdt

Laplacetransform

1π

−i∞∫0

e−(it)αe−xtidt 1π

∞∫0

e−(it)αe−xtidt

Table I. Summary of Fourier and Laplace representationsfor the stable and the trans-stable functions

0 10x

10-2

100

S(x;

)-Fo

urie

r

=0.90=0.70=0.50=0.40

0 10x

10-2

100

T(x

;)-

Lap

lace

=0.90=0.70=0.50=0.40

100 101

x

10-2

100

S(x;

)-Fo

urie

r

=0.90=0.70=0.50=0.40

100 101

x

10-2

100

T(x

;)-

Lap

lace

=0.90=0.70=0.50=0.40

Figure 2. Comparison of numerical integration 0 < α < 1between Fourier and Laplace transform of the stable S(x;α)and the trans-stable T (x;α) functions using recursive adap-tive Simpson quadrature method [37]. The absolute errortolerance of the method is ξ = 3.5 × 10−8. Top plots areshown in semi-logarithmic scale and in logarithmic scale.

2. For 1 < α < 2

Here we will shown that for 1 < α < 2 the stableand trans-stable functions have the same asymptotic be-haviour on large events if the integrals are appropriatelytruncated.

Let us recall Eq. (16) for the attenuation factor,

g(x, r, θ;α) = −rα cos (θα)− rx sin θ.

In the previous section, it was shown that cos(θα) is al-ways positive in the first quadrant of the complex planeif 0 < α ≤ 1. Otherwise, if α > 1, then cos(θα) < 0 whenθ = π/2. Consequently, limr→∞I(x, r, θ;α) = ∞ in thisrange, so that the right hand side of Eq. (14) can not beneglected. Therefore S(x) 6= T (x) if α > 1.

We can find an approximation between these two func-tions if the τ value in the contour of Figure 1 is kept largebut finite (τ <∞). Thus, Eq. (13) becomes:

S(x;α, τ)− T (x;α, τ) = −3∑i=2

∫Si

I (x, z;α) dz, (20)

where S(x;α, τ) and T (x;α, τ) are the truncated inte-grals in Eqs. (10) and (11) respectively:

5

S(x;α, τ) =1

π

τ∫0

e−tα

eixtdt, (21)

T (x;α, τ) =1

π

iτ∫0

e−tα

eixtdt. (22)

Now, the right hand of Eq. (20) can be evaluated inthe limit where x→∞. First, notice that in the contourof integration in Figure 1 the magnitude of r is boundedby the condition 0 < r <

√2 τ and sin(θ) > 0 in the first

quadrant, thus:

limx→∞g(x, r, θ;α) = −∞.

Consequently, limx→∞I (x, z;α) = 0 so that the integralon the right of Eq. (20) vanishes at x → ∞. Therefore,the asymptotic behavior is obtained for 1 < α < 2,

S(x;α, τ) ∼ T (x;α, τ) as x→∞. (23)

This demostrates that both functions are asymptoticallyequivalent when the integrals are truncated.

The next step is to find the truncantion value τ thatleads to the best approximation of these functions. Thevalue of τ should be chosen to minimize the truncationerror and at the same time to make the domain of integra-tion as small as possible. With this aim, the trans-stablefunction T (x;α, τ) in Eq. (11) is expressed in its Laplacerepresentation by usying the change of variable t→ −it.Thus,

T (x;α, τ) =1

π

∫ τ

0

I (x, t;α) dt, (24)

where I corresponds to Laplace transform integrandshown in Eq. (19) and Table I,

I (x, t;α) = e(−it)αe−xti. (25)

Then, considering Euler’s representation for a complexexponential function eiθ = cos (θ)+isin (θ), the followingequations are obtained to express Eq. (25):

I (x, t;α) = eg(x,t;α)+ih(x,t;α), (26)

g(x, t;α) = −tα cos(πα

2

)− xt, (27)

h(x, t;α) = −tα sin(πα

2

)+π

2. (28)

The instantaneous amplitude |I| = eg will be determinedby the attenuation factor in Eq. (27). For that reason,

an analysis of g(x, t;α) was made in Figure 3. The curveg(x, t;α) = 0 divides two regions, one with exponentialgrowth (g > 0) and the other with exponential decay(g < 0).

In Figure 3, two sub-regions can be recognized in g <0. The first one, “Zone A” which contains negative gvalues with downward trend ∂g/∂t < 0 that is faster asx → ∞. The second sub-region is “Zone B”, it containssmaller negative g values that follow an upward trendand ∂g/∂t > 0 displaying an increase behaviour whenx → 0. Considering these sub-regions, the truncation τin Eq. (24) will depend on x value as follows:

• For x→ 0, The integration must avoid zone B. Thevalues of g(x, t;α) in this zone lead to an exponen-tial growth due to an upward trend ∂g/∂t > 0,consequently |I|9 0.

• For x → ∞, the integration should be restrictedto zone A. The downward trend ∂g/∂t < 0 leadsto obtain g(x, t;α)→ 0. Consequently, the conver-gence of |I| → 0 occurs faster as t→∞.

zoneB

t

x

0( , ; ) 0g x t

( , ; ) 0g x t

zoneA

g(x,t; )=0

x=αt

Positive values with upward trend

Negative values with downward trendNegative values with upward trend

Figure 3. Curve g = 0 separates two regions with g >0 and g < 0. In the latter region, two zones can be distin-guished: zone A with ∂g/∂t < 0 and zone B with ∂g/∂t >0. The equation x = αt is an estimation of the boundary be-tween zones A and B.

For x→ 0, the cut off τ1 which avoids most of zone Bis defined by x = αt. This equation is an estimation ofthe boundary between zones A and B for all range of αvalues.

The cut off τ1 obeys a linear equation and is obtainedfrom the following equations:

eg(x,τ1;α) = |I| = ε and x = ατ1, (29)

where the tolerance ε represents a negligible instanta-neous amplitude

∣∣I∣∣.

6

0 1 2 3 4 5 6 7 8 9t

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5x

1e-07

1e-07

1e-07

1e-06

1e-06

1e-06

1e-05

1e-05

1e-05

1e-05

0.0001

0.0001

0.0001

0.0001

0.0010.001

0.001

0.001

0.001

0.00250.0025

0.0025

0.0025

0.0025

0.010.010.01

0.01

0.01

0.01

0.050.050.050.05

0.05

0.05

0.05

0.40.4

0.40.40.4

0.4

0.40.4

0.750.75

0.750.75

0.750.75

0.750.75

11

11

1

1

11

11

22

2

2

2

44

4

4

4

10

10

10

10

40

40

40

40

200

200

200

500

500 1000

1000

(tc,x

c)

x=-t -1cos( /2)

1 =x/ if x<x

c

1=t

c if x>x

c

0 3 6 9t-1

0

1x=3.75 1=2.55

-1

0

11-truncation

-wavelength

0 3 6 9t

-1

0

1

x=2.9 1=2.07

-1

0

11-truncation

-wavelength

0 3 6 9t-2

-1

0

1

2x=1.6 1=1.14

-2

-1

0

1

21-truncation

-wavelength

0 3 6 9t-1000

0

1000x=0.5 1=0.36

-1000

0

1000

1-truncation

-wavelength

Figure 4. Contour plot of eg that represents the instantaneous amplitude |I| for α = 1.4 and tolerance ε = 10−3 in Eq. (26)and (27). The red line x was drawn by considering g = 0 which represents the limit between the positive and negative valuesof the attenuation factor g. The truncation τ1 is applied following Eq. (30). Downward trend before τ1 and upward trendafter τ1 of |I| can be observed in the right part of the figure for x = 0.5 and x = 1.6. For x = 2.9 and x = 3.8 the truncationis made after reaching a small value on the modulation I.

For x→∞, the cut off τ1 will restrict the integrationof I on a closed interval [0, tc]. This occurs due to a fasterdownward trend ∂g/∂t < 0. The tc value represents thepoint where the instantaneous amplitude can be consid-ered a negligible quantity |I| = ε. Thus, the cut off τ1obeys an equation of a vertical line τ1 = tc.

Notice that there are two different definitions for τ1.Each one corresponds to a particular sub-regions A (x→∞) or B (x→ 0). Consequently, the truncation τ1 for thetrans-stable function is defined by two equations whichdepend on the x and ε values. These two equation havetheir intersection point at (tc, xc):

τ1(ε, x) =

{tc(ε) if x > xcx/α if x < xc

for α > 1, (30)

where tc(ε) and xc are given by the implicit form of thefollowing equations:

αtc2 + tc

α cos (πα/2) + ln(ε) = 0,xc = αtc.

(31)

Figure 4 illustrates the contour plot of the instanta-neous amplitude |I| for α = 1.4. The truncation τ1 ispresented as a cut-off made when a negligible value ofinstantaneous amplitude is achieved |I| = ε = 10−3. Thepoint (xc, tc) is located at the intersection between thecontour line of the given tolerance ε and the equationτ1 = x/α. The truncation τ1 avoids zone B which con-tains negative values for g with ∂g/∂t > 0. One canobserve that there is an abrupt upward trend in |I| forx→ 0. So, the truncation τ1 allows us to make a perfectcut off before this upward trend starts. It is noticeablethat with a small tolerance ε the intersection will occurin the rightmost part of the figure, consequently the in-terval of integration will be wider and a more accurateresult can be obtained.

Figure 5 shows how the solutions of trans-stable andstable distribution functions are quite similar after xcvalue, which depends on the tolerance ε. For a smallerε the similarity of both asymptotic series is expected toimprove due to a wider interval of integration. However,the value xc will be higher and the similarity will startat the rightmost part of the axis.

7

0 10x

10-4

10-3

10-2

10-1

100S(

x;)-

Four

ier

=1.20=1.40=1.60=1.90

0 10x

10-4

10-2

100

T(x

;)-

Lap

lace

xc( )

=1.20=1.40=1.60=1.90

100 101

x

10-4

10-3

10-2

10-1

100

S(x;

)-Fo

urie

r

=1.20=1.40=1.60=1.90

100 101

x

10-4

10-2

100T

(x;

)-L

apla

ce

xc( )

=1.20=1.40=1.60=1.90

Figure 5. Comparison of numerical integration 1 < α < 2between Fourier and Laplace transform of stable S(x;α) andtrans-stable T (x;α) functions using recursive adaptive Simp-son quadrature method [37]. The absolute error toleranceof the method is ξ = 3.5 × 10−8. Top plots are shown insemi-logarithmic scale and in logarithmic scale.

IV. ASYMPTOTIC EXPANSIONS

Asymptotic expansions are developed to obtain closed-form representations for the stable distribution functionS(x, α). These expansions are based on the Taylor seriesof the complex exponential function,

ez ∼n∑k=0

zk

k!as n→∞. (32)

Two different asymptotic expansions will be performed.The first one corresponds to the ‘inner expansion’. To getthis solution the stable distribution function is evaluatedby expanding eixt of Eq. (10) and (21) around x = 0.The second one refers to the ‘outer expansion’, whichis the asymptotic series expansion for x → ∞. Whenx >> 1, the oscillations of the integrands in Eq. (10)and (21) are large. Consequently, there are importantcancellations due to factor eixt in the integral. Thus,we focus our integration in the region with the majorcontribution in the integral, that is around t = 0. Inconsequence, the amplitude of the integral et

αis replaced

by its Taylor expansion around t = 0. To guaranteethe convergence of the series expansion, the improperintegrals are truncated. The truncation occurs because ofthe sufficient conditions for Riemann integral existence.

These conditions are that the integrand must be boundedand the domain of integration is a closed interval [38, 39].

A. Inner Expansion

The inner expansion is obtained making a substitutionof eixt by its Taylor series expansion given by Eq. (32)in the integrand of the stable distribution I. After thissubstitution, the integrals in Eq. (10) and (21) can beanalytically solved. The difference between these twoequations are the truncation on the interval of integra-tion.

For α ≤ 1 the convergence of the series is slow, de-manding a large value of order n in Eq. (32) to reach anacceptable similarity with the original integrand I. Forthis reason, the improper integral is truncated after asmall enough amplitude of I is obtained. For α > 1 theconvergence occurs faster and truncation is not needed.

1. For 0 < α ≤ 1

The inner expansion is obtained by substituting eixt inEq. (21) by its Taylor expansion using Eq. (32). Then:

Si(x;α, ε) =1

π

τ2(x,ε)∫0

e−tα

eixtdt ∼ 1

π

τ2(x,ε)∫0

Indt as n→∞,

(33)

where In is given by:

In(x; t, α) =

n∑k=0

e−tα (ixt)k

k!. (34)

The upper limit τ2 is given by the following equation:

τ2(x, ε) = − ln(ε)

x. (35)

This truncation results from the equation eixτ2 = ε,where ε represents the tolerance that needs to be small toensure a cut-off when negligible quantities of |I| and |In|are obtained. Consequently, the area under the curve ofboth functions are similar.

The convergence of In to I demands a large value oforder n in Eq. (32), as it can be observed in Fig (6). Thisoccurs because of slow decay of e−t

α

value for α < 1. Thisis the reason to evaluate the integral in the closed interval[0,τ2], where the original integrand I and its Taylor seriesapproximation In are similar.

The integrals in Eq. (33) can be solved without diffi-culty. Then, the inner expansion si is given by the realpart of this solution,

si(x;α, ε) = Re(Si(x;α, ε)).

8

0 1 2 3 4 5t-2

-1

0

1

2

I

x=4.5 n=20 =0.75

IIn

0 1 2 3 4 5t10-20

100

1020

I- I n

I - In

2

0 1 2 3 4 5t-2

-1

0

1

2

I

x=4.5 n=30 =0.75

IIn

0 1 2 3 4 5t10-20

100

1020 I-

I n

I - In

2

0 1 2 3 4 5t-2

-1

0

1

2

I

x=4.5 n=50 =0.75

IIn

0 1 2 3 4 5t

10-20

100

1020

I- I n

I - In

2

0 1 2 3 4 5t-2

-1

0

1

2

I

x=4.5 n=20 =1.8

IIn

0 1 2 3 4 5t10-20

100

1020

I- I n

I- In

2

0 1 2 3 4 5t-2

-1

0

1

2

I

x=4.5 n=30 =1.8

IIn

0 1 2 3 4 5t10-20

100

1020

I- I n

I- In

2

0 1 2 3 4 5t-2

-1

0

1

2

I

x=4.5 n=50 =1.8

IIn

0 1 2 3 4 5t10-20

100

1020

I- I n

I- In

2

Figure 6. Comparison of I = etα

eixt and In = etα n∑k=0

(ixt)k

k!of Eq. (33) and (38), where In is obtained by replacing eixt in

I by its Taylor expansion. The plots are for α = 0.75 (left) and α = 1.8 (right). For α ≤ 1, the truncation is required to en-sure a cut-off when negligible quantities of |I| and |In| are obtained. The truncated error of the Taylor expansion is measuredby using the absolute value of the difference |I − In|. For α > 1, since the convergence is fast the truncation is unnecessary.For these particular examples, the integrand I is evaluated at x = 4.5 for three cases of n = 20, 30, 50 with ε = 10−9.

Consequently,

si(x;α, ε) =1

πα

∞∑k=0

xk

k!γ

(k + 1

α, τ2(x; ε)

α

)cos

(πk

2

),

(36)where γ represents the incomplete gamma function [40],

γ(z, b) =

∫ b

0

xz−1e−xdx. (37)

Due to a computation of the incomplete gamma func-tion γ, Eq. (36) was modified for numerical analysis inMatlab2 [41].

2Note: Matlab defines the incomplete gamma function as γ∗

γ∗(b, z) =1

Γ(z)

∫ b

0xz−1e−xdx

where Γ(z) is the gamma function.

2. For 1 < α < 2

Here it is derived the inner expansion si for α > 1 fromthe non-truncated form of stable distribution function.This derivation is made by substituting eixt in Eq. (10)by its Taylor expansion in Eq. (32), then:

Si(x;α) =1

π

∞∫0

e−tα

eixtdt ∼ 1

π

∞∫0

Indt as n→∞.

(38)For α > 1, the convergence of integrand I and the in-tegrand after the substitution In occurs faster than forα < 1. This feature is observed in Figure (6), wherean acceptable convergence between I and In is obtainedwith a small n value. Consequently, the integral is eval-uated without truncation or taking the limit ε → 0 inEq. (33).

Then, it is only considered the real part of the solutionof Eq. (38),

si(x;α) = Re(Si(x;α)).

9

Consequently,

si(x;α) =1

πα

∞∑k=0

xk

k!Γ

(k + 1

α

)cos

(πk

2

), (39)

where Γ represents the gamma function [40],

Γ(b) =

∫ ∞0

xb−1e−xdx.

Examples for α = 0.75 and α = 1.80 are shown onFigures 7 and 8 respectively. In Figure 7, for α ≤ 1the truncation τ2 is needed, otherwise the convergenceto stable distribution function will be ultraslow as n →∞. This is evident when a comparison is made betweensubfigure 7a and 7b. They represent a non-truncated andtruncated stable solution respectively. The subfigure 7bdisplays an acceptable convergence with a smaller ordern. In Figure 8, for α > 1 the convergence to the stabledistribution function occurs faster and no truncation isneeded. For both cases the inner expansion si behaveswell because it converges to s(x;α).

0 1 2 3 4 5 6 7x

10-2

10-1

100

s in (x;

,n)

1

3

5100

(a)

Stable Function s(x; )

Inner Expansion sin(x; ,n)

0 1 2 3 4 5 6 7x

10-2

10-1

100

s in (x;

,,n

) 2 32

3840

42

44

50 56 62 68 (b)

Stable Function s(x; )

Inner Expansion sin(x; , ,n)

Figure 7. Inner expansion of the stable distribution func-tion for α = 0.75. This is obtained by applying Taylor ex-pansion around t = 0. The subfigure (a) is a non-truncatedintegral. The subfigure (b) is the truncated integral withtolerance ε = 10−9 in Eq. (36), which displays a fast conver-gence due to integral’s truncation.

0 1 2 3 4 5 6 7x

10-2

100

s in (x;

,n) 2 8

14 20

2632

5056

7480

98104

Stable Function s(x; )

Inner Expansion sin(x; ,n)

Figure 8. Inner expansion of the stable distribution forα = 1.80 as a result of applying a Taylor expansion in the‘exponential of the phase’ of the integrand. The figure illus-trates the non-truncated stable solution in Eq. (39). Thisfigure displays a fast convergence so that no truncation isneeded.

B. Outer Expansion

The outer expansion is obtained making a substitutionof the amplitude e−t

αin the integrand of the truncated

stable distribution function I in Eq. (21) by its Taylor se-ries expansion around t = 0. Then, the following relationis obtained:

So(x;α, ε) =1

π

τ3(ε)∫0

e−tα

eixtdt ∼ 1

π

τ3(ε)∫0

Gndt as n→∞,

(40)

where Gn is given by:

Gn(x;α) =

n∑k=0

(−tα)k

k!eixt. (41)

The upper limit τ3 is given by the following equation:

τ3(ε) = [− ln(ε)]1/α

. (42)

This truncation is calculated from e−τ3α

= ε, where ε isdefined as tolerance and represents a negligible instanta-neous amplitude when ε is small. The truncation allowsa faster convergence of Gn to I and reduces the errorof integration due to an accurate approximation on theinterval [0, τ3]. The original integrand I and the new in-tegrand after applying Taylor series Gn in Eq. (40) wereevaluated in Figure 9. Since the convergence of Gn to Iis slow, the truncation τ3 is considered to define the newinterval of integration [0, τ3].

To obtain the outer solution so a change of variableafter the series expansion is applied in Eq. (40). Thechange of variable is −u = ixt, so −du = ixdt. Thisgives us an approximation of the form:

10

0 20 40 60

t

-1

-0.5

0

0.5

1

I

x=4.5 n=20 =0.75

IG

n

0 20 40 60

t

10-40

10-20

100

1020

1040

I- G

n

I- Gn

3

0 20 40 60

t

-1

-0.5

0

0.5

1

I

x=4.5 n=30 =0.75

IG

n

0 20 40 60

t

10-40

10-20

100

1020

1040

I- G

n

I- Gn

3

0 20 40 60

t

-1

-0.5

0

0.5

1

I

x=4.5 n=60 =0.75

IG

n

0 20 40 60

t

10-40

10-20

100

1020

1040

I- G

n

I- Gn

3

0 2 4 6

t

-1

-0.5

0

0.5

1

I

x=4.5 n=20 =1.8

IG

n

0 2 4 6

t

10-40

10-20

100

1020

1040

I- G

n

I- Gn

3

0 2 4 6

t

-1

-0.5

0

0.5

1

I

x=4.5 n=30 =1.8

IG

n

0 2 4 6

t

10-40

10-20

100

1020

1040

I- G

n

I- Gn

3

0 2 4 6

t

-1

-0.5

0

0.5

1

I

x=4.5 n=60 =1.8

IG

n

0 2 4 6

t

10-40

10-20

100

1020

1040

I- G

n

I- Gn

3

Figure 9. Comparison of I = etα

eixt and Gn = (−tα)k

k!eixtdt of Eq. (40), where Gn is the Taylor expansion of I around

t = 0. The plots correspond to α = 0.75 (left) and α = 1.8 (right). Truncation of the integral is required for both cases. Thereason of that is to reach an accurate approximation between the original integrand I and the one after the series expansionGn. The error is measured by the absolute value of the difference |I −Gn|. For these particular examples, the integrand I isevaluated at x = 4.5 for three cases of n = 20, 30, 50 with ε = 10−9.

So(x;α, ε) ∼ 1

π

n∑k=0

(−1)k

k!

(−1

ix

)kα+1−ixτ3(ε)∫

0

ukαe−udu.

(43)

To solve the integral, the incomplete gamma functionof imaginary argument γ(v, iz) is used [40, 42]. The fol-lowing solution is presented by Barak as a special case ofconfluent hypergeometric function [42],

γ(v, iz) =

∫ iz

0

tv−1e−tdt

= (iz)vv−11F1(v, 1 + v,−iz),

(44)

where 1F1(v, 1 + v,−iz) represents the Confluent Hyper-geometric function. Then, comparing Eq. (43) and (44),we obtain the following relation betwen the variables,v = kα+ 1, z = −xτ3 and t = u.

Finally, the real part of the solution is:

so(x;α, ε) = Re(So(x;α, ε)).

Consequenlty,

so(x;α, ε) =− 1

π

∞∑k=1

(−1)k

k!

(cos (παk)

kα+ 1

)...

...(−τ3(ε))kα+1

1F1(kα+ 1, kα+ 2, ixτ3(ε)).

(45)

Figure 10 shows the calculation of Eq. 45 for α = 1.8.In this figure is evident that the outer expansion so con-verges slowly. This occurs due to computation of the con-fluent hypergeometric function 1F1 which demands con-siderable computational time. The series that define thefunction 1F1 do not have a trivial structure, this createsnumerical issues which makes the calculation computa-tional inefficient [43]. The approximation in Figure 10shows how the convergence demands a large value of or-der n to obtain an accurate approximation at the tail.The convergence resembles waves that slowly start to de-crease from the tails to the peak of stable distribution.The series until n = 30 does not show an acceptableapproximation. Only an approximation on tails are ob-tained after n = 40. For x→ 0, the series of so convergesto a specific value different of the stable distribution func-tion. The convergence to the stable distribution functionis observed only for x→∞.

11

0 1 2 3 4 5 6 7x

-500

0

500

s on (x;

,,n

)

(a)

Stable Function s(x; )

Outer Expansion son(x; , ,n)n=10

0 1 2 3 4 5 6 7x

-0.2

-0.1

0

0.1

0.2

0.3

s on (x;

,,n

)

(b)

Stable Function s(x; )

Outer Expansion son(x; , ,n)

n=30

n=40

n=100

Figure 10. Outer expansion so for α = 1.80 with toleranceε = 10−6 in Eq. (45). The subfigure (a) and (b) correspondto n = 10, 30, 40, 100 respectively.

C. Outer expansion by Trans-Stable distribution

Because of the slow convergence of the outer expansionso and its wave-like behavior, an alternative approxima-tion is obtained using the trans-stable function T (x;α).As it was previously explained in section III, the solu-tions of trans-stable T (x;α) and stable S(x;α) functionsare identical for 0 < α ≤ 1 and similar for 1 < α < 2 af-ter the xc value. Consequently, the improper integralin Eq. (19) is used to calculate the series expansions for0 < α ≤ 1 and the truncated trans-stable integral in itsLaplace representation in Eq. (24) for 1 < α < 2.

This outer expansion to is given by the analytical solu-tion of the trans-stable function after applying the Taylorseries of e−(it)α around t = 0 in the trans-stable integrandI using Eq. (32) and (19). Then, the following equationis shown:

To(x;α, ε) =1

π

τ1(x,ε)∫0

e−(it)αe−xtidt ∼ 1

π

τ1(x,ε)∫0

Kndt,

(46)

where Kn is given by:

Kn(x) =

n∑k=0

(−(it)α)k

k!e−xti. (47)

Due to a slow convergence of Kn to I the cut-off τ1 isapplied. The truncation τ1 has two different expressions.For α ≤ 1, the truncation τ1 depends on the tolerance εand for α > 1 it depends on the tolerance ε and x values.These expressions will be explained in the following sub-sections.

To solve the integral in Eq. (46), the following changeof variable is applied: xt = u and xdt = du. This leadsto the following series expansion:

To(x;α, ε) ∼ 1

π

∞∑k=0

(−1)k

k!

(−1

ix

)kα+1xτ1(x,ε)∫

0

u(kα+1)−1e−udu.

(48)

The upper limit of the integral changes from τ1 to xτ1,but still remains on the real axis. The integral above canbe solved using the incomplete gamma function definedin Eq. (37). Then, the real part of the result is obtained,

to(x;α, ε) = Re(To(x;α, ε)).

Consequently,

to(x;α, ε) =

− 1

π

∞∑k=1

(−1)k

k!

(1

x

)kα+1

sin

(παk

2

)γ (kα+ 1, xτ1(x, ε)).

(49)

The determination of τ1 for α ≤ 1 and α > 1 is presentedin the following subsections.

1. For 0 < α ≤ 1

For α ≤ 1, the cut-off τ1 in Eq. (49) is given by thefollowing equation:

τ1(ε) = [− ln(ε)]1/α

for α ≤ 1. (50)

This truncation is obtained from e−τ1α

= ε, where thetolerance ε represents a negligible instantaneous ampli-tude for the integrands in Eq. (46).

2. For 1 < α < 2

The truncation τ1 in Eq. (49) for 1 < α < 2 was alreadyobtained in subsection III-2 and defined by Eq. (30) as:

τ1(x, ε) =

{tc(ε) if x > xc,x/α if x < xc,

for α > 1,

where tc and xc were defined by Eq.(31). As indicatedin subsection III-2, the value of τ1 is used to minimizethe truncation error and at the same time to make thedomain of integration as small as possible.

12

The outer expansion by the trans-stable function con-verges to the original trans-stable function. Examplesare shown in Figure 11 for α ≤ 1 and Figure 12 forα > 1. Note that in both cases the truncation τ1 allowsa faster and more accurate convergence to the real partof the trans-stable distribution t(x;α). Consequently theouter solution to shows an identical solution as s(x;α) forα ≤ 1 and the same asymptotic behaviour for α > 1. Fora smaller ε the convergence of these outer expansions tothe trans-stable function will occur faster. Also in Figure11 the non-truncated trans-stable expansion is shown asan expansion that converges extremely slowly requiringa higher order n than truncated trans-stable expansionto obtain an acceptable convergence. In Figure 12 thenon-truncated trans-stable expansion does not convergeto trans-stable function at all.

0 1 2 3 4 5 6 7x

10-2

10-1

100

t on (x;

,n)

(a)

1

2

8

Stable Function s(x; )Trans-stable Function t(x; )

Outer Expansion ton(x; ,n)

10027

503225

0 1 2 3 4 5 6 7x

10-2

10-1

100

t on (x;

,,n

)

27

32

50

(b)

2

825

27

Stable Function s(x; )Trans-stable Function t(x; )

Outer Expansion ton(x; , ,n)

Figure 11. Outer expansion of the trans-stable functionfor α = 0.75. This result is obtained from the Taylor ex-pansion of the integrand around t = 0 in Eq. (49) and (50).The subfigure (a) is the non-truncated integral that showsslow convergence. The subfigure (b) corresponds to truncatedintegral with tolerance ε = 10−6. The subfigure (b) displaysa faster convergence to the trans-stable function as a resultof the truncation of the integral.

0 1 2 3 4 5 6 7x

10-2

100

t on (x;

,n)

(xc)

(a)

1

35

7 Stable Function s(x; )Trans-stable Function t(x; )

Outer Expansion ton(x; ,n)

0 1 2 3 4 5 6 7x

10-2

100

t on (x;

,,n

)

3

7

(xc)

(b) Stable Function s(x; )Trans-stable Function t(x; )

Outer Expansion ton(x; , ,n)

Figure 12. Outer expansion of the trans-stable for α = 1.80as a result of applying Taylor expansion of the integrandaround t = 0 in Eq. (30) and (49). The subfigure (a) showsthat the non-truncated integral does not converge to thetrans-stable function. The subfigure (b) corresponds to thetruncated integral with tolerance ε = 10−6. The subfigure (b)displays a fast convergence as a result of the truncation ofthe integral.

V. UNIFORM SOLUTION

The uniform solution is presented as the combinationof the inner solution and the outer solution to constructan approximation valid for all x ∈ [−∞,∞]. To constructthe uniform solution an asymptotic matching methodbased on boundary-layer theory is applied [44, 45]. Thismethod is based on superposing the inner and outer so-lution and substracting the overlap between them,

su(x) = yout(x) + yin(x)− yoverlap(x). (51)

The overlap is defined as the limit of the rightmost edgeof yin and the leftmost edge of yout,

yoverlap = limx→0

yout = limx→∞

yin. (52)

For this case, our proposed uniform solution su is con-structed based on our inner expansion si and our outerexpansion to. These previous solutions were already de-fined in section IV.

For a better understanding of our uniform solution su,two sub-sections A and B are presented. Sub-section A

13

Range of α 0 < α ≤ 1 1 < α < 2

Normalized stable

distribution (s)

s(x;α) = 1π

∞∫0

e−tα

cos(tx)dt

Normalized trans-stable

distribution (t)

t(x;α) = 1π

∞∫0

e−tα cos(πα2 )−xtsin(tα sin(πα

2))dt

Inner expansion (sni )

sni (x;α, ε) =1

πα

n∑k=0

xk

k!γ

(k + 1

α, τ2

α

)cos

(πk

2

)

τ2 =− ln(ε)

x

sni (x;α) =1

πα

n∑k=0

xk

k!Γ

(k + 1

α

)cos

(πk

2

)

Outer expansion (sno ) sno (x;α, ε) = − 1

π

n∑k=1

(−1)k

k!

(cos (παk)

kα+ 1

)(−τ3)kα+1

1F1(kα+ 1, kα+ 2, ixτ3)

τ3 = [− ln(ε)]1/α

Outer solution 3 (tno ) tno (x;α, ε) = − 1

π

n∑k=1

(−1)k

k!

(1

x

)kα+1

γ (kα+ 1, xτ1) sin

(παk

2

)

τ1 = (− ln(ε))1/ατ1 =

{tc if x > xcx/α if x < xc

Complete and incomplete

gamma functions (Γ) & (γ)

Γ(z) =

∫ ∞0

xz−1e−xdx γ(z, b) =

∫ b

0

xz−1e−xdx

Table II. Summary of Inner and Outer Solutions

contains a summary of inner and outer expansions pre-viously obtained. In sub-section B the steps taken toobtain su are explained.

A. Summary of inner and outer expansions

Table II contains the normalized stable and trans-stable distribution and the summary of previous resultsobtained from stable and trans-stable functions by ap-plying Taylor expansions. The series refers to one innerexpansion si and two outer expansions so and to.

3Note: Refer to equation Eq. (30) to obtain tc and xc value for1 < α < 2

For the inner expansion si, the solution for α ≤ 1corresponds to a truncated stable solution which al-lows a faster convergence. For α > 1 the series isobtained from the non-truncated stable solution. Theonly difference between them is the use of the incom-plete gamma function γ in the solution for α ≤ 1, whereΓ(z) = limb→∞ γ(z, b). Consequently, for both cases thetruncated series can provide a good approximation. How-ever, in the case of α ≤ 1 we must take the limit as:

s(x;α) = limε→0

(limn→∞

sni (x;α, ε))

for x <∞.

In general the order how we apply the limits cannot beexchanged. However, in the case of α > 1 the order ofthe limits does not affect the convergence. Taking a smallvalue of ε ensures a faster convergence.

For the outer expansion two expressions were derived.

14

The first outer espansion so is obtained by performing theTaylor expansion around t = 0 on the truncated stabledistribution. This solution displays a slow convergencefor n → ∞. The second outer expansion to is obtainedby applying the Taylor expansion on the truncated trans-stable function for x→∞. The truncation of to dependson α and there are two different cases. For α ≤ 1 itconverges to the exact solution of s(x;α) and for α > 1it converges to the same solution at the tails of s(x;α).To guarantee convergence, we need to take the limit as,

t(x;α) = limε→0

(limn→∞

tno (x;α, ε))

for x > 0.

Exchanging the order of the limits will affect the con-vergence. The outer expansion that will be used is to,because it displays a faster convergence and it does notexhibit wavelike behavior.

B. Steps to obtain the uniform solution

To obtain the uniform solution su the condition inEq. (52) needs to be satisfied. The inner expansion siand the outer expansion to have to be multiplied withan appropriate coefficient A(x) to obtain the asymptoticsolutions with a common matching value ym. These oper-ations will allow us to obtain yout and yin. Consequently,Eq. (51) will be applied to obtain the closed-form solutionof the stable distribution function.

Below the steps are explained to obtain the location ofthe matching between the inner and the outer solutions(xm, ym), the coefficient A(x), and the uniform solutionsu.

1. Finding inner and outer limit (xm, ym)

Considering si and to as good approximations to thestable distribution function, we must require that theinner and the outer expansions will be close enough be-fore matching them [46]. Consequently, the point wherethe matching between si and to takes place is (xm, ym)and it represents the location where the minimal verticaldistance between the inner si and the outer solution tooccurs.

The distance function between si and s is defined as δiand the distance function between to and s is δo. Con-sequently, (xm, ym) is the point where the Pythagoreanaddition of these distances is minimal Eq. (53):

δi2(x;α, ε) = (s(x;α)− si(x;α, ε))

2,

δo2(x;α, ε) = (s(x;α)− to(x;α, ε))

2,

δ2(x;α, ε) = δo2(x;α, ε) + δi

2(x;α, ε),

d(δ2(x;α, ε)

)dx

∣∣∣∣∣xm

= 0. (53)

The xm value is obtained from the previous equation.Then, ym is defined by the equidistant point betweenboth functions,

ym =si(xm) + to(xm)

2. (54)

2. Defining the inner and the outer solutions yinand yout

To obtain the uniform solution su, the asymptoticmatching method based on boundary layer theorem [44]is applied. Consequenlty, the inner solution yin and theouter solution yout must have a matching asymptotic be-haviour. More precisely, the limit of the outer solutionyout when x → 0 should correspond to the limit of theinner solution yin when x → ∞. To obtain yin and youtsolutions, the series expansions si and to are multipliedby an appropiate coefficients to meet the requirements ofmatching asymptotic expansions, so the yin and yout aredefined as follows:

yin(x;α, ε, µ) = (si(x)− ym) (1−A(x;µ)) + ym, (55)

yout(x;α, ε, µ) = (to(x)− ym)A(x;µ) + ym, (56)

where the overlapping factor A(x) is defined as:

A(x;µ) =1

2

(1 + tanh

(x− xmµ

)). (57)

The A(x;µ) is used to smooth si and to and providesthem with a symmetric overlap section around xm andgives yin and yout an asymptotic behaviour. The variableµ determines the width of the overlap between yin andyout.

It is easy to see that Eq. (55) and Eq. (56) satisfyEq. (52), where the limits of yout and yin converge to aconstant value ym.

3. Defining the Uniform Solution su

The inner solution yin Eq.(55) and the outer solutionyout in Eq.(56) were defined to fulfill the requirementsfor matching asymptotic expansions. Then, Eq.(51) isapplied to obtain the uniform solution su,

sni,nou (x;α, ε, µ) =tnoo (x;α, ε)

2+snii (x;α, ε)

2....

....+ tanh

(x− xmµ

)(tnoo (x;α, ε)

2− snii (x;α, ε)

2

).

(58)

15

4. Find the best su by choosing the most appropiate µ

The width of the overlap between yin and yout can beoptimized to obtain the closest solution su of the stabledistribution function. The most appropriate value of µneeds to be obtained for each particular value of α. Forthat, the least square method will be applied between theoriginal s(x;α) and the new closest solution su(x;α, ε, µ).Applying Eq. (7) and (58) the following equation is ob-tained:

L(µ) =

N∑i=1

(su(xi;α, ε, µ)− s(xi;α))2, (59)

where the N value represents the length of the sampleused to minimize L.

The similarity between the exact solution of s(x;α)and the uniform solution su(x;α, ε, µ) is observed in Fig-ure 13 and 14 for α = 0.75 and α = 1.80 respectively.For α < 1, a good approximation between s(x;α) andsu(x;α, ε, µ) is obtained in the tails after mixing two dif-ferent orders. The order for the inner solution is ni = 6,which makes the solution concave upward. The order forthe outer solution is no = 17, which makes the solutionconcave downward. This combination of orders will en-sure a good matching asymptotic behaviour. For α > 1,the uniform solution works well, and a good uniform so-lution is obtained quickly with a lower order n = 6.

Lower orders can be used for both cases, where themost important aspect to consider is the different con-cavity between yin and yout for the matching asymptoticbehaviour. The concavity of the inner and outer solutionis defined by the trigonometric element in each solution.

0 1 2 3 4 5 6 7x

10-2

10-1

100

s(x;

)

Stable Function sInner Solution y

inOuter Solution y

out

Uniform Solution su

(xm

,ym

)

Figure 13. Uniform solution su for α = 0.75 as a result ofjoining the inner solution yin with the outer solution yout.The tolerance ε = 10−6, µ = 0.052 and ni = 6 and no = 17.

0 1 2 3 4 5 6 7x

10-3

10-2

10-1

100

s(x;

)

Stable Function sInner Solution y

inOuter Solution y

out

Uniform Solution su

(xm

,ym

)

Figure 14. Uniform solution su for α = 1.80 as a result ofjoining the inner solution yin with the outer solution yout.The tolerance ε = 10−6, µ = 0.4 and n = 8.

VI. CONCLUSIONS

In this paper we presented a uniform solution of thestable distribution. This solution converges to the sta-ble distribution function in the full range of x values−∞ < x < ∞. This condition makes our uniform so-lution more robust than previous analytical expressionsthat were only applicable for extreme values x → 0 orx → ∞. Also, our uniform solution removes the neg-ative values obtained in previous numerical solutions ofthe stable distribution function for all α values, whichmakes this solution more reliable because a probabilitydensity function must be always positive.

The uniform solution is the result of an asymptoticmatching between the inner and outer expansions. Theinner expansion results from the Taylor series expansionof the characteristic function of the stable distributionaround x = 0. The outer expansion is obtained fromthe Taylor expansion of the integrand of the trans-stablefunction around t = 0. The convergence of these expan-sions is guaranteed if the integrands are truncated, andthe speed of convergence depends on how is the trunca-tion implemented.

For α ≤ 1, the uniform solution provides a good ap-proximation for all the range of x values. Also, the nu-merical integration of the trans-stable function consti-tutes a second option which allows us to obtain a robustnumerical solution of the stable distribution function andremoves the oscillations. For α > 1, the uniform solutionprovides an analytical solution of the stable distributionfunction based on fast converging series.

ACKNOWLEDGMENTS

F. Alonso-Marroquin thanks Hans J. Herrmann foruseful discussions and his hospitality in ETHZ.

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