# q-Gaussian Distributions: Simplifications and Simulations

Authors: Paweł J. Szabłowski
Journal: Journal of Probability and Statistics
 (2009)

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## Abstract

We present some properties of measures (q-Gaussian) that orthogonalize the set of q-Hermite polynomials. We also present an algorithm for simulating i.i.d. sequences of random variables having q-Gaussian distribution.


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## Body

## 1. Introduction

The paper is devoted to recollection of known and presentation of some new properties of a distribution calledq-Gaussian. We propose also a method of simulation of i.i.d. sequences drawn from it.q-Gaussian is in fact a family of distributions indexed by a parameter q∈[-1,1]. It is defined as follows.Forq=-1, it is a discrete 2-point distribution, which assigns values 1/2 to -1 and 1.Forq∈(-1,1), it has density given by
(1.1)fH(x∣q)=1-q2π4-(1-q)x2∏k=0∞((1+qk)2-(1-q)x2qk)∏k=0∞(1-qk+1),
for |x|≤2/1-q. In particular fH(x∣0)=(1/2π)4-x2, for |x|≤2. Hence it is Wigner distribution with radius 2.Forq=1,q-Gaussian distribution is the Normal distribution with parameters 0 and 1.In Figure1, we present plots of fH(x∣-.4) in blue,fH(x∣.1) in orange,fH(x∣.8) in red,  and standard normal density   in black.Figure 1
Plots offH(·∣q) for different q.This family of distributions was defined first in the paper of Bożejko et al., in 1997 [1] in noncommutative probability context. Later ([2, 3]) it appeared in quite classical context namely as a stationary distribution PH of discrete time random field X={Xn}n∈ℤ defined by the following relationships: 𝔼(Xi)=0,𝔼(Xi2)=1,i∈ℤ,(1.2)∃a∈ℝ;∀n∈ℤ:𝔼(Xn∣ℱ≠n)=a(Xn-1+Xn+1),a.s.∃A,B,C∈ℝ;∀n∈ℤ:𝔼(Xn2∣ℱ≠n)=A(Xn-12+Xn+12)+BXn-1Xn+1+C,a.s.,
where ℱ≠m:=σ(Xk:k≠m). It turns out that parameters a,A,B,C are related to one another in such a way that there are two parameters q≥-1 and 0<|ρ|<1 and all others can be expressed through them:
(1.3)a=ρ1+ρ2,A=ρ2(1-qρ2)(ρ2+1)(1-qρ4),B=ρ2(1-ρ2)(1+q)(ρ2+1)(1-qρ4),C=(1-ρ2)21-qρ4.
Then, one proves that
(1.4)∀n∈ℤ,k,i≥1:𝔼(Hk(Xn∣q)∣ℱ≤n-i)=ρkiHk(Xn-i∣q),a.s.,
where ℱ≤m:=σ(Xk:k≤m), (similarly one defines ℱ≥m:=σ(Xk:k≥m)) and Hk(x∣q):k≥-1 are q-Hermite polynomials defined below. It turns out that for q>1 the one-dimensional distribution of the process X is not defined by moments. This case is treated separately (e.g., in [4]).As mentioned earlier, here we will consider only the case|q|≤1. We will denote family of q-Gaussian distributions by PH(q) or simply PH.It turns out that there is quite large literature where this distribution appears and is used to model different phenomena. See, for example, [5–9]. Random field defined above is a model of notions that first appeared in noncommutative context and hence establishes a link between noncommutative and classical probability theories.Remark 1.1.
In literature there exists another family of distributions under the same name. It appears in the context of (Boltzmann-Gibbs)-statistical mechanics. See, for example, [10] for applications and review.
Both families are indexed by basically one parameterq∈[-1,1], and for q=1 both include ordinary N(0,1) distribution.In the sequel we will use the following traditional notation used in so-called “q-series theory” [0]q=0,[n]q=1+q+⋯+qn-1;n≥1,[0]q!=1,[n]q!=∏i=1n[i]q, [nk]q=[n]q!/[k]q![n-k]q!, for n≥0,k=0,…,n and 0 otherwise, and (a∣q)n=∏k=0n-1(1-aqk), for n=1,2,…,∞ (so-called Pochhammer symbol). Sometimes (a∣q)n will be abbreviated to (a)n, if it will not cause misunderstanding. Notice that (q)n=(1-q)n[n]q!,[nk]q=(q)n/(q)k(q)n-k and that [n]q,[n]q! and [nk]q tend to n,n!, and (nk) (Newton's symbol), respectively, as q->1.Remark 1.2.
Introducing new variablez defined by the relationship 1-(1-q)x2/2=cos2πz we can express q-Gaussian density through Jacobi θ functions defined, for example, in [11]. Namely, we have for z∈[-1/2,1/2](1.5)fH(2sin(πz)1-q∣q)=Cqθ3(z∣q)θ2(z∣q),
with Cq=1-q(q∣q2)/2πq1/4(q2∣q2) where θ3(z∣q) and θ2(z∣q) are so-called third and second Jacobi Theta functions.Let us introduce family of polynomials (calledq-Hermite) satisfying the following three-term recurrence relationship:
(1.6)Hn+1(x∣q)=xHn(x∣q)-[n]qHn-1(x∣q),
with H-1(x∣q)=0,H0(x∣q)=1. Notice that Hn(x∣0)=Un(x/2),n≥-1, where Un(x) are Chebyshev polynomials of the second kind defined by(1.7)Un(cosθ)=sin((n+1)θ)sinθ,
and that Hn(x∣1)=Hn(x),n≥-1 where Hn(x) are (“probabilist”) Hermite polynomials, that is, polynomials orthogonal with respect to Gaussian N(0,1) measure.It turns out thatq-Gaussian is the distribution with respect to which q-Hermite polynomials are orthogonal. This fact can be easily deduced from (1.4).Thus in particular using the condition(1.8)∀n≥1;q∈(-1,1):∫-2/1-q2/1-qHn(x∣q)fH(x∣q)dx=0,
we can get all moments of q-Gaussian distribution. Hence in particular we have 𝔼X2n+1=0,n≥0,𝔼X2=1,𝔼X4=2+q,𝔼X6=5+6q+3q2+q3 if only X~PH(q).The aim of this paper is to makeq-Gaussian distribution more friendly by presenting an alternative form of the density fH for q∈(-1,1), more easy to deal with (in particular we find the c.d.f. of PH), and suggest a method of simulation of i.i.d. sequences having density fH.


## 2. Expansion offH

In this section we will prove the following expansion theorem.Theorem 2.1.
For all|q|<1,|x|≤2/1-q one has
(2.1)fH(x∣q)=1-q2π4-(1-q)x2∑k=1∞(-1)k-1q(k2)U2k-2(x1-q2),
where (nk)=n!/k!(n-k)!.As a corollary we get expression for the c.d.f. function ofPH.Corollary 2.2.
The distribution function ofq-Gaussian distribution is given by
(2.2)FH(y∣q)=12+1πarcsin(y1-q2)+12π4-(1-q)y2∑k=1∞(-1)k-1q(k2)(1+qk)U2k-1(y1-q/2)2k.Identity (2.1) can be a source of many interesting identities, which may not be widely known outside the circle of researchers working in special functions.Corollary 2.3.
For allq∈(-1,1)(i)
forallx∈[-2/1-q,2/1-q](2.3)(q)∞∏k=1∞((1+qk)2-(1-q)x2qk)=∑k=1∞(-1)k-1q(k2)U2k-2(x1-q2),
where the polynomials Uk are defined by (1.7).In particular, one has(ii)
(2.4)(-q)∞(q2∣q2)∞=∑k=1∞q(k2),
a particular case of so-called Jacobi's “triple product identity”.(iii)
(2.5)(q3∣q3)∞=1+∑k=1∞(-1)3k(q(3k2)+q(3k+12)),(iv)
(2.6)(q)∞3=1+∑k=2∞(-1)k+1(2k-1)q(k2),(v)
(2.7)∀n∈ℕ:[2nn]q=∑k=1n(-1)k-1(1+qk)q(k2)[2nn-k]q,Corollary 2.4.
fH(x∣q) is bimodal for q∈(-1,q0), where q0(≅-.107) is the largest real root of the equation ∑k=0∞(2k+1)2qk(k+1)/2=0.Lemma 2.5.
For allq∈(-1,1)and n≥4 one has(i)
(2.8)sup|x|<2/1-q|fH(x∣q)-1-q2π4-(1-q)x2(∑k=1n-1(-1)k-1q(k2)U2k-2(x1-q2))|≤n|q|(n-1)(n-2)/2π(1-q2)2,(ii)
(2.9)sup|x|<2/1-q|FH(y∣q)-12-1πarcsin(y1-q2)-12π4-(1-q)y2∑k=1n-1(-1)k-1q(k2)(1+qk)U2k-1(y1-q/2)2k|≤|q|n(n-1)/2+|q|n(n+1)/22π(1-|q|n).Remark 2.6.
Using the assertion of the above corollary one can approximate the densityfH as well as function FH(y∣q)-1/2-1/πarcsin(y1-q/2) of PH by expressions of the type 4-(1-q)x2×polynomialinx with great accuracy. This expression is simple to analyze, simulate, and calculate interesting characteristics. Of course one should be aware that for small values of n(1-q/2π)4-(1-q)x2×(∑k=1n(-1)k-1q(k2)U2k-2(x1-q/2) is not nonnegative for all (1-q)x2≤4! To give a scent of how many n's are needed to obtain the given accuracy we solved numerically (using program Mathematica) the equation
(2.10)n|q|(n-1)(n-2)/2π(1-q2)2=ε,
for several values q and ε. Let us denote by N(q,ε) the solution of this equation:
(2.11)wehaveε=∖q=.1.4.7.9.99..013.594.977.7114.9356.73.0014.045.678.7316.5360.86.00014.766.269.6117.9764.70.
We also performed similar calculations for equation
(2.12)|q|n(n-1)/2+|q|n(n+1)/22π(1-|q|n)=ε,obtaining:ε=∖q=.1.4.7.9.99..012.33.35.19.533.0012.84.26.311.539.00013.24.737.313.344.


## 3. Simulation

There is an interesting problem of quick simulation of i.i.d. sequences drawn fromq-Gaussian distribution, using few realizations of i.i.d. standard uniform variates. One possibility is the rejection method  (see, e.g., [12]). It is not optimal in the sense that it uses least realizations of independent, uniform on [0,1] variates. But, as one can see below, it works.To apply this method one has to compare density of the generated variates with another density that has the property of being “easy generated” or another words i.i.d. sequences of variables having this control density are easily obtainable. In the case of densityfH such natural candidate is (1-q)(4-(1-q)x2)/2π. However this density is unimodal, while the densities fH for q below certain negative value are bimodal. This would lead to inefficient simulation method requiring many trial observations to be generated from (1-q)(4-(1-q)x2)/2π to obtain one observation from fH for sufficiently small q. That is why we decided to take as “easy generated” density the following one:
(3.1)fE(x∣q)=(1-q)(4-(1-q)x2)∏j=13((1+qj)2-(1-q)qjx2)2π[9]q[5]q,
defined for x∈[-2/1-q,2/1-q]. However to be sure that this distribution can be used one has to prove the following inequalities presented by the following Lemma.Lemma 3.1.
For-1<q<1 and x∈[-2/1-q,2/1-q] one has
(3.2)fH(x∣q)fE(x∣q)≤M(q),
where
(3.3)M(q)={(1+q)(1-q3)(1-q5)(1-q9)∏k=4∞(1-q2k)(1+qk)ifq∈(0,1),(1+q)(1-q3)(1-q5)(1-q9)∏k=4∞(1+|q|k)2(1-qk)ifq∈(-1,0).FunctionM(q) has the plot in Figure 2Figure 2
M(q)(0÷5) versus q(-1÷1).Now following [12] we can simulate sequences of independent random variables with q-Gaussian  distribution. If q=±1 then such simulation is trivial.Forq∈(-1,1) we use Lemma 3.1 and program Mathematica. We generated sequence of independent random variables from density fE by inversion method (see [12]), since fE(x∣q) can be integrated leading to cumulative distribution function (c.d.f.):
(3.4)∫-2/1-qxfE(y∣q)dy=D(q)x(1-q)(4-(1-q)x2)Q2(x3,q)+C(q)arcsin(x1-q2)=defFE(x∣q),
for |x|≤2/1-q, where Q2 denotes quadratic polynomial in x3 with coefficients depending on q, while the constants D and C are known functions of q. Recall that the inversion method requires solving numerically the sequence of equations FE(x∣q)=ri, where ri are observations drawn from standard uniform distribution.Since the functionFE(x∣q) is strictly increasing on its support and its derivative is known, there are no numerical problems in solving this equation. Due to efficient procedure “FindRoot” of Mathematica solving this equation is quick.Now let us recall how rejection method works in caseq∈(-1,1).Applying algorithm described in [12], the rule to get one realization of random variable having density fH is as follows.(1)
We generate two variables:X~fE(.) and Y~U(0,1).(2)
SetT=M(q)fE(X∣q)/fH(X∣q).(3)
IfYT>1 then set Z=X otherwise repeat (1) and (2).For details ofMathematica program realizing the above mentioned algorithm, see Appendix.To see how this algorithm works, we present two simulation results performed (consisting of2000 simulations) with q=-.8 (red dots) and q=.8 (green dots) in Figure 3.Figure 3
Simulation of i.i.d. sequences fromPH.Unfortunately this algorithm turns out to be very inefficient forq close to -1, more practically less than say -.85. One can see this by examining Figure 2. Values of M(q) are very large then, showing that one needs very large number of observations from density fE to obtain one observation from fH. Thus there is still an open question to generate efficiently observations from fH for values close to -1.One might be inclined to use formula (2.2) and inversion method applied to its finite approximation and again use procedure “FindRoot”. Well we applied this idea to simulate 2000 observations from PH for q=-.95. It worked giving the results in Figure 4:Figure 4
i.i.d. sequence fromPH by inversion method for q=-.95.We used procedure “Findroot” of Mathematica. It worked, as one can see, however it lasted quite a time to get the result.Besides, when we tried to get2000 observations from PH for q=-.97, numerical errors seemed to play an important role as one can notice judging from black dots that appeared between levels 0 and -.5 on the picture in Figure 5.Figure 5
i.i.d. sequence fromPH by inversion method for q=-.97.


## 4. Proofs

Proof of Theorem2.1.
Let us denotez=x1-q/2. Hence |z|<1. We have
(4.1)fH(z∣q)=(q)∞1-q4π1-z2∏k=0∞((1+qk)2-4z2qk).
Now let us notice that
(4.2)(1+qk)2-4z2qk=(1+qk-2zqk/2)(1+qk+2zqk/2)=((1-z2+iz)2+qk)((1-z2-iz)2+qk).
Now notice that since |z|<1, we see that |1-z2+iz|=1. Thus we can write 1-z2+iz=exp(iθ) where θ=arcsinz and also 1-z2=cosθ. Hence we can write
(4.3)(1+qk)2-4z2qk=(1+e2iθqk)(1+e-2iθqk),
and consequently
(4.4)fH(z∣q)=(q)∞1-q4πcosθ(-e2iθ)∞(1+e-2iθ)(-qe-2iθ)∞.
We will now use so-called “triple product identity” (see [13, Theorem 10.4.1., page 497]) that states in our setting, that
(4.5)(q)∞(-e2iθ)∞(-qe-2iθ)∞=∑k=-∞∞(-1)kq(k2)(-e2iθ)k=∑k=-∞∞qk(k-1)/2e2ikθ=1+∑k=1∞qk(k-1)/2e2ikθ+∑j=1∞qj(j+1)/2e-2ijθ=1+∑k=1∞q(k2)e2ikθ+∑k=2∞q(k2)e-2i(k-1)θ.
Now notice that
(4.6)(1+e-2iθ)cosθ=2cosθeiθcosθ=2e-iθ.
Hence,
(4.7)fH(z∣q)=1-q2πe-iθ(1+∑k=1∞q(k2)e2ikθ+∑k=2∞q(k2)e-2i(k-1)θ)=1-q2πe-iθ×(1+e2θ+∑k=2∞q(k2)(cos2kθ+cos2(k-1)θ)+i∑k=2∞q(k2)(sin2kθ-sin2(k-1)θ))=1-qπcosθ+1-qπ∑k=2∞q(k2)cos(2k-1)θ.
To return to variable z we have to recall definition of Chebyshev polynomials. Namely, we have
(4.8)cos((2k-1)θ)=cos((2k-1)arcsinz)=cos((2k-1)(π2-arccosz))=(-1)k+1sin((2k-1)arccosz)=(-1)k+11-z2U2k-2(z),
where Un(z) is the Chebyshev polynomial of the second kind. More precisely we have here
(4.9)Un(z)=sin((n+1)arccosz)1-z2.
It is well known, that sequence {Un} satisfies three-term recurrence equation
(4.10)Un+1(z)-2xUn(z)+Un-1(z)=0,
with U-1(z)=0,U0(z)=1, and can be calculated directly (see [14, Theorem 7.2, page 188]) as in (1.7). Thus we have shown that
(4.11)fH(z∣q)=1-qπ1-z2(∑k=1∞(-1)k-1q(k2)U2k-2(z)),
or equivalently
(4.12)fH(x∣q)=1-q2π4-(1-q)x2(∑k=1∞(-1)k-1q(k2)U2k-2(x1-q/2)).Proof of Corollary2.2.
We haveFH(y|q)=∫-2/1-qyfH(x)dx. Now we change variable to z=x1-q/2 and use (2.1). Thus
(4.13)FH(y∣q)=2π(∫-1y1-q/21-z2dz+∑k=2∞(-1)k-1q(k2)∫-1y1-q/21-z2U2k-2(z)dz).
We use now the following, easy to prove, formulae:  ∫-1y1-z2U2n(z)dz=1-y2(U2n+1(y)/(4n+4)-U2n-1(y)/(4n)) and ∫-1y1-x2dx=(π/2)+(1/2)arcsiny+(1/2)y1-y2 and after rearranging terms get (2.2).Proof of Corollary2.3.
LetfU(x∣q)=(1-q/2π)4-(1-q)x2 for |x|1-q≤2. Assertion (i) is obtained directly after noting that
(4.14)fH(x∣q)=fU(x∣q)∏k=1∞((1+qk)2-(1-q)x2qk)(1-qk).
Following (2.1), we get
(4.15)fH(x∣q)=fU(x∣q)(∑k=1∞(-1)k-1q(k2)U2k-2(x1-q2)).(ii) and (iii) are obtained by inserting x=0 and x=1/1-q in (2.1) and canceling out common factors. From (2.1) it follows also that values U2n(0) and U2n(1/2) will be needed. Keeping in mind (4.9) we see that U2n(0)=cos(nπ)=(-1)n and
(4.16)U2n(12)={1ifn=3m,0ifn=3m+1,-1ifn=3m+2.
On the other hand we see that
(4.17)∏k=1∞((1+qk)2-(1-q)x2qk)∏k=0∞(1-qk+1)|x=0=∏k=1∞(1-q2k)(1+qk),∏k=1∞((1+qk)2-(1-q)x2qk)∏k=0∞(1-qk+1)|x=1/1-q=∏k=1∞((1+qk)2-qk)∏k=1∞(1-qk)=∏k=1∞(1+qk+q2k)∏k=1∞(1-qk)=∏k=1∞(1-q3k).
(iv) Putting x=±2/1-q in (2.3) we get
(4.18)∏k=1∞(1-qk)3=∑k=1∞(-1)k-1q(k2)U2k-2(1).
Now recall thatU2k(1)=2k+1.
(v) To see this notice thatq-Hermite polynomials are orthogonal with respect to the measure with density fH. Thus we have
(4.19)∀m>0:∫-2/1-q-2/1-qHm(x∣q)fH(x∣q)dx=0.
Using (2.1) know that ∀m>0(4.20)∫-2/1-q-2/1-qHm(x∣q)4-(1-q)x2(1+∑k=2∞(-1)k+1q(k2)U2k-2(x1-q2))dx=0.
Observing that function fH is symmetric and q-Hermite polynomials of odd order are odd functions, we deduce that above mentioned identities are trivial for odd m. Thus, let us concentrate on even m. Introducing new variable z=x1-q/2, and multiplying both sides of this identity by (1-q)m/2 we get
(4.21)∀m>1:∫-11hm(z∣q)(1+∑k=2∞(-1)k+1q(k2)U2k-2(z))1-z2dz=0,
where hm(z∣q)=(1-q)m/2Hm(2z/1-q∣q). Polynomials hm are called continuous q-Hermite polynomials. It can be easily verified (following (1.6) that they satisfy the following three-term recurrence equation:
(4.22)hn+1(t∣q)=2thn(t∣q)-(1-qn)hn-1(t∣q),
with h-1=0,h0(z∣q)=1. Moreover, it is also known that (see, e.g., [13]),
(4.23)hn(cosθ∣q)=∑k=0n[nk]cos((n-2k)θ).
Let us change once more variables in (4.21) and put z=cosτ. Then, for all m>1,(4.24)∫0πhm(cosτ∣q)×(1+∑k=2∞(-1)k+1q(k2)U2k-2(cosτ))×sinτdτ=0,
or for all m>1,(4.25)∫0πhm(cosτ∣q)(∑k=1∞(-1)k-1q(k2)sin(2k-1)τsinτ)dτ=0.
Keeping in mind that 2sin(2k-1)τsinτ=cos(2k-2)τ-cos(2k)τ, we see that
(4.26)∀m>1:∫0πh2m(cosτ∣q)(∑k=1∞(-1)k-1q(k2)(cos(2k-2)τ-cos(2k)τ))dτ=0.
Now keeping in mind that ∫0πcos2mτdτ=0 for m>0 we see that
(4.27)∀m>1:∫0πh2m(cosτ∣q)(∑k=m+1∞(-1)k+1q(k2)(cos(2k-2)τ-cos(2k)τ))dτ=0.
On the other hand taking into account (4.23) we see that
(4.28)∫0πh2m(cosτ∣q)cos2kτdτ=π[2mm-k]q,
for k=0,1,…,m,. Hence we have (2.7).Proof of Corollary2.4.
Keeping in mind thatfH is symmetric with respect to x we deduce that the point of change of modality of fH must be characterized by the condition fH′′(0∣q0)=0. Calculating second derivative of the right hand side of (2.1) and remembering that (4-(1-q)x2)′|x=0=0 we end up with an equation0=-(1-q)/2∑k=0∞qk(k+1)/2+2(1-q)×∑k=1∞(-1)kk(k+1)qk(k+1)/2(-1)k+1. Now since 4k(k+1)+1=(2k+1)2 we get equation in Corollary 2.4 defining q0.To prove Lemma2.5 we need the following lemma.Lemma 4.1.
Suppose0<r<1 and n≥3. Then
(4.29)∑k≥n(2k-1)r(k2)≤2nrn(n-1)/2(1-r2)2.Proof.
Recall that for|ρ|<1 we have: ∑i≥1iρi-1=(1/(1-ρ))′=1/(1-ρ)2 and that ∑i=0mρi=(1-ρm+1)/(1-ρ). Thus we have
(4.30)∑k≥n(2k-1)r(k2)=r(n-1)(n-4)/2∑k≥n(2k-1)rk(k-1)/2-(n-1)(n-4)/2=r(n-1)(n-4)/2×∑k≥n(2k-1)r2k-2rk(k-1)/2-(n-1)(n-4)/2-2(k-1).
Now notice that k(k-1)/2-(n-1)(n-4)/2-2(k-1)=(1/2)(k-n)(k+n-5)≥0 for k≥n≥3. Hence
(4.31)∑k≥n(2k-1)r(k2)≤r(n-1)(n-4)/2∑k≥n(2k-1)r2k-2≤2r(n-1)(n-4)/2∑k≥nk(r2)k-1=2r(n-1)(n-4)/2ddr2(11-r2-1-r2n1-r2)=2r(n-1)(n-4)/2ddr2((r2)n1-r2)=2r(n-1)(n-4)/2n(r2)n-1-(n-1)(r2)n(1-r2)2≤2nrn(n-1)/2(1-r2)2.Proof of Lemma2.5.
(i) We have(4.32)Un(cosθ)=sin((n+1)θ)sinθ.
From this fact we deduce that sup|x|≤1|Un(x)|=(n+1). Now using Lemma (4.1) we have
(4.33)|fH(x∣q)-1-q2π4-(1-q)x2(1+∑k=2n-1(-1)k+1q(k2)U2k-2(x1-q2))|=|1-q2π4-(1-q)x2∑k≥n(-1)k+1q(k2)U2k-2(x1-q2)|≤1-q2π4-(1-q)x2∑k≥n|q|(k2)(2k-1)≤1-qπ2n|q|n(n-1)/2(1-q2)2.(ii)(4.34)|FH(y∣q)-12-1πarcsin(y1-q2)-4-(1-q)y22π∑k=1n-1(-1)k-1q(k2)U2n-2(y1-q2)|≤4-(1-q)y22π∑k=n∞|q|(k2)|U2n-2(y1-q2)|≤12π∑k=n∞|q|(k2)(1+|q|k),
since sup|x|≤11-x2|Un(x)|=1. Now to get (ii) we use routine transformations and sum two geometric series.Proof of Lemma3.1.
Notice that comparing definitions offH and fE we have
(4.35)fH(x∣q)=fE(x∣q)×1-q91-q1-q51-q×∏k=4∞((1+qk)2-(1-q)x2qk)×∏k=1∞(1-qk)=(1+q)(1-q3)(1-q5)(1-q9)×∏k=4∞((1+qk)2-(1-q)x2qk)(1-qk).
Now if q∈[0,1) we have
(4.36)((1+qk)2-(1-q)x2qk)(1-qk)≤(1+qk)2(1-qk)=(1-q2k)(1+qk).
If q∈(-1,0) then
(4.37)((1+qk)2-(1-q)x2qk)(1-qk)≤(1-qk){((1+qk)2-4qk)ifkisodd(1+qk)2ifkiseven=(1-qk)(1+|q|k)2,
since then sup|x|≤2/1-q((1+qk)2-(1-q)x2qk)=((1+qk)2-4qk)=(1-qk)2.



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*Source: 752430-2009-08-23.xml*