# On two parameter bivariate kernel built of $q-$ultraspherical polynomials
and other Lancaster type expansions of bivariate distribution

Paweł J. Szabłowski  Emeritus in Department of Mathematics and Information
Sciences,
Warsaw University of Technology
ul Koszykowa 75, 00-662 Warsaw, Poland<EMAIL_ADDRESS>
(Date: September, 2022)

###### Abstract.

Our most important result concerns the positivity of certain kernels built of
the so-called $q-$ultraspherical polynomials. Since this result apears at
first sight as primarily important for those who are working in orthogonal
polynomials, $q-$series theory and the so-called quantum polynomials, it might
have a limited number of interested researchers. That is why, we put our
result into a broader context. We recall the theory of Hilbert-Schmidt
operators, Lancaster expansions and their applications in Mathematical
statistics, or bivariate distributions absolutely continuous with respect to
the product of their marginal distributions leading to the generation of
Markov process with polynomial conditional moments (the main representative of
such processes is a famous Wiener process).

###### Key words and phrases:

bivariate kernels, Lancaster expansions, orthogonal polynomials, $q-$series,
$q-$ultraspherical , Hermite, $q-$Hermite, Laguerre polynomials

###### 2020 Mathematics Subject Classification:

Primary 47B07, 62H05; Secondary 47B32, 46N30

##  1\. Introduction

As stated in the Abstract our main result concerns the positivity of certain
bivariate kernels built of the so-called $q-$ultraspherical polynomials. The
proof is long and difficult involving certain identities from the so-called
$q-$series. That is why, we will present our results in a broader context that
includes, Hilbert-Schmidt operators, kernels built of orthogonal polynomials,
Markov processes with polynomial conditional moments and bivariate
distributions that are absolutely continuous with respect to the products of
their marginals and their applications in Mathematical statistics.

We will be dealing mostly with expressions of the type

$K(x,y)=\sum_{j=0}^{\infty}c_{j}a_{j}(x)b_{j}(y),$

where $\left\\{\alpha_{j}(x)\right\\}$ and $\left\\{\beta_{j}(y)\right\\}$ are
the sets of polynomials orthogonal with respect to some finite measures
respectively $d\alpha(x)$ and $d\beta(y).$ Since the terminology found in the
literature is somewhat confusing, we will call such expressions kernels,
symmetric when the two measures are the same (and consequently families
$\left\\{\alpha_{j}(x)\right\\}$ and $\left\\{\beta_{j}(y)\right\\}$ are the
same) or non-symmetric when the two measures are different. The most general
application of such kernels are the so-called Hilbert-Schmidt operators
considered in functional analysis. Namely, imagine that we have two Hilbert
spaces $L_{2}(d\alpha(x))$ and $L_{2}(d\beta(y))$ of functions that are square
integrable with respect to respectively $d\alpha$ and $d\beta$. Let
$\left\\{a_{j}(x)\right\\}$ and $\left\\{b_{j}(y)\right\\}$ be the orthonormal
bases of these spaces. Let us take a function $h$ say, from
$L_{2}(d\alpha(x)).$ It can be presented in the following form
$h(x)\allowbreak=\allowbreak\sum_{n\geq 0}h_{n}a_{n}(x),$ with $\sum_{n\geq
0}\left|h_{n}\right|^{2}<\infty.$ Then

$f(y)=\int K(x,y)h\left(x\right)d\alpha(x)=\sum_{n\geq 0}c_{n}h_{n}b_{n}(y),$

We can observe that if only, $\sum_{n\geq 0}\left|c_{n}\right|^{2}<\infty$
then $f\left(y\right)\in L_{2}(d\beta(y)).$ Moreover, $K(x,y)$ is a
Hilbert–Schmidt operator with the norm equal $\sum_{n\geq
0}\left|c_{n}\right|^{2}$.

If $K(x,y)$ is non-negative for almost all $x$ mod $d\alpha$ and $y$ mod
$d\beta$ then we talk about positive kernels, it is worth to mention that the
positive symmetric kernels are the source of stationary Markov processes with
all its conditional moments being polynomials in the conditioning random
variable. How to do it see, e.g., [29], [30] and [31]. The nice introduction
and the list of literature treating this subject was done in the recently
published in Stochastics paper [35]. How it is done will be explained
partially below, when we will be discussing the so-called Lancaster
expansions.

Another problem is to present kernels in the compact, simple forms. The
process of doing so is called summing the kernels. It turns out to be a very
hard problem. Only a few kernels are summed. A sample of a few positive,
summed kernels will be presented in Section 5.

We are going to present several, including, compactly supported, bivariate
distribution that allows the so-called Lancaster expansion. In the series of
papers [9], [11], [10], and [12], H. Lancaster analysed the family of
bivariate distributions which allow a special type of expansion of its Radon-
Nikodym derivative of these distributions with respect to the products of
their marginal measures. More precisely, if $d\mu(x,y)$ is the distribution in
question and $d\alpha(x)$ and $d\beta(x)$ are its marginal distribution and
moreover the following condition is satisfied

(1.1)
$\int\int\left(\frac{\partial^{2}\mu(x,y)}{\partial\alpha(x)\partial\beta(y)}\right)^{2}d\alpha(x)d\beta(y)<\infty,$

where the integration is over the support of the product measure $d\alpha
d\beta$, then the following expansion is convergent at least in mean-squares
with respect to the product measure:

(1.2) $d\mu(x,y)=d\alpha(x)d\beta(y)\sum_{j=0}^{\infty}c_{j}a_{j}(x)b_{j}(y).$

From the theory of orthogonal series, it follows, that if the condition (1.1)
is satisfied and we deal with expansion (1.2) convergent in mean-square, then
we have to have

$\sum_{n\geq 0}c_{n}^{2}<\infty,$

that is $\left\\{c_{n}\right\\}_{n\geq 0}\in l_{2},$ the space of square
summable sequences. When equipped in the following norm
$\left\|\mathbf{c}\right\|\allowbreak=\allowbreak\sum_{n\geq 0}c_{n}^{2}$
where: $\mathbf{c}\allowbreak=\allowbreak\left\\{c_{n}\right\\}_{n\geq 0},$
then $l_{2}$ becomes a Banach space.

In the expansion (1.2), $\left\\{a_{j}\right\\}$ and $\left\\{b_{j}\right\\}$
are orthonormal sequences with respect to respectively $d\alpha$ and $d\beta$.
We will call expansions of the form (1.2) satisfying a condition (1.1)
Lancaster expansion, briefly LE. One has to Notice, that expressions of the
type (1.2), more precisely of the form

$\sum_{j=0}^{\infty}c_{j}a_{j}(x)b_{j}(y),$

where $\left\\{a_{j}\right\\}$ and $\left\\{b_{j}\right\\}$ are two families
of polynomials are kernels, discussed above.

Obviously, for probabilists the most interesting are those kernels that are
nonnegative for certain ranges of $x$ and $y.$ Nonnegative kernels on some
subsets of $\mathbb{R}^{2}$ will be called Lancaster kernels briefly LK. One
can expect to be able to construct with the help of such a kernel a positive,
bivariate measure by the proper scaling. The term LE will be used as the most
precise and suitable for the probabilistic context of this paper.

Probably the first LE expansion was the following one:

$\displaystyle\exp(-(x^{2}+y^{2})/2)\sum_{j\geq
0}^{\infty}\frac{\rho^{j}}{j!}H_{j}(x)H_{j}(y)=$
$\displaystyle\exp(-(x^{2}-2\rho
xy+y^{2})/(2(1-\rho^{2}))/(2\pi\sqrt{1-\rho^{2}}),$

that is convergent for all $x,y\in\mathbb{R}$ and $\left|\rho\right|<1.$ (See
[13]).

As mentioned earlier, in Section 5, we will list all known to the author such
LE.

Why such expansions are important? Lancaster himself a long time ago pointed
out possible applications in mathematical statistics, hence we will not repeat
these arguments. In this paper, we will concentrate on applications in
distribution theory by indicating bivariate distributions having simple
structure. As shown in [34] bivariate distribution that allows LE and satisfy
the condition (1.1) constitute a family of bivariate distributions that have
the property that all its conditional moments of degree say $n$ are
polynomials of order $n$ in the conditioning random variable. More precisely,
for all random variables $(X,Y)$ having bivariate distribution given by (1.2)
we have for all $n\geq 0$

$E(a_{n}(X)|Y=y)=c_{n}b_{n}(y).$

We have immediate observation concerning coefficients
$\left\\{c_{n}\right\\}_{n\geq 0}$ that appear in the definitions of LK.
Namely, if we raise the restriction that LK of LE must lead to probability
(i.e. integrating to $1$) distributions we see that the coefficients
$\left\\{c_{n}\right\\}_{n\geq 0}$ form a convex cone in the space $L_{2}$ of
square summable sequences.

Other possible applications are in the theory of Markov processes. Namely
recall, that every Markov process $\left\\{X_{t}\right\\}_{t\in\mathbb{R}}$ is
completely defined by two families of measures. The first one is the family of
the so-called marginal distributions, i.e. family (indexed by the time
parameter $t)$ of one-dimensional distributions of a random variable $X_{t}.$
The other family of distributions is the family of the conditional
distributions of $X_{t+\tau}|X_{\tau}=x.$ One could utilize symmetric LE
almost immediately by finding positive constants $\gamma_{n}$ such that
$\exp(-t\gamma_{n})\allowbreak=\allowbreak c_{n}$ as it was done e.g. in [35]
and define a stationary Markov process.

There are many examples of LE that stem from the application of the
polynomials from the so-called Askey-Wilson family of polynomials. These
polynomials involve notions of the so-called $q-$ series theory, so in the
next section we will present basic notions and facts from this theory. The
traditional terminology calls polynomial appearing within $q-$series theory,
quantum polynomials, see, e.g., an excellent monograph [7] of Ismail.

The paper is organized as follows. In the next Section 2, we present notation
used traditionally within the $q-$series theory and also present a list of
polynomials mostly from the so-called Askey–Wilson (AW) scheme. It is
important to present these polynomials and their relationship to the
$q-$ultraspherical that is the main subject of the paper. In the proof of our
result, we will need the so-called connection coefficients between
$q-$ultraspherical and the other members of the AW scheme. The next Section 4
is dedicated to the proof of our main result i.e. summing and proving the
positivity of certain kernel built of $q-$ultraspherical polynomials, The next
Section 5 lists simple, known to the author, summed kernels both symmetric and
non-symmetric kernels. Finally, the last Section 6 contains longer, requiring
tedious calculations, proofs and other auxiliary results from $q-$series
theory.

##  2\. Notation, definition and some methods of obtaining LE

$q$ is a parameter. We will assume that $-1<q\leq 1$ unless otherwise stated.
The case $q\allowbreak=\allowbreak 1$ may not always be considered directly,
but sometimes as left-hand side limit ( i.e., $q\longrightarrow 1^{-}$). We
will point out these cases.

We will use traditional notations of the $q-$series theory, i.e.,

$\left[0\right]_{q}\allowbreak=\allowbreak
0,~{}\left[n\right]_{q}\allowbreak=\allowbreak
1+q+\ldots+q^{n-1}\allowbreak,\left[n\right]_{q}!\allowbreak=\allowbreak\prod_{j=1}^{n}\left[j\right]_{q},\text{with
}\left[0\right]_{q}!\allowbreak=1\text{,}$
$\genfrac{[}{]}{0.0pt}{}{n}{k}_{q}=\left\\{\begin{array}[c]{ccc}\frac{\left[n\right]_{q}!}{\left[n-k\right]_{q}!\left[k\right]_{q}!}&,&n\geq
k\geq 0\\\ 0&,&\text{otherwise}\end{array}\right.\text{.}$

$\binom{n}{k}$ will denote the ordinary, well-known binomial coefficient.

It is useful to use the so-called $q-$Pochhammer symbol for $n\geq 1$

$\left(a|q\right)_{n}=\prod_{j=0}^{n-1}\left(1-aq^{j}\right),~{}~{}\left(a_{1},a_{2},\ldots,a_{k}|q\right)_{n}\allowbreak=\allowbreak\prod_{j=1}^{k}\left(a_{j}|q\right)_{n}\text{,}$

with $\left(a|q\right)_{0}\allowbreak=\allowbreak 1$.

Often $\left(a|q\right)_{n}$ as well as
$\left(a_{1},a_{2},\ldots,a_{k}|q\right)_{n}$ will be abbreviated to
$\left(a\right)_{n}$ and
$\left(a_{1},a_{2},\ldots,a_{k}\right)_{n},$ if it will not cause
misunderstanding.

We will also use the following symbol $\left\lfloor n\right\rfloor$ to denote
the largest integer not exceeding $n$.

It is worth to mention the following four formulae, that are well-known, see,
e.g., [8](Introduction). Namely, the following formulae are true for
$\left|t\right|<1,$ $\left|q\right|<1$ (derived already by Euler, see [3]
Corollary 10.2.2)

(2.1) $\displaystyle\frac{1}{(t)_{\infty}}\allowbreak$
$\displaystyle=\allowbreak\sum_{k\geq 0}\frac{t^{k}}{(q)_{k}}\text{,}$ (2.2)
$\displaystyle(t)_{\infty}\allowbreak$ $\displaystyle=\allowbreak\sum_{k\geq
0}(-1)^{k}q^{\binom{k}{2}}\frac{t^{k}}{(q)_{k}},$ (2.3)
$\displaystyle\frac{1}{(t)_{n+1}}$ $\displaystyle=\sum_{j\geq
0}\genfrac{[}{]}{0.0pt}{}{n}{j}_{q}t^{j},$ (2.4)
$\displaystyle\left(t\right)_{n}$
$\displaystyle=\sum_{j}^{n}\genfrac{[}{]}{0.0pt}{}{n}{j}_{q}q^{\binom{j}{2}}(-t)^{j}\text{.}$

It is easy to Notice, that
$\left(q\right)_{n}=\left(1-q\right)^{n}\left[n\right]_{q}!$ and that

$\genfrac{[}{]}{0.0pt}{}{n}{k}_{q}\allowbreak=\allowbreak\allowbreak\left\\{\begin{array}[c]{ccc}\frac{\left(q\right)_{n}}{\left(q\right)_{n-k}\left(q\right)_{k}}&,&n\geq
k\geq 0\\\ 0&,&\text{otherwise}\end{array}\right.\text{.}$

The above-mentioned formula is just an example where direct setting
$q\allowbreak=\allowbreak 1$ is senseless however, the passage to the limit
$q\longrightarrow 1^{-}$ makes sense.

Notice, that, in particular,

(2.5) $\left[n\right]_{1}\allowbreak=\allowbreak
n,~{}\left[n\right]_{1}!\allowbreak=\allowbreak
n!,~{}\genfrac{[}{]}{0.0pt}{}{n}{k}_{1}\allowbreak=\allowbreak\binom{n}{k},~{}(a)_{1}\allowbreak=\allowbreak
1-a,~{}\left(a|1\right)_{n}\allowbreak=\allowbreak\left(1-a\right)^{n}$

and

(2.6)
$\left[n\right]_{0}\allowbreak=\allowbreak\left\\{\begin{array}[c]{ccc}1&\text{if}&n\geq
1\\\
0&\text{if}&n=0\end{array}\right.,~{}\left[n\right]_{0}!\allowbreak=\allowbreak
1,~{}\genfrac{[}{]}{0.0pt}{}{n}{k}_{0}\allowbreak=\allowbreak
1,~{}\left(a|0\right)_{n}\allowbreak=\allowbreak\left\\{\begin{array}[c]{ccc}1&\text{if}&n=0\\\
1-a&\text{if}&n\geq 1\end{array}\right..$

$i$ will denote imaginary unit, unless otherwise clearly stated. Let us define
also:

(2.7) $\displaystyle\left(ae^{i\theta},ae^{-i\theta}\right)_{\infty}$
$\displaystyle=\prod_{k=0}^{\infty}v\left(x|aq^{k}\right)\text{,}$ (2.8)
$\displaystyle\left(te^{i\left(\theta+\phi\right)},te^{i\left(\theta-\phi\right)},te^{-i\left(\theta-\phi\right)},te^{-i\left(\theta+\phi\right)}\right)_{\infty}$
$\displaystyle=\prod_{k=0}^{\infty}w\left(x,y|tq^{k}\right)\text{,}$ (2.9)
$\displaystyle\left(ae^{2i\theta},ae^{-2i\theta}\right)_{\infty}$
$\displaystyle=\prod_{k=0}^{\infty}l\left(x|aq^{k}\right)\text{,}$

where,

(2.10) $\displaystyle v(x|a)\allowbreak$ $\displaystyle=\allowbreak
1-2ax+a^{2},$ (2.11) $\displaystyle l(x|a)$ $\displaystyle=(1+a)^{2}-4x^{2}a,$
(2.12) $\displaystyle w(x,y|a)$
$\displaystyle=(1-a^{2})^{2}-4xya(1+a^{2})+4a^{2}(x^{2}+y^{2})$

and, as usually in the $q-$series theory, $x\allowbreak=\allowbreak\cos\theta$
and $y=\cos\phi$.

We will use also the following denotation:

$S(q)\overset{df}{=}\left\\{\begin{array}[c]{ccc}[-2/\sqrt{1-q},2/\sqrt{1-q}]&\text{if}&\left|q\right|<1\\\
\mathbb{R}&\text{if}&q=1\end{array}\right..$

###  2.1. Method of expansion of the ratio of densities

We will use through the paper the following way of obtaining infinite
expansions of the type

$\sum_{j\geq 0}a_{n}p_{n}(x),$

that are convergent almost everywhere on some subset of $\mathbb{R}$. Namely,
in sight of [19] the following setting is considered. Say we have two measures
on $\mathbb{R}$ both having densities $f$ and $g$. Suppose that, we know that
$\int(f/g)^{2}gdx$ is finite. Further suppose also that we know two families
of orthogonal polynomials $\left\\{\alpha_{n}\right\\}$ and
$\left\\{\beta_{n}\right\\}$, such that the first one is orthogonal with
respect to the measure having the density $f$ and the other is orthogonal with
respect to the measure having the density $g$. Then we know that $f/g$ can be
expanded in an infinite series

(2.13) $\sum_{n\geq 0}a_{n}\beta_{n}(x),$

that is convergent in $L^{2}(\mathbb{R},g)$. We know in particular that
$\sum_{n\geq 0}\left|a_{n}\right|^{2}<\infty$. If additionally

$\sum_{n\geq 0}\left|a_{n}\right|^{2}\log^{2}(n+1)<\infty,$

then by the Rademacher–Meshov theorem, we deduce that the series in question
converges not only in $L^{2},$ but also almost everywhere with respect to $g$.

Thus, we will get the condition $\sum_{n\geq 0}\left|a_{n}\right|^{2}<\infty$
satisfied for free. Moreover, in many cases we will have
$\left|a_{n}\right|^{2}\leq r^{n}$ for some $r<1$. Hence the condition
$\sum_{n\geq 0}\left|a_{n}\right|^{2}\log^{2}(n+1)<\infty$ is also naturally
satisfied. If one knows the connection coefficients between the families
$\left\\{\alpha_{n}\right\\}$ and $\left\\{\beta_{n}\right\\}$,i.e., a set of
coefficients $\left\\{c_{k,n}\right\\}_{n\geq 1,0\leq k\leq n}$ satisfying

$\beta_{n}\allowbreak(x)=\allowbreak\sum_{k=0}^{n}c_{k,n}\alpha_{k}(x),$

then $a_{n}\allowbreak=\allowbreak c_{0,n}/\int\beta_{n}^{2}gdx$. We will
refer to this type of reasoning as D(ensity) E(expansion) I(idea) (*,*) (that
is DEI(*,*)), where the stars point out to the connection coefficient formulae
and the formula for $\int\beta_{n}^{2}gdx$.

##  3\. Families of polynomials appearing in the paper including those forming
part of the Askey-Wilson scheme

All families of polynomials listed in this section are described in many
positions of literature starting from [1], [3], [7]. However, as it was
noticed by the author in [22], by changing the parameters to complex conjugate
and changing the usual range of all variables from $[-1,1]$ to $S(q)$, we
obtain polynomials from the AW scheme suitable for probabilistic applications.
Recently, in the review paper [33] and a few years earlier in [24] the author
described and analyzed the polynomials of this scheme with conjugate complex
parameters. Thus, we will refer to these two papers for details.

The families of orthogonal polynomials will be identified by their three-term
recurrences. Usually, the polynomials mentioned in such a three-term
recurrence will be monic (i.e., having a coefficient of the highest power of
the variable equal to $1$). The cases when the given three-term recurrence
leads to non-monic polynomials will be clearly pointed out. Together with the
three-term recurrence we will mention the measure, usually having density,
that makes a given family of polynomials orthogonal.

In order not to allow the paper to be too large, we will mention only basic
the properties of the polynomials involved. More properties and relationships
between used families of polynomials could be found in already mentioned
fundamental positions of literature like [1], [3], [7] or [8]. One has to
remark that the polynomials of the AW scheme are presented in their basic
versions where all ranges of the variables are confined to the segment
$[-1,1]$ in these positions in the literature. As mention before, for the
probabilistic applications more useful are versions where variables range over
$S(q).$ Then it is possible to pass with $q$ to $1$ and compare the results
with the properties of Hermite polynomial and Normal distribution which are
the reference points to all distribution comparisons in probability theory.

We recall these families of polynomials for the sake of completeness of the
paper.

###  3.1. Chebyshev polynomials

They are of two types denoted by $\left\\{T_{n}\right\\}$ and
$\left\\{U_{n}\right\\}$ called respectively of the first and second kind
satisfying the same three-term recurrence, for $n\geq 1$

$2xU_{n}(x)=U_{n+1}(x)+U_{n-1}(x),$

with different initial conditions $T_{0}(x)=1=U_{0}(x)$ and $T_{1}(x)=x$ and
$U_{1}(x)=2x.$ Obviously, they are not monic. They are orthogonal respectively
with respect to arcsine distribution with the density
$f_{T}(x)\allowbreak=\allowbreak\frac{1}{\pi\sqrt{1-x^{2}}}$ and to the
semicircle or Wigner distribution with the density
$f_{U}(x)\allowbreak=\allowbreak\frac{2}{\pi}\sqrt{1-x^{2}}.$ Besides we have
also the following orthogonal relationships:

$\displaystyle\int_{-1}^{1}T_{n}(x)T_{m}(x)f_{T}(x)dx\allowbreak$
$\displaystyle=\allowbreak\left\\{\begin{array}[c]{ccc}0&\text{if}&m\neq n\\\
1&\text{if}&m=n=0\\\ 2&\text{if}&m=n>0\end{array}\right.,$
$\displaystyle\int_{-1}^{1}U_{n}(x)U_{m}(x)f_{U}(x)dx\allowbreak$
$\displaystyle=\allowbreak\left\\{\begin{array}[c]{ccc}0&\text{if}&m\neq n\\\
1&\text{if }&m=n\end{array}\right..$

More about their properties one can read in [14].

###  3.2. Hermite polynomials

We will consider here only the so-called probabilistic Hermite polynomials
namely the ones satisfying the following three-term recurrence:

$H_{n+1}(x)=xH_{n}(x)-nH_{n-1}(x),$

with initial conditions $H_{0}(x)=1,$ $H_{1}(x)\allowbreak=\allowbreak x.$
They are monic and orthogonal with respect to the Normal $N(0,1)$ distribution
with the well-known density
$f_{N}(x)\allowbreak=\allowbreak\frac{1}{\sqrt{2\pi}}\exp(-x^{2}/2).$ They
satisfy the following orthogonal relationship:

(3.1)
$\int_{-\infty}^{\infty}H_{n}(x)H_{m}(x)f_{N}(x)dx=\left\\{\begin{array}[c]{ccc}0&\text{if
}&m\neq n\\\ n!&\text{if}&m=n\end{array}\right..$

###  3.3. $q-$Hermite polynomials

They will be denoted $H_{n}(x|q),$ called $q-$Hermite polynomials are defined
by the following three-term recurrence:

(3.2)
$xH_{n}\left(x|q\right)=H_{n+1}\left(x|q\right)\allowbreak+\left[n\right]_{q}H_{n-1}\left(x\right),$

for $n\geq 1$ with $H_{-1}\left(x|q\right)\allowbreak=\allowbreak 0$,
$H_{1}\left(x|q\right)\allowbreak=\allowbreak 1$. Notice, that now polynomials
$H_{n}(x|q)$ are monic and also that

$\lim_{q\rightarrow 1^{-}}H_{n}(x|q)=H_{n}(x).$

Let us define the following nonnegative function where we denoted

(3.3)
$f_{h}\left(x|q\right)=\frac{2\left(q\right)_{\infty}\sqrt{1-x^{2}}}{\pi}\prod_{k=1}^{\infty}l\left(x|q^{k}\right),$

with, as before, $l(x|a)\allowbreak=\allowbreak(1+a)^{2}-4x^{2}a$ and let us
define a new density

$f_{N}\left(x|q\right)\allowbreak=\allowbreak\left\\{\begin{array}[c]{ccc}\sqrt{1-q}f_{h}(x\sqrt{1-q}/2|q)/2&\text{if}&\left|q\right|<1\\\
\exp\left(-x^{2}/2\right)/\sqrt{2\pi}&\text{if}&q=1\end{array}\right..$

Note, that $f_{N}$ is non-negative if only $x\in S(q).$ It turns out that we
have the following orthogonal relationship:

(3.4)
$\int_{S(q)}H_{n}(x|q)H_{m}(x|q)f_{N}(x|q)dx=\left\\{\begin{array}[c]{ccc}0&\text{if
}&n\neq m\\\ \left[n\right]_{q}!&\text{if}&n=m\end{array}\right..$

Notice, that if $X\sim f_{N}(x|q)$, then as $H_{1}(x|q)\allowbreak=\allowbreak
x,$ $H_{2}(x|q)\allowbreak=\allowbreak x^{2}-1,$ we deduce that
$EX\allowbreak=\allowbreak 0$ and $EX^{2}\allowbreak=\allowbreak 1.$

###  3.4. Big $q-$Hermite polynomials

More on these polynomials can be read in [33], [28] or [24]. Here, we will
only mention that these polynomials are, for example, defined by the
relationship:

$H_{n}(x|a,q)\allowbreak=\sum_{j=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{j}_{q}q^{\binom{j}{2}}(-a)^{j}H_{n-j}\left(x|q\right),$

where $H_{n}$ denotes a continuous $q-$Hermite polynomials, defined above and
$a\in(-1,1)$. They satisfy the following three-term recurrence:

$xH_{n}(x|a,q)=H_{n+1}(x|a,q)+aq^{n}H_{n}(x|a,q)+[n]_{q}H_{n-1}(x|a,q),$

with $H_{-1}(x|a,q)\allowbreak=\allowbreak 0$ and
$H_{0}(x|a,q)\allowbreak=\allowbreak 1.$ It is known in particular that the
characteristic function of the polynomials $\left\\{H_{n}\right\\}$ for
$\left|q\right|<1$ is given by the formula:

$\sum_{n=0}^{\infty}\frac{t^{n}}{\left[n\right]_{q}!}H_{n}\left(x|q\right)=\varphi\left(x|t,q\right),$

where

(3.5)
$\varphi\left(x|t,q\right)\allowbreak=\allowbreak\frac{1}{\prod_{k=0}^{\infty}\left(1-(1-q)xtq^{k}+(1-q)t^{2}q^{2k}\right)}.$

Notice, that it is convergent for $t$ such that $\left|t\sqrt{1-q}\right|<1$
and $x\in S(q)$. These polynomials satisfy the following orthogonality
relationship:

$\int_{S\left(q\right)}H_{n}\left(x|a,q\right)H_{m}\left(x|a,q\right)f_{bN}\left(x|a,q\right)dx\allowbreak=\allowbreak\left[n\right]_{q}!\delta_{m,n},$

where

$f_{bN}(x|a,q)=f_{N}(x|q)\varphi(x|a,q).$

There exists one more interesting relationship between $q-$Hermite and big
$q-$Hermite polynomials. Namely, following paper of Carlitz [5] we get

$H_{n}(x|a,q)\sum_{j\geq
0}\frac{a^{j}}{\left[j\right]_{q}!}H_{j}(x|q)=\sum_{j\geq
0}\frac{a^{j}}{\left[j\right]_{q}!}H_{j+n}(x|q).$

Note that this expansion nicely compliments and generalizes the results of
Proposition 3.

###  3.5. Continuous $q-$ultraspherical polynomials

These polynomials first were considered by Rogers in 1894 (see [17], [16],
[18]). They were defined for $\left|x\right|\leq 1$ by the three-term
recurrence e.g. given in [8](14.10.19). We have the celebrated connection
coefficient formula for the Rogers polynomials see [7],(13.3.1), that will be
of use in the sequel, of course after proper rescaling.

(3.6)
$C_{n}\left(x|\gamma,q\right)\allowbreak=\allowbreak\sum_{k=0}^{\left\lfloor
n/2\right\rfloor}\frac{\beta^{k}\left(\gamma/\beta\right)_{k}\left(\gamma\right)_{n-k}\left(1-\beta
q^{n-2k}\right)}{(q)_{k}\left(\beta
q\right)_{n-k}\left(1-\beta\right)}C_{n-2k}\left(x|\beta,q\right).$

We will consider them, with a different scaling of the variable $x\in S(q)$
and parameter $\beta\in[-1,1]$. We have

(3.7)
$R_{n}(x|\beta,q)=\left[n\right]_{q}!C_{n}(x\sqrt{1-q}/2|\beta,q)(1-q)^{n/2},$

Then, their three-term recurrence becomes

(3.8) $\left(1-\beta
q^{n}\right)xR_{n}\left(x|\beta,q\right)=R_{n+1}\left(x|\beta,q\right)+\left(1-\beta^{2}q^{n-1}\right)\left[n\right]_{q}R_{n-1}\left(x|\beta,q\right)\text{.}$

with $R_{-1}(x|\beta,q)\allowbreak=\allowbreak 0,$
$R_{0}(x|\beta,q)\allowbreak=\allowbreak 1.$ Let us define the following
density

(3.9)
$f_{C}(x|\beta,q)\allowbreak=\allowbreak\frac{(\beta^{2})_{\infty}}{(\beta,\beta
q)_{\infty}}f_{h}\left(x|q\right)/\prod_{j=0}^{\infty}l\left(x|\beta
q^{j}\right),$

where $f_{h}$ is define by (3.3). Let us modify it by considering

(3.10) $f_{R}\left(x|\beta,q\right)=\sqrt{1-q}f_{C}(x\sqrt{1-q}/2|\beta,q)/2.$

Then we have the following orthogonal relationship satisfied by polynomials
$\left\\{R_{n}\right\\}.$

(3.11)
$\displaystyle\int_{S(q)}R_{n}\left(x|\beta,q\right)R_{m}\left(x|\beta,q\right)f_{R}\left(x|\beta,q\right)dx$
(3.14) $\displaystyle=\left\\{\begin{array}[c]{ccc}0&\text{if}&m\neq n\\\
\frac{\left[n\right]_{q}!(1-\beta)\left(\beta^{2}\right)_{n}}{\left(1-\beta
q^{n}\right)}&\text{if}&m=n\end{array}\right..$

Obviously, polynomials $\left\\{R_{n}\right\\}$ are not monic. We can easily
Notice, that the coefficient by $x^{n}$ in $R_{n}$ is $(\beta)_{n}.$ Hence by
defining a new sequence of polynomials

(3.15) $V_{n}(x|\beta,q)\allowbreak=\allowbreak R_{n}(x|\beta,q)/(\beta)_{n},$

we get the sequence of monic versions of the polynomials
$\left\\{R_{n}\right\\}.$

Since our main result concerns summing certain kernel built of polynomials
$\left\\{R_{n}\right\\},$ we will need the following Lemma where the
relationships between polynomials $\left\\{R_{n}\right\\}$ and
$\left\\{H_{n}\right\\}$ are exposed.

######  Lemma 1.

1)

(3.16) $\displaystyle R_{n}(x|r,q)$ $\displaystyle=\sum_{k=0}^{\left\lfloor
n/2\right\rfloor}\frac{\left[n\right]_{q}!}{\left[k\right]_{q}!\left[n-2k\right]_{q}!}q^{\binom{k}{2}}(-r)^{k}\left(r\right)_{n-k}H_{n-2k}(x|q),$
(3.17) $\displaystyle H_{n}(x|q)$ $\displaystyle=\sum_{k=0}^{\left\lfloor
n/2\right\rfloor}\frac{\left[n\right]_{q}!(1-rq^{n-2k})}{\left[k\right]_{q}!\left[n-2k\right]_{q}!(1-r)(1-rq)_{n-k}}r^{k}R_{n-2k}(x|r,q).$
(3.18)
$\displaystyle\int_{S\left(q\right)}R_{n}\left(x|\beta,q\right)R_{m}\left(x|\beta,q\right)f_{R}\left(x|\beta,q\right)dx$
(3.21) $\displaystyle=\left\\{\begin{array}[c]{ccc}0&\text{if}&m\neq n\\\
\frac{\left[n\right]_{q}!\left(\beta^{2}\right)_{n}(1-\beta)}{\left(1-\beta
q^{n}\right)}&\text{if}&m=n\end{array}\right.,$

(3.22) $\displaystyle\int_{S(q)}H_{n}(x)R_{m}(x|\beta,q)f_{N}(x|q)dx=$ (3.25)
$\displaystyle\left\\{\begin{array}[c]{ccc}0&\text{if}&n>m\text{ or }n+m\text{
is odd}\\\
q^{\binom{(m-n)/2}{2}}\frac{\left[m\right]_{q}!(-\beta)^{(m-n)/2}}{\left[(m-n)/2\right]_{q}!}\left(\beta\right)_{(m+n)/2}&\text{if}&\text{otherwise}\end{array}\right..$

and

(3.26) $\displaystyle\int_{S(q)}H_{n}(x)R_{m}(x|\beta,q)f_{R}(x|\beta,q)dx=$
(3.29) $\displaystyle\left\\{\begin{array}[c]{ccc}0&\text{if}&m>n\text{ or
}\left|n-m\right|\text{ is odd}\\\
\frac{\beta^{(n-m)/2}\left(\beta^{2}\right)_{m}\left[n\right]_{q}!}{(1-\beta)\left[(n-m)/2\right]_{q}!\left(\beta
q\right)_{(n+m)/2}}&\text{if}&\text{otherwise}\end{array}\right..$

where:

(3.30) $\displaystyle f_{R}(x|\beta,q)=(1-\beta)f_{CN}(x|x,\beta,q)=$
$\displaystyle
f_{N}(x|q)\frac{\left(\beta^{2}\right)_{\infty}}{\left(\beta\right)_{\infty}\left(\beta
q\right)_{\infty}\prod_{j=0}^{\infty}\left((1+\beta
q^{j}\right)^{2}-(1-q)\beta q^{j}x^{2}))}=$
$\displaystyle(1-\beta)f_{N}(x|q)\sum_{n\geq
0}\frac{\beta^{n}H_{n}^{2}(x|q)}{\left[n\right]_{q}!}=(1-\beta)f_{N}(x|q)\sum_{n\geq
0}\frac{\beta^{n}H_{2n}(x|q)}{\left[n\right]_{q}!\left(\beta\right)_{n+1}}.$

(3.31) $f_{N}(x|q)=f_{R}(x|\gamma,q)\sum_{n\geq
0}(-\gamma)^{n}q^{\binom{n}{2}}\frac{(\gamma)_{n}(1-\gamma
q^{2n})}{\left[n\right]_{q}!(1-\gamma)(\gamma^{2})_{2n}}R_{2n}(x|\gamma,q).$

2) We have

(3.32) $\displaystyle R_{n}(x|r,q)R_{m}(x|r,q)$
$\displaystyle=\sum_{k=0}^{\min(n,m)}\genfrac{[}{]}{0.0pt}{}{m}{k}_{q}\genfrac{[}{]}{0.0pt}{}{n}{k}_{q}\left[k\right]_{q}!\frac{\left(r\right)_{m-k}\left(r\right)_{n-k}\left(r\right)_{k}\left(r^{2}\right)_{n+m-k}\left(1-rq^{n+m-2k}\right)}{\left(1-r\right)\left(rq\right)_{n+m-k}\left(r^{2}\right)_{m+n-2k}}\times$
$\displaystyle R_{n+m-2k}(x|r,q),$

(3.33) $\displaystyle H_{m}(x|q)R_{n}(x|r,q)=$
$\displaystyle\sum_{s=0}^{\left\lfloor(n+m)/2\right\rfloor}\genfrac{[}{]}{0.0pt}{}{n}{s}_{q}\left[s\right]_{q}!H_{n+m-2s}(x|q)\sum_{k=0}^{s}\genfrac{[}{]}{0.0pt}{}{m}{s-k}_{q}\genfrac{[}{]}{0.0pt}{}{n-s}{k}_{q}(-r)^{k}q^{\binom{k}{2}}(r)_{n-k},$
(3.34) $\displaystyle H_{m}(x|q)R_{n}(x|r,q)=$
$\displaystyle\sum_{u=0}^{\left\lfloor(n+m)/2\right\rfloor}\frac{\left[n\right]_{q}!\left[m\right]_{q}!(1-rq^{n+m-2u})}{\left[u\right]_{q}!\left[n+m-2u\right]_{q}!(1-r)}R_{n+m-2u}(x|r,q)\sum_{s=0}^{u}\genfrac{[}{]}{0.0pt}{}{u}{s}_{q}\frac{r^{u-s}}{\left(rq\right)_{n+m-u-s}}\times$
$\displaystyle\sum_{k=0}^{s}\genfrac{[}{]}{0.0pt}{}{s}{k}_{q}\genfrac{[}{]}{0.0pt}{}{m+m-2s}{m+k-s}_{q}q^{\binom{k}{2}}(-r)^{k}\left(r\right)_{n-k}.$

###### Proof.

1) (3.17) and (3.16) are adaptations of (3.6) with either
$\beta\allowbreak=\allowbreak 0$ or $\gamma\allowbreak=\allowbreak 0.$ When we
consider $\beta\allowbreak=\allowbreak 0,$ one has to careful and notice, that

$\beta^{n}(\gamma/\beta)_{n}=\prod_{j=0}^{n-1}(\beta-\gamma
q^{j})\rightarrow(-\gamma)^{n}q^{\binom{n}{2}},$

where the limit is taken when $\beta\rightarrow 0.$ When rescaling. to $S(q)$
we use formula (3.7). Formulae (3.22) and (3.26) follow almost directly
expansions respectively (3.16), (3.17) and the fact that

$\int_{S(q)}H_{n}(x|q)H_{m}(x|q)f_{N}(x|q)dx=\left[n\right]_{q}!\delta_{nm}.$

in the first case and (3.18) in the second. Formula (3.18) is given in
[21](Proposition 1(3.3)). To get (3.31) we use DEI(3.16,3.1).

2) Again (3.32) is an adaptation of the well-known formulae derived by Rogers
himself by the end of the nineteenth century concerning polynomial $C_{n}$
related to polynomials $R_{n}$ by the formula (3.7). Formula (3.33) appeared
in the version for polynomials $h$ and $C$ in [33](8.3) but in fact, it was
proved by Al-Salam and Ismail. in [2]. Formula (3.34) is obtained directly by
inserting (3.17) into (3.33). ∎

###  3.6. Al-Salam–Chihara polynomials.

Originally, they were defined by the $3$ parameters $q$ and two others $a$ and
$b$ both from the segment $[-1,1],$ for $\left|x\right|\leq 1$. We will
consider $a$ and $b$ complex conjugate such that
$ab\allowbreak=\allowbreak\rho^{2}$ and
$a+b\allowbreak=\allowbreak\frac{y}{\sqrt{1-q}}\rho$ and $x,y\in
S\left(q\right).$ Then they will be denoted $P_{n}(x|y,\rho,q).$ It is known
see, e.g., [24], that they satisfy the following three-term recurrence:

(3.35) $P_{n+1}(x|y,\rho,q)=(x-\rho
yq^{n})P_{n}(x|y,\rho,q)-(1-\rho^{2}q^{n-1})[n]_{q}P_{n-1}(x|y,\rho,q),$

with $P_{-1}(x|y,\rho,q)\allowbreak=\allowbreak 0$ and
$P_{0}(x|y,\rho,q)\allowbreak=\allowbreak 1.$

These polynomials are orthogonal with respect to the measure with the
following density:

(3.36)
$f_{CN}\left(x|y,\rho,q\right)=f_{N}(x|q)\frac{(\rho^{2})_{\infty}}{W(x,y|\rho,q)},$

where

(3.37)
$W(x,y|\rho,q)\allowbreak=\allowbreak\prod_{k=0}^{\infty}W\left(x,y|\rho
q^{k}\right)\allowbreak,$

and
$W(x,y|\rho)\allowbreak=\allowbreak(1-\rho^{2})^{2}\allowbreak-\allowbreak(1-q)\rho
xy(1+\rho^{2})\allowbreak+\allowbreak\rho^{2}(1-q)(x^{2}+y^{2}).$ One can show
easily that

(3.38) $f_{CN}(x|y,\rho,q)\rightarrow\exp\left(-\frac{(x-\rho
y)^{2}}{2(1-\rho^{2})}\right)/\sqrt{2\pi(1-\rho^{2})},$

as $q\rightarrow 1,$ that is it tends to the density of the normal $N(\rho
y,1-\rho^{2})$ distribution. That is why $f_{CN}$ is called conditional
$q-$Normal.

The orthogonal relation for these polynomials are the following:

(3.39)
$\int_{S\left(q\right)}P_{n}(x|y,\rho,q)P_{m}\left(x|y,\rho,q\right)f_{CN}\left(x|y,\rho,q\right)dx=\left\\{\begin{array}[c]{ccc}0&\text{if}&m\neq
n\\\
\left[n\right]_{q}!\left(\rho^{2}\right)_{n}&\text{if}&m=n\end{array}\right..$

Another fascinating property of the distribution $f_{CN}$ is the following
Chapman-Kolmogorov property:

(3.40)
$\int_{S\left(q\right)}f_{CN}\left(z|y,\rho_{1},q\right)f_{CN}\left(y|x,\rho_{2},q\right)dy=f_{CN}\left(x|z,\rho_{1}\rho_{2},q\right).$

It has been noticed in [24] that the following relationships combine
ultraspherical and other mentioned above families of polynomials.

The relationship between the two families of polynomials
$\left\\{H_{n}\right\\}$ and $\left\\{P_{n}\right\\}$ are the following It has
been shown in [4],

(3.41) $\displaystyle P_{n}\left(x|y,\rho,q\right)$
$\displaystyle=\sum_{j=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{j}_{q}\rho^{n-j}B_{n-j}\left(y|q\right)H_{j}\left(x|q\right),$
(3.42) $\displaystyle H_{n}\left(x|q\right)$
$\displaystyle=\sum_{j=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{j}_{q}\rho^{n-j}H_{n-j}\left(y|q\right)P_{j}\left(x|y,\rho,q\right),$

where polynomials $\left\\{B_{n}\right\\}$ satisfy the following three-term
recurrence

$B_{n+1}(x|q)=-xq^{n}B_{n}(x|q)+q^{n-1}\left[n\right]_{q}B_{n-1}(x|q),$

with $B_{-1}(x|q)\allowbreak=\allowbreak 0,$
$B_{0}(x|q)\allowbreak=\allowbreak 1$. It has been noticed in [24] or [28]
that the following particular case are true.

######  Proposition 1.

For $n\geq 0$ and, remembering that the polynomials $H_{n}$, $T_{n}$ and
$U_{n}$ defined above, we have

i) $R_{n}\left(x|0,q\right)\allowbreak=\allowbreak H_{n}\left(x|q\right)$,

ii)
$R_{n}\left(x|q,q\right)\allowbreak=\allowbreak\left(q\right)_{n}U_{n}\left(x\sqrt{1-q}/2\right),$

iii)
$\lim_{\beta->1^{-}}\frac{R_{n}\left(x|\beta,q\right)}{\left(\beta\right)_{n}}\allowbreak=\allowbreak
2\frac{T_{n}\left(x\sqrt{1-q}/2\right)}{(1-q)^{n/2}}$,

iv)
$R_{n}(x|\beta,1)\allowbreak=\allowbreak\left(\frac{1+\beta}{1-\beta}\right)^{n/2}H_{n}\left(\sqrt{\frac{1-\beta}{1+\beta}}x\right)$,

v)
$R_{n}(x|\beta,0)\allowbreak=\allowbreak(1-\beta)U_{n}(x/2)-\beta(1-\beta)U_{n-2}(x/2)$.

vi) $R_{n}(x|\beta,q)=P_{n}(x|x,\beta,q)$, where $P_{n}$ is defined by its
three-term recurrence (3.35).

Having done all those preparations we are ready to present our main result.

##  4\. Positive, summable kernel built of $q-$ultraspherical polynomials

One can find in the literature attempts to sum the kernels for
$q-$ultraspherical polynomials like, e.g., [6] or [15]. However, the kernel we
will present has a simple sum and moreover depends on two parameters, more
precisely on the symmetric function of two parameters, not on only one.

Let us denote by $w_{n}(m,r_{1},r_{2},q)$ the following symmetric polynomial
of degree $2n$ and by $\phi_{n}(r_{1},r_{2},q)$ the following rational,
symmetric function, both in $r_{1}$ and $r_{2}$ that will often appear in the
sequel:

(4.1) $\displaystyle w_{n}(m,r_{1},r_{2},q)$
$\displaystyle=\sum_{s=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{s}_{q}r_{1}^{s}\left(q^{m}r_{2}^{2}\right)_{s}r_{2}^{n-s}\left(q^{m}r_{1}^{2}\right)_{n-s},$
(4.2) $\displaystyle\phi_{n}(r_{1},r_{2},q)$
$\displaystyle=\frac{w_{n}(0,r_{1},r_{2},q)}{\left(r_{1}^{2}r_{2}^{2}\right)_{n}}.$

Let us immediately notice, that

$w_{n}(m,r_{1},r_{2},q)=q^{-nm/2}w_{n}(0,r_{1}q^{m/2},r_{2}q^{m/2},q).$

######  Theorem 1.

The following symmetric bivariate kernel is nonnegative on $S(q)\times S(q)$

$\sum_{n\geq
0}\phi_{n}(r_{1},r_{2},q)\frac{(1-r_{1}r_{2}q^{n})}{\left[n\right]_{q}!(r_{1}^{2}r_{2}^{2})_{n}(1-r_{1}r_{2}q)}R_{n}(x|r_{1}r_{2},q)R_{n}(y|r_{1}r_{2},q),$

where $\left\\{R_{n}(x|\beta,q\right\\}_{n\geq 0}$ are the so-called
$q-$ultraspherical polynomials. Functions $\left\\{w_{n}\right\\}$ are given
by (4.1) and $r_{1},r_{2}\in(-1,1)$.

Moreover, we have

(4.3) $\displaystyle f_{R}(x|r_{1}r_{2},q)f_{R}(y|r_{1}r_{2},q)\sum_{n\geq
0}\frac{\phi_{n}(r_{1},r_{2},q)(1-r_{1}r_{2}q^{n})}{\left[n\right]_{q}!(r_{1}^{2}r_{2}^{2})_{n}(1-r_{1}r_{2}q)}R_{n}(x|r_{1}r_{2},q)R_{n}(y|r_{1}r_{2},q)$
$\displaystyle=(1-r_{1}r_{2})f_{CN}\left(y|x,r_{1},q\right)f_{CN}\left(x|y,r_{2},q\right).$

where $f_{CN}\left(x|y,r_{2},q\right)$ denotes the so-called conditional
$q-$Normal distribution, defined above by (3.36).

Denoting by $\hat{R}_{n}(x|r_{1}r_{2},q)\allowbreak=\allowbreak
R_{n}(x|r_{1}r_{2},q)\sqrt{1-r_{1}r_{2}q^{n}}/\sqrt{\left[n\right]_{q}!\left(r_{1}^{2}r_{2}^{2}\right)_{n}(1-r_{1}r_{2})}$
the orthonormal version of the polynomials $R_{n}$ we get more friendly
version of our result

(4.4)
$\displaystyle(1-r_{1}r_{2})f_{CN}\left(y|x,r_{1},q\right)f_{CN}\left(x|y,r_{2},q\right)$
$\displaystyle=f_{R}(x|r_{1}r_{2},q)f_{R}(y|r_{1}r_{2},q)\sum_{n\geq
0}\phi_{n}(r_{1},r_{2},q)\hat{R}_{n}(x|r_{1}r_{2},q)\hat{R}_{n}(y|r_{1}r_{2},q).$

######  Corollary 1.

i) Setting $r_{1}\allowbreak=\allowbreak\rho$ and
$r_{2}\allowbreak=\allowbreak 0$ we have the following expansion which is true
for all $x,y\in S(q),\left|\rho\right|<1,$ $-1<q\leq 1.$

$f_{N}(x|q)f_{N}(y|q)\sum_{n\geq
0}\frac{\rho^{n}}{\left[n\right]_{q}!}H_{n}(x|q)H_{n}(y\\}q)=f_{N}(y|q)f_{CN}(x|y,\rho,q).$

In other words, the well-known Poisson-Mehler expansion formula is a
particular case of (4.3).

ii) Let us set $r_{1}\allowbreak=\allowbreak r\allowbreak=\allowbreak-r_{2}.$
We have then

$\phi_{n}(r,-r,q)=\left\\{\begin{array}[c]{ccc}0&\text{if}&n\text{ is odd}\\\
r^{2k}\left(q|q^{2}\right)_{k}/\left(r^{4}q|q^{2}\right)_{k}&\text{if}&n=2k\end{array}\right.,$

and consequently we get

$\displaystyle(1+r^{2})f_{CN}\left(y|x,r,q\right)f_{CN}\left(x|y,-r,q\right)=$
$\displaystyle f_{R}(x|-r^{2},q)f_{R}(y|-r^{2},q)\sum_{k\geq
0}r^{2k}\left(q|q^{2}\right)_{k}/\left(r^{4}q|q^{2}\right)_{k}\hat{R}_{n}(x|-r^{2},q)\hat{R}_{n}(y|-r^{2},q).$

iii)

$\displaystyle\frac{1+r_{1}r_{2}}{2\pi(1-r_{1}r_{2})}\exp(-\frac{(1-r_{1}r_{2})x^{2}}{2(1+r_{1}r_{2})}-\frac{(1-r_{1}r_{2})y^{2}}{2(1+r_{1}r_{2})})\times$
$\displaystyle\sum_{n\geq
0}\left(\frac{r_{1}+r_{2}}{\left(1+r_{1}r_{2}\right)(1-r_{1}^{2}r_{2}^{2})}\right)^{n}\left(\frac{1+r_{1}r_{2}}{1-r_{1}r_{2}}\right)^{n}H_{n}\left(x\sqrt{\frac{1-r_{1}r_{2}}{1+r_{1}r_{2}}}\right)H_{n}\left(y\sqrt{\frac{1-r_{1}r_{2}}{1+r_{1}r_{2}}}\right)$
$\displaystyle=(1-r_{1}r_{2})\exp\left(-\frac{\left(x-r_{1}y\right)^{2}}{2\left(1-r_{1}\right)^{2}}--\frac{\left(y-r_{2}x\right)^{2}}{2\left(1-r_{2}\right)^{2}}\right)/(2\pi(1-r_{1})(1-r_{2})).$

###### Proof.

i) We use the fact that $R_{n}(x|0,q)\allowbreak=\allowbreak H_{n}(x|q)$ and
$f_{N}(x|0,q)\allowbreak=\allowbreak f_{N}(x|q)$ and also that
$\sum_{j=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{j}_{q}r_{1}^{j}\left(r_{2}^{2}\right)_{j}r_{2}^{n-j}\left(r_{1}^{2}\right)_{n-j}$
$\allowbreak=\allowbreak r_{1}^{n}$ when $r_{2}\allowbreak=\allowbreak 0.$

ii) We have

$\displaystyle\phi_{n}(r,-r,q)=\frac{1}{\left(-r^{2}\right)_{n}}r^{n}\sum_{j=0}^{n}(-1)^{j}(r^{2})_{j}\left(r^{2}\right)_{n-j}$
$\displaystyle=\left\\{\begin{array}[c]{ccc}0&\text{if}&n\text{ is odd}\\\
r^{2k}\left(q|q^{2}\right)_{k}\left(r^{2}|q^{2}\right)_{k}/\left(-r^{2}\right)_{2k}&\text{if}&n=2k\end{array}\right..$

iii) We use (3.38) with $x\allowbreak=\allowbreak y,$ Proposition (1), the
fact that in this case
$\phi_{n}(r_{1},r_{2},1)\allowbreak=\allowbreak(\frac{r_{1}+r_{2}}{1+r_{1}r_{2}})^{n}$
and the fact that polynomials $H_{n}(\sqrt{a}x)$ are orthogonal with respect
to the measure with the density$\exp(-\alpha x^{2}/2)/\sqrt{2\pi\alpha}.$ ∎

Before we present a somewhat complicated proof of this theorem let us
formulate and/or prove some auxiliary results.

######  Theorem 2.

For all $x,y\in S(q),$ $r_{1}r_{2}\in(-1,1)$ and $-1<q\leq 1$ we have

(4.5) $\displaystyle 0$
$\displaystyle\leq(1-r_{1}r_{2})f_{CN}\left(y|x,r_{1},q\right)f_{CN}\left(x|y,r_{2},q\right)$
$\displaystyle=f_{N}(x|q)f_{R}(y|r_{1}r_{2},q)\sum_{n\geq
0}\frac{1}{\left[n\right]_{q}!}H_{n}(x|q)D_{n}(y|r_{1},r_{2},q),$

where

(4.6)
$D_{n}(y|r_{1},r_{2},q)=\sum_{j=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{j}_{q}r_{1}^{n-j}r_{2}^{j}\left(r_{1}\right)_{j}H_{n-j}(y|q)R_{j}(y|r_{1}r_{2},q)/\left(r_{1}^{2}r_{2}^{2}\right)_{j}.$

We also have

(4.7)
$\int_{S(q)}(1-r_{1}r_{2})f_{CN}\left(y|x,r_{1},q\right)f_{CN}\left(x|y,r_{2},q\right)dx=f_{R}(y|r_{1}r_{2},q).$

###### Proof.

This Theorem is composed of the results that appeared in [22] and, e.g., [21].
Namely, in [22] the following result was proved (Thm.3 (3.4))

$\displaystyle\allowbreak\frac{f_{CN}\left(z|x,\rho_{2},q\right)f_{CN}\left(x|y,\rho_{1},q\right)f_{N}\left(y|q\right)}{f_{CN}\left(z|y,\rho_{1}\rho_{2},q\right)f_{N}\left(y|q\right)}$
$\displaystyle=f_{N}\left(x|q\right)\sum_{j=0}^{\infty}\frac{1}{\left[j\right]_{q}!}H_{j}\left(x|q\right)C_{j}\left(y,z|\rho_{1},\rho_{2},q\right),$

where

$C_{n}\left(y,z|\rho_{1},\rho_{2},q\right)=\sum_{s=0}^{n}\allowbreak\genfrac{[}{]}{0.0pt}{}{n}{s}_{q}\rho_{1}^{n-s}\rho_{2}^{s}\left(\rho_{1}^{2}\right)_{s}H_{n-s}\left(y|q\right)P_{s}\left(z|y,\rho_{1}\rho_{2},q\right)/(\rho_{1}^{2}\rho_{2}^{2})_{s}.$

Now in [21] has been noticed (Proposition 1(3.1)) that
$(1-r)f_{CN}(y|y,r,q)\allowbreak=\allowbreak f_{R}(y|r,q)$ where $f_{R}$ is
the density of the measure that makes polynomials $\left\\{R_{n}\right\\}$
orthogonal. Thus, by replacing $\rho_{1}$ and $\rho_{2}$ by $r_{1}$ and
$r_{2}$ and identifying $y$ and $z$ we get (4.5). Let us denote now

$D_{n}(y|\rho_{1},\rho_{2},q)=C_{n}(y,y|\rho_{1},\rho_{2},q),$

∎

######  Remark 1.

$(1-r_{1}r_{2})f_{CN}\left(y|x,r_{1},q\right)f_{CN}\left(x|y,r_{2},q\right)$
is a symmetric function with respect both $x$ and $y$ as well as $r_{1}$ and
$r_{2}.$

To see this, let us refer to the definition of the density $f_{CN}$. Namely,
we have

$\displaystyle(1-r_{1}r_{2})f_{CN}\left(y|x,r_{1},q\right)f_{CN}\left(x|y,r_{2},q\right)$
$\displaystyle=(1-r_{1}r_{2})f_{N}(x|q)f_{N}(y|q)/(W(x,y|r_{1},q)W(x,y|r_{2},q))$

where $W(x,y|r_{1},q)$ is defined by (3.37).

The next several partial results require very special and tedious calculations
that can be interested only for those who are working on $q-$series theory.
That is why we will shift all such results, although important, to the last
section containing some proofs and any auxiliary material from $q-$series
theory, in order not to blur the logic of the arguments leading to the final
result.

Our following result presents the functions $\left\\{D_{n}\right\\}$ given by
(4.6) as a combination of only polynomials $\left\\{R_{n}\right\\}$ or in
other words expands each $D_{n}$ in the basis of $\left\\{R_{n}\right\\}.$ The
proof is simple ideologically, but externally hard in terms of specialized
calculations, which is why we will shift it to the last section.

######  Proposition 2.

We have for all $n\geq 0,$ $r_{1},r_{2}\in(-1,1)$ and $-1<q\leq 1$:

(4.8) $D_{n}(y|r_{1},r_{2},q)=\sum_{u=0}^{\left\lfloor
n/2\right\rfloor}\frac{\left[n\right]_{q}!(1-r_{1}r_{2}q^{n-2u})(r_{1}r_{2})^{u}}{\left[u\right]_{q}!\left[n-2u\right]_{q}!\left(r_{1}r_{2}\right)_{n-u+1}}R_{n-2u}(y|r_{1}r_{2},q)\gamma_{n,u}(r_{1},r_{2},q),$

where

(4.9) $\displaystyle\gamma_{n,u}(r_{1},r_{2},q)$
$\displaystyle=\frac{1}{(r_{1}^{2}r_{2}^{2})_{2n}}\sum_{m=0}^{u}\genfrac{[}{]}{0.0pt}{}{u}{m}_{q}\left(r_{2}^{2}\right)_{m}\left(r_{1}r_{2}q^{n-u-m+1}\right)_{m}$
$\displaystyle\times\left(r_{1}^{2}\right)_{m}\left(r_{1}r_{2}\right)_{m}w_{n-2m}(m,r_{1}r_{2},q),$

is well defined for all $n\geq 0$ and $u\leq 0\leq\left\lfloor
n/2\right\rfloor.$

###### Proof.

The long, tedious proof is shifted to Section 6. ∎

Before we formulate the next result let us formulate the following simple
result.

######  Proposition 3.

For all $x\in S(x),$ $\left|r\right|<1,$ $\left|q\right|\leq 1,$ $m\geq 0$ we
have

(4.10) $\displaystyle(1-r)f_{N}(x|q)\sum_{j\geq
0}H_{j}(x|q)H_{j+m}(x|q)=f_{R}(x|q,q)R_{m}(x|r,q)/\left(r^{2}\right)_{m}=$
(4.11) $\displaystyle
f_{R}(x|r,q)\sum_{k=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{k}_{q}(-r)^{k}q^{\binom{k}{2}}H_{m-j}(x|q)R_{j}(x|r,q)/\left(r^{2}\right)_{j}=$
(4.12) $\displaystyle(1-r)f_{N}(x|q)\sum_{s\geq
0}\frac{r^{s}}{\left[s\right]_{q}!(r)_{m+s+1}}H_{2s+m}(x|q).$

###### Proof.

We start with the result of [22](Lemma 3i)) where we set
$x\allowbreak=\allowbreak y$ and make other adjustments and get directly
(4.10) and (4.11).

Now, to get (4.12) we proceed as follows

$\displaystyle f_{N}(x|q)\sum_{j\geq
0}\frac{r^{j}}{\left[j\right]_{q}!}H_{j}(x|q)H_{j+m}(x|q)$
$\displaystyle=f_{N}(x|q)\sum_{j\geq
0}\frac{r^{j}}{\left[j\right]_{q}!}\sum_{k}^{j}\genfrac{[}{]}{0.0pt}{}{j}{k}_{q}\genfrac{[}{]}{0.0pt}{}{j+m}{k}_{q}\left[k\right]_{q}!H_{2j+m-2k}(x|q)$
$\displaystyle=f_{N}(x|q)\sum_{j\geq
0}\frac{r^{j}}{\left[j\right]_{q}!}\sum_{k}^{j}\genfrac{[}{]}{0.0pt}{}{j}{k}_{q}\genfrac{[}{]}{0.0pt}{}{j+m}{k}_{q}\left[k\right]_{q}!H_{2j+m-2k}(x|q)$
$\displaystyle=f_{N}(x|q)\sum_{j\geq
0}\frac{r^{j}}{\left[j\right]_{q}!}\sum_{s=0}^{j}\genfrac{[}{]}{0.0pt}{}{j}{s}_{q}\genfrac{[}{]}{0.0pt}{}{j+m}{m+s}_{q}\left[j-s\right]_{q}!H_{2s+m}(x|q)$
$\displaystyle=f_{N}(x|q)\sum_{s\geq
0}\frac{r^{s}}{\left[s\right]_{q}!}H_{2s+m}(x|q)\sum_{j\geq
s}r^{j-s}\genfrac{[}{]}{0.0pt}{}{j-s+(m+s)}{m+s}_{q}$
$\displaystyle=f_{N}(x|q)\sum_{s\geq
0}\frac{r^{s}}{\left[s\right]_{q}!(r)_{m+s+1}}H_{2s+m}(x|q).$

∎

######  Lemma 2.

We have $n\geq 0:$

$\displaystyle\int_{S(q)}R_{n}(x|r_{1}r_{2},q)(1-r_{1}r_{2})f_{CN}\left(y|x,r_{1},q\right)f_{CN}\left(x|y,r_{2},q\right)dx$
$\displaystyle=\phi_{n}(r_{1},r_{2},q)R_{n}(y|r_{1}r_{2},q),$
$\displaystyle\int_{S(q)}R_{n}(y|r_{1}r_{2},q)(1-r_{1}r_{2})f_{CN}\left(y|x,r_{1},q\right)f_{CN}\left(x|y,r_{2},q\right)dx$
$\displaystyle=\phi_{n}(r_{1},r_{2},q)R_{n}(x|r_{1}r_{2},q).$

Further, for all $n\geq 0$ and $k\leq\left\lfloor n/2\right\rfloor$

$\gamma_{n,k}(r_{1},r_{2},q)=\phi_{n-2k}(r_{1},r_{2},q).\allowbreak$

In particular we have

$\displaystyle\gamma_{2k,k}(r_{1},r_{2},q)$ $\displaystyle=1,$
$\displaystyle\gamma_{n,0}(r_{1},r_{2},q)$
$\displaystyle=\phi_{n}(r_{1},r_{2},q).$

###### Proof.

The long, detailed proof by induction is shifted to Section 6. ∎

###### Proof of the Theorem 1.

Now, it is enough to recall results of say [34] to get our expansion. ∎

Now we can apply our results and complement the results of [21]. Let us
recall, that in this paper the following $3-$dimensional distribution

$f_{3D}(x,y,z|\rho_{12},\rho_{13},\rho_{23},q)=(1-r)f_{CN}(x|y,\rho_{12},q)f_{CN}(y|z,\rho_{23},q)f_{CN}(z|x,\rho_{13},q),$

where we denoted $r\allowbreak=\allowbreak\rho_{12}\rho_{23}\rho_{13},$ has
the property that all its conditional moments are the polynomials in the
conditioning random variable(s). Hence, it is a compactly supported
generalization of the Normal distribution. Let us recall also, that the one-
dimensional marginals are the same and equal to $f_{R}(.|r,q)$. Moreover, the
two-dimensional distributions are the following,, e.g., the marginal
distribution of $(Y,Z)$ is equal
$(1-r)f_{CN}(y|z,\rho_{23},q)f_{CN}(z|y,\rho_{12}\rho_{13},q).$ Consequently,
we can now expand it in the following way:

$\displaystyle(1-r)f_{CN}(y|z,\rho_{23},q)f_{CN}(z|y,\rho_{12}\rho_{13},q)$
$\displaystyle=f_{R}(x|r,q)f_{R}(y|r,q)\sum_{n\geq
0}\phi_{n}(\rho_{23},\rho_{13}\rho_{12},q)\hat{R}_{n}(x|r,q)\hat{R}_{n}(y|r,q),$

and consequently simplify the formula for the conditional moment $E(.|Z),$
namely, we have

(4.13)
$E(\hat{R}_{n}(Y|r,q)|Z)\allowbreak=\allowbreak\phi_{n}(\rho_{23},\rho_{13}\rho_{12},q)\hat{R}_{n}(Z|r,q),$

as it follows from our main result.

As a corollary, we get the following nice identities that seem to be of their
own interest.

######  Corollary 2.

i) For all $r_{1},r_{2},q\in\mathbb{C}$, $n\geq 0,$ $0\leq u\leq\left\lfloor
n/2\right\rfloor$ we have

$\displaystyle\sum_{m=0}^{u}\genfrac{[}{]}{0.0pt}{}{u}{m}_{q}\left(r_{2}^{2}\right)_{m}\left(r_{1}r_{2}q^{n-u-m+1}\right)_{m}\left(r_{1}^{2}\right)_{m}\left(r_{1}r_{2}\right)_{m}w_{n-2m}(m,r_{1},r_{2},q)$
$\displaystyle=\left(r_{1}^{2}r_{2}^{2}q^{n-2u}\right)_{2u}w_{n-2u}(0,r_{1},r_{2},q),$

since we have polynomials on both sides of the equations.

ii) For all $\left|r_{1}\right|,\left|r_{2}\right|<1,q\in S(q)$, $n\geq 0$

$\displaystyle\left|\phi_{n}(r_{1},r_{2},q)\right|$ $\displaystyle\leq 1,$
$\displaystyle\sum_{n\geq 0}\left|\phi_{n}(r_{1},r_{2},q)\right|^{2}$
$\displaystyle<\infty.$

###### Proof.

i) The proof follows directly the definition of $\phi_{n}$ and its proved
properties. ii) The fact that $\sum_{n\geq
0}\left|\phi_{n}(r_{1},r_{2},q)\right|^{2}<\infty,$ follows directly the fact
that we are dealing with the mean-square expansion. The fact that
$\left|\phi_{n}(r_{1},r_{2},q)\right|\leq 1$, follows the probabilistic
interpretation of our result. Namely recall, (4.13) with
$r_{1}\allowbreak=\allowbreak\rho_{23},$
$r_{2}\allowbreak=\allowbreak\rho_{13}\rho_{12}$ and the fact that
$E\hat{R}_{n}(Y|r_{1}r_{2},q)\allowbreak=\allowbreak 0$ and
$E\hat{R}_{n}^{2}(Y|r_{1}r_{2},q)\allowbreak=\allowbreak 1$. Now recall, the
well-known fact that the variance of the conditional expectation of a random
variable, does not exceed its variance. Consequently, we have

$\displaystyle 1=E\hat{R}_{n}^{2}(r_{1}r_{2},q)\geq
E(E(\hat{R}_{n}(Y|r,q)|Z))^{2}=$
$\displaystyle\phi_{n}(r_{1},r_{2},q)^{2}E\hat{R}_{n}^{2}(r_{1}r_{2},q)=\phi_{n}(r_{1},r_{2},q)^{2}.$

∎

##  5\. List of Lancaster kernels

###  5.1. Symmetric kernels

Below, we list symmetric Lancaster kernels that can be easily obtained using,
mentioned in the Introduction, the idea of expansion of the ratio of two
densities. The list below has simple sums and moreover sometimes lead to
smooth stationary Markov processes.

1\. We start with the following LK called the Poisson-Mehler kernel.

(5.1) $f_{CN}(x|y,\rho,q)/f_{N}(x|q)=\sum_{n\geq
0}\frac{\rho^{n}}{\left[n\right]_{q}!}H_{n}(x|q)H_{n}(y|q).$

Its justification is IDE(3.42,3.4). It leads to the so-called $q-$Ornstein-
Uhlenbeck process, for details see, e.g., [20].

2\. One should mention the following particular case of the above-mentioned
formula, that is $q\mathbb{\allowbreak=\allowbreak}0$. Then
$H_{n}(x|0)\allowbreak=\allowbreak U_{n}(x/2)$ and
$\left[n\right]_{0}!\allowbreak=\allowbreak 1$ and finally we get for all
$\left|x\right|,\left|y\right|\leq 2,\left|\rho\right|<1$:

(5.2) $\sum_{n\geq
0}\rho^{n}U_{n}(x/2)U_{n}(y/2)\allowbreak=\allowbreak\frac{1-\rho^{2}}{(1-\rho^{2})^{2}-\rho(1+\rho^{2})xy+\rho^{2}(x^{2}+y^{2})}.$

Recently, in [27] (5.2) formula has been proven by other means together with
the following one:

3.

$\sum_{n\geq
0}\rho^{n}T_{n}(x/2)T_{n}(y/2)=\frac{4(1-\rho^{2})-\rho(3+\rho^{2})xy+2\rho^{2}(x^{2}+y^{2})}{4((1-\rho^{2})^{2}-\rho(1+\rho^{2})xy+\rho^{2}(x^{2}+y^{2}))}.$

4\. The following expansion appeared recently in [35](4.7):

$\displaystyle 1+2\sum_{n\geq 1}\rho^{n^{2}}T_{n}(x)T_{n}(y)=$
$\displaystyle\frac{1}{2}\left(\theta(\rho,\arccos(x)-\arccos(y)\right)+\theta(\rho,\arccos(x)+\arccos(y)),$

where $\theta$ denotes the Jacobi Theta function.

5\. The following expansion appeared recently in [35](unnamed formula):

$\displaystyle\frac{4}{\pi^{2}}\sqrt{(1-x^{2})(1-y^{2})}\sum_{n\geq
0}\rho^{n(n+2)}U_{n}(x)U_{n}(y)=$
$\displaystyle\frac{1}{\rho\pi^{2}}\left(\theta(\rho,\arccos(x)-\arccos(y)\right)-\theta(\rho,\arccos(x)+\arccos(y)),$

where, as before, $\theta$ denotes Jacobi Theta function.

6\. It is well-known that the following LE is also true

$\displaystyle(1-\rho)\sum_{n\geq
0}\frac{n!}{\Gamma(n+\alpha)}L_{n}^{\alpha}(x)L_{n}^{\alpha}(y)\rho^{n}=$
$\displaystyle\exp\left(-\rho\frac{x+y}{1-\rho}\right)I_{\alpha-1}\left(\frac{2(x,y\rho)^{1/2}}{1-\rho}\right)/(xy\rho)^{(\alpha-1)/2},$

where $L_{n}^{(\alpha)}(x)$ denotes generalized Laguerre polynomials i.e. the
ones orthogonal with respect to the measure with the density
$\exp(-x)x^{\alpha-1}/\Gamma(\alpha)$ for $x\geq 0$ and $\alpha\geq 1.$
$I_{\alpha-1}(x)$ denotes a modified Bessel function of the first kind. This
kernel also leads to a smooth, stationary Markov process, as shown in [35].

Let us remark that, following [35], the kernels mentioned at points 1., 4.,
5., 6. allow to generate stationary Markov processes with polynomial
conditional moments that allow continuous path modification.

###  5.2. Non-symmetric kernels

It has to be underlined that the list of non-symmetric kernels presented below
is far from being complete. There is nice paper, namely [25] mentioning more
of them. However, they are in a very complicated form, interesting only to
specialists in $q-$series theory.

1\. We start with the following, known, but recently presented with general
setting in [27]:

$\sum_{n\geq
0}\rho^{n}U_{n}(x/2)T_{n}(y/2)=\frac{2(1-\rho^{2})+\rho^{2}x^{2}-\rho
xy}{2((1-\rho^{2})^{2}-\rho(1+\rho^{2})xy+\rho^{2}(x^{2}+y^{2}))}.$

2\. In [24](3.12) we have the kernel that after adjusting it to ranges of
$x,y,z$ equal to $S\left(q\right)$ and utilizing and the fact that

$\displaystyle
f_{CN}(y|x,\rho_{1},q)f_{CN}(x|z,\rho_{2},q)/(f_{CN}(y|z,\rho_{1}\rho_{2},q)f_{CN}(x|y,\rho_{1},q))\allowbreak$
$\displaystyle=\allowbreak f_{CN}(z|y,\rho_{2},q)/f_{CN}(z|x,\rho_{2},q),$

we get finally the following non-symmetric kernel:

$\sum_{j\geq
0}\frac{\rho_{2}^{j}}{\left[j\right]_{q}!\left(\rho_{1}^{2}\rho_{2}^{2}\right)_{j}}P_{j}\left(z|y,\rho_{1}\rho_{2},q\right)P_{j}\left(x|y,\rho_{1},q\right)=\frac{f_{CN}\left(z|x,\rho_{2},q\right)}{f_{CN}(z|y,\rho_{1}\rho_{2},q)},$

true for all $\left|x\right|,\left|y\right|,\left|z\right|\in
S(q),\left|\rho_{1}\right|,\left|\rho_{1}\right|<1$ and $\left|q\right|\leq
1,$ where $P_{j}(x|y,\rho,q)$ and $f_{CN}(x|y,\rho,q)$ are defined
respectively by (3.35) and (3.36).

2s. Using the fact that

$P_{n}(x|y,\rho,1)\allowbreak=\allowbreak(1-\rho^{2})^{n/2}H_{n}\left((x-\rho
y)/\sqrt{1-\rho^{2}}\right),$

and

$f_{CN}(x|y,\rho,1)\allowbreak=\frac{1}{\sqrt{2\pi(1-\rho^{2})}}\exp\left(-\frac{\left(x-\rho
y\right)^{2}}{2(1-\rho^{2})}\right),$

by [33](8.24) and [33](8.32) we get the following kernel

$\displaystyle\sum_{j\geq
0}\frac{\rho_{2}^{j}(1-\rho_{1}^{2})^{j/2}}{j!(1-\rho_{1}^{2}\rho_{2}^{2})^{j/2}}H_{j}(\frac{z-\rho_{1}\rho_{2}y}{\sqrt{1-\rho_{1}^{2}\rho_{2}^{2}}}H_{j}(\frac{x-\rho_{1}y}{\sqrt{1-\rho_{1}^{2}}})$
$\displaystyle=\sqrt{\frac{1-\rho_{1}^{2}\rho_{2}^{2}}{1-\rho_{2}^{2}}}\exp\left(\frac{(z-\rho_{1}\rho_{2}y)^{2}}{2(1-\rho_{1}^{2}\rho_{2}^{2})}-\frac{(z-\rho_{2}x)^{2}}{2(1-\rho_{2}^{2})}\right).$

3\. The following non-symmetric kernel was presented in [23]. For
$\left|b\right|>\left|a\right|,\left|q\right|<1,$ $x,y\in S\left(q\right)$

$\sum_{n\geq
0}\frac{a^{n}}{\left[n\right]_{q}!b^{n}}H_{n}(x|a,q)H_{n}\left(y|b,q\right)\allowbreak=\allowbreak\left(a^{2}/b^{2}\right)_{\infty}\prod_{k=0}^{\infty}\frac{\left(1-(1-q)xaq^{k}+(1-q)a^{2}q^{2k}\right)}{w\left(x,y,q^{k}a/b|q\right)},$

where $H_{n}(x|a,q)$ denotes the so-called big $q-$Hermite polynomials. Now
let us change a bit this expansion by renaming its parameters. Let us denote
$a/b\allowbreak=\allowbreak\rho.$ Then we can recognize that
$\prod_{k=0}^{\infty}(1-(1-q)axq^{k}+(1-q)a^{2}q^{2k})\allowbreak=\allowbreak
1/\varphi(x|a,q)$ and

$\left(a^{2}/b^{2}\right)_{\infty}\prod_{k=0}^{\infty}\frac{1}{w\left(x,y,q^{k}a/b|q\right)}=f_{CN}(x|y,a/b,q)/f_{N}(x|q).$

See also [26].

##  6\. Complicated proofs and auxiliary facts from $q-$series theory.

We start with the series of some auxiliary, simplifying formulae.

######  Lemma 3.

The generating function of the sequence
$\left\\{w_{n}(0,r_{1},r_{2},q)\right\\}$ is the following:

$\sum_{n\geq
0}\frac{t^{n}}{\left(q\right)_{n}}w_{n}(0,r_{1},r_{2},q)=\frac{\left(tr_{1}r_{2}^{2}\right)_{\infty}\left(tr_{2}r_{1}^{2}\right)_{\infty}}{\left(tr_{1}\right)_{\infty}\left(tr_{2}\right)_{\infty}}.$

Consequently, we have, for all $n\geq 0,$ $r_{1},r_{2}\in(-1,1)$ and $-1<q\leq
1$:

(6.1) $\displaystyle w_{n}(m,r_{1},r_{2},q)$
$\displaystyle=q^{-nm/2}w_{n}(0,r_{1}q^{m/2},r_{2}q^{m/2})$
$\displaystyle=\sum_{s=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{s}_{q}r_{1}^{n-s}r_{2}^{s}\left(q^{m}r_{1}r_{2}\right)_{s}\left(q^{m}r_{1}r_{2}\right)_{n-s}.$

and

$\sum_{n\geq
0}\frac{t^{n}}{\left(q\right)_{n}}w_{n}(m,r_{1},r_{2},q)=\frac{\left(tq^{m}r_{1}r_{2}^{2}\right)_{\infty}\left(tq^{m}r_{2}r_{1}^{2}\right)_{\infty}}{\left(tr_{1}\right)_{\infty}\left(tr_{2}\right)_{\infty}}$

###### Proof.

We have

$\sum_{n\geq
0}\frac{t^{n}}{\left(q\right)_{n}}w_{n}(0,r_{1},r_{2},q)=\sum_{n\geq
0}\sum_{s=0}^{n}\frac{\left(tr_{1}\right)^{s}}{\left(q\right)_{s}}\left(r_{2}^{2}\right)_{s}\frac{\left(tr_{2}\right)^{n-s}}{\left(q\right)_{n-s}}\left(r_{1}^{2}\right)_{n-s}.$

Now, changing the order of summation we get

$\sum_{n\geq
0}\frac{t^{n}}{\left(q\right)_{n}}w_{n}(0,r_{1},r_{2},q)=\sum_{s\geq
0}\frac{\left(tr_{1}\right)^{s}}{\left(q\right)_{n}}\left(r_{2}^{2}\right)_{s}\sum_{n\geq
s}\frac{\left(tr_{2}\right)^{n-s}}{\left(q\right)_{n}}\left(r_{1}^{2}\right)_{n-s}=\frac{\left(tr_{1}r_{2}^{2}\right)_{\infty}\left(tr_{2}r_{1}^{2}\right)_{\infty}}{\left(tr_{1}\right)_{\infty}\left(tr_{2}\right)_{\infty}}.$

Now, we Notice, that

$\frac{\left(tr_{1}r_{2}^{2}\right)_{\infty}\left(tr_{2}r_{1}^{2}\right)_{\infty}}{\left(tr_{1}\right)_{\infty}\left(tr_{2}\right)_{\infty}}=\frac{\left(tr_{1}r_{2}^{2}\right)_{\infty}\left(tr_{2}r_{1}^{2}\right)_{\infty}}{\left(tr_{2}\right)_{\infty}\left(tr_{1}\right)_{\infty}},$

from which follows directly (6.1). ∎

######  Lemma 4.

The following identity is true for all $n\geq 0,$ $r_{1},r_{2}\in(-1,1)$ and
$-1<q\leq 1$:

$\sum_{s=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{s}_{q}r_{1}^{n-s}r_{2}^{s}\left(r_{1}^{2}\right)_{s}\left(r_{1}r_{2}\right)_{s}/\left(r_{1}^{2}r_{2}^{2}\right)_{s}=\frac{1}{\left(r_{1}^{2}r_{2}^{2}\right)_{n}}w_{n}(0,r_{1},r_{2},q)=\phi_{n}(r_{1},r_{2},q).$

###### Proof.

We will prove it by the generating function method. Namely, we will show that
the generating function of the sequence
$\left\\{\sum_{s=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{s}_{q}r_{1}^{n-s}r_{2}^{s}\left(r_{1}^{2}\right)_{s}\left(r_{1}r_{2}\right)_{s}\left(r_{1}^{2}r_{2}^{2}q^{s}\right)_{n-s}\right\\}_{n\geq
0}$ is equal to
$\frac{\left(tr_{1}r_{2}^{2}\right)_{\infty}\left(tr_{2}r_{1}^{2}\right)_{\infty}}{\left(tr_{1}\right)_{\infty}\left(tr_{2}\right)_{\infty}}.$
We have taking

$\displaystyle\sum_{n\geq
0}\frac{t^{n}}{\left(q\right)_{n}}\sum_{s=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{s}_{q}r_{1}^{n-s}r_{2}^{s}\left(r_{1}^{2}\right)_{s}\left(r_{1}r_{2}\right)_{s}\left(r_{1}^{2}r_{2}^{2}q^{s}\right)_{n-s}$
$\displaystyle=\sum_{s\geq
0}\frac{\left(tr_{2}\right)^{s}}{\left(q\right)_{s}}\left(r_{1}^{2}\right)_{s}\left(r_{1}r_{2}\right)_{s}\sum_{n\geq
s}\frac{\left(tr_{1}\right)^{n-s}}{\left(q\right)_{n-s}}\left(r_{1}^{2}r_{2}^{2}q^{s}\right)_{n-s}$
$\displaystyle=\sum_{s\geq
0}\frac{\left(tr_{2}\right)^{s}}{\left(q\right)_{s}}\left(r_{1}^{2}\right)_{s}\left(r_{1}r_{2}\right)_{s}\frac{\left(tr_{1}^{3}r_{2}^{2}q^{s}\right)_{\infty}}{\left(tr_{1}\right)_{\infty}}$
$\displaystyle=\frac{\left(tr_{1}^{3}r_{2}^{2}\right)_{\infty}}{\left(tr_{1}\right)_{\infty}}\sum_{s\geq
0}\frac{\left(tr_{2}\right)^{s}}{\left(q\right)_{s}}\frac{\left(r_{1}^{2}\right)_{s}\left(r_{1}r_{2}\right)_{s}}{\left(tr_{1}^{3}r_{2}^{2}\right)_{s}}=\frac{\left(tr_{1}^{3}r_{2}^{2}\right)_{\infty}}{\left(tr_{1}\right)_{\infty}}~{}_{2}\phi_{1}\left(\begin{array}[c]{cc}r_{1}^{2}&r_{1}r_{2}\\\
tr_{1}^{3}r_{2}^{2}&r\end{array};q,tr_{2}\right),$

where ${}_{2}\phi_{1}$ denotes the so-called basic hypergeometric function
(different from the function defined by (4.2)). Reading its properties, in
particular, the so-called reduction formulae we observe that:

$tr_{2}=\frac{tr_{1}^{3}r_{2}^{2}}{r_{1}^{2}r_{1}r_{2}}.$

So now we use the so-called $q$-Gauss sum (one of the reduction formulae for
${}_{2}\phi_{1}$) and get

${}_{2}\phi_{1}\left(\begin{array}[c]{cc}r_{1}^{2}&r_{1}r_{2}\\\
tr_{1}^{3}r_{2}^{2}&r\end{array};q,tr_{2}\right)=\frac{\left(tr_{1}^{3}r_{2}^{2}/r_{1}^{2}\right)_{\infty}\left(tr_{1}^{3}r_{2}^{2}/r_{1}r_{2}\right)_{\infty}}{\left(tr_{1}^{3}r_{2}^{2}\right)_{\infty}\left(tr_{2}\right)_{\infty}}.$

Hence, indeed we have

$\frac{\left(tr_{1}^{3}r_{2}^{2}\right)_{\infty}}{\left(tr_{1}\right)_{\infty}}~{}_{2}\phi_{1}\left(\begin{array}[c]{cc}r_{1}^{2}&r_{1}r_{2}\\\
tr_{1}^{3}r_{2}^{2}&r\end{array};q,tr_{2}\right)=\frac{\left(tr_{1}r_{2}^{2}\right)_{\infty}\left(tr_{2}r_{1}^{2}\right)_{\infty}}{\left(tr_{1}\right)_{\infty}\left(tr_{2}\right)_{\infty}}.$

∎

######  Lemma 5.

The following identity is true for all $n\geq 0,$ $r_{1},r_{2}\in(-1,1)$ and
$-1<q\leq 1$:

$\sum_{j=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{j}_{q}(-1)^{m-j}q^{\binom{m-j}{2}}\frac{\left(aq^{j}\right)_{m}}{1-aq^{j}}=\left\\{\begin{array}[c]{ccc}1/(1-a)&if&m=0\\\
0&if&m>0\end{array}\right..$

###### Proof.

It is obvious that when $m\allowbreak=\allowbreak 0$ the identity is true.
Hence let us assume that $m\geq 1$. We have, after changing the index of
summation from $j$ to $t\allowbreak=\allowbreak m-j$ and applying (2.4)

$\displaystyle\sum_{j=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{j}_{q}(-1)^{j}q^{\binom{j}{2}}\frac{\left(aq^{m-j}\right)_{m}}{1-aq^{m-j}}=\sum_{j=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{j}_{q}(-1)^{j}q^{\binom{j}{2}}(aq^{m-j})_{m-1}$
$\displaystyle=\sum_{j=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{j}_{q}(-1)^{j}q^{\binom{j}{2}}\sum_{k=0}^{m-1}\genfrac{[}{]}{0.0pt}{}{m-1}{k}_{q}(-aq^{m-j})^{k}q^{\binom{k}{2}}$
$\displaystyle=\sum_{k=0}^{m-1}\genfrac{[}{]}{0.0pt}{}{m-1}{k}_{q}(-a)^{k}q^{m-j})^{k}q^{\binom{k}{2}}\sum_{j=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{j}_{q}(-1)^{j}q^{\binom{j}{2}}q^{k(m-j)}$
$\displaystyle=\sum_{k=0}^{m-1}\genfrac{[}{]}{0.0pt}{}{m-1}{k}_{q}(-a)^{k}q^{m-j})^{k}q^{\binom{k}{2}}q^{km}\sum_{j=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{j}_{q}(-1)^{j}q^{\binom{j}{2}}q^{-kj}$
$\displaystyle=\sum_{k=0}^{m-1}\genfrac{[}{]}{0.0pt}{}{m-1}{k}_{q}(-a)^{k}q^{m-j})^{k}q^{\binom{k}{2}}q^{km}(q^{-k})_{m}=0.$

This is so since $\left(q^{-k}\right)_{m}\allowbreak=\allowbreak 0$ for every
$k\allowbreak=\allowbreak 0,\ldots,m-1.$ ∎

######  Lemma 6.

The following identity is true for all $n,t\geq 0,$ $r_{1},r_{2}\in(-1,1)$ and
$-1<q\leq 1$:

(6.2)
$\sum_{k=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{k}_{q}(-r_{2}^{2})^{k}q^{\binom{k}{2}}\frac{\left(r_{1}^{2}\right)_{t+m+k}}{\left(r_{1}^{2}r_{2}^{2}\right)_{t+m+k}}=\frac{\left(r_{2}^{2}\right)_{m}\left(r_{1}^{2}\right)_{t+m}}{\left(r_{1}^{2}r_{2}^{2}\right)_{t+2m}}.$

###### Proof.

Recall, that in [22] (Lemma 1ii) ) the following identity has been proved for
all $-1<q\leq 1,$ $a,b\in\mathbb{R}$ and $n\geq 0$:

(6.3)
$\sum_{j=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{j}_{q}(-b)^{j}q^{\binom{j}{2}}\left(a\right)_{j}\left(abq^{j}\right)_{n-j}\allowbreak=\allowbreak\left(b\right)_{n}.$

So now let us transform a bit the identity that we have to prove. Namely, we
have after multiplying both sides by $\left(r_{1}^{2}r_{2}^{2}\right)_{t+m}$
and dividing again both sides by $\left(r_{1}^{2}\right)_{t+m}$ we get

$\sum_{k=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{k}_{q}(-r_{2}^{2})^{k}q^{\binom{k}{2}}\left(r_{1}^{2}q^{t+m}\right)_{k}\left(r_{1}^{2}r_{2}^{2}q^{t+m}\right)_{m-k}.$

Now we apply (6.3) with $a\allowbreak=\allowbreak r_{1}^{2}q^{t+m}$ and
$b\allowbreak=\allowbreak r_{2}^{2}$ and get (6.2). ∎

###### Proof of Proposition 2.

We start with inserting (3.34) into (4.6) getting:

$\displaystyle
D_{n}(y|r_{1},r_{2},q)=\sum_{s=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{s}_{q}r_{1}^{n-s}r_{2}^{s}\frac{\left(r_{1}^{2}\right)_{s}}{\left(r_{1}^{2}r_{2}^{2}\right)_{s}}\times$
$\displaystyle\sum_{u=0}^{\left\lfloor
n/2\right\rfloor}\frac{\left[s\right]_{q}!\left[n-s\right]_{q}!(1-r_{1}r_{2}q^{n-2u})}{\left[u\right]_{q}!\left[n-2u\right]_{q}!}R_{n-2u}(y|r_{1}r_{2},q)\times$
$\displaystyle\sum_{m=0}^{u}\genfrac{[}{]}{0.0pt}{}{u}{m}_{q}\frac{(r_{1}r_{2})^{u-m}}{(1-r_{1}r_{2})\left(r_{1}r_{2}q\right)_{n-m-u}}\sum_{k=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{k}_{q}\genfrac{[}{]}{0.0pt}{}{n-2m}{n+k-s-m}_{q}(-r_{1}r_{2})^{k}q^{\binom{k}{2}}\left(r_{1}r_{2}\right)_{s-k}$

First we change the order of summation getting:

$\displaystyle D_{n}(y|r_{1},r_{2},q)=\sum_{u=0}^{\left\lfloor
n/2\right\rfloor}\frac{\left[n\right]_{q}!(1-r_{1}r_{2}q^{n-2u})}{\left[u\right]_{q}!\left[n-2u\right]_{q}!\left(r_{1}r_{2}\right)_{n-u+1}}R_{n-2u}(y|r_{1}r_{2},q)\sum_{s=0}^{n}r_{1}^{n-s}r_{2}^{s}\frac{\left(r_{1}^{2}\right)_{s}}{\left(r_{1}^{2}r_{2}^{2}\right)_{s}}$
$\displaystyle\times\sum_{m=0}^{u}\genfrac{[}{]}{0.0pt}{}{u}{m}_{q}\left(r_{1}r_{2}\right)^{u-m}\left(r_{1}r_{2}q^{n+1-u-m}\right)_{m}\times$
$\displaystyle\sum_{k=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{k}_{q}\genfrac{[}{]}{0.0pt}{}{n-2m}{n+k-s-m}_{q}(-r_{1}r_{2})^{k}q^{\binom{k}{2}}\left(r_{1}r_{2}\right)_{s-k}.$

On the way we made use of the well-known property of the $q-$Pochhammer symbol
(see, e.g., [8]):

$\left(a\right)_{n+m}=\left(a\right)_{n}\left(aq^{n}\right)_{m}.$

Now Notice, that for $\genfrac{[}{]}{0.0pt}{}{n-2m}{n-s-(m-k)}_{q}$ to be
nonzero we have to have $n-s\geq m-k$ and $n-2m-n+s+m-k\geq\ $which leads to
$s\geq m+k.$ Hence we have further:

$\displaystyle D_{n}(y|r_{1},r_{2},q)=\sum_{u=0}^{\left\lfloor
n/2\right\rfloor}\frac{\left[n\right]_{q}!(1-r_{1}r_{2}q^{n-2u})}{\left[u\right]_{q}!\left[n-2u\right]_{q}!\left(r_{1}r_{2}\right)_{n-u+1}}R_{n-2u}(y|r_{1}r_{2},q)\times$
$\displaystyle\sum_{m=0}^{u}\genfrac{[}{]}{0.0pt}{}{u}{m}_{q}(r_{1}r_{2})^{u-m}\left(r_{1}r_{2}q^{n-u-m+1}\right)_{m}\times$
$\displaystyle\sum_{k=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{k}_{q}(-r_{1}r_{2})^{k}q^{\binom{k}{2}}\sum_{s=m+k}^{n-(m-k)}\genfrac{[}{]}{0.0pt}{}{n-2m}{n-s-(m-k)}_{q}r_{1}^{n-s}r_{2}^{s}\frac{\left(r_{1}^{2}\right)_{s}}{\left(r_{1}^{2}r_{2}^{2}\right)_{s}}\left(r_{1}r_{2}\right)_{s-k}$
$\displaystyle=\sum_{u=0}^{\left\lfloor
n/2\right\rfloor}\frac{\left[n\right]_{q}!(1-r_{1}r_{2}q^{n-2u})}{\left[u\right]_{q}!\left[n-2u\right]_{q}!\left(r_{1}r_{2}\right)_{n-u+1}}R_{n-2u}(y|r_{1}r_{2},q)\times$
$\displaystyle\sum_{m=0}^{u}\genfrac{[}{]}{0.0pt}{}{u}{m}_{q}(r_{1}r_{2})^{u-m}\left(r_{1}r_{2}q^{n-u-m+1}\right)_{m}\sum_{k=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{k}_{q}(-r_{1}r_{2})^{k}q^{\binom{k}{2}}$
$\displaystyle\times\sum_{t=0}^{n-2m}\genfrac{[}{]}{0.0pt}{}{n-2m}{t}_{q}r_{1}^{n-t-
m-k}r_{2}^{t+m+k}\frac{\left(r_{1}^{2}\right)_{t+m+k}}{\left(r_{1}^{2}r_{2}^{2}\right)_{t+m+k}}\left(r_{1}r_{2}\right)_{t+m}.$

Further, we have

$\displaystyle D_{n}(y|r_{1},r_{2},q)=\sum_{u=0}^{\left\lfloor
n/2\right\rfloor}\frac{\left[n\right]_{q}!(1-r_{1}r_{2}q^{n-2u})}{\left[u\right]_{q}!\left[n-2u\right]_{q}!\left(r_{1}r_{2}\right)_{n-u+1}}R_{n-2u}(y|r_{1}r_{2},q)\times$
$\displaystyle\sum_{m=0}^{u}\genfrac{[}{]}{0.0pt}{}{u}{m}_{q}(r_{1}r_{2})^{u-m}\left(r_{1}r_{2}q^{n-u-m+1}\right)_{m}\times$
$\displaystyle\sum_{k=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{k}_{q}(-r_{1}^{2})^{k}q^{\binom{k}{2}}\sum_{t=0}^{n-2m}\genfrac{[}{]}{0.0pt}{}{n-2m}{t}_{q}r_{1}^{n-t-2m}r_{2}^{t}\frac{\left(r_{1}^{2}\right)_{t+m+k}}{\left(r_{1}^{2}r_{2}^{2}\right)_{t+m+k}}\left(r_{1}r_{2}\right)_{t+m}.$

Now we change the order of summation in the last two sums and we get

$\displaystyle D_{n}(y|r_{1},r_{2},q)=\sum_{u=0}^{\left\lfloor
n/2\right\rfloor}\frac{\left[n\right]_{q}!(1-r_{1}r_{2}q^{n-2u})}{\left[u\right]_{q}!\left[n-2u\right]_{q}!\left(r_{1}r_{2}\right)_{n-u+1}}R_{n-2u}(y|r_{1}r_{2},q)\times$
$\displaystyle\sum_{m=0}^{u}\genfrac{[}{]}{0.0pt}{}{u}{m}_{q}(r_{1}r_{2})^{u-m}\left(r_{1}r_{2}q^{n-u-m+1}\right)_{m}\times$
$\displaystyle\sum_{t=0}^{n-2m}\genfrac{[}{]}{0.0pt}{}{n-2m}{n-2m-t}_{q}r_{1}^{n-t-2m}r_{2}^{t}\left(r_{1}r_{2}\right)_{t+m}\sum_{k=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{k}_{q}(-r_{1}^{2})^{k}q^{\binom{k}{2}}\frac{\left(r_{1}^{2}\right)_{t+m+k}}{\left(r_{1}^{2}r_{2}^{2}\right)_{t+m+k}}.$

Now, Notice, that after applying Lemma 6 we get

$\displaystyle D_{n}(y|r_{1},r_{2},q)=\sum_{u=0}^{\left\lfloor
n/2\right\rfloor}\frac{\left[n\right]_{q}!(1-r_{1}r_{2}q^{n-2u})(r_{1}r_{2})^{u}}{\left[u\right]_{q}!\left[n-2u\right]_{q}!\left(r_{1}r_{2}\right)_{n-u+1}}R_{n-2u}(y|r_{1}r_{2},q)\times$
$\displaystyle\sum_{m=0}^{u}\genfrac{[}{]}{0.0pt}{}{u}{m}_{q}\left(r_{2}^{2}\right)_{m}\left(r_{1}r_{2}q^{n-u-m+1}\right)_{m}$
$\displaystyle\times\sum_{t=0}^{n-2m}\genfrac{[}{]}{0.0pt}{}{n-2m}{n-2m-t}_{q}r_{1}^{n-t-2m}r_{2}^{t}\left(r_{1}r_{2}\right)_{t+m}\frac{\left(r_{1}^{2}\right)_{t+m}}{\left(r_{1}^{2}r_{2}^{2}\right)_{t+2m}}.$

Now, notice, that

$\displaystyle\sum_{t=0}^{n-2m}\genfrac{[}{]}{0.0pt}{}{n-2m}{n-2m-t}_{q}r_{1}^{n-t-2m}r_{2}^{t}\left(r_{1}r_{2}\right)_{t+m}\frac{\left(r_{1}^{2}\right)_{t+m}}{\left(r_{1}^{2}r_{2}^{2}\right)_{t+2m}}\allowbreak=$
$\displaystyle\allowbreak\sum_{s=0}^{n-2m}\genfrac{[}{]}{0.0pt}{}{n-2m}{s}_{q}r_{1}^{s}r_{2}^{n-2m-s}\left(r_{1}r_{2}\right)_{n-m-s}\frac{\left(r_{1}^{2}\right)_{n-m-s}}{\left(r_{1}^{2}r_{2}^{2}\right)_{n-s}}=$
$\displaystyle\sum_{s=0}^{n-2m}\genfrac{[}{]}{0.0pt}{}{n-2m}{s}_{q}r_{1}^{s}r_{2}^{n-2m-s}\left(r_{1}r_{2}\right)_{n-m-s}\frac{\left(r_{1}^{2}\right)_{n-m-s}}{\left(r_{1}^{2}r_{2}^{2}\right)_{n-s}}=$
$\displaystyle\frac{\left(r_{1}^{2}\right)_{m}\left(r_{1}r_{2}\right)_{m}}{\left(r_{1}^{2}r_{2}^{2}\right)_{2m}}\sum_{s=0}^{n-2m}\genfrac{[}{]}{0.0pt}{}{n-2m}{s}_{q}r_{1}^{s}r_{2}^{n-2m-s}\frac{\left(r_{1}r_{2}q^{m}\right)_{n-2m-s}\left(r_{1}^{2}q^{m}\right)_{n-m-s}}{\left(r_{1}^{2}r_{2}^{2}q^{2m}\right)_{n-s}}$

and further

$\displaystyle\frac{\left(r_{1}^{2}\right)_{m}\left(r_{1}r_{2}\right)_{m}}{\left(r_{1}^{2}r_{2}^{2}\right)_{2m}}q^{-m(n-2m)/2}\times$
$\displaystyle\sum_{s=0}^{n-2m}\genfrac{[}{]}{0.0pt}{}{n-2m}{s}_{q}(q^{m/2}r_{1})^{s}(q^{m/2}r_{2})^{n-2m-s}\frac{\left(r_{1}r_{2}q^{m}\right)_{n-2m-s}\left(r_{1}^{2}q^{m}\right)_{n-m-s}}{\left(r_{1}^{2}r_{2}^{2}q^{2m}\right)_{n-s}}$
$\displaystyle=\frac{\left(r_{1}^{2}\right)_{m}\left(r_{1}r_{2}\right)_{m}}{\left(r_{1}^{2}r_{2}^{2}\right)_{2m}}\frac{1}{\left(r_{1}^{2}r_{2}^{2}q^{2m}\right)_{n-2n}}\sum_{s=0}^{n-2m}\genfrac{[}{]}{0.0pt}{}{n-2m}{s}_{q}r_{1}^{s}r_{2}^{n-s}\left(r_{2}^{2}q^{m}\right)_{s}\left(r_{1}^{2}q^{m}\right)_{n-s}$
$\displaystyle=\frac{\left(r_{1}^{2}\right)_{m}\left(r_{1}r_{2}\right)_{m}}{\left(r_{1}^{2}r_{2}^{2}\right)_{n}}w_{n-2m}(m,r_{1},r_{2},q).$

On the last stage we applied Lemma 4 and of course many times identity

$\left(a\right)_{n+m}\allowbreak=\allowbreak\left(a\right)_{n}\left(aq^{n}\right)_{m}.$

Concluding we have

$\displaystyle
D_{n}(y|r_{1},r_{2},q)=\frac{1}{\left(r_{1}^{2}r_{2}^{2}\right)_{n}}\sum_{u=0}^{\left\lfloor
n/2\right\rfloor}\frac{\left[n\right]_{q}!(1-r_{1}r_{2}q^{n-2u})(r_{1}r_{2})^{u}}{\left[u\right]_{q}!\left[n-2u\right]_{q}!\left(r_{1}r_{2}\right)_{n-u+1}}R_{n-2u}(y|r_{1}r_{2},q)\times$
$\displaystyle\sum_{m=0}^{u}\genfrac{[}{]}{0.0pt}{}{u}{m}_{q}\left(r_{2}^{2}\right)_{m}\left(r_{1}r_{2}q^{n-u-m+1}\right)_{m}\left(r_{1}^{2}\right)_{m}\left(r_{1}r_{2}\right)_{m}w_{n-2m}(m,r_{1}r_{2},q).$

Or after applying the definition of $\gamma_{n,k}(r_{1},r_{2},q)$ given by
(4.9)

$D_{n}(y|r_{1},r_{2},q)=\sum_{u=0}^{\left\lfloor
n/2\right\rfloor}\frac{\left[n\right]_{q}!(1-r_{1}r_{2}q^{n-2u})(r_{1}r_{2})^{u}}{\left[u\right]_{q}!\left[n-2u\right]_{q}!\left(r_{1}r_{2}\right)_{n-u+1}}R_{n-2u}(y|r_{1}r_{2},q)\gamma_{n,u}(r_{1},r_{2},q).$

∎

###### Proof.

The idea of the proof is the following. Recall, that the function
$K(x.y)\allowbreak=\allowbreak(1-r_{1}r_{2})f_{CN}\left(y|x,r_{1},q\right)f_{CN}\left(x|y,r_{2},q\right),$
as shown in the Remark 1, is a symmetric function of $x$ and $y$. Now imagine
that we multiply function $K$ by any function $g(x)$, and integrate the
product over $S(q)$ with respect to $x$. Let us call the result $h(y).$ Now
imagine that we multiply $K(x,y)$ by $g(y)$ and integrate with respect to $y$
over $S(q).$ We should get $h(x)$.

The proof will be by induction with respect to $s\allowbreak=$$n-2u$. The
induction assumption is the following:

$\displaystyle\gamma_{n,k}(r_{1},r_{2},q)\allowbreak$
$\displaystyle=\allowbreak\phi_{n-2k}(r_{1},r_{2},q)\text{ and }$
$\displaystyle\int_{S(q)}K(x,y)R_{n}(x|r_{1}r_{2},q)dx$
$\displaystyle=\phi_{n}(r_{1},r_{2},q)R_{n}(y|r_{1}r_{2},q).$

So, now to start induction, let us set $s\allowbreak=\allowbreak 0$.
Integrating the left-hand side of (4.5) gives $f_{R}(y|r_{1}r_{2},q).$ Now
integrating the right-hand side of (4.5) with respect to $y$ we get

$\displaystyle f_{N}(x|q)\sum_{n\geq
0}\frac{H_{2n}(x|q)}{\left[2n\right]_{q}!}\int_{S(q)}D_{2n}(y|r_{1},r_{2},q)f_{R}(y|r_{1}r_{2},q)dy=$
$\displaystyle f_{N}(x|q)\sum_{n\geq
0}\frac{H_{2n}(x|q)}{\left[2n\right]_{q}!}\frac{(1-r_{1}r_{2})(r_{1}r_{2})^{n}\left[2n\right]_{q}!}{\left[n\right]_{q}!(r_{1}r_{2})_{n+1}}\gamma_{2n,n}(r_{1},r_{2},q)$
$\displaystyle=(1-r_{1}r_{2})f_{N}(x|q)\sum_{n\geq
0}\frac{H_{2n}(x|q)}{\left[n\right]_{q}!}\frac{(r_{1}r_{2})^{n}}{(r_{1}r_{2})_{n+1}}\gamma_{2n,n}(r_{1},r_{2},q).$

Hence, using Proposition 6 and uniqueness of the expansion we must have

$\gamma_{2n,n}(r_{1},r_{2},q)=\phi_{0}(r_{1},r_{2},1)=1,$

for all $n\geq 0$.

So now let us take $s\allowbreak=\allowbreak m$ and make the induction
assumption that $\gamma_{n,k}\left(r_{1},r_{2},q\right)=$
$\phi_{n-2k}(r_{1},r_{2},q)$ whenever $n-2k<m$.

So let us multiply the left-hand side of (4.5) by $R_{m}(x|r_{1}r_{2},q)$ and
integrate with respect to $x$ over $S\left(q\right).$ Since (3.22) is zero for
$n>m$, we have

$\displaystyle f_{R}(y|r_{1}r_{2},q)\sum_{s=0}^{\left\lfloor
m/2\right\rfloor}\frac{D_{m-2s}(y|r_{1},r_{2},q)}{\left[m-2s\right]_{q}!}\int
H_{m-2s}(x|q)R_{m}(x|r_{1}r_{2},q)f_{N}(x|q)dx=$ $\displaystyle
f_{R}(y|r_{1}r_{2},q)\sum_{s=0}^{\left\lfloor
m/2\right\rfloor}(-r_{1}r_{2})^{s}q^{\binom{s}{2}}\frac{\left[m\right]_{q}!\left(r_{1}r_{2}\right)_{m-s}}{\left[s\right]_{q}!}\frac{1}{\left[m-2s\right]_{q}!}\times$
$\displaystyle\sum_{u=0}^{\left\lfloor
m/2\right\rfloor-s}\frac{\left[m-2s\right]_{q}!(1-r_{1}r_{2}q^{m-2s-2u})(r_{1}r_{2})^{u}}{\left[u\right]_{q}!\left[m-2s-2u\right]_{q}!(r_{1}r_{2})_{m-2s-u+1}}R_{m-2s-2u}(r_{1},r_{2},q)\gamma_{m-2s,u}(r_{1},r_{2},q)=$
$\displaystyle f_{R}(y|r_{1}r_{2},q)\sum_{k=0}^{\left\lfloor
m/2\right\rfloor}\frac{\left[m\right]_{q}!(1-r_{1}r_{2}q^{m-2k})(r_{1}r_{2})^{k}}{\left[k\right]_{q}!\left[m-2k\right]_{q}!}R_{m-2k}(r_{1},r_{2},q)\gamma_{m-2k,0}(r_{1},r_{2},q)$
$\displaystyle\times\sum_{s=0}^{k}\genfrac{[}{]}{0.0pt}{}{k}{s}_{q}(-1)^{s}q^{\binom{s}{2}}\frac{\left(r_{1}r_{2}\right)_{m-s}}{(r_{1}r_{2})_{m-k-s+1}}=$

and further

$\displaystyle f_{R}(y|r_{1}r_{2},q)\sum_{k=0}^{\left\lfloor
m/2\right\rfloor}\frac{\left[m\right]_{q}!(1-r_{1}r_{2}q^{m-2k})(r_{1}r_{2})^{k}}{\left[k\right]_{q}!\left[m-2k\right]_{q}!}R_{m-2k}(r_{1},r_{2},q)\gamma_{m-2k,0}(r_{1},r_{2},q)$
$\displaystyle\times\sum_{u=0}^{k}\genfrac{[}{]}{0.0pt}{}{k}{u}_{q}(-1)^{k-u}q^{\binom{k-u}{2}}\frac{\left(r_{1}r_{2}\right)_{m-k+u}}{(r_{1}r_{2})_{m-2k+u+1}}=$
$\displaystyle f_{R}(y|r_{1}r_{2},q)\sum_{k=0}^{\left\lfloor
m/2\right\rfloor}\frac{\left[m\right]_{q}!(1-r_{1}r_{2}q^{m-2k})(r_{1}r_{2})^{k}}{\left[k\right]_{q}!\left[m-2k\right]_{q}!}R_{m-2k}(r_{1},r_{2},q)\gamma_{m-2k,0}$
$\displaystyle\times\sum_{u=0}^{k}\genfrac{[}{]}{0.0pt}{}{k}{u}_{q}(-1)^{k-u}q^{\binom{k-u}{2}}\frac{\left(r_{1}r_{2}q^{m-2k+u}\right)_{k}}{(1-r_{1}r_{2}q^{m-2k+u+1})}=$
$\displaystyle
f_{R}(y|r_{1}r_{2},q)R_{m}(r_{1},r_{2},q)\phi_{m}(r_{1},r_{2},q).$

We applied here induction assumption as well as Lemma 5 with
$a\allowbreak=\allowbreak r_{1}r_{2}q^{m-2k}$.

Now let us multiply the (4.5) by $R_{m}(y|r_{1}r_{2},q)$ and integrating with
respect to $y$ over $S(q),$ We get then

$\displaystyle f_{N}(x|q)\sum_{n\geq
0}\frac{1}{\left[n\right]_{q}!}H_{n}(x|q)\int_{S(q)}D_{n}(y|r_{1},r_{2},q)R_{m}f_{R}(y|r_{1}r_{2},q)dy=$
$\displaystyle f_{N}(x|q)\sum_{u\geq
0}\frac{H_{2u+m}(x|q)\left[2u+m\right]_{q}!(1-r_{1}r_{2}q^{m})\left[m\right]_{q}!(r_{1}^{2}r_{2}^{2})_{m}(1-r_{1}r_{2})}{\left[2u+m\right]_{q}!\left[u\right]_{q}!\left[m\right]_{q}!\left(r_{1}r_{2}\right)_{m+u+1}(1-r_{1}r_{2}q^{m})}=$
$\displaystyle(1-r_{1}r_{2})f_{N}(x|q)(r_{1}^{2}r_{2}^{2})_{m}\sum_{u\geq
0}\frac{H_{2u+m}(x|q)}{\left[u\right]_{q}!\left(r_{1}r_{2}\right)_{m+u+1}}\gamma_{2u+m,u}(r_{1},r_{2},q).$

Now, in order to have this expression to be equal to
$R_{m}(x|r_{1}r_{2},q)f_{R}(x|r_{1}r_{2},q)$, in the face of Proposition1(3)
of [21], we must have

$\gamma_{2u+m,u}(r_{1},r_{2},q)=\phi_{m}(r_{1},r_{2},q).$

∎

## References

  * [1] Richard Askey and James Wilson, Some basic hypergeometric orthogonal polynomials that generalize Jacobi polynomials. _Mem. Amer. Math. Soc_. 54(319) (1985) iv+55 pp.
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