\section{Introduction and notation}

Let us first make some remarks concerning notation. $\alpha ,$ $\beta ,$
\ldots $ will denote positive measures on the real line. We will assume that
all of these measures have infinite supports. In order to be able to use
sometimes probabilistic notation we will assume that all considered measures
are normalized. Integrals of integrable function $f$ with respect to measure
say $\alpha $ will be denoted by either of the following notations
\begin{equation*}
\int f(x)d\alpha (x),~~\int fd\alpha ,Ef,~~Ef(Z),~~E_{\alpha }f(Z),
\end{equation*
depending on the context and the need to specify details. In above formulae
Z$ denotes random variable with distribution $\alpha .$ Probability theory
assures that $Z$ always exist.

Matrices and vectors (always columns) will be generally denoted by bold type
letters. The most important vector and matrix are $\mathbf{X}_{n}\allowbreak
=\allowbreak (1,x,\ldots ,x^{n})^{T}$ ($T-$transposition) and
\begin{equation}
\mathbf{M}_{n}(\alpha )\allowbreak =\allowbreak \left[ m_{i+j}(\alpha
\right] _{j,i=0,\ldots ,n},  \label{matrix}
\end{equation
where $m_{n}(\alpha )\allowbreak =\allowbreak \int x^{n}d\alpha (x).$ In
other words $\mathbf{M}_{n}(\alpha )\allowbreak =\allowbreak E_{\alpha
\mathbf{X}_{n}\mathbf{X}_{n}^{T}.$ Matrices of this form i.e. having the
same elements on counter diagonals are called Hankel matrices.

\begin{definition}
We will say that the moment problem is determinate if moment sequence
\left\{ m_{n}\left( \alpha \right) \right\} _{n\geq 0}$ is defined by only
one measure measure $\alpha .$ Otherwise we say that the moment problem is
indeterminate.
\end{definition}

\begin{remark}
There exist necessary criteria allowing to check if the moment problem is
determinate or not. E.g. Carleman's criterion states that if $\sum_{n\geq
0}m_{2n}^{1/2n}<\infty $ then the problem is indeterminate. Or if $\int \exp
\left( \left\vert x\right\vert \right) d\alpha \left( x\right) <\infty $
then the problem is determinate.
\end{remark}

In the sequel we will assume generally that our moment problem is
determinate.

$\left( \mathbf{A}\right) _{j,k}$ will denote $(j,k)-$th entry of matrix
\mathbf{A.}$

Infinite support assumption assures that for every $n$ one can always find
n+1$ linearly independent vectors of the form $(1,x_{k},\ldots
,x_{k}^{n})^{T},$where $x_{k}\in \limfunc{supp}\alpha ,$ $k\allowbreak
=\allowbreak 1,\ldots ,n+1$. Besides we know that if $\limfunc{supp}\alpha $
is infinite then matrices $\mathbf{M}_{n}(\alpha )$ are non-singular for
every $n.$ Let us remark immediately that the matrix $\mathbf{M}_{n}$ is the
main submatrix of the matrix $\mathbf{M}_{n+1}.$ Let us introduce sequence
\begin{equation}
\Delta _{n}(\alpha )\allowbreak =\allowbreak \det \mathbf{M}_{n}(\alpha ),
\label{det}
\end{equation
$n\geq 1$, of determinants of matrices $\mathbf{M}_{n}(\alpha ).$

Let us also introduce vectors consisting of successive moments
\begin{equation*}
\mathbf{m}_{n}^{T}(\alpha )\allowbreak =\allowbreak (1,\ldots ,m_{n}(\alpha
)).
\end{equation*
Vector $\mathbf{m}_{n}(\alpha )$ is the first column of the matrix $\mathbf{
}_{n}(\alpha ).$

In order to avoid repetition of assumption we will assume that matrices
\mathbf{M}_{n}(\alpha )$ exist for all $n\geq 0.$ In other words we assume
that all moments of the measure $\alpha $ exist. Obviously $(0,0)$ entry of
the matrix $\mathbf{M}_{n}$ is equal to $1.$

We know that given measure $\alpha $ such that all moments exist one can
define the set of polynomials $\left\{ p_{n}(x,\alpha )\right\} _{n\geq -1}$
with $p_{-1}(x,\alpha )\allowbreak =\allowbreak 0,$ $p_{0}(x,\alpha
)\allowbreak =\allowbreak 1$ and such that $p_{n}$ is of degree $n$ and for
n+m\neq -2:
\begin{equation*}
\int p_{n}\left( x,\alpha \right) p_{m}(x,\alpha )d\alpha (x)\allowbreak
=\allowbreak \delta _{n,m},
\end{equation*
where $\delta _{n,m}$ denoted Kronecker's delta. Moreover if we declare that
all leading coefficients of polynomials $p_{n}(x,\alpha )$ are positive then
coefficients $\pi _{n,i}(\alpha )$ of the expansio
\begin{equation}
p_{n}\left( x,\alpha \right) \allowbreak =\allowbreak \sum_{i=0}^{n}\pi
_{n,i}\left( \alpha \right) x^{i},  \label{p_o}
\end{equation
are defined uniquely by the measure $\alpha .$ According to our convention,
later we will drop dependence on $\alpha ,$ if measure $\alpha $ is clearly
specified.

Let us define vectors $\mathbf{P}_{n}(x)\allowbreak =\allowbreak
(p_{0}(x),\ldots ,p_{n}(x))^{T}$ and the lower triangular matrix $\mathbf
\Pi }_{n}$ with entries $\pi _{i,j}.$ Of course we set $\pi
_{i,j}\allowbreak =\allowbreak 0$ for $j>i.$ We obviously have
\begin{equation}
\mathbf{P}_{n}(x)=\mathbf{\Pi }_{n}\mathbf{X}_{n}.  \label{_p}
\end{equation
To continue introduction of notation let $\lambda _{n,i}(\alpha )$ denote
coefficients in the following expansions
\begin{equation}
x^{n}\allowbreak =\allowbreak \sum_{i=0}^{n}\lambda _{n,i}(\alpha
)p_{i}\left( x,\alpha \right) .  \label{inv_p}
\end{equation
Consequently let us introduce lower triangular matrices $\mathbf{\Lambda
_{n}$ with entries $\lambda _{i,j}$ if $i\geq j$ and $0$ otherwise.

We obviously have
\begin{equation}
\mathbf{X}_{n}\allowbreak =\allowbreak \mathbf{\Lambda }_{n}\mathbf{P
_{n}(x),~~\mathbf{\Pi }_{n}\mathbf{\Lambda }_{n}\allowbreak =\allowbreak
\mathbf{\Lambda }_{n}\mathbf{\Pi }_{n}=\mathbf{I}_{n},  \label{_x}
\end{equation
where $\mathbf{I}_{n}$ denotes $(n+1)\times (n+1)$ identity matrix.

Since polynomials $\left\{ p_{n}\right\} $ are orthonormal then there exist
two number sequences $\left\{ a_{n}\right\} ,$ $\left\{ b_{n}\right\} $ such
that polynomials $\left\{ p_{n}\right\} $ satisfy the following 3-term
recurrence
\begin{equation}
xp_{n}(x)\allowbreak =\allowbreak
a_{n+1}p_{n+1}(x)+b_{n}p_{n}(x)+a_{n}p_{n-1}(x),  \label{3tr}
\end{equation
with $a_{0}\allowbreak =\allowbreak 0$ and $n\geq 0.$ We know also that
\begin{equation}
a_{n}\allowbreak =\allowbreak \frac{\pi _{n-1,n-1}}{\pi _{n,n}
,~~b_{n}\allowbreak =\allowbreak \int xp_{n}^{2}\left( x\right) d\alpha
\left( x\right) ,  \label{coeff}
\end{equation
consequently that $b_{0}\allowbreak =\allowbreak m_{1}.$ For details see
e.g. \cite{Akhizer65} or \cite{Nev86}. \

Combining (\ref{3tr}) and (\ref{p_o}) we get the following set of recursive
equations to be satisfied by coefficients $\pi _{n,j}$.
\begin{eqnarray}
a_{n+1}\pi _{n+1,0}+b_{n}\pi _{n,0}+a_{n}\pi _{n-1,0}\allowbreak
&=&\allowbreak 0,  \label{_f} \\
a_{n+1}\pi _{n+1,j}+b_{n}\pi _{n,j}+a_{n}\pi _{n-1,j}\allowbreak
&=&\allowbreak \pi _{n,j-1},  \label{_s}
\end{eqnarray
for $n\geq 0,$ $j\allowbreak =\allowbreak 1,\ldots ,n$ , remembering that
\pi _{n,j}\allowbreak =\allowbreak 0$ for $j>n.$ Further combining (\ref{3tr
) and (\ref{inv_p}) we get the following set of equations to be satisfied by
coefficients $\lambda _{n,i}.$
\begin{eqnarray}
\lambda _{n+1,n+1}\allowbreak &=&\allowbreak \lambda _{n,n}a_{n+1},
\label{lam1} \\
\lambda _{n+1,0}\allowbreak &=&\allowbreak \lambda _{n,0}b_{0}+\lambda
_{n,1}a_{1}  \label{lam2} \\
\lambda _{n+1,i} &=&\lambda _{n,i-1}a_{i}+\lambda _{n,i}b_{i}+\lambda
_{n,i+1}a_{i+1}  \label{lam3}
\end{eqnarray}
with, $\lambda _{0,0}\allowbreak =\allowbreak 1,$ so $\lambda
_{n,n}\allowbreak =\allowbreak \prod_{j=1}^{n}a_{j},$ $n\geq 1.$

\begin{remark}
As it follows from formula (2.1.6) of \cite{IA} coefficients $\pi _{n,i}$
can be expressed as determinants of certain submatrices built of moment
matrix $\mathbf{M}_{n}.$ In particular denoting by $D_{n}^{(i,j)}$ the
determinant of a submatrix obtained by removing row number $i+1$ and column
number $j+1$ of the matrix $\mathbf{M}_{n}.$ We have $\pi _{n,i}\allowbreak
=\allowbreak (-1)^{n-i}D_{n}^{(i,n)}/\sqrt{\Delta _{n}\Delta _{n-1}}$ the so
called Heine representation of orthogonal polynomials (see formula (2.2.6)
in \cite{IA}).
\end{remark}

Let us also consider the family of associated polynomials $\left\{
q_{n}(x)\right\} _{n\geq -1}.$ As it follows say \cite{Akhizer65} or \cit
{Sim05} they satisfy the same 3-term recurrence but with different initial
values. Namely we assume that $q_{-1}(x)\allowbreak =\allowbreak -1$ and
q_{0}(x)\allowbreak =\allowbreak 0.$ Following (\ref{3tr}) we see that then
q_{1}(x)\allowbreak =\allowbreak 1/a_{1}.$ One knows also (see e.g. \cit
{Sim05}) that
\begin{equation*}
q_{n}(x)=\int \frac{p_{n}(x)-p_{n}(y)}{x-y}d\alpha (y).
\end{equation*
Now since $(x^{k}-y^{k})/(x-y)\allowbreak =\allowbreak x^{k-1}\allowbreak
+x^{k-2}y\allowbreak +\allowbreak \allowbreak \ldots \allowbreak
+\allowbreak y^{k-1},$ $p_{n}(x)\allowbreak =\allowbreak \sum_{j=0}^{n}\pi
_{n,j}x^{j}$ and $\int y^{k}d\alpha \left( y\right) \allowbreak =\allowbreak
m_{k}$ we deduce that
\begin{equation}
q_{n}\left( x\right) =\sum_{j=1}^{n}\pi
_{n,j}\sum_{k=0}^{j-1}m_{j-1-k}x^{k}\allowbreak =\allowbreak
\sum_{k=0}^{n-1}x^{k}\sum_{j=k+1}^{n}\pi _{n,j}m_{j-1-k}.  \label{ass_pol}
\end{equation
Let us define also $(n+1)-$vector $\mathbf{Q}_{n}(x)\allowbreak =\allowbreak
(0,q_{1}(x),\ldots ,q_{n}(x))^{T}.$

It should be stressed that the presented above approach of treating first $n$
elements of the sequence $\left\{ p_{j}\left( x\right) \right\} _{j\geq -1}$
as a $(n+1)-$vector a result of certain matrix multiplication is very
fruitful although not original. Traces of it appear in \cite{Bre80} or \cit
{Chih79} and can be traced even earlier. Its relation to Choleski
decomposition of the moment matrix and inverse of moment matrix are also not
original. However such view appears in the literature as 'yet another
possibility' of looking on the main result. In this paper this approach is a
basic tool to get known results and new ones mostly concerning connection
and linearization formulae.

Most of our results concern the so called "truncated moment problem" that is
we in fact assume that we know finite number (say $2n+1$ including moment or
order $0)$ of moments of some distribution. That is in many cases the
assumption that the matrix $\mathbf{M}_{n}$ exists for all $n$ will not be
needed. Then we derive $n+1$ polynomials $\{p_{i}\}_{i=0}^{n}$ that are
mutually orthogonal, we find coefficients of expansion of $x^{i}$ in terms
of these polynomials as well as we derive all of the so called linearization
coefficients i.e. coefficients of the expansions $p_{i}(x)p_{j}(x)$ in
polynomials $\{p_{i}\}_{i=0}^{n}$ for all $i+j\leq n.$

Given two distributions (say $\alpha $ and $\delta )$ and $2$ respective
moment sequences we are able to derive all so called "connection
coefficients" i.e. coefficients of the expansion of say $p_{j}\left(
x,\delta \right) $ in $\{p_{i}(x,\alpha )\}_{i=0}^{j}$ and conversely. Due
to very efficient numerical algorithms of Cholesky decomposition and
inversion of lower triangular matrices all these calculations can be done
within seconds using today's computers.

Of course we present also results that require existence of all moments.
These are some limit properties of arithmetic averages of orthogonal
polynomials and more importantly results concerning expansions of
Radon--Nikodym derivatives of one distribution with respect to the other
(see (\ref{rozkl})).

The paper is organized as follows. In the next Section \ref{chol} we present
consequences of are our approach and derive with its help mostly known
results. We do this basically to illustrate the usefulness of our approach.
By the end of this Section in Subsection \ref{recurr} we relate coefficients
of the power series expansion of polynomials $\left\{ p_{n}\right\} $ to the
coefficients of the 3-term recurrence satisfied by these polynomials. More
precisely we partially solve systems of equations (\ref{_f}), (\ref{_s}) and
(\ref{lam1}), (\ref{lam2}), (\ref{lam3}) . The solution is exact in the case
of symmetric measures $\alpha .$

As we think particularly interesting and new are the results concerning
connection coefficients presented in Section \ref{con-coef} containing not
only formula for the connection coefficients between two sets of orthogonal
polynomials related to two measures but also expansion of Radon--Nikodym
derivative of one measure with respect to the other in a Fourier series of
orthogonal polynomials related to one of the measures.

Interesting and new seems also Section \ref{Line} presenting general formula
for the linearization coefficients. Longer and uninteresting proofs are
shifted to Section \ref{dow}.

\section{Cholesky decomposition and its consequences\label{chol}}

Our basic tool in what follows is the so called Cholesky decomposition of
the symmetric, positive definite matrix. Below we collect some of the
properties of Cholesky decomposition in the following simple proposition.

\begin{proposition}
\label{Cholesky}Suppose positive normalize measure $\alpha $ has support of
infinite cardinality and $\int x^{2N}d\alpha <\infty $ for some $N\geq 0$.
Then

i) there exists unique real, non-singular lower triangular matrix $\mathbf{L
_{N}(\alpha )$ such that $\mathbf{M}_{N}(\alpha )\allowbreak =\allowbreak
\mathbf{L}_{N}(\alpha )\mathbf{L}_{N}^{T}(\alpha ),$

ii) entries of matrix $\mathbf{L}_{N}$ can be calculated recursively
\begin{equation}
l_{n,n}=\sqrt{m_{2n}-\sum_{j=0}^{n-1}l_{n,j}^{2}},~~l_{n+1,k}=(m_{n+k+1}
\sum_{j=0}^{k-1}l_{n+1,j}l_{k,j})/l_{k,k},  \label{gen}
\end{equation
with $l_{0,0}\allowbreak =\allowbreak 1$ for $n\allowbreak =\allowbreak
0,\ldots ,N.$ Entries $l_{n,n}$ have also the following interpretation
\begin{equation}
l_{n,n}^{2}\allowbreak =\allowbreak \frac{\Delta _{n}}{\Delta _{n-1}},
\label{l2n}
\end{equation
where sequence $\left\{ \Delta _{n}\right\} $ is defined by (\ref{det}). In
particular we have:
\begin{eqnarray}
l_{1,1}\allowbreak &=&\allowbreak \sqrt{m_{2}-m_{1}^{2}},~~l_{2,2}=\sqrt
m_{4}\allowbreak -\allowbreak m_{2}^{2}-\frac{(m_{3}-m_{1}m_{2})^{2}}
(m_{2}-m_{1}^{2})}}  \label{_odch} \\
l_{i,0}\allowbreak &=&\allowbreak m_{i},~~l_{i,1}\allowbreak =\allowbreak
\frac{(m_{i+1}-m_{i}m_{1})}{l_{1,1}},  \label{mom} \\
l_{i,2}\allowbreak &=&\allowbreak \frac{1}{l_{2,2}}(m_{i+2}-m_{i}m_{2
\allowbreak -\allowbreak \frac{(m_{3}-m_{2}m_{1})(m_{i+1}-m_{i}m_{1})}
(m_{2}-m_{1}^{2})}),  \label{li2}
\end{eqnarray
$i\allowbreak =\allowbreak 1,\ldots ,N,\allowbreak $

iii) $\forall ~0\leq i,j\leq N
\begin{equation*}
m_{i+j}\allowbreak =\allowbreak \sum_{k=0}^{\min (i,j)}l_{i,k}l_{j,k}.
\end{equation*}
\end{proposition}

\begin{proof}
i) Follows the existence and uniqueness of the Cholesky decomposition (see
e.g. Theorem 8.2.1 of \cite{Serre}) and the fact that if the support of a
positive measure is infinite then matrix $\mathbf{M}_{N}$ exists, is
symmetric and positive definite. Besides by the Cauchy Theorem we have
\Delta _{n}\allowbreak =\allowbreak \left( \det L_{n}\right) ^{2}\allowbreak
=\allowbreak \prod_{i=0}^{n}l_{i,i}^{2}.$ Since $\Delta _{n-1}\allowbreak
=\allowbreak \prod_{j=0}^{n-1}l_{i,i}^{2}$ we get our assertion. ii) Follows
one of the algorithms of obtaining Cholesky decomposition (so called
Cholesky--Banachiewicz algorithm that can be found in. e.g. \cite{Fox64}).
\end{proof}

We have obvious observations that we collect in the next proposition:

\begin{proposition}
\label{interpretacja}Let $\mathbf{M}_{n}$ and $\mathbf{M}_{n}^{-1},$ $n\geq
0 $ be respectively sequence of moment matrices and the sequence of its
inverses of some measure $\alpha .$ Let us denote $\mathbf{M
_{n}^{-1}\allowbreak =\allowbreak \lbrack \mu _{i,j}^{(n)}]_{0\leq i,j\leq
n} $ i.e. that $\mu _{i,j}^{(n)}$ is $(i,j)$ entry of the matrix $\mathbf{M
_{n}^{-1}.$ Let $\mathbf{L}_{n}$ be defined by the sequence of lower
triangular matrices forming Cholesky decomposition of matrices $\mathbf{M
_{n}$, then

i) $\forall n\geq 0,$ $\mathbf{\Pi }_{n}=\mathbf{L}_{n}^{-1},$ $\mathbf
\Lambda }_{n}=\mathbf{L}_{n}.$ That is $\mathbf{\Lambda }_{n}\mathbf{\Lambda
}_{n}^{T}\allowbreak =\allowbreak \mathbf{M}_{n}$ and $\mathbf{\Pi }_{n}^{T
\mathbf{\Pi }_{n}\allowbreak =\allowbreak \mathbf{M}_{n}^{-1}$ in particular
$\sum_{k=\max (i,j)}^{n}\pi _{k,i}\pi _{k,j}\allowbreak =\allowbreak \mu
_{i,j}^{(n)}$ and $\sum_{k=0}^{\min (i,j)}\lambda _{i,k}\lambda
_{j,k}\allowbreak =\allowbreak m_{i+j}.$

ii) $\mathbf{P}_{n}^{T}(x)\mathbf{P}_{n}(y)\allowbreak =\allowbreak
\sum_{i=0}^{n}p_{i}(x)p_{i}(y)\allowbreak =\allowbreak \mathbf{X}_{n}^{T
\mathbf{M}_{n}^{-1}\mathbf{Y}_{n}$, thus $\mathbf{X}_{n}^{T}\mathbf{M
_{n}^{-1}\mathbf{Y}_{n}$ is the reproducing kernel and $1/\mathbf{X}_{n}^{T
\mathbf{M}_{n}^{-1}\mathbf{X}_{n}$ is the Christoffel function of the
measure $\alpha .$ Consequently
\begin{eqnarray*}
\left\vert \mathbf{X}_{n}^{T}\mathbf{M}_{n}^{-1}\mathbf{Y}_{n}\right\vert
&\leq &\frac{1}{\xi _{0,n}}\sqrt{(1+\ldots +x^{2n})(1+\ldots +y^{2n})}, \\
\frac{\xi _{0,n}}{1+\ldots +x^{2n}}\, &\leq &\frac{1}{\mathbf{X
_{n}^{T}M_{n}^{-1}\mathbf{X}_{n}}\leq \frac{\xi _{n,n}}{1+\ldots +x^{2n}},
\end{eqnarray*
where $\xi _{0,n}\leq \xi _{1,n}\leq \ldots \leq \xi _{j,n}\leq \ldots \leq
\xi _{n,n}$ denote eigenvalues of the matrix $\mathbf{M}_{n}$ in
non-decreasing order.$.$

iii) $\int \mathbf{P}_{n}^{T}(x)\mathbf{M}_{n}\mathbf{P}_{n}(x)\allowbreak
d\alpha (x)=\allowbreak \sum_{i=0}^{n}m_{2i}\allowbreak =\allowbreak \xi
_{0,n}+\ldots +\xi _{n,n}.$

iv) $\frac{1}{2\pi }\int_{0}^{2\pi }\mathbf{P}_{n}(e^{it})\mathbf{P
_{n}^{T}(e^{-it})dt\allowbreak =\allowbreak \mathbf{\Pi }_{n}\mathbf{\Pi
_{n}^{T}$, consequently $\frac{1}{2\pi }\sum_{j=0}^{n}\int_{0}^{2\pi
}\left\vert p_{j}(e^{it})\right\vert ^{2}dt\allowbreak =\allowbreak \func{tr
(\mathbf{M}_{n}^{-1})\allowbreak =\allowbreak \sum_{j=0}^{n}1/\xi _{j,n},$

v) $\frac{1}{\xi _{n,n}}\leq \sum_{j\geq 0}^{n}\left\vert
p_{j}(0)\right\vert ^{2}\allowbreak =\allowbreak \mu _{0,0}^{(n)}\allowbreak
\leq \allowbreak \frac{1}{\xi _{0,n}},$

vi) $(\sum_{j=1}^{n-1}m_{j}^{2})/\xi _{n,n}\allowbreak \leq \allowbreak
\sum_{j=1}^{n}\left\vert q_{j}\left( 0\right) \right\vert ^{2}\allowbreak
=\allowbreak \sum_{1\leq i,j\leq n}\mu _{i,j}^{(n)}m_{i-1}m_{j-1}\allowbreak
\leq \allowbreak (\sum_{j=1}^{n-1}m_{j}^{2})/\xi _{0,n},$ and
\sum_{j=1}^{n}q_{j}(0)p_{j}(0)\allowbreak =\allowbreak
\sum_{j=1}^{n}m_{j-1}\mu _{0,j}^{(n)}$,

vii) $\frac{1}{\sqrt{n+1}\log ^{2}(n+2)}\sum_{i=0}^{n}p_{i}(x,\alpha
)\longrightarrow 0,$ $\alpha -$a.s. as $n\longrightarrow \infty .$
\end{proposition}

\begin{proof}
Is shifted to Section \ref{dow}.
\end{proof}

\begin{remark}
Part of assertion i) namely the statement $\mathbf{\Pi }_{n}^{T}\mathbf{\Pi
_{n}\allowbreak =\allowbreak \mathbf{M}_{n}^{-1}$ and assertion iv) were
shown in \cite{Berg11}. We presented these statements for completeness of
the paper.
\end{remark}

\begin{remark}
Notice that from vi) it follows that if the moment problem is indeterminate
i.e. when $\sum_{j=1}^{\infty }\left\vert q_{j}\left( 0\right) \right\vert
^{2}<\infty $ (see e.g. Theorem 2.17 of \cite{Sim98}) we get estimate of the
speed of the divergence of $\xi _{n,n}$ to infinity.
\end{remark}

\begin{remark}
Assertion vii) of Proposition \ref{interpretacja} gives in fact an estimate
of the speed of convergence in Law of Large Numbers that sequence of
orthogonal polynomials satisfies. Namely this assertion can be written in
the form $\frac{\sqrt{n+1}}{\log ^{2}(n+2)}\frac{1}{n+1
\sum_{i=0}^{n}p_{i}(x,\alpha )\longrightarrow 0,\alpha -a.s.$as
n\longrightarrow \infty .$ This result is in the spirit of \cite{Morg55} and
his followers.
\end{remark}

As a corollary we have the following observations:

\begin{corollary}
Coefficients $a_{n}$ and $b_{n}:$ $n\geq 0$ defining the 3-term recurrence
are related to the moment matrix by the formulae
\begin{equation}
a_{n}^{2}\allowbreak =\allowbreak \frac{\Delta _{n}\Delta _{n-2}}{\Delta
_{n-1}^{2}},~\text{~}b_{n}\allowbreak =\allowbreak \frac{\Delta _{n-1}}
\Delta _{n}}l_{n+1,n}l_{n,n}-\frac{\Delta _{n-2}}{\Delta _{n-1}
l_{n,n-1}l_{n-1,n-1},  \label{wsp}
\end{equation
for $n\geq 2$ with $a_{0}=0,$ $a_{1}^{2}\allowbreak =\allowbreak \Delta
_{2}\allowbreak =\allowbreak m_{2}-m_{1}^{2}.$
\end{corollary}

\begin{proof}
Following (\ref{coeff}) and Proposition \ref{interpretacja} i) we deduce
a_{n}^{2}\allowbreak =\allowbreak \pi _{n-1,n-1}^{2}/\pi _{n,n}^{2}.$ Since
\pi _{n,n}\allowbreak =\allowbreak l_{n,n}^{-1}$ we apply (\ref{l2n}). To
get formula for $b_{n}$ first we observe that $(i,i-1)$ entry of the the
inverse of the lower triangular matrix $\mathbf{L}_{n}\allowbreak
=\allowbreak \lbrack l_{i,j}]_{i=0,\ldots ,n,j=0,\ldots ,i}$ is equal to $
\frac{l_{i,i-1}}{l_{i,i}l_{i-1,i-1}}.$ Besides dividing both sides of (\re
{_s}) with $j\allowbreak =\allowbreak n$ by $\pi _{n,n}$ we get
\begin{equation*}
\frac{\pi _{n+1,n}}{\pi _{n+1,n+1}}+b_{n}=\frac{\pi _{n,n-1}}{\pi _{n,n}}.
\end{equation*
Now we have $\frac{\pi _{n+1,n}}{\pi _{n+1,n+1}}\allowbreak =\allowbreak
\frac{l_{n+1,n}l_{n+1,n+1}}{_{l_{n+1,n+1}l_{n,n}}}\allowbreak =\allowbreak
\frac{l_{n+1,n}}{l_{n,n}}$ and similarly $\frac{\pi _{n,n-1}}{\pi _{n,n}
\allowbreak =\allowbreak -\frac{l_{n,n-1}}{l_{n-1,n-1}}.$ Finally we use
\ref{l2n}).
\end{proof}

\begin{remark}
Notice that if $\alpha $ is a symmetric measure, then its moments of odd
order are equal to zero consequently following (\ref{gen}) and (\ref{wsp})
we deduce that coefficients $b_{n}$ are all equal to zero. On the other hand
coefficients $a_{n}$ can be expressed as functions of determinants of the
moment matrix as it follows from (\ref{wsp}). Coefficients $a_{n}$ determine
completely in this case orthogonal polynomials $p_{i}(x),$ $i\allowbreak
=\allowbreak 1,\ldots ,n$ by (\ref{_f}), (\ref{_s}) and (\ref{_p}) and
consequently matrices $\mathbf{L}_{i}(\alpha ),$ $i\allowbreak =\allowbreak
1,\ldots ,n$ which determine moment matrix $\mathbf{M}_{n}(\alpha )$ as
shown by Proposition \ref{interpretacja},i). In other words we have an
algorithm for regaining moments from the sequence of determinants of the
leading principal submatrices of the moment matrix. This is a particular
property of special Hankel matrices with zero as $(i,j)-$entries for $i+j$
being odd.
\end{remark}

\subsection{Coefficients of the 3-term recurrence\label{recurr}}

Below we will formulate a sequence of observations concerning two systems of
equations (\ref{_f})-(\ref{lam3}) which relate coefficients of 3-term
recurrence to coefficients $\pi _{n,i}$ and $\lambda _{n,j}.$

\begin{proposition}
\label{3t-r}i) $\forall n\geq 0:a_{n}>0.$

ii) Let us denote $\eta _{n,i}\allowbreak =\allowbreak \pi
_{n,i}\prod_{j=1}^{n}a_{j}$, $\tau _{n,i}\allowbreak =\allowbreak \lambda
_{n,i}/\prod_{j=1}^{i}a_{j}\allowbreak =\allowbreak \lambda _{n,i}/\lambda
_{i,i}$ and by $\tilde{p}_{n}(x)$ denote the monic version of polynomial
p_{n}(x)$ the
\begin{eqnarray*}
\tilde{p}_{n}(x) &=&\sum_{k=0}^{n}\eta _{n,k}x^{k}, \\
x^{n}\allowbreak  &=&\allowbreak \sum_{k=0}^{n}\tau _{n,k}\tilde{p}_{k}(x).
\end{eqnarray*}

Coefficients $\left\{ \eta _{n,j},\tau _{n,j}\right\} _{n\geq 0,0\leq j\leq
n}$ satisfy the following system of equations
\begin{eqnarray}
\eta _{n+1,0} &=&-b_{n}\eta _{n,0}\allowbreak -a_{n}^{2}\eta _{n-1,0},
\label{_1} \\
\eta _{n+1,j} &=&\eta _{n,j-1}-b_{n}\eta _{n,j}-a_{n}^{2}\eta _{n-1,j},
\label{_2} \\
\tau _{n+1,0}\allowbreak  &=&\allowbreak b_{0}\tau _{n,0}+\tau
_{n,1}a_{1}^{2},  \label{_3} \\
\tau _{n+1,j} &=&\tau _{n,j-1}+b_{j}\tau _{n,j}+a_{j+1}^{2}\tau _{n,j+1},
\label{_4}
\end{eqnarray
$n\geq 0,$ $j\leq n,$ with $\eta _{0,0}\allowbreak =\allowbreak 1,\eta
_{n,n}\allowbreak =\allowbreak 1,$ and $\tau _{0,0}\allowbreak =\allowbreak
\tau _{n,n}\allowbreak =\allowbreak 1$ for $n>0.$

iii) $\forall i>j:\sum_{k=i}^{j}\eta _{j,k}\tau _{k,i}\allowbreak
=\allowbreak 0\allowbreak =\allowbreak \sum_{k=i}^{j}\tau _{j,k}\eta _{k,i}.$
\end{proposition}

\begin{proof}
Is shifted to Section \ref{dow}.
\end{proof}

\begin{proposition}
\label{aux}Let us consider $4$ auxiliary number sequences $\left\{ \xi
_{n,j}^{\left( i\right) }\right\} _{n,j\geq 0},$ $\left\{ \zeta
_{n,j}^{\left( i\right) }\right\} _{n,j\geq 0}$, $i\allowbreak =\allowbreak
1,2$ satisfying the following systems of recurrences for $n\geq 0$,
\begin{eqnarray}
\xi _{n+1,0}^{(1)}\allowbreak  &=&\allowbreak -a_{n}^{2}\xi
_{n-1,0}^{(1)},~\xi _{n+1,0}^{(2)}=-b_{n}\xi _{n,0}^{(2)},  \label{_11} \\
\xi _{n+1,j}^{(1)}\allowbreak  &=&\allowbreak \xi
_{n,j-1}^{(1)}-a_{n}^{2}\xi _{n-1,j}^{(1)},~\xi _{n+1,j}^{(2)}=\xi
_{n,j-1}^{(2)}-b_{n}\xi _{n,j}^{(2)},  \label{_22} \\
\zeta _{n+1,0}^{(1)}\allowbreak  &=&\allowbreak a_{1}^{2}\zeta
_{n,1}^{(1)},~\zeta _{n+1,0}^{(2)}=b_{0}\zeta _{n,0}^{(2)},  \label{_33} \\
\zeta _{n+1,j}^{(1)}\allowbreak  &=&\allowbreak \zeta
_{n,j-1}^{(1)}+a_{j+1}^{2}\zeta _{n,j+1}^{(1)},~\zeta _{n+1,j}^{(2)}=\zeta
_{n,j-1}^{(2)}+b_{j}\zeta _{n,j}^{(2)}.  \label{_44}
\end{eqnarray
with $\allowbreak \allowbreak \xi _{n,j}^{(i)}\allowbreak =0$ when
j>n\allowbreak ,$ for $i=1,2$ and $\xi _{0,0}^{(1)}\allowbreak =\allowbreak
\zeta _{0,0}^{(2)}\allowbreak =\allowbreak 1$. Then
\begin{gather}
\xi _{n+j,n}^{(1)}=\left\{
\begin{array}{ccc}
0 & if & j=2k+1 \\
(-1)^{k}\sum_{\substack{ 1\leq j_{1}<\ldots <j_{k}\leq n+j-1 \\
j_{m+1}-j_{m}\geq 2,m=1,\ldots ,k-1}}\prod_{m=1}^{k}a_{j_{m}}^{2} & if & j=2
\end{array
\right. ,  \label{sol11} \\
\xi _{n+j,n}^{(2)}=(-1)^{j}\sum_{0\leq k_{1}<\ldots <k_{j}\leq
n+j-1}\prod_{m=1}^{j}b_{k_{m}},  \label{sol12} \\
\zeta _{n+l,n}^{(1)}\allowbreak =\allowbreak \left\{
\begin{array}{ccc}
0 & if & l=2k+1 \\
\sum_{j_{1}=1}^{n+1}a_{j_{1}}^{2}\sum_{j_{2}=1}^{j_{1}+1}a_{j_{1}}^{2}\ldots
\sum_{j_{k}=1}^{j_{k-1}+1}a_{j_{k}}^{2} & if & l=2
\end{array
\right. ,  \label{sol21} \\
\zeta _{n+j,n}^{(2)}\allowbreak =\allowbreak
\sum_{k_{1}=0}^{n}b_{k_{1}}\sum_{k_{2}=k_{1}}^{n}b_{k_{2}}\ldots
\sum_{k_{j}=k_{j-1}}^{n}b_{k_{j}}.  \label{sol22}
\end{gather}
\end{proposition}

\begin{proof}
Is shifted to Section \ref{dow}
\end{proof}

\begin{proposition}
\label{part} i) Let us denote $\hat{\eta}_{n,k}\allowbreak =\allowbreak \eta
_{n,k}-\xi _{n,k}^{(1)}-\xi _{n,k}^{(2)}$ and $\hat{\tau}_{n,k}\allowbreak
=\allowbreak \tau _{n,k}-\zeta _{n,k}^{(1)}-\zeta _{n,k}^{(2)},$ for $n\geq
k\geq 0.$ We have
\begin{eqnarray}
\hat{\eta}_{n+1,k} &=&\hat{\eta}_{n,k-1}-b_{n}\hat{\eta}_{n,k}-a_{n}^{2}\hat
\eta}_{n-1,k}-b_{n}\xi _{n,k}^{(1)}-a_{n}^{2}\xi _{n-1,k}^{(2)},
\label{aux1} \\
\hat{\tau}_{n+1,k} &=&\hat{\tau}_{n,k-1}+b_{k}\hat{\tau}_{n,k}+a_{k+1}^{2
\hat{\tau}_{n,k+1}+a_{k+1}^{2}\zeta _{n,k+1}^{(2)}+b_{k}\zeta _{n,k}^{(1)},
\label{aux2}
\end{eqnarray
with $\hat{\eta}_{0,0}\allowbreak =\allowbreak \hat{\tau}_{0,0}\allowbreak
=\allowbreak -1$. In particular we have:

ii) $\eta _{n+1,n}\allowbreak =\allowbreak \xi _{n+1,n}^{(2)}\allowbreak
=\allowbreak -\tau _{n+1,n}$ for $n\geq 0.$

iii) $\eta _{n+2,n}\allowbreak =\allowbreak \xi _{n+2,n}^{(2)}\allowbreak
+\xi _{n+2,n}^{(1)}\allowbreak ,$ $\tau _{n+2,n}\allowbreak =\allowbreak
\zeta _{n+2,n}^{(1)}+\zeta _{n+2,n}^{(2)}$ for $n\geq 0.$

iv) $\tau _{n+3,n}\allowbreak =\allowbreak \zeta _{n+3,n}^{(2)}+\zeta
_{n+2,n}^{(1)}\zeta _{n+1,n}^{(2)}+\sum_{j=1}^{n+1}a_{j}^{2}(b_{j-1}+b_{j});$

$\eta _{n+3,n}\allowbreak =\allowbreak \xi
_{n+3,3}^{(2)}+\sum_{j=1}^{n+2}a_{j}^{2}\sum_{k=0,k\neq j,j-1}^{n+2}b_{k},$
\newline
$\eta _{n+4,n}\allowbreak =\allowbreak \xi _{n+4,n}^{(1)}+\xi
_{n+4,n}^{(2)}+\sum_{k=1}^{n+3}a_{i}^{2}\sum_{0\leq i<j\leq n+3,i,j\neq
k.k-1}b_{i}b_{j},$ \newline
$\tau _{n+4,n}\allowbreak =\allowbreak -\eta _{n+4,n}-\eta _{n+4,n+1}\tau
_{n+1,n}-\eta _{n+4,n+2}\tau _{n+2,n}-\eta _{n+4,n+3}\tau _{n+3,n}.$

Assume that $\forall n\geq 0:b_{i}\allowbreak =\allowbreak 0,$ then:

v) $\eta _{n,0}\allowbreak =\allowbreak \left\{
\begin{array}{ccc}
0 & if & n=2k-1 \\
(-1)^{k}\prod_{j=1}^{k}a_{2j-1}^{2} & if & n=2
\end{array
\right. ,$ $k\allowbreak =\allowbreak 1,2,\ldots .$

\begin{equation}
\eta _{n+l,n}\allowbreak =\xi _{n+l,n}^{(1)},\tau _{n+l,n}\allowbreak
=\allowbreak \zeta _{n+l,n}^{(1)}.  \label{sol2}
\end{equation}

for $l\allowbreak n\geq 0.$
\end{proposition}

\begin{proof}
is shifted to Section \ref{dow}
\end{proof}

\begin{remark}
As pointed in Proposition \ref{3t-r},ii) coefficients $\eta _{i,j}$ are the
power coefficients of monic orthogonal polynomials i.e. orthogonal
polynomials with the leading coefficient equal to $1.$ The similar formula
to (\ref{sol2}) for orthogonal polynomials on the unit circle was proved by
in \cite{Gol07}. Formulae given in assertion ii) and iii) were given in \cit
{Chih79} (Thm. 4.2 (d) and ibidem Exercise 4.1, p24).We present them for
completeness of the paper.
\end{remark}

\begin{remark}
Notice that $(-1)^{k}\sum_{\substack{ 1\leq j_{1}<\ldots <j_{k}\leq n-1  \\
j_{m+1}-j_{m}\geq 2,m=1,\ldots ,k-1}}\prod_{m=1}^{k}a_{j_{m}}^{2}$ can also
be written as
\begin{equation*}
(-1)^{k}\sum_{j_{1}=1}^{n-2k+1}a_{j_{1}}^{2
\sum_{j_{2}=j_{1}+2}^{n-2k+3}a_{j_{2}}^{2}\ldots
\sum_{j_{k}=j_{k-1}+2}^{n-1}a_{j_{k}}^{2}.
\end{equation*}
\end{remark}

As a corollary we get also the following recursive formula expressing
moments in terms of the coefficients $a_{n}$ and $b_{n}$ of the 3-term
recurrence.

\begin{proposition}
\label{moments}i) $m_{j}\allowbreak =\allowbreak -\sum_{k=1}^{j-1}\eta
_{j-1,k-1}m_{k},$

If we assume that all coefficients $b_{n}\allowbreak =\allowbreak 0$ $n\geq
0,$ then we have simplified version of the previous statement:

ii) $m_{2k-1}\allowbreak =\allowbreak 0$ $k\allowbreak =\allowbreak
1,2,\ldots ,$ $m_{4}\allowbreak =\allowbreak a_{1}^{2}(a_{1}^{2}+a_{2}^{2}),$
$m_{2k}\allowbreak =\allowbreak
(\sum_{j=1}^{2k-2}a_{j}^{2})m_{2k-2}-\sum_{j=2}^{k-1}\eta
_{2k-1,2k-1-2j}m_{2k-2j},$ $k\geq 3.$
\end{proposition}

\begin{proof}
i) We use the (\ref{_x}) and (\ref{mom}) which leads to the identity
\forall j\geq 1
\begin{equation*}
\sum_{k=0}^{j}\eta _{j,k}m_{k}\allowbreak =\allowbreak 0.
\end{equation*
Consequently $m_{j}\allowbreak =\allowbreak -\sum_{k=0}^{j-1}\eta
_{j,k}m_{k}.$ Now we utilize (\ref{_2}) and get:
\begin{eqnarray*}
m_{j}\allowbreak &=&\allowbreak -\sum_{k=0}^{j-1}(-b_{j-1}\eta
_{j-1,k}-a_{j-1}^{2}\eta _{j-2,k}+\eta _{j-1,k-1})m_{k}\allowbreak \\
&=&b_{j-1}\sum_{k=0}^{j-1}\eta _{j-1,k}m_{k}+a_{j-1}^{2}\sum_{k=0}^{j-2}\eta
_{j-2,k}m_{k}-\sum_{k=0}^{j-1}\eta _{j-1,k-1}m_{k} \\
&=&-\sum_{k=1}^{j-1}\eta _{j-1,k-1}m_{k}.
\end{eqnarray*}

ii) By i) we have $m_{2j}\allowbreak =\allowbreak \allowbreak
-\sum_{k=1}^{2j-1}\eta _{2j-1,k-1}m_{k}\allowbreak =\allowbreak
-\sum_{n=1}^{j-1}\eta _{2j-1,2n-1}m_{2n}$ since $m_{j}$ with odd $j$ are
equal to zero. Now we recall that $\eta _{2j-1,2j-3}\allowbreak =\allowbreak
-\sum_{i=1}^{2j-2}a_{i}^{2}$ by Proposition \ref{3t-r}, iv).
\end{proof}

\section{Connection coefficients and Radon--Nikodym derivatives.\labe
{con-coef}}

In this subsection we will express the so called connection coefficients
between two sets of $N-$orthogonal polynomials. So let us assume that we
have two moment matrices $\mathbf{M}_{N}\left( \alpha \right) $ and $\mathbf
M}_{N}\left( \delta \right) .$ Let $\mathbf{L}_{N}\left( \alpha \right) $
and $\mathbf{L}_{N}(\delta )$ be their Cholesky decomposition matrices and
\left\{ \mathbf{P}_{N}\left( x,\alpha \right) \right\} $ and $\left\{
\mathbf{P}_{N}(x,\delta )\right\} $ respective sets of $N-$orthogonal
polynomials. Then we have

\begin{lemma}
\label{con-c}We have
\begin{equation*}
\mathbf{P}_{N}(x,\delta )\allowbreak =\allowbreak \mathbf{L}_{N}^{-1}(\delta
)\mathbf{L}_{N}(\alpha )\mathbf{P}_{N}\left( x,\alpha \right) ,
\end{equation*
or more precisely for all $n\allowbreak =\allowbreak 1,\ldots ,N:$
\begin{equation*}
p_{n}(x,\delta )\allowbreak =\allowbreak \sum_{k=0}^{n}\gamma _{n,k}(\delta
,\alpha )p_{k}\left( x,\alpha \right) ,
\end{equation*
where
\begin{equation}
\gamma _{n,k}(\delta ,\alpha )\allowbreak =\allowbreak \sum_{j=k}^{n}\pi
_{n,j}\left( \delta \right) \lambda _{j,k}(\alpha ).  \label{conn}
\end{equation
Moreover, if we assume that polynomials $\left\{ \tilde{p}_{n}(x,\delta )
\tilde{p}_{n}(x,\alpha )\right\} _{n\geq 0}^{N}$ are assumed to be monic
then we have the same formula with coefficients $\pi $ replaced by $\eta $
and $\lambda $ by $\tau $ both defined in Proposition \ref{3t-r}.
\end{lemma}

\begin{proof}
This formula follows simple observation that
\begin{equation*}
\mathbf{X}_{n}=\mathbf{L}_{n}(\alpha )\mathbf{P}_{n}(x,\alpha ).
\end{equation*
Then we apply Proposition \ref{interpretacja} i).

The fact that the same formula is satisfied by $\eta ^{\prime }s$ and $\tau
^{\prime }s$ instead by $\pi ^{\prime }s$ and $\lambda ^{\prime }s$ follows
the fact that we have $\mathbf{X}_{n}\allowbreak =\allowbreak \mathbf{\tilde
L}}_{n}(\alpha )\mathbf{\tilde{P}}_{n}(x,\alpha )$ where $\mathbf{\tilde{P}
_{n}(x,\alpha )$ denotes the vector $(1,\tilde{p}_{1}(\alpha ),\ldots
\tilde{p}_{n}(\alpha ))^{T}$ while $\mathbf{\tilde{L}}_{n}(\alpha )$ denotes
lower triangular matrix with $(i,j)$ entry equal to $\tau _{i,j}(\alpha ).$
\end{proof}

\begin{corollary}
\label{f_few}Let $\left\{ b_{n}(\iota ),a_{n+1}(\iota )\right\} _{n\geq 0},$
$\iota \allowbreak =\allowbreak \delta ,\alpha $ denote coefficients of
3-term recurrences of polynomials respectively $\left\{ \tilde{p
_{n}(x,\delta )\right\} _{n\geq -1}$ and $\left\{ \tilde{p}_{n}(x,\alpha
)\right\} _{n\geq -1}.$ Then

i)
\begin{equation*}
\gamma _{n,n-1}(\delta ,\alpha )\allowbreak =\allowbreak
\sum_{k=0}^{n-1}b_{k}(\alpha )-b_{k}(\delta ),
\end{equation*}

ii)
\begin{gather*}
\gamma _{n,n-2}(\delta ,\alpha )\allowbreak =\allowbreak
\sum_{k=1}^{n-1}(a_{k}^{2}\left( \alpha \right) -a_{k}^{2}\left( \delta
\right) )\allowbreak +\allowbreak \frac{1}{2}(\sum_{j=0}^{n-2}(b_{j}(\alpha
)-b_{j}(\delta )))^{2}\allowbreak  \\
+\allowbreak \frac{1}{2}\sum_{j=0}^{n-2}(b_{j}^{2}(\alpha )-b_{j}^{2}(\delta
))\allowbreak -\allowbreak b_{n-1}(\delta )\sum_{j=0}^{n-2}(b_{j}(\alpha
)-b_{j}(\delta )).
\end{gather*}
\end{corollary}

\begin{proof}
i) We use (\ref{conn}) with $\pi $ replaced by $\eta $ and $\lambda $
replaced by $\tau $ and get $\gamma _{n,n-1}\left( \delta ,\alpha \right)
\allowbreak =\allowbreak \eta _{n,n-1}(\delta )\tau _{n,n}\left( \alpha
\right) \allowbreak +\allowbreak \eta _{n,n}\left( \delta \right) \tau
_{n,n-1}\allowbreak =\allowbreak \sum_{k=0}^{n-1}(b_{k}(\alpha
)-b_{k}(\delta ))$ by Proposition \ref{part}, ii)

ii) $\gamma _{n,n-2}(\delta ,\alpha )\allowbreak =\allowbreak \eta
_{n,n-2}(\delta )\tau _{n-2,n-2}\left( \alpha \right) \allowbreak
+\allowbreak \eta _{n,n-1}\left( \delta \right) \tau _{n-1,n-2}\left( \alpha
\right) \allowbreak +\allowbreak \eta _{n,n}\left( \delta \right) \tau
_{n,n-2}\left( \alpha \right) \allowbreak =\allowbreak \eta _{n,n-2}(\delta
)\allowbreak +\allowbreak \tau _{n,n-2}\left( \alpha \right) \allowbreak
+\allowbreak \eta _{n,n-1}\left( \delta \right) \tau _{n-1,n-2}\left( \alpha
\right) $. Now we apply Proposition \ref{part}, iii) and do some algebra.
\end{proof}

\begin{corollary}
\label{con_symm}Let us assume that both distributions $\alpha $ and $\delta $
are symmetric and coefficients of the 3-term recurrences satisfied by monic
polynomials orthogonal with respect to distributions $\alpha $ and $\delta $
are respectively $\left\{ a_{n}\left( \alpha \right) \right\} _{n\geq 0}$
and $\left\{ a_{n}\left( \delta \right) \right\} _{n\geq 0}$ then
\begin{equation*}
\tilde{p}_{n}\left( x,\delta \right) =\sum_{k=0}^{\left\lfloor
n/2\right\rfloor }\gamma _{n,n-2k}(\delta ,\alpha )\tilde{p}_{n-2k}(x,\alpha
),
\end{equation*
where
\begin{gather*}
\gamma _{n,n-2k}\left( \delta ,\alpha \right) \allowbreak =\allowbreak  \\
\sum_{m=0}^{k}(-1)^{m}\sum_{\substack{ 1\leq j_{1}\leq j_{2}\ldots
,j_{m}\leq n-1, \\ j_{k+1}-j_{k}\geq 2, \\ k=1,\ldots ,m-1}
\prod_{i=1}^{m}a_{j_{1}}^{2}\left( \alpha \right) \ldots a_{j_{m}}^{2}\left(
\alpha \right) \sum_{i_{1}=1}^{k-m}a\left( \delta \right) _{j_{1}}^{2}\ldots
\sum_{i_{k-m}=1}^{i_{k-m-1}+1}a\left( \delta \right) _{i_{k-m}}^{2}.
\end{gather*}
In particular
\begin{eqnarray*}
\gamma _{n,n}\left( \delta ,\alpha \right) \allowbreak  &=&\allowbreak 1, \\
\gamma _{n,0}(\delta ,\alpha )\allowbreak  &=&\allowbreak \left\{
\begin{array}{ccc}
0 & if & n\text{ is odd,} \\
\chi _{k} & if & n\allowbreak =\allowbreak 2k
\end{array
\right.  \\
\gamma _{n,n-2}\left( \delta ,\alpha \right)
&=&\sum_{k=1}^{n-1}(a_{k}^{2}\left( \alpha \right) -a_{k}^{2}\left( \delta
\right) ),
\end{eqnarray*
where
\begin{gather*}
\chi _{k}\allowbreak =\allowbreak  \\
\sum_{m=0}^{k}(-1)^{m}\sum_{\substack{ 1\leq j_{1}\leq j_{2}\ldots
,j_{m}\leq n-1, \\ j_{k+1}-j_{k}\geq 2, \\ k=1,\ldots ,m-1}
\prod_{i=1}^{m}a\left( \alpha \right) _{j_{1}}^{2}\ldots a\left( \alpha
\right) _{j_{m}}^{2}\sum_{i_{1}=1}^{k-m}a\left( \delta \right)
_{j_{1}}^{2}\ldots \sum_{i_{k-m}=1}^{i_{k-m-1}+1}a\left( \delta \right)
_{i_{k-m}}^{2}.
\end{gather*}
\end{corollary}

\begin{remark}
It turns out that pairs of systems of orthogonal polynomials with the
property that the connection coefficients between them are nonnegative are
important. Basing on Corollaries \ref{f_few} and \ref{con_symm} we see that
a necessary conditions for coefficients $\gamma _{n,j}\left( \delta ,\alpha
\right) $ be nonnegative that $\forall n\geq 0:\sum_{j=0}^{n}b_{j}\left(
\delta \right) \geq \sum_{j=0}^{n}b_{j}\left( \alpha )\right) .$ If the
measures that orthogonalize those systems of polynomials are such that
\forall n\geq 0:b_{n}\left( \delta \right) \allowbreak =\allowbreak
b_{n}\left( \alpha \right) $ then the necessary condition for coefficients
\gamma _{n,j}\left( \delta ,\alpha \right) $ to be nonnegative is $\forall
n\geq 0:\sum_{j=1}^{n}a_{j}^{2}\left( \delta \right) \geq
\sum_{j=1}^{n}a_{j}^{2}\left( \alpha \right) .$ The discussion why the
nonnegativity of connection coefficients is important and what are the
consequences of this fact is given in \cite{Szwarc92}.
\end{remark}

Following slight modification (ratio of densities is substituted by the
Radon--Nikodym derivative of respective measures) of Proposition 1 iii) of
\cite{Szab10} we deduce the following general statement concerning :

\begin{corollary}
If $\frac{d\alpha }{d\delta }(x)\allowbreak =\allowbreak 1/Q_{r}(x)$ where
Q_{r}$ is a polynomial of order $r$ (positive on $\limfunc{supp}\delta )$
then for $N\geq r+1$ the symmetric matrix
\begin{equation*}
\mathbf{L}_{N}^{-1}(\alpha )\mathbf{M}_{N}(\delta )\left( \mathbf{L
_{N}^{-1}(\alpha )\right) ^{T}
\end{equation*}
is a '$r-$ribbon' matrix i.e. its $(i,j)$ entries such that $\left\vert
i-j\right\vert >r$ are zeros.
\end{corollary}

\begin{proof}
By the above mentioned Proposition we deduce that the lower triangular
matrix $\mathbf{L}_{N}^{-1}(\alpha )\mathbf{L}_{N}(\delta )$ is a '$r-
ribbon' matrix. Then we have $\mathbf{L}_{N}^{-1}(\alpha )\mathbf{L
_{N}(\delta )(\mathbf{L}_{N}^{-1}(\alpha )\mathbf{L}_{N}(\delta
))^{T}\allowbreak =\allowbreak \mathbf{L}_{N}^{-1}(\alpha )\mathbf{M
_{N}(\delta )\left( \mathbf{L}_{N}^{-1}(\alpha )\right) ^{T}$ . Then we use
the fact that if $\mathbf{A}$ is a lower triangular '$r-$ribbon' matrix then
$\mathbf{AA}^{T}$ is also a '$r-$ribbon' matrix.
\end{proof}

As a more interesting consequence of Lemma \ref{con-c} we have an important
expansion of the Radon--Nikodym derivative of two measures $\alpha <<\delta
. $

\begin{theorem}
\label{expansion}Let the two measures $\alpha $ and $\delta $ both having
all moments be such that $\alpha <<\delta $ and $\int (\frac{d\alpha }
d\delta }(x))^{2}d\delta (x)\allowbreak <\allowbreak \infty ,$ where $\frac
d\alpha }{d\delta }(x)$ denotes their Radon--Nikodym derivative. The
\begin{equation}
\frac{d\alpha }{d\delta }(x)\allowbreak =\allowbreak \sum_{j=0}^{\infty
}E_{\alpha }p_{j}(Z,\delta )p_{j}(x,\delta ),  \label{rozkl}
\end{equation
in $L_{2}(\limfunc{supp}\delta ,\mathcal{F},d\delta ),$ where $\mathcal{F}$
denotes Borel sigma field of $\limfunc{supp}\delta .$ In particular we have
(Parseval's formula)
\begin{equation}
\int (\frac{d\alpha }{d\delta }(x))^{2}d\delta (x)=\sum_{j\geq 0}(E_{\alpha
}p_{j}(Z,\delta ))^{2}.  \label{bessel}
\end{equation
Additionally when $\sum_{j\geq 0}(E_{a}p_{j}(Z,\delta ))^{2}\ln
(j+1)^{2}\allowbreak <\allowbreak \infty ,$ we have $\delta $ almost
everywhere convergence.
\end{theorem}

\begin{proof}
Although the idea of this simple in fact theorem appeared in \cite{Szab10}
where also its numerous nontrivial applications were presented we will give
its simple proof for completeness of the paper. \newline
Radon--Nikodym derivative $\frac{d\alpha }{d\delta }(x)$ is square
integrable with respect to the measure $\delta $ i.e. hence it can be
expanded in a Fourier series with respect to the system of orthogonal
polynomials $\left\{ p_{j}\left( x,\delta \right) \right\} _{j\geq 0}
\begin{equation*}
\frac{d\alpha }{d\delta }(x)=\sum_{j\geq 0}\omega _{j}p_{j}\left( x,\delta
\right) .
\end{equation*
Now let us multiply both sides of this expansion by $p_{k}\left( x,\delta
\right) $ and integrate with respect to $\delta \left( dx\right) .$ On the
right hand side we will get $\omega _{k}$ while on the left hand side $\int
p_{k}\left( x,\delta \right) \alpha \left( dx\right) \allowbreak
=\allowbreak E_{\alpha }p_{k}\left( Z,\delta \right) .$ (\ref{bessel})
follows Bessel equality of orthogonal series. If $\sum_{j\geq
0}(E_{a}p_{j}(Z,\delta ))^{2}\ln (j+1)^{2}\allowbreak <\allowbreak \infty $
then we apply Rademacher--Menshov Theorem (see e.g. \cite{Alexits}) and get
almost everywhere convergence.
\end{proof}

\begin{remark}
Let us notice that if we write $p_{n}(x,\delta )\allowbreak =\allowbreak
\sum_{i=0}^{n}\gamma _{n,i}(\delta ,\alpha )p_{i}(x,\alpha )$ then $\gamma
_{n,0}(\delta ,\alpha )\allowbreak =\allowbreak E_{\alpha }p_{n}(Z,\delta )$
after integrating both sides with respect to $\alpha \left( dx\right) .$
\end{remark}

\begin{example}
As a corollary we will get the famous Poisson--Mehler expansion formula (
\ref{P-M}), below). In order not to repeat too many known details we refer
the reader to \cite{Szab10}, \cite{SzablAW} as far as the ideas and
calculations are concerned and to \cite{IA} in order to get more properties
of the mentioned below families of orthogonal polynomials.

Namely we will consider the so called $q-$Hermite polynomials defined for
\left\vert q\right\vert <1$ as $H_{n}(x|q)/\sqrt{[n]_{q}!},$ where
H_{n}(x|q)$ are monic polynomials satisfying 3-term recurrence given by
(2.3) of \cite{SzablAW}.

We used here traditional notation common in the so called $q-$series theory:
$[n]_{q}\allowbreak =\allowbreak (1-q^{n})/(1-q),$ for $\left\vert
q\right\vert <1$ and $[n]_{1}\allowbreak =\allowbreak n,$
[n]_{q}!\allowbreak =\allowbreak \prod_{j=1}^{n}[j]_{q},$ with $[0]_{q}!=1$
(a)_{n}\allowbreak =\allowbreak \prod_{i=0}^{n-1}(1-aq^{i}),$ (the so called
$q-$Pochhammer symbol).

One can consider also the case $q\allowbreak =\allowbreak 1$ obtaining
similar results but for the sake of simplicity let us consider only the case
$\left\vert q\right\vert <1.$ Let us mention only that for $q\allowbreak
=\allowbreak 1$ $q-$Hermite polynomials are in fact equal to classical
Hermite polynomials, more precisely the ones that are orthogonal with
respect to measure with the density $\exp \left( -x^{2}/2\right) /\sqrt{2\pi
}.$

It is known that $q-$Hermite polynomials are orthogonal for $\left\vert
q\right\vert <1,$ $x\in S\left( q\right) \allowbreak =\{x\in \mathbb{R
:\left\vert x\right\vert \leq 2/\sqrt{1-q}\}\allowbreak $ with respect to
measure with the density $f_{N}(x|q)$ whose exact formula is not very
important and which is given e.g. in \cite{SzablAW} (formula (2.10)). The
measure with the density $f_{N}(x|q)$ is our measure $\delta .$ It is also
known (see same references) that the measure with the density $:
\begin{equation*}
f_{CN}\left( x|y,\rho ,q\right) =f_{N}\left( x|q\right) \prod_{k=0}^{\infty
\frac{(1-\rho ^{2}q^{k})}{w_{k}\left( x,y|\rho ,q\right) },
\end{equation*
where
\begin{equation*}
w_{k}\left( x,y|\rho ,q\right) =(1-\rho ^{2}q^{2k})^{2}-(1-q)\rho
q^{k}(1+\rho ^{2}q^{2k})xy+(1-q)\rho ^{2}(x^{2}+y^{2})q^{2k},
\end{equation*
for $x,y\in S(q),$ $\left\vert \rho \right\vert <1$ for $\left\vert
q\right\vert <1$ has orthonormal polynomials equal to the so called
Al-Salam--Chihara polynomials $P_{n}(x|y,\rho ,q)$ satisfying the 3-term
recurrence given by formula (2.6) of \cite{SzablAW} divided by $\sqrt{(\rho
^{2})_{n}[n]_{q}!}$ as it follows from Proposition 1,iii) of \cite{SzablAW}
(to get orthonormality).

Measure with density the $f_{CN}$ it is our measure $\alpha .$ Following
formula (4.7) in \cite{IRS99} we deduce that
\begin{equation*}
E_{\alpha }H_{n}\left( Z|q\right) \allowbreak =\allowbreak \rho
^{n}H_{n}(y|q).
\end{equation*
where $y\in S(q)$ and $\left\vert \rho \right\vert <1$ are some parameters.
Details are in \cite{SzablAW} but the can be traced to earlier works of
Bryc, Matysiak, Szab\l owski \cite{bms}.
\begin{equation*}
\frac{d\alpha }{d\delta }\left( x\right) \allowbreak =\allowbreak
\prod_{k=0}^{\infty }\frac{(1-\rho ^{2}q^{k})}{w_{k}\left( x,y|\rho
,q\right) }I_{S\left( q\right) }\left( x\right) .
\end{equation*
Notice also that this function is bounded from above and as such square
integrable with respect to any finite measure on $S(q).$ Again details of
the proof of this simple fact are in \cite{SzablAW}. Now following (\re
{rozkl}) we get
\begin{equation}
\prod_{k=0}^{\infty }\frac{(1-\rho ^{2}q^{k})}{w_{k}\left( x,y|\rho
,q\right) }=\sum_{j\geq 0}\frac{\rho ^{j}}{[j]_{q}!}H_{j}(x|q)H_{j}(y|q),
\label{P-M}
\end{equation
for every $y\in S\left( q\right) $ and almost all $x\in S\left( q\right) .$
Notice that for $q=1$ (\ref{P-M}) is also true but it requires some more
properties of Hermite polynomials.
\end{example}

\begin{remark}
Situation described above is an illustration of the situation often met in
the theory of Markov processes. Namely suppose that we have process $\mathbf
X}\allowbreak =\allowbreak \{X_{t}:t\in T\},$ where $T$ is some ordered set
of infinite cardinality and $\forall t\in T:$ $X_{t}$ is a random variable
with support of infinite cardinality. Suppose $dP_{t}$ is the distribution
of $X_{t}$ and that $E_{t}\left\vert X_{t}\right\vert ^{n}$ is finite for
all $t$ and $n.$ Suppose also that $\left\{ p_{n}^{(t)}\right\} $ are
polynomials orthogonal with respect to $dP_{t}.$ Further suppose that the
conditional distribution of $X_{s}$ given $X_{t}\allowbreak =\allowbreak x$
for $s>t$ i.e. $dC_{s,t}$ is absolutely continuous with respect to $dP_{s}$
and that $\frac{dC_{s,t}}{dP_{s}}(x)$ is square integrable with respect to
dP_{s}$ for every $s>t$ and $y\in \limfunc{supp}X_{t}.$ Then as it follows
from Theorem \ref{expansion} in $L_{2}(\limfunc{supp}X_{s},\mathcal{F
,dP_{s})$ we have
\begin{equation*}
dC_{s,t}\allowbreak =\allowbreak (\sum_{j\geq
0}E_{s,t}p_{j}^{(s)}(X_{s})p_{j}^{(s)}(x))dP_{s},
\end{equation*
where as usually in the theory of Markov processes
E_{s,t}(p_{j}^{(s)}(X_{s}))$ denotes expectation with respect to
distribution $C_{s,t}$ i.e. it denotes conditional expectation of
p_{j}^{(s)}(X_{s})$ given $X_{t}\allowbreak =\allowbreak x.$ In other words
we get expansion of the transfer function of our process.
\end{remark}

\section{Linearization coefficients\label{Line}}

Notice that Propositions \ref{Cholesky} and \ref{interpretacja} allow us to
formulate an algorithm to get so called 'linearization coefficients'. Let us
recall that linearization formula is popular name for the expansions of the
for
\begin{equation*}
p_{n}\left( x\right) p_{m}\left( x\right) \allowbreak =\allowbreak
\sum_{j=0}^{m+n}c_{n,m,j}p_{j}(x).
\end{equation*
The problem is to find coefficients $c_{n,m,j}$ for all $n,$ $m\geq 1$. We
have the following lemma.

\begin{lemma}
\label{Lin_c}For $\forall n,m\geq 0$ and $s=0,\ldots ,m+n$
\begin{equation*}
c_{n,m,s}=\left( \sum_{\substack{ 0\leq j\leq n,  \\ 0\leq k\leq m,j+k\geq s
}}\pi _{n,j}\pi _{m,k}\lambda _{j+k,s}\right) .
\end{equation*}
\end{lemma}

\begin{proof}
For $N\geq \max (m,n)$ we have:
\begin{gather*}
p_{n}(x)p_{m}(x)\allowbreak =\allowbreak (\mathbf{P}_{N}(x)\mathbf{P
_{N}^{T}(x))_{n,m}=(\mathbf{\Pi }_{N}\mathbf{X}_{N}\mathbf{X}_{N}^{T}\mathbf
\Pi }_{N}^{T})_{n,m} \\
\sum_{j,k=0}^{N}\left( \mathbf{\Pi }_{N}\right) _{n,j}\left( \mathbf{X}_{N
\mathbf{X}_{N}^{T}\right) _{j,k}\left( \mathbf{\Pi }_{N}^{T}\right)
_{k,m}=\sum_{j,k=0}^{N}\pi _{n,j}x^{j+k}\pi _{m,k}= \\
\sum_{s=0}^{2N}p_{s}(x)\left( \sum_{j,k=0}^{N}\pi _{n,j}\pi _{m,k}\lambda
_{j+k,s}\right) .
\end{gather*
We now use the fact that $\pi _{n,j}\allowbreak =\allowbreak 0$ for $n<j$
and $\lambda _{k,j}\allowbreak =\allowbreak 0$ for $k<j.$
\end{proof}

\begin{remark}
Following general properties of orthogonal polynomials we deduce that
\forall k<|n-m|:\left( \sum_{\substack{ 0\leq i\leq n,  \\ 0\leq j\leq
m,j+i\geq k}}\pi _{n,i}\pi _{m,j}\lambda _{i+j,k}\right) \allowbreak
=\allowbreak 0.$ More precisely $c_{n,m,s}\allowbreak =\allowbreak 0$ for
s=0,\ldots ,\left\vert n-m\right\vert -1.$
\end{remark}

\begin{remark}
By Proposition \ref{3t-r} we deduce that for monic versions of polynomials
p_{n}$ we have similar formula. More precisely let $\tilde{p}_{n}(x)$ be the
monic version of polynomial $p_{n}(x)$ then
\begin{equation*}
\tilde{p}_{n}(x)\tilde{p}_{n}\left( x\right) \allowbreak =\allowbreak
\sum_{s=0}^{n+m}\tilde{c}_{n,m,s}\tilde{p}_{s}(x),
\end{equation*
where
\begin{equation}
\tilde{c}_{n,m,s}\allowbreak =\allowbreak \left( \sum_{\substack{ 0\leq
j\leq n,  \\ 0\leq k\leq m,j+k\geq s}}\eta _{n,j}\eta _{m,k}\tau
_{j+k,s}\right) .  \label{lin}
\end{equation}
This is so since $\left( \prod_{j=1}^{n}a_{j}\right) \pi _{n,k}\left(
\prod_{j=1}^{m}a_{j}\right) \pi _{m,l}\frac{1}{\prod_{j=1}^{s}a_{j}}\lambda
_{l+k,s}\allowbreak =\allowbreak \eta _{n,k}\eta _{m,l}\tau _{k+l,s}.$
\end{remark}

\begin{corollary}
We have i)
\begin{equation*}
c_{n,m,m+m-1}\allowbreak =\allowbreak \sum_{j=\max
(n,m)}^{n+m-1}(b_{j}-b_{j-\max (n,m)}),
\end{equation*}

ii)
\begin{gather*}
c_{n,m,n+m-2}\allowbreak =\allowbreak \sum_{j=\max
(m,n)}^{m+n-1}a_{j}^{2}\allowbreak -\allowbreak \sum_{j=1}^{\min
\{n,m)-1}a_{j}^{2}\allowbreak -\allowbreak \frac{1}{2}(\sum_{j=\max
(n,m)}^{m+n-2}b_{j}\allowbreak  \\
-\allowbreak \sum_{j=0}^{\min (n,m)-1}b_{j})^{2}\allowbreak -\allowbreak
\frac{1}{2}(\sum_{j=\max (n,m)}^{m+n-2}b_{j}^{2}\allowbreak -\allowbreak
\sum_{j=0}^{\min (n,m)-1}b_{j}^{2}).
\end{gather*}
\end{corollary}

\begin{proof}
i) By (\ref{lin}) we have $c_{n,m,n+m-1}\allowbreak =\allowbreak \eta
_{n,n}\eta _{m,m}\tau _{n+m,n\_m-1}\allowbreak +\allowbreak \eta
_{n,n-1}\eta _{m,m}\tau _{n+m-1,n+n-1}\allowbreak +\allowbreak \eta
_{n,n}\eta _{m,m-1}\tau _{n+m-1,n+n-1}\allowbreak =\allowbreak
\sum_{k=0}^{n+m-1}b_{k}\allowbreak \allowbreak -\allowbreak
\sum_{k=0}^{n-1}b_{k}\allowbreak -\allowbreak \sum_{k=0}^{m-1}b_{k}.$

ii) By (\ref{lin}) we have $c_{n,m,m+m-2}\allowbreak =\allowbreak \eta
_{n,n}\eta _{m,m}\tau _{n+m,n\_m-2}\allowbreak +\allowbreak \eta
_{n,n-2}\eta _{m,m}\tau _{n+m-2,n+m-2}+\allowbreak \eta _{n,n}\eta
_{m,m-2}\tau _{n+m-2,n+m-2}+\allowbreak \eta _{n,n-1}\eta _{m,m}\tau
_{n+m-1,n+m-2}\allowbreak +\allowbreak \eta _{n,n}\eta _{m,m-1}\tau
_{n+m-1,n+m-2}\allowbreak \allowbreak +\allowbreak \eta _{n,n-1}\eta
_{m,m-1}\tau _{n+m-2,m+n-2}\allowbreak \allowbreak =\allowbreak
\sum_{k=1}^{n+m-1}a_{k}^{2}\allowbreak -\allowbreak
\sum_{k=1}^{n-1}a_{k}^{2}-\sum_{k=1}^{m-1}a_{k}^{2}\allowbreak -\allowbreak
(\sum_{j=0}^{n-1}b_{j}\allowbreak +\allowbreak
\sum_{j=0}^{m-1}b_{j})\sum_{j=0}^{n+m-2}b_{j}\allowbreak +\allowbreak
\sum_{j=0}^{n-1}b_{j}\sum_{j=0}^{m-1}b_{j}.$ After little algebra we get the
desired form.
\end{proof}

\begin{remark}
As in the case of the connection coefficients the fact that linarization
coefficients are nonnegative is important. Why it is so, what are the
strightforward consequeces of this fact and in what particular situation it
happens is again given in \cite{Szwarc92}. From the the above mentioned
Corollary one can derive in fact necessary condition for linearization
coefficients to be nonnegative.
\end{remark}

\section{Proofs\label{dow}}

\begin{proof}[Proof of Proposition \protect\ref{interpretacja}]
i) Follows uniqueness of both Cholesky decomposition and orthonormal
polynomials provided sign of the leading coefficient is selected.

ii) We
\begin{eqnarray*}
\int \mathbf{P}_{n}(x,\alpha )\mathbf{P}_{n}^{T}(x,\alpha )d\alpha
(x)\allowbreak &=&\allowbreak \mathbf{L}_{n}^{-1}\int \mathbf{X}_{n}\mathbf{
}_{n}^{T}d\alpha (x)\left( \mathbf{L}_{n}^{-1}\right) ^{T}\allowbreak \\
&=&\allowbreak \mathbf{L}_{n}^{-1}\mathbf{M}_{n}\left( \mathbf{L
_{n}^{-1}\right) ^{T}\allowbreak =\allowbreak \mathbf{I}_{n}.
\end{eqnarray*
Further we have
\begin{equation*}
\mathbf{P}_{n}^{T}(x,\alpha )\mathbf{P}_{n}(y,\alpha )\allowbreak
\allowbreak =\allowbreak \mathbf{X}_{n}^{T}\allowbreak \left( \mathbf{L
_{n}^{-1}\right) ^{T}\mathbf{L}_{n}^{-1}\mathbf{Y}_{n}\allowbreak
=\allowbreak \left( \mathbf{X}_{n}\right) ^{T}(\mathbf{L}_{n}\mathbf{L
_{n}^{T})^{-1}\mathbf{Y}_{n}.
\end{equation*
Thus obviously we have
\begin{equation*}
\left\vert \mathbf{X}_{n}\right\vert ^{2}/\xi _{n,n}\allowbreak \leq
\allowbreak \mathbf{X}_{n}^{T}\mathbf{M}_{n}^{-1}\mathbf{Y}_{n}\allowbreak
\leq \allowbreak \left\vert \mathbf{X}_{n}\right\vert ^{2}/\xi _{0,n},~
\text{and }\left\vert \mathbf{X}_{n}\right\vert ^{2}\allowbreak =\allowbreak
\sum_{i=0}^{n}x^{2i}.
\end{equation*
iii)
\begin{eqnarray*}
\int \mathbf{P}_{n}^{T}(x)\mathbf{M}_{n}\mathbf{P}_{n}(x)d\alpha \allowbreak
&=&\allowbreak \int \func{tr}(\mathbf{M}_{n}\mathbf{P}_{n}(x)\mathbf{P
_{n}^{T}(x))d\alpha \\
&=&\allowbreak \func{tr}\mathbf{M}_{n}\mathbf{L}_{n}^{-1}\mathbf{M
_{n}\left( \mathbf{L}_{n}^{-1}\right) ^{T}\allowbreak =\allowbreak \func{tr
\mathbf{M}_{n}.
\end{eqnarray*}

iv) Denote $\mathbf{e}_{n}^{T}(t)\allowbreak =\allowbreak (1,e^{it},\ldots
,e^{int}).$ We have by Proposition \ref{interpretacja}, i) $\mathbf{P
_{n}(e^{it})\allowbreak =\allowbreak \mathbf{\Pi }_{n}\mathbf{e
_{n}^{T}(t)\allowbreak ,$ hence $\frac{1}{2\pi }\int_{0}^{2\pi }\mathbf{P
_{n}(e^{it})\mathbf{P}_{n}^{T}(e^{-it})dt\allowbreak \allowbreak
=\allowbreak \mathbf{\Pi }_{n}(\frac{1}{2\pi }\int_{0}^{2\pi }\mathbf{e
_{n}(t)\mathbf{e}_{n}^{T}(-t)dt)\mathbf{\Pi }_{n}^{T}.$ Secondly notice that
$\left( k,j\right) -$th entry of the matrix $\mathbf{e}_{n}(t)\mathbf{e
_{n}^{T}(-t)$ is equal to $e^{it(k-j)}$ consequently $(\frac{1}{2\pi
\int_{0}^{2\pi }\mathbf{e}_{n}(t)\mathbf{e}_{n}^{T}(-t)dt)$ is equal to an
identity matrix. Second statement follows the fact that $\frac{1}{2\pi
\sum_{j=0}^{n}\int_{0}^{2\pi }\left\vert p_{j}(e^{it})\right\vert ^{2}dt$ is
the trace of $\frac{1}{2\pi }\int_{0}^{2\pi }\mathbf{P}_{n}(e^{it})\mathbf{P
_{n}^{T}(e^{-it})dt.$ But $\func{tr}(\mathbf{\Pi }_{n}\mathbf{\Pi
_{n}^{T})\allowbreak =\allowbreak \func{tr}(\mathbf{\Pi }_{n}^{T}\mathbf{\Pi
}_{n})\allowbreak =\allowbreak \func{tr}\mathbf{M}_{n}^{-1}.$

v) By Proposition \ref{interpretacja}, ii) considered for $x\allowbreak
=\allowbreak y\allowbreak =\allowbreak 0.$ We get $\sum_{i=0}^{n}\left\vert
p_{i}(0)\right\vert ^{2}\allowbreak =\allowbreak \mathbf{0}_{n}^{T}\mathbf{M
_{n}^{-1}\mathbf{0}_{n}\mathbf{,}$ where $\mathbf{0}_{n}^{T}\allowbreak
=\allowbreak (1,0,\ldots ,0),$ which means that $\sum_{i=0}^{n}\left\vert
p_{i}(0)\right\vert ^{2}$ is $(0,0)$ entry of $\mathbf{M}_{n}^{-1}.$

vi) Following (\ref{ass_pol}) we see that $q_{n}(0)\allowbreak =\allowbreak
\sum_{j=1}^{n}\pi _{n,j}m_{j-1}.$ In other words $\mathbf{Q
_{n}(0)\allowbreak =\allowbreak \mathbf{\Pi }_{n}\mathbf{\mu }_{n}.$ Now
\sum_{j=1}^{n}\left\vert q_{j}\left( 0)\right) ^{2}\right\vert \allowbreak
=\allowbreak \mathbf{Q}_{n}^{T}(0)\mathbf{Q}_{n}(0).$ On the way we utilize
the fact that $\mathbf{\Pi }_{n}^{T}\mathbf{\Pi }_{n}\allowbreak
=\allowbreak \mathbf{M}_{n}^{-1}.$ Following similar arguments we have
\sum_{j=1}^{n}q_{j}(0)p_{j}(0)\allowbreak =\allowbreak (1,0,\ldots ,0
\mathbf{M}_{n}\mathbf{\mu }_{n}.$

vii) Let us denote $\bar{p}_{n}(x,\alpha )\allowbreak =\allowbreak \frac{1}
\sqrt{n+1}\log ^{2}(n+2)}\sum_{i=0}^{n}p_{i}(x,\alpha ).$ It satisfies
recursion
\begin{equation*}
\bar{p}_{n+1}(x,\alpha )\allowbreak =\allowbreak \sqrt{\frac{n+1}{n+1}}\frac
\log ^{2}(n+2)}{\log ^{2}(n+3)}\bar{p}_{n}(x)+p_{n+1}(x)/\sqrt{n+2}\log
^{2}(n+3).
\end{equation*
Since we have $\sum_{n\geq 0}\frac{\log ^{2}(n+2)}{(n+2)\log ^{4}(n+2)
<\infty $ we deduce by Rademacher--Menshov theorem that series $\sum_{n\geq
0}\frac{p_{n}(x)}{\sqrt{n+1}\log ^{2}(n+2)}$ converges $\alpha -$a.s.
Further we apply \cite{Szab87} (Thm. 5).
\end{proof}

\begin{proof}[Proof of Proposition \protect\ref{3t-r}]
Multiplying both sides of (\ref{_f}) and (\ref{_s}) by $\prod_{i=1}^{n}a_{i}$
and dividing both sides of (\ref{lam1}) and (\ref{lam3}) by
\prod_{i=1}^{n}a_{i}$ we see that quantities $\eta $ $\tau $ satisfy the
following system of equations:) Follows immediately formulae (\ref{wsp}).

iii) Follows the fact that $j>i:$ $\sum_{k=i}^{j}\pi _{j,k}\lambda
_{k,j}\allowbreak =\allowbreak \sum_{k=i}^{j}\lambda _{j,k}\pi
_{k,j}\allowbreak =\allowbreak 0$ and the fact that $\lambda _{j,k}\pi
_{k,i}\allowbreak =\allowbreak \tau _{j,k}\eta _{k,i}$ and similarly for the
product $\eta _{j,k}\tau _{k,i}.$
\end{proof}

\begin{proof}[Proof of Proposition \protect\ref{aux}]
First let us consider sequences with upper indices $\left( 1\right) .$ We
have $\xi _{n+1,0}^{(1)}\allowbreak =\allowbreak -a_{n}^{2}\xi
_{n-1,0}^{(1)} $ $.$ Recall that then $\xi _{0,0}^{(1)}\allowbreak
=\allowbreak 1$ and $\xi _{1,0}^{(1)}\allowbreak =\allowbreak 0.$ So we see
that $\eta _{n,0}$ with odd $n$ must be equal to zero.

To see $\xi _{n,n-2k+1}^{(1)}\allowbreak =\allowbreak 0,$ and $\zeta
_{n,n-2k-1}^{(1)},$ $k\allowbreak =\allowbreak 1,2,\ldots ,$ $n\geq 2k+1$ is
easy since then our formulae (\ref{_22}) and (\ref{_44}) become now:
\begin{eqnarray}
\xi _{n+1,n+1-2k-1}^{(1)} &=&-a_{n}^{2}\xi
_{n-1,n-1-(2k-1)}^{(1)}\allowbreak +\allowbreak \xi _{n,n-2k-1}^{(1)},
\label{pom} \\
\zeta _{n+1,n+1-(2k+1)}^{(1)} &=&\zeta
_{n,n-(2k+1)}^{(1)}+a_{n-1-(2k-1)}^{2}\zeta _{n-1,n-1-(2k-1).}^{(1)}
\label{pom1}
\end{eqnarray
We argue in case of (\ref{pom}) by induction assuming $\eta
_{n-1,n-2k}\allowbreak =\allowbreak 0$ and having $\eta _{2k+1,0}\allowbreak
=\allowbreak 0$ as shown above. In the case of (\ref{pom1}) firstly we
notice that from (\ref{_33}) with $n\allowbreak =\allowbreak 0$ we deduce
that $\zeta _{1,0}\allowbreak =\allowbreak 0$. Then taking in (\ref{pom1})
n\allowbreak =\allowbreak 2$ and $k\allowbreak =\allowbreak 1$ we deduce
that $\zeta _{3,0}\allowbreak =\allowbreak 0.$ We use induction in the
similar way and deduce that $\zeta _{2k+1,0}\allowbreak =\allowbreak 0,$
k\allowbreak =\allowbreak 0,\ldots $ . Now taking $k\allowbreak =\allowbreak
0$ we get
\begin{equation*}
\zeta _{n+1,n}\allowbreak =\allowbreak \zeta _{n,n-1}+a_{n-2}\zeta
_{n-1,n}\allowbreak =\allowbreak \zeta _{n,n-1},
\end{equation*
from which we deduce that $\zeta _{n,n-1}\allowbreak =\allowbreak 0$ for
n\geq 1.$ Now take $k\allowbreak =\allowbreak 1$ we get
\begin{equation*}
\zeta _{n+1,n-2}\allowbreak =\allowbreak \zeta _{n,n-3}+a_{n-2}\zeta
_{n-1,n-2}\allowbreak =\allowbreak \zeta _{n,n-3},
\end{equation*
from which we deduce that $\zeta _{n,n-3}\allowbreak =\allowbreak 0$ for all
$n\geq 3.$ In the similar way we show that $\zeta _{n,n-2k-1}$ $\allowbreak
=\allowbreak 0$ for all $n\geq 2k+1.$

Hence let us consider the case of even differences in indices $i$ and $j$ in
$\xi _{i,j}^{(1)}.$

The proofs will be by induction. Let us prove (\ref{sol11}) first. We will
prove it for indices $(n,n-2k).$ Recursive formula (\ref{_2}) becomes now
\begin{equation}
\eta _{n+1,n+1-2k}=-a_{n}^{2}\eta _{n-1,n-1-2(k-1)}+\eta _{n,n-2k}
\label{eqn}
\end{equation
First notice that since sign of $\eta _{n,n-2k}$ is $(-1)^{k}$ and of $\eta
_{n-1,n-2k+1}$ is $(-1)^{k-1}$ by induction assumption we deduce that the
sign of $\eta _{n+1,n+1-2k}$ is $(-1)^{k}$ as claimed. Secondly notice (\re
{sol11}) can be interpreted as a sum of products of elements of $k-
combinations drawn from the set $\left\{ a_{1}^{2},\ldots
,a_{n-1}^{2}\right\} $ such that distance between numbers of chosen elements
is greater than $1.$ For example from $3$ elements we can select only one
such $2-$combinations. Alter little reflection one sees that one there are
\binom{n-k}{k}$ combinations consequently that $\eta _{n,n-2k}$ contains
\binom{n-k}{k}$ products. Equation (\ref{eqn}) states that sum of such
products of $k-$combinations chosen form the set $\{1,,\ldots ,n\}$ can be
decomposed on the sum of such products chosen from the set set with indices
\left\{ 1,\ldots ,n-1\right\} $ and a sum of products containing element
a_{n}^{2}$ times products of similarly chosen $(k-1)-$combinations but from
the set $\left\{ 1,\ldots ,n-2\right\} .$ There are $\binom{n-k}{k}$
summands of the first type and $\binom{n-1-(k-1)}{k-1}$ summand of the
second type (i.e. containing $a_{n}^{2}).$ The total number of summands in
\eta _{n+1,n+1-2k}$ is jus
\begin{equation*}
\binom{n-1-(k-1)}{k-1}+\binom{n-k}{k}=\binom{n+1-k}{k},
\end{equation*
by the well know property of the Pascal triangle as it should be.

The proof of (\ref{sol21}). Let us denote by $\beta _{n,l}$ the right hand
side of (\ref{sol2}). We have:
\begin{eqnarray*}
\beta _{n+2k,n}-\beta _{n-1+2k,n-1}\allowbreak &=&\allowbreak
a_{n+1}^{2}\sum_{j_{2}=1}^{n+2}a_{j_{1}}^{2}\ldots \sum_{j_{k}=1}^{j_{k-1}+1}
\\
&=&a_{n+1}^{2}\beta _{n+1+2k-2,n+1}.
\end{eqnarray*
Further we have $\beta _{n+2,n}\allowbreak =\allowbreak
\sum_{j_{1}=1}^{n+1}a_{j_{1}}^{2}$ by direct calculation. Now notice that
sequences $\zeta ^{(1)}$ and $\beta $ satisfy the same difference equations
and have the same initial conditions. Hence they are identical.

Now let us consider sequences with upper index $(2).$ First of all notice
that by (\ref{sol22}) can be also written
\begin{equation}
\zeta _{n+j,n}^{(2)}\allowbreak =\allowbreak
\sum_{k_{1}=0}^{n}b_{k_{1}}\sum_{k_{2}=0}^{k_{1}}b_{k_{2}}\ldots
\sum_{k_{j}=0}^{k_{j-1}}b_{k_{j}}.  \label{pp}
\end{equation}
Let us denote by $\gamma _{n+j,n}$ the left hand side of (\ref{pp}). From
this form we can easily deduce that
\begin{equation*}
\gamma _{n+1+j,n+1}-\gamma _{n+j,n}\allowbreak =\allowbreak b_{n+1}\gamma
_{n+j,n+1}\allowbreak .
\end{equation*
Thus $\gamma _{n+j,n}$ satisfies the same recurrence as $\zeta
_{n+j,n}^{(2)} $ with the same initial condition. Consequently $\zeta
_{n+j,n}^{(2)}\allowbreak =\allowbreak \gamma _{n+j,n}.$

Now it remained to prove (\ref{sol12}) . First of all notice that left hand
side of (\ref{sol12}) is a sum of products of all of the sets $\left\{
b_{0},\ldots ,b_{n+j-1}\right\} $ of the size $j$. Let us denote it by
\delta _{n+j,n}.$ From what was stated earlier it follows that $\delta
_{n+1+j,n+1}-\delta _{n+j,n}$ is equal to the sum of product of subsets of
the set $\left\{ b_{0},\ldots ,b_{n+j-1}\right\} $ of the size $j$ that
contain element $b_{n+j-1}.$ Another word it is equal to
-(-1)^{j-1}b_{n+j-1}\delta _{n+j-1,n}.$ Hence $\delta _{n+j,n}$ and $\xi
_{n+j,n}^{(2)}$ satisfy the same recurrence with the same initial condition.
\end{proof}

\begin{proof}[Proof of Proposition \protect\ref{part}]
i) Combining (\ref{_1}), (\ref{_2}) with (\ref{_11}) and (\ref{_22}) we get
\hat{\eta}_{n+1,j}\allowbreak =\allowbreak \eta _{n+1,j}-\xi
_{n+1,j}^{(1)}-\xi _{n+1,j}^{(2)}\allowbreak =\allowbreak \eta
_{n,j-1}-b_{n}\eta _{n,j}-a_{n}^{2}\eta _{n-1,j}\allowbreak -\allowbreak
(\xi _{n,j-1}^{(1)}-a_{n}^{2}\xi _{n-1,j}^{(1)})\allowbreak -\allowbreak
(\xi _{n,j-1}^{(2)}-b_{n}\xi _{n,j}^{(2)})\allowbreak =\allowbreak \hat{\eta
_{n,j-1}\allowbreak -\allowbreak b_{n}(\eta _{n,j}-\xi _{n,j}^{(2)}-\xi
_{n,j}^{(1)})\allowbreak \allowbreak -\allowbreak \allowbreak b_{n}\xi
_{n,j}^{(1)}-a_{n}^{2}(\eta _{n-1,j}-\xi _{n-1,j}^{(1)}-\xi
_{n-1,j}^{(2)})\allowbreak -a_{n}^{2}\allowbreak \xi _{n-1,j}^{(2)}$ which
is first of the equations in i). Now let us consider (\ref{_3}), (\ref{_4}),
(\ref{_33}) and (\ref{_44}). We get $\allowbreak \hat{\tau
_{n+1,j}=\allowbreak \tau _{n+1,j}\allowbreak -\allowbreak \zeta
_{n+1,j}^{(1)}\allowbreak -\allowbreak \zeta _{n+1,j}^{(2)}\allowbreak
=\allowbreak \tau _{n,j-1}\allowbreak +\allowbreak b_{j}\tau
_{n,j}\allowbreak +\allowbreak a_{j+1}^{2}\tau _{n,j+1}\allowbreak
\allowbreak -\allowbreak (\allowbreak \zeta _{n,j-1}^{(1)}+a_{j+1}^{2}\zeta
_{n,j+1}^{(1)})\allowbreak -\allowbreak (\zeta _{n,j-1}^{(2)}+b_{j}\zeta
_{n,j}^{(2)})\allowbreak =\allowbreak \hat{\tau}_{n,j-1}\allowbreak
+\allowbreak a_{j+1}^{2}(\tau _{n,j+1}\allowbreak -\allowbreak \zeta
_{n,j+1}^{(1)}\allowbreak -\allowbreak \zeta _{n,j+1}^{(2)})\allowbreak
+\allowbreak a_{j+1}^{2}\zeta _{n,j+1}^{(2)}\allowbreak +\allowbreak
b_{j}(\tau _{n,j}\allowbreak -\allowbreak \zeta _{n,j}^{(2)}\allowbreak
-\allowbreak \zeta _{n,j}^{(1)})\allowbreak +\allowbreak b_{j}\zeta
_{n,j}^{(1)}.$

ii) Consider (\ref{aux1}) and (\ref{aux2}) with $k\allowbreak =\allowbreak
n. $ We get then $\hat{\eta}_{n+1,n}\allowbreak =\allowbreak \hat{\eta
_{n,n-1}$ and $\hat{\tau}_{n+1,n}\allowbreak =\allowbreak \hat{\tau
_{n,n-1}. $ Since for $n\allowbreak =\allowbreak 1$ we have $\hat{\eta
_{1,0}\allowbreak =\allowbreak \hat{\tau}_{1,0}\allowbreak =\allowbreak 0$
we get the assertion.

iii) Take $k\allowbreak =\allowbreak n-1$ in (\ref{aux1}) and (\ref{aux2}).
We get then $\hat{\eta}_{n+1,n-1}\allowbreak =\allowbreak \hat{\eta
_{n,n-2}\allowbreak -\allowbreak b_{n}\hat{\eta}_{n,n-1}\allowbreak
-\allowbreak a_{n}^{2}\hat{\eta}_{n-1,n-1}\allowbreak -\allowbreak b_{n}\xi
_{n,n-1}^{(1)}\allowbreak -\allowbreak a_{n}^{2}\xi
_{n-1,n-1}^{(2)}\allowbreak =\allowbreak \hat{\eta}_{n,n-2},$ since $\hat
\eta}_{n,n-1}\allowbreak =\allowbreak \xi _{n,n-1}^{(1)}\allowbreak
=\allowbreak 0$ and $\hat{\eta}_{n-1,n-1}\allowbreak =\allowbreak -\xi
_{n-1,n-1}^{(2)}\allowbreak =\allowbreak -1.$ Further since $\hat{\eta
_{2,0}\allowbreak =\allowbreak 0$ we get the assertion.

iv) As before we take $k\allowbreak =\allowbreak n-2$ in (\ref{aux1}) and
\ref{aux2}). We get then $\hat{\tau}_{n+1,n-2}\allowbreak =\allowbreak \hat
\tau}_{n,n-3}\allowbreak \allowbreak +\allowbreak b_{n-2}\hat{\tau
_{n,n-2}\allowbreak +\allowbreak +a_{n-1}^{2}\hat{\tau}_{n,n-1}\allowbreak
+\allowbreak a_{n-1}^{2}\zeta _{n,n-1}^{(2)}\allowbreak +\allowbreak
b_{n-2}\zeta _{n,n-2}^{(1)}\allowbreak =\allowbreak \hat{\tau
_{n,n-3}\allowbreak +\allowbreak \allowbreak a_{n-1}^{2}\zeta
_{n,n-1}^{(2)}\allowbreak +\allowbreak b_{n-2}\zeta _{n,n-2}^{(1)},$ since
\hat{\tau}_{n,n-2}\allowbreak =\allowbreak \hat{\tau}_{n,n-1}\allowbreak
=\allowbreak 0$ as shown above. Besides $\zeta _{n,n-1}^{(2)}\allowbreak
=\allowbreak \sum_{k=0}^{n-1}b_{k}$ and $\zeta _{n,n-2}^{(1)}\allowbreak
=\allowbreak \sum_{k=1}^{n-1}a_{k}^{2}$ as shown in (\ref{sol21}) and (\re
{sol22}). Now it a matter of algebra. We reason in the similar way in case
of $\eta _{n+3,n}.$

So now let us consider $\eta _{n+4,n}.$ By taking $k\allowbreak =\allowbreak
n-3$ in (\ref{aux1}) we get: $\hat{\eta}_{n+1,n-3}\allowbreak =\allowbreak
\hat{\eta}_{n,n-4}\allowbreak -\allowbreak b_{n}\hat{\eta
_{n,n-3}\allowbreak -\allowbreak a_{n}^{2}\hat{\eta}_{n-1,n-3}\allowbreak
-\allowbreak b_{n}\xi _{n,n-3}^{(1)}\allowbreak -\allowbreak a_{n}^{2}\xi
_{n-1,n-3}^{(2)}\allowbreak =\allowbreak \hat{\eta}_{n,n-4}\allowbreak
-\allowbreak b_{n}\hat{\eta}_{n,n-3}\allowbreak \allowbreak -\allowbreak
a_{n}^{2}\sum_{0\leq k_{1}<k_{2}\leq n-2}b_{k_{1}}b_{k_{2}}$ since $\hat{\et
}_{n-1,n-3}\allowbreak =\allowbreak \xi _{n,n-3}^{(1)}\allowbreak
=\allowbreak 0$ as shown above and by (\ref{sol11}). Further we use (\re
{sol12})and some algebra.

v) To see that (\ref{sol2}) holds true it is enough to apply (\ref{aux1})
and (\ref{aux2}) with $b_{k}\allowbreak =\allowbreak 0,$ $k\geq 0$ which
results in $\xi _{n,k}^{(2)}\allowbreak =\allowbreak \zeta
_{n,k}^{(2)}\allowbreak =\allowbreak 0,$ for $n>0,$ $k\allowbreak \geq 0$
and which leads to relationships $\hat{\eta}_{n+1,k}=\hat{\eta
_{n,k-1}\allowbreak -\allowbreak a_{n}^{2}\hat{\eta}_{n-1,k}$ and $\hat{\tau
_{n+1,k}=\hat{\tau}_{n,k-1}\allowbreak +\allowbreak a_{k+1}^{2}\hat{\tau
_{n,k+1}$ with $\hat{\eta}_{n,n}\allowbreak =\allowbreak \hat{\tau
_{n,n}\allowbreak =\allowbreak 0,$ for $n>0$ and $\hat{\eta
_{i,0}\allowbreak =\allowbreak \hat{\tau}_{i,0}\allowbreak =\allowbreak 0$
for $i\allowbreak =\allowbreak 1,2.$ Now it is elementary to see that we
must have $\hat{\eta}_{n,k}\allowbreak =\allowbreak \hat{\tau
_{n,k}\allowbreak =\allowbreak 0$ for all $n>0,$ $k\geq 0.$
\end{proof}