text:
\section{Introduction}
\setcounter{equation}{0}
Any stochastic process must be realized in a probability space, a triple made up of a sample space  $\Omega$,
a sigma-algebra
$\mathscr B$, and a choice of probability measure $P$ defined on
$(\Omega, \mathscr B)$. In research papers so far, covering a
general family of stationary-increment processes, the
choice of sample space  $\Omega$ has typically been   $\mathcal
S^\prime$ (the space of tempered distributions in the sense of
Laurent Schwartz). However for applications, a better choice for
$\Omega$ (the sample space) is clearly the continuous functions,
$C(\mathbb R)$  if the process is indexed by time
(sample paths for the process.)  In a setting more general than
$\mathcal S^\prime$ versus $C(\mathbb R)$  we present an
explicit measure-isomorphism between the two choices. It will be
given by an explicit formula, will be well suited for
computations; and our formulas are new even in the case of
$\mathcal S^\prime$ versus $C(\mathbb R)$. In addition to this
isomorphism, we offer a number of applications to stochastic
integration.\\

In our discussion below and in our proofs, we will be making use
of the following earlier papers
\cite{MR0622034,MR0370681,MR2793121}.   The literature on
families of Gaussian processes is vast, both in pure and in
applied areas each with a different viewpoint. Our present motivation stems
from a certain harmonic analysis, from spectral theoretic
computations, and from applications in quantum theory. In our
analysis, we study a particular family of Gaussian processes
(see sections \ref{sec2} and \ref{sec4}) and, in our proofs, we
utilize of tools from a host of areas. Specifically, we utilize families of
operators in Hilbert spaces, with emphasis on explicit reproducing
kernel Hilbert spaces. To refer the reader to useful
references, we mention the following
papers/books whose theme is close to that adopted here:
\cite{ajnfao,MR2793121,MR2788708,MR2444857,MR2397796,MR1810995,MR2805533,MR2459226},
and for papers/books stressing the Hilbert space, see e.g.,
\cite{MR2521338,MR2105321,MR2918388,MR2797301,MR2766605,MR2683408,MR3081915,MR2920083,
MR2662722,MR2480780,MR2518590}, and the papers cited therein. We make use of basic tools from harmonic analysis, and Gaussian
processes in Hilbert space; for background references, see e.g.,
\cite{MR1216518,lifs,MR1943282,MR2932674,MR2728698,MR2512800}.\\

The paper is organized as follows: In the first half of the paper
(sections \ref{sec2}-\ref{sec4}), we introduce a general class of
Gaussian processes indexed by a sigma-algebra of a general but fixed
measure space. This material will aid us in two ways, in the second half of the paper,
sections \ref{sec6}-\ref{sec7}.  First it will unify our approach to
generalized Ito-integrals, subsequently studied. Secondly, our general theory will be
used in the proofs of our results covering our main theme, i.e., setting up a
measure-theoretic equivalence between the two approaches to the formulation of Gaussian
processes outlined above. This thread culminates with Theorem \ref{thm82}.

\section{Preliminaries}
\setcounter{equation}{0}
\label{sec2}
 Below we briefly sketch the framework for the particular class of Gaussian processes to be studied in detail.
We first gather some notations and definitions needed in the sequel.
We already mentioned the Schwartz space $\mathcal S$ and its dual $\mathcal S^\prime$. The duality between the two spaces
will be denoted as
\begin{equation}
\label{pairing}
\langle \xi,\varphi\rangle,\quad \varphi\in\mathcal S\quad\text{\and}\quad \xi\in\mathcal S^\prime,\,\,\text{the distribution $\xi$ applied to
$\varphi$.}
\end{equation}
\begin{Dn}
\label{def21}
Let $M=\mathbb R$ and let $\mathscr F=\mathscr
B(\mathbb R)$ denote the Borel sigma-algebra. A measure $\sigma$ on $\mathscr B(\mathbb R)$ is called
tempered if it satisfies
\begin{equation}
\label{8} \int_{\mathbb R}\frac{d\sigma(u)}{(u^2+1)^p}<\infty
\end{equation}
for some $p\in\mathbb N_0$.\\
The measure $\sigma$ will be called symmetric if
\begin{equation}
\label{neww2}
\sigma(A)=\sigma(-A),\quad\forall A\in\mathscr B.
\end{equation}
\end{Dn}

\begin{Dn} A cylinder set in $\mathcal S^\prime$ is a subset of $\mathcal S^\prime$ of the form
\begin{equation}\label{eqdefcyl}
\left\{\xi\in\mathcal S^\prime\,;\, (\langle \xi,\varphi_1\rangle,\ldots, \langle \xi,\varphi_n\rangle)\in A\right\}
\end{equation}
where $\varphi_1,\ldots,\varphi_n$ are preassigned in the Schwartz space $\mathcal S$, and  $A\subset \mathbb R^n$ is an
open subset of $\mathbb R^n$.
\label{defcyl}
\end{Dn}

We denote the Fourier transform by
\begin{equation}
\widehat{f}(\lambda)=\int_{\mathbb R}e^{-i\lambda x}f(x)dx,\quad \text{for}\quad f\in\mathbf L_2(\mathbb R).
\end{equation}

\begin{Tm}
\label{tm32}
Let $\sigma$ be fixed. There is a real-valued Gaussian process
$X^{(\sigma)}$ indexed by the space of Schwartz functions
$\mathcal S$ and realized on a probability space $(\mathcal
S^\prime,\mathscr C,P^{(\sigma)})$, where $\mathscr C$ denotes the
cylinder algebra, such that, with $\mathbb
E^{(\sigma)}[u]=\int_{\mathcal S^\prime}udP^{(\sigma)}$ and with the notation \eqref{pairing},
\[
X_\varphi^{(\sigma)}(\xi)=\langle\xi,\varphi\rangle,
\]
we have\smallskip

$(1)$
\[
\mathbb
E^{(\sigma)}\left[X^{(\sigma)}_\varphi\right]=0,\quad\forall
\varphi\in\mathcal S.
\]

and\smallskip

$(2)$
\begin{equation}\label{090714}
\mathbb
E^{(\sigma)}\left[e^{iX^{(\sigma)}_\varphi}\right]=e^{-\frac{1}{2}\int_{\mathbb
R}|\widehat{\varphi}(u)|^2d\sigma(u)}.
\end{equation}
\end{Tm}

We will return to a detailed construction    of this process in Sections
\ref{sec4}-\ref{sec7}, but we first need some technical
preparation.
\begin{proof}[Proof of Theorem \ref{tm32}] In the construction of the Gaussian process
\begin{equation}
\label{gauss090714}
(X_\varphi^{(\sigma)})_{\varphi\in\mathcal S}\quad\text{on}\quad (\mathcal S^\prime,\mathscr C, P^{(\sigma)})
\end{equation}
in Theorem \ref{tm32}, we "invert" a transform. Specifically, we
consider the right hand side in \eqref{090714} as a continuous
and positive definite function on $\mathcal S$, and we apply
Minlos' theorem to verify the existence of the Gaussian process
\eqref{gauss090714}.
\end{proof}
\begin{Rk}
\label{rknew1}
{\rm By this approach, we do not get
explicit formulas for \eqref{gauss090714} that can be computed on
sample-paths. This situation is remedied in Theorems \ref{tm61}
and \ref{thm82} below; see especially formulas
\eqref{new17}--\eqref{new19}.}
\end{Rk}
\section{ Gaussian processes}
\setcounter{equation}{0} \label{sec3}

This section is divided into four parts. In the first, our starting point is a given sigma-finite measure space $(M,\mathscr F, \sigma)$, and we outline properties of a Gaussian process  $W^{(\sigma)}$  which is indexed canonically by the sets from the sigma-algebra of $M$. In subsection \ref{sec3.2},
we prove an inversion-formula: We show that if $\sigma$ is refinable (Definition \ref{newdf1}), then this measure $\sigma$  may be recovered as a quadratic variation computed from $W^{(\sigma)}$ . Moreover, with this, we formulate and prove a generalized Ito-lemma for $W^{(\sigma)}$. And, as a result, we obtain a general stochastic integration.
We use this in the subsequent Section
\ref{sec3.3} to introduce a coordinate system (Theorem \ref{thm9}) on the probability space  $\Omega$ carrying the process.
Moreover, we obtain (Corollary \ref{cor12}) a generalized Fourier transform in $\mathbf L_2(\Omega, P)$. We write the
proofs in the case of real valued functions, and real valued
random variables, but the arguments apply {\it mutatis mutandis}
for the complex case as well.

\subsection{Measure space associated Gaussian processes}

\begin{Dn}
\label{def1} Let $(M,\mathscr F,\sigma)$ be a sigma-finite measure
space, meaning that there exists an increasing sequence
$A_1\subset A_2\subset\cdots$ of elements of $\mathscr F$, such
that $\cup_{n=1}^\infty A_n=M$ and $\sigma(A_n)<\infty$ for all
$n\in\mathbb N$.
We say that the Gaussian process $W=(W_A^{(\sigma)})_{A\in\mathscr F}$ is {\rm associated with $(M,\mathscr F, \sigma)$} if there is a probability space $(\Omega,\mathcal B,P)$ such that the following two conditions hold:\\
$(i)$ Setting $\mathbb E[X]=\int_\Omega X(\w)dP(\w)$, we have
\begin{eqnarray}
\label{1}
\mathbb E[W_A]&=&0,\quad\forall A\in\mathscr F,\\
\label{2}
\mathbb E[W_AW_B]&=&\sigma(A\cap B),\quad \forall A,B\in\mathscr F.
\end{eqnarray}
$(ii)$ For every finite sequence $A_1,A_2,\ldots,A_n$ of elements
of $\mathscr F$, the collection of random variables
\[
W_{A_1},W_{A_2}, \ldots,W_{A_n}
\]
is jointly Gaussian with
covariance
\begin{equation}
\label{3}
\mathbb E[W_{A_i}W_{A_j}]=\sigma(A_i\cap A_j),\quad i,j=1,\ldots, n.
\end{equation}
\end{Dn}

\begin{La}
For every sigma-finite measure space $(M,\mathscr F,\sigma)$, an
associated Gaussian process exists. \label{lem1}
\end{La}

\begin{proof}The function $K(A,B)=\sigma(A\cap B)$ is positive definite on $\mathscr F$;  and so it is
the covariance function of a zero-mean Gaussian process with covariance function $\sigma(A\cap B)$,
as follows from \cite[pp. 466-467]{MR58:31324b}.
\end{proof}

We note that in the present setting,  reference \cite{MR0622034} is also helpful for a general discussion of positive definite kernels and Gaussian processes.

\begin{Ex} {\rm The Cameron-Martin space revisited:}
Let $(M,\mathscr F,\sigma)=(\mathbb R_+,\mathscr B, \lambda)$,
where $\mathbb R_+$ is the half-line $[0,\infty)$, $\mathscr B$
denotes the Borel subsets, and $\lambda$ is the restriction of the
Lebesgue measure to $[0,\infty)$. Then the reproducing kernel
Hilbert space associated with the positive definite function
\[
K(A,B)=\lambda(A\cap B),\quad A,B\in\mathscr B,
\]
is the Cameron-Martin Hilbert space $\mathcal H_1^{(\lambda)}$ where
\[
\mathcal H_1^{(\lambda)}=\left\{f\in\mathbf L_2(\mathbb R_+)\,|\, f^\prime\in \mathbf L_2(\mathbb R_+),\,\, and\,\, f(0)=0\right\}
\]
and
\[
\|f\|^2_{\mathcal H_1^{(\lambda)}}=\int_0^1|f^\prime(x)|^2dx,\quad f\in \mathcal H_1^{(\lambda)}.
\]
\label{cameron}
\end{Ex}
\begin{proof}[Discussion of Example \ref{cameron}:]
For $s,t\in[0,\infty)$ set $A=[0,s]$ and $B=[0,t]$. Then
\[
\lambda (A\cap B)=s\wedge t=k(s,t)=k_s(t),
\]
and
\[
\langle f,k_s\rangle_{\mathcal H_1^{(\lambda)}}=f(s),\quad \forall f\in \mathcal H_1^{(\lambda)}\quad \text{and}\quad s\in[0,\infty)
\]
\end{proof}

\begin{Rk}
\label{rkvu160714}
{\rm The conclusion in the example above is still valid if the measure $\sigma$ for $(\mathbb R_+,\mathscr B,\sigma)$ is
tempered; see Definition \ref{def21}. The modifications are as follows:\\
$(i)$ With $\sigma$ given, we get $W^{(\sigma)}$ and its covariance kernel. The corresponding reproducing kernel Hilbert
space $RK(\sigma)$ is described as follows:
\[
RK(\sigma)=\left\{F\,\,:\,\, F(t)=\int_0^tf(x)d\sigma(x)\,\,\text{for some $f\in\mathbf L_2(\sigma)$}\right\}
\]
with norm
\[
\|F\|^2_{RK(\sigma)}=\int_0^\infty |f(x)|^2d\sigma(x),
\]
and\\
$(ii)$ the reproducing kernel for $RK(\sigma)$ is
\[
k^{(\sigma)}(t,s)=\sigma([0,t\wedge s]).
\]
}
\end{Rk}
\begin{La}
\label{lemma3} If $(M,\mathscr F,\sigma)$ is a sigma-finite
measure space, and $(W_A^{(\sigma)})_{A\in\mathscr F}$ is an
associated Gaussian process on the probability space
$(\Omega,\mathcal B,P)$, then the Ito integral
\[
\int_Mf(x)dW^{(\sigma)}(x)\,\,\in\,\,\mathbf L_2(\Omega,\mathcal
B,P)
\]
exits for all $f\in\mathbf L_2(M,\mathscr F,\sigma)$, and the isometry
\begin{equation}
\label{4} \mathbf
E\left[\big|\int_Mf(x)dW^{(\sigma)}(x)\big|^2\right]=\int_M|f(x)|^2d\sigma(x)
\end{equation}
holds.
\end{La}

\begin{proof}
Let $f_s$ be a generic simple function, that is
\begin{equation}\label{5}
f_s(x)=\sum_{k=1}^nc_k\chi_{A_k}(x),\quad n\in\mathbb N,\quad A_1,\ldots,A_n\in\mathscr F\quad\text{and}\quad c_1,\ldots,
c_n\in\mathbb R,
\end{equation}
where $\chi_{A_k}$ denotes the appropriate indicator function with
the sets $A_1,\ldots,A_n$ being such that $A_j\cap A_k=\emptyset$ for
$j\not= k$, and $c_1,\ldots,c_n\in\mathbb R$. Set
\begin{equation}
\label{6}
\int_Mf_s(x)dW^{(\sigma)}(x)=\sum_{k=1}^nc_kW_{A_k}^{(\sigma)}.
\end{equation}
We claim that \eqref{4} holds. The desired conclusion \eqref{4} follows
for all $f\in\mathbf L_2(M,\mathscr F,\sigma)$ as the simple functions
are dense in this latter space and since \eqref{4} is a densely defined isometry between Hilbert spaces, and thus has a unique everywhere
and continuous isometric extension. Indeed (and with the limit meaning approximation by simple functions),
\[
\begin{split}
\mathbb E\left[\big|\int_Mf(x)dW^{(\sigma)}(x)\big|^2\right)&=\lim\sum_{j,k=1}^nc_jc_k{\mathbb E}\left[W_{A_j}^{(\sigma)}
W_{A_k}^{(\sigma)}\right] \\
&=\lim\sum_{j,k=1}^nc_jc_k\sigma(A_j\cap A_k)\\
&=\lim\sum_{k=1}^nc_k^2\sigma(A_k)\\
&=\int_M|f(x)|^2d\sigma(x)
\end{split}
\]
as claimed.
\end{proof}

\begin{Rk}
\label{newrk}
{\rm A refinement of the proof of Lemma \ref{lemma3}, for deterministic functions, also yields the case of Ito-integral and Ito-isometry, for
stochastic adapted process.
The purpose of the extension is to set the stage for the Ito lemma (Corollary \ref{cy322}) where the reasoning relies on the generality of stochastic adapted processes, indexed by a given measure space.
But for pedagogical reasons, we have chosen to first state the result in the "easier" special case of the Wiener integration for deterministic functions.  Below we now point out how the proof of this case carries over, with suitable modifications,  to the case of Ito-integration, and we establish the Ito-isometry for the case of a family of processes which we call $\mathcal F$-adapted. See Definition \ref{lalaldo} and Lemma \ref{lalala}.
}
\end{Rk}

\begin{Cy}
\label{cy3b} Let $\left(W_A^{(\sigma)}\right)_{A\in\mathscr F}$
be a Gaussian process defined on the probability space
$(\Omega,\mathscr B,P)$, and let $I=(a,b]$ be a finite interval.
Take an $A\in\mathscr F$ such that $0<\sigma(A)<\infty$. Then
\[
P\left(\left\{\w\in\Omega\,|\, W^{(\sigma)}_A(\w)\in
I\right\}\right)=P\left(\left\{a<W_A^{(\sigma)}\le
b\right\}\right)=\frac{1}{\sqrt{2\pi}}\int_{\frac{a}{\sqrt{\sigma(A)
}}}^{\frac{b}{\sqrt{\sigma(A)}}}e^{-\frac{x^2}{2}}dx.
\]
\end{Cy}

\begin{Cy}
\label{cyvu}
Let $W^{(\sigma)}(f)$, $f\in\mathbf L_2(\sigma)$ be as in
\eqref{7}. Then,
\[
\mathbb E_{P_\sigma}\left[W^{(\sigma)}(f_1)W^{(\sigma)}(f_2)
\right]=\langle f_1,f_2\rangle_{\sigma}\left(=\int_Mf_1(u)f_2(u)
d\sigma(u)\right),\quad\forall f_1,f_2\in\mathbf L_2(\sigma).
\]
\end{Cy}
\begin{proof}
This is the polarization of
\[
\mathbb E_{P_\sigma}\left[\left(W^{(\sigma)}(f)\right)^2\right]
=\|f\|^2_\sigma.
\]
\end{proof}

\begin{Nn} (The Wiener integral)
With $(M,\mathscr F,\sigma)$ and $(W_A^{(\sigma)})_{A\in\mathscr
F}$ as above, we set for $f\in\mathbf L_2(M,\mathscr F,\sigma)$
\begin{equation}
\label{7} W^{(\sigma)}(f)=\int_MfdW^{(\sigma)}.
\end{equation}
\label{n4}
\end{Nn}

Let $(M,\mathscr F,\sigma)$ be as above with $W^{(\sigma)}$ being the associated Gaussian process. Let $H_0,H_1,\ldots$ denote the Hermite
polynomials, defined by the generating function
\begin{equation}
e^{zx-\frac{z^2}{2}}=\sum_{n=0}^\infty \frac{z^n}{n!}H_n(x).
\label{te1}
\end{equation}
Let $\psi\in\mathcal S$ be a (real-valued) Schwartz function, with Hermite expansion
\begin{equation}
\label{te2}
\psi(x)=\sum_{n=0}^\infty c_nH_n(x),\quad x\in\mathbb R,
\end{equation}
and set
\begin{equation}
\label{te3}
[\psi](x)\stackrel{\rm def.}{=}\sum_{n=0}^\infty n!c_n^2x^n,\quad x\in\mathbb R.
\end{equation}
With these notations we can now state:

\begin{Cy}
With $W^{(\sigma)}$, $\psi$ and $[\psi]$ as above we have the following: Let $f_1,f_2\not=0$ be in $\mathbf L_2(\sigma)$. Then,
\begin{equation}
\label{te4}
\mathbb E_{P_\sigma}\left[\psi\left(\frac{1}{\|f_1\|_\sigma}  W^{(\sigma)}(f_1)\right)
\left(\psi\left(\frac{1}{\|f_2\|_\sigma}  W^{(\sigma)}(f_2)\right)\right)\right]=[\psi]\left(\frac{\langle f_1,f_2\rangle_\sigma}{
\|f_1\|_\sigma\|f_2\|_\sigma}\right).
\end{equation}
\end{Cy}
\begin{proof} By Lemma \ref{lemma3} applied to the $\mathcal N(0,1)$ random variables $\frac{1}{\|f_1\|_\sigma}  W^{(\sigma)}(f_1)$ and $\frac{1}{\|f_2\|_\sigma}  W^{(\sigma)}(f_2)$ we have
\begin{equation}
\label{tset}
\mathbb E_{P_\sigma}\left[\left(H_n\left(\frac{1}{\|f_1\|_\sigma}  W^{(\sigma)}(f_1)\right)\right)
\left(H_k\left(\frac{1}{\|f_2\|_\sigma}  W^{(\sigma)}(f_2)\right)\right)\right]=n!\delta_{k,n}\left(\frac{\langle f_1,f_2\rangle_\sigma}{
\|f_1\|_\sigma\|f_2\|_\sigma}\right)^n.
\end{equation}
Consider now the left hand side of \eqref{te4}, and replace $\psi$ by its expansion \eqref{te2}. Taking into
account \eqref{tset} we get the right hand side of \eqref{te4}.
\end{proof}
In the computation of \eqref{tset}, we use the following fact about Gaussian vectors in $\mathbb R^2$.

\begin{La}
Let $c\in\mathbb R$, $|c|<1$, and let $\gamma_2^{(c)}$ be the $\mathbb R^2$-Gaussian joint density with covariance matrix
$\begin{pmatrix}1&c\\c&1
\end{pmatrix}$, then for the Hermite functions $H_n$ and $H_k$ we have:
\begin{equation}
\iint_{\mathbb R^2}H_n(x)H_k(y)d\gamma_2^{(c)}(x,y)=\delta_{n,k}n!c^n,\quad n,k\in\mathbb N_0.
\end{equation}
\end{La}
The proof is by direct computation, making use of the Ornstein-Uhlenbeck semigroup $P_t$ given by
\begin{equation}
(P_th)(x)=\int_{\mathbb R}h(e^{-t}x+\sqrt{1-e^{-2t}}y)d\gamma_1(y),\quad h\in\mathbf L_2(\mathbb R,d\gamma_1).
\end{equation}
For details see for instance \cite[pp. 9-28]{MR2962301}.
\begin{Cy}
Let $W^{(\sigma)}$, $\psi$ and $[\psi]$ as above, and let $A,B\in\mathscr F$ be such that $0<\sigma(A)<\infty$ and
$0<\sigma(B)<\infty$. Then,
\begin{equation}
\label{te4b}
\mathbb E_{P_\sigma}\left[\left(\psi\left(\frac{1}{\sqrt{\sigma(A)}}  W^{(\sigma)}_A\right)\right)
\left(\psi\left(\frac{1}{\sqrt{\sigma(B)}}  W^{(\sigma)}_B\right)\right)\right]=[\psi]\left(\frac{\sigma(A\cap B)}{
\sqrt{\sigma(A)\sigma(B)}}\right).
\end{equation}
\end{Cy}

\begin{Dn}
Let $(M,\mathscr F,\sigma)$ and  $(W_A^{(\sigma)})_{A\in\mathscr
F}$ be as in Definition \ref{def1}. We say that a (finite or)
countable family $A_1,A_2,\ldots\in\mathscr F$ is a partition of
$A$ if $A=\cup_{k\in\mathbb N}A_k$, and $A_k\cap A_n=\emptyset$
for $k\not= n$.
\end{Dn}

\begin{Cy}
Let $\sigma$ be a tempered Borel measure on $\mathbb R$, and let $(X_\varphi^{(\sigma)})_{\varphi\in\mathcal S}$ be the corresponding Gaussian process, namely $X_\varphi^{(\sigma)}(\xi)=\langle\xi,\varphi\rangle$. Then,
\begin{equation}
\label{090714b} X_{\varphi}^{(\sigma)}=\int_{\mathbb
R}\widehat{\varphi}(u)dW_u^{(\sigma)},
\end{equation}
where $W^{(\sigma)}$ in \eqref{090714b} is the process in Definition \ref{def1}, and the integral is the generalized Ito
integral of Lemma \ref{lemma3}.
\end{Cy}
\begin{proof}
Pick a finite partition $\left\{x_j\right\}$ of $\mathbb R$ such that the integral in \eqref{090714b} is approximated by
\[
\sum_j\widehat{\varphi}(x_j)W^{(\sigma)}_{[x_j,x_{j+1})}.
\]
The approximation is in the $L_2$ norm in $\mathbf L_2(\mathcal S^\prime,P_\sigma)$. Using
\[
\prod_j\mathbb E_{P_\sigma}\left[e^{i\widehat{\varphi}(x_j)W^{(\sigma)}_{[x_j,x_{j+1})}}\right]=e^{-\frac{1}{2}\sum_j|\widehat{\varphi}
(x_j)|^2\sigma\left([x_j,x_{j+1)}\right)},
\]
we conclude, upon passing to the limit of refinements, that
\begin{equation}
\label{newnew2}
\mathbb E_\sigma\left[e^{iX_{\varphi}^{(\sigma)}}\right]=e^{-\frac{1}{2}\int_{\mathbb R}|\widehat{\varphi}(u)|^2d\sigma(u)}.
\end{equation}
Equivalently, the Gaussian process defined in \eqref{090714b},
satisfies equation $(2)$ of Theorem \ref{tm32}. By
Minlos' theorem,  the process in $(2)$ is uniquely determined
by the characteristic function $e^{-\frac{1}{2}\int_{\mathbb
R}|\widehat{\varphi}(u)|^2d\sigma(u)}$.
\end{proof}

\subsection{Quadratic variation}
\label{sec3.2}
\begin{La}
Let $(M,\mathscr F,\sigma)$ be a measure space with $\sigma $ being sigma-finite, and let $(W_A^{(\sigma)})_{A\in\mathscr F}$
be the Gaussian process of Definition \ref{def1}. Then for its moments we have:\\
$(1)$ The odd moments vanish,
\[
E_{P_\sigma}\left[(W_A^{(\sigma)})^{2k+1}\right]=0,\quad k\in\mathbb N_0,
\]
and\\
$(2)$ the even moments are given by:
\[
E_{P_\sigma}\left[(W_A^{(\sigma)})^{2k}\right]=(2k-1)!!(\sigma(A))^k,\quad k\in\mathbb N,
\]
for all $A\in\mathscr F$ such that $\sigma(A)<\infty$ where,
\[
(2k-1)!!=\frac{(2k)!}{2^kk!}.
\]
\label{toto}
\end{La}
We note that in particular we have:
\[
\begin{split}
\mathbb E_{P_\sigma}[W_A^{(\sigma)}]&=0,\\
\mathbb E_{P_\sigma}[(W_A^{(\sigma)})^2]&=\sigma(A),\\
\mathbb E_{P_\sigma}[(W_A^{(\sigma)})^4]&=3(\sigma(A))^2.\\
\end{split}
\]
\begin{proof}[Proof of Lemma \ref{toto}] The asserted moment expressions in the lemma follow from comparing
powers in the moment generating function
\[
\mathbb E_{P_\sigma}\left[e^{itW_A^{(\sigma)}}\right]=e^{-\frac{t^2\sigma(A)}{2}},
\]
valid for all $A\in\mathscr F$ such that $\sigma(A)<\infty$.
\end{proof}

\begin{Pn}
\label{prop37} Let the process $(W_A^{(\sigma)})_{A\in\mathscr F}$
be realized on $\mathbf L_2(\Omega,\mathscr B,P)$ as outlined in
Lemma \ref{lemma3}, and let $\mathbb E$ be the expectation operator
defined by $P$. Then for any $A\in\mathscr F$ and every
partition $(A_k)_{k\in\mathbb N}$ of $A$, the following identity
holds:
\begin{equation}
\label{new2} \mathbb
E\left[|\sigma(A)-\sum_{k}\left(W_{A_k}^{(\sigma)}\right)^2|^2\right]=
2\sum_{k}(\sigma(A_k))^2=2\sum_{k}
\left(\mathbb E\left[(W_{A_k}^{(\sigma)})^2\right]\right)^2.
\end{equation}
\end{Pn}
\begin{proof}
We compute the left hand side in \eqref{new2} with the use of
Lemma \ref{lemma3} as follows: We first note that the random
variables $W^{(\sigma)}_{A_k}$ and $W^{(\sigma)}_{A_n}$ are
independent when $k\not=n$. This follows from \eqref{3} together with the
fact that they are Gaussian; see Corollary \ref{cy3b}. Set
$s=\sigma(A)$ and $s_k=\sigma(A_k)$ (so that $s=\sum_{k=1}^\infty
s_k$). When $k\not=n$ we have (using independence)
\[
\mathbb
E\left[\left(W_{A_k}^{(\sigma)}\right)^2\left(W_{A_n}^{(\sigma)}\right)^2\right]=s_ks_n,
\]
This is utilized so as to compute the right hand side of \eqref{new2}. We get
\[
\begin{split}
\mathbb
E\left[|\sigma(A)-\sum_{k=1}^\infty\left(W_{A_k}^{(\sigma)}\right)^2|^2\right]&=
s^2-2s^2+\sum_{k=1}^\infty \mathbb
E\left[\left(W_{A_k}^{(\sigma)}\right)^4\right]+2\sum_{k<n}s_ks_n\\
&=-s^2+3\sum_{k=1}^\infty s_k^2+2\sum_{k<n}s_ks_n\\
&=2\sum_{k=1}^\infty s_k^2,
\end{split}
\]
which is the desired right hand side of \eqref{new2}.
\end{proof}

\begin{Dn}
\label{defnew2} Let $(M,\mathscr F,\sigma)$ and
$(W_A^{(\sigma)})_{A\in\mathscr F}$ be as above. For
$A\in\mathscr F$,  we denote by ${\rm PAR}(A)$ the set of all
$\mathscr F$-partition of $A$, and we denote by
\begin{equation}
\label{new3} {\rm Var}^{(\sigma)}_{-}(A)=\inf_{(A_k)_{k\in\mathbb
N}\in{\rm PAR}(A)}\sum_{k=1}^\infty (\sigma(A_k))^2
\end{equation}
the lower variation of sum of squares.
\end{Dn}

\begin{Cy}
\label{cornew3} Let $A\in\mathscr F$, and consider the sum of
random variables squares
\[
\sum_{k=1}^\infty \left(W_{A_k}^{(\sigma)}\right)^2
\]
on $\Omega$ for all partitions of $A$. It is $\chi^2$-distributed
on the probability space $(\Omega,\mathscr B,P)$ of Lemma
\ref{lemma3}. Then the following two conditions are equivalent.
Given $\e>0$,\\
$(i)$\hspace{1cm} $(A_k)\in{\rm PAR}(A)$ satisfies
${\rm Var}^{(\sigma)}_{-}(A)=0$ and $\sum_{k}(\sigma(A_k))^2\le \epsilon$,\\ and\\
$(ii)$\hspace{1cm}$\mathbb
E\left[|\sigma(A)-\sum_{k}\left(W_{A_k}^{(\sigma)}\right)^2|^2\right]\le
2\e$. \label{cor39}
\end{Cy}

\begin{Rk}\label{newrk}
{\rm \mbox{}\\
$(a)$ The meaning of $(ii)$ is the assertion that, when ${\rm
Var}^{(\sigma)}_{-}=0$, the random variable
\[
\sum_{k=1}^\infty \left(W_{A_k}^{(\sigma)}\right)^2
\]
is constant $P$-a.e. on $\Omega$. (It is called the quadratic
variation of $W_A^{(\sigma)}$).\\
$(b)$ The conditions in the Corollary are satisfied if
$(M,\mathscr F,\sigma)$ is taken to be the real line with the
Borel sets and the Lebesgue measure. For example if $A=[0,T]$ we
consider a sequence of partitions consisting of dyadic intervals
and
\[
{\rm Var}^{(\sigma)}_{-}([0,T])\le
T^2\sum_{k=1}^{2^n}\left(\frac{1}{2^n}\right)^2=\frac{T^2}{2^n}\rightarrow 0
\]
as $n\rightarrow\infty$.\\
$(c)$ More generally, consider $(M,\mathscr F,\sigma)$ as above,
and let $A\in\mathscr F$ be such that $0<\sigma(A)<\infty$.
Suppose that for all $n\in\mathbb N$ there is a partition
$\left\{A_1,\ldots, A_n\right\}$ such that
\begin{equation}
\label{new4} \sigma(A_k)=\frac{1}{n}\sigma(A),\quad k=1,\ldots, n.
\end{equation}
Then
\begin{equation}
\label{new5} {\rm Var}^{(\sigma)}_{-}(A)=0.
\end{equation}
Indeed, if \eqref{new4} holds, then
\[
\sum_{k=1}^n\sigma(A_k)^2=\frac{\sigma(A)^2}{n}\rightarrow 0,
\]
so \eqref{new5} follows.}
\end{Rk}

\begin{Dn}
\label{newdf1} Let $(M,\mathscr F,\sigma)$ be specified as above.
We say that it is \underline{refinable} if for every
$A\in\mathscr F$ and every $\e>0$ there exists $(A_k)\in{{\rm
PAR}}(A)$ such that
\begin{equation}
\label{newn1} |(A_k)|=\sup_{k}\sigma(A_k)<\e.
\end{equation}
\end{Dn}

\begin{Cy} Let $(M,\mathscr F,\sigma)$ and $(W_A^{(\sigma)})_{A\in\mathscr
F}$ be as above, and assume that $(M,\mathscr F,\sigma)$ is
refinable. Then for every $A\in\mathscr F$ and every
$(A_k^{(n)})\in{\rm PAR}(A)$ such that
\[
\lim_{n\rightarrow\infty}|A_k^{(n)}|=0,
\]
we have
\begin{equation}
\label{newn2} \lim_{n\rightarrow\infty}\sum_k
\left(W^{(\sigma)}_{A_k^{(n)}}\right)^2=\sigma(A),\quad P\, a.e.
\end{equation}
namely, in the limit, the left hand side random variable reduces to the constant $\sigma(A)$.
\label{cy314}
\end{Cy}
\begin{proof} Using Proposition \ref{prop37} and Corollary
\ref{cor39} we only need to show that
\begin{equation}
\label{newn3}
\lim_{n\rightarrow\infty}\sum_{k}\left(\sigma(A_k^{(n)})\right)^2=0.
\end{equation}
But the following holds for the left hand side of \eqref{newn3}:
\[
\sum_{k}\left(\sigma(A_k^{(n)})\right)^2\le|A_k^{(n)}|\sum_{k}\sigma(A_k^{(n)}).
\]
We now use that, by assumption,
\[
\lim_{n\rightarrow\infty}|A_k^{(n)}|=0,
\]
and that $\sum_{k}\sigma(A_k^{(n)})=\sigma(A)<\infty$ since
$(A_k^{(n)})\in{\rm PAR}(A)$ for all $n\in\mathbb N$. The conclusion  \eqref{newn3} now follows.
\end{proof}
\begin{Rk}{\rm It is obvious that if $(M,\mathscr F,\sigma)$ is
refinable, it has to be \underline{non-atomic}.} \end{Rk}

\begin{Dn}
\label{lalaldo}
The stochastic process $Y$ defined on $M$ with values in $\mathbf L_2(\Omega,\mathcal B,P)$ is called $\mathcal F$-adapted if the following condition
holds for every $A,B\in \mathcal F$ be such that $A\cap B=\emptyset$. Set $\mathcal F_B$ to be the sigma-algebra generated by the random variables $W^{(\sigma)}_C$, where $C$ runs through all subsets of $B$ which belong to $\mathcal F$.
For all $x\in A$, $Y_x$ is  $\mathcal F_A$-measurable,
and, in addition, is independent of $\mathcal F_B$.
\end{Dn}

\begin{La}
\label{lalala}
Let $Y$ be a $\mathcal F$-adapted process such that the function $x\mapsto E\left(|Y(x)|^2\right)$ is measurable and $\int_M\mathbb E\left(|Y(x)|^2\right)d\sigma(x)<\infty$. Then the random variables
\[
\sum_{k=1}^nY(x_k)W^{(\sigma)}_{A_k}
\]
where $\left\{A_k\right\}_{k=1}^n$ is a covering of $M$ by pairwise disjoint measurable sets and $x_k\in A_k$, $k=1,\ldots, n$,
converges to an element in $\mathbf L_2(\Omega,\mathcal B,P)$, which we denote $\int_M Y(x)dW_x^{(\sigma)}$. Furthermore,
we have the Ito-isometry property
\begin{equation}
\label{chloe}
\mathbb E\left(\big|\int_M Y(x)dW_x^{(\sigma)}\big|^2\right)=\int_M\mathbb E\left(|Y(x)|^2\right)d\sigma(x).
\end{equation}
\end{La}

\begin{proof}
When computing the difference between two such sums, say  $\left\{A_k\right\}_{k=1}^n$ (with points $\left\{x_k\right\}_{k=1}^n$)
and $\left\{B_k\right\}_{k=1}^m$ (with points $\left\{y_k\right\}_{k=1}^m$)
we build a covering of pairwise disjoint measurable sets from the two given covering, say
$\left\{C_k\right\}_{k=1}^p$ (with points $\left\{z_k\right\}_{k=1}^p$)
The integral
\[
\mathbb E\left(
\big|\sum_{k=1}^nY(x_k)W^{(\sigma)}_{A_k}-\sum_{k=1}^mY(y_k)W^{(\sigma)}_{B_k}\big|^2\right)
\]
can then be divided into two groups of terms: The sum
\[
\mathbb E\left(
\big|\sum_{k=1}^m(Y(z_{i_k})-Y(z_{j_k})W^{(\sigma)}_{C_k}\big|^2\right)=\sum_{k=1}^p\mathbb E\left(|Y(z_{i_k})-
Y(z_{j_k})|^2\right)\sigma(C_k)
\]
which goes to $0$ since $\int_M\left(\mathbb E\left(|Y(x)|^2\right)\right)d\sigma(x)<\infty$
and the cross-products
\[
\sum_{k,\ell=1}^p
\mathbb E\left(Y(z_k)W_{C_k}^{(\sigma)}\overline{Y(z_\ell)}W_{C_\ell}^{(\sigma)}\right)=
\big|\mathbb E\left(\sum_{k=1}^pY(z_k)W_{C_k}^{(\sigma)}\right)\big|^2
\]
which goes to $0$ since $\sigma$ is refinable, and hence non-atomic.
\end{proof}

\begin{Cy}
\label{cy322}
Let $(M,\mathscr F,\sigma)$ be a sigma-finite measure space and assume it is refinable (see Definition \ref{newdf1}). Let
$(W_A^{(\sigma)})_{A\in\mathscr F}$ be the corresponding Gaussian process (see Definition \ref{def1}). Let $f:\,\mathbb R\,\longrightarrow\,\mathbb R$ be a given $C^2$-function. Then for all $A\in\mathscr F$ such that $0<\sigma(A)<\infty$ we have:
\begin{equation}\label{itoformula}
f(W^{(\sigma)}_A)-f(0)=\int_Af^\prime(W^{(\sigma)}_x)dW_x^{(\sigma)}+\frac{1}{2}\int_Af^{\prime\prime}
(W^{(\sigma)}_x)d\sigma(x).
\end{equation}
\end{Cy}

\begin{proof}[A sketch of the proof] First note that all the terms in \eqref{itoformula} are random variables. Given $W^{(\sigma)}_A$,
by $f(W_A^{(\sigma)})$ we mean the composition of the function $W_A^{(\sigma)}$ from $\Omega$ to $\mathbb R$ with $f$, and similarly for the terms under the integrals on the right hand side of \eqref{itoformula}. Further, we stress that the first term
$\int_Af^\prime(W^{(\sigma)}_x)dW_x^{(\sigma)}$ is an Ito integral in the sense of Lemma \ref{lalala} and Remark \ref{newrk}, but now with the random process
$f^\prime(W^{(\sigma)}_x)$ occuring under the integral. By the arguments of Lemma \ref{lemma3}, we have
\[
\mathbb E_{P_\sigma}\left[\big|\int_Af^\prime(W^{(\sigma)}_x)dW_x^{(\sigma)} \big|^2\right]=\int_A\mathbb E_{P_\sigma}
\left[|f^\prime(W^{(\sigma)}_x)|^2\right]d\sigma(x).
\]
With that we note that the proof of \eqref{itoformula} is concluded through the use of the same arguments utilized in the
 proof of the classical Ito formula, see e.g. \cite{Revuz91}. Specifically, these
include $(i)$ integration by parts; $(ii)$ a stopping/truncation argument, enabling to prove the result
for $W_A$ restricted to compacts; $(iii)$ then, on compacts, any $C^2$
function may be written as a limit of
polynomial functions; and, finally, $(iv)$ the use of standard convergence
together with Proposition \ref{prop37} and
a probabilistic Dominated Convergence allows to complete the proof.
\end{proof}

\begin{Rk}{\rm We refer to \cite[Theorem 8.2]{MR2793121} for a Ito formula where the stochastic term is computed as a Wick product integral, in the setting of an associated Gelfand triple.}
\end{Rk}

\begin{Rk}
{\rm
We note that Ito integration is done with stochastic integrands, say $Z_t$,
being adapted with respect to an underlying filtration, namely, an increasing
sequence of sub-sigma algebras ${\mathcal F_t}$. This is to say that for all $t\ge 0$, $Z_t$ is $\mathcal F_t$-measurable.
Given the fact that no natural order may be invoked in the present general setting, such terminology
obviously becomes irrelevant here. This is where Definition \ref{lalaldo} is called for.}
\end{Rk}
\subsection{Independent standard Gaussian summands}
\label{sec3.3}
\begin{Ex}
\label{ex5} Take $M=\mathbb R$ and let $\mathscr F=\mathscr
B(\mathbb R)$ denote the Borel sigma-algebra. Let $\sigma$ be a
tempered measure, that is, subject to \eqref{8},
so that we have the Gelfand triple with continuous inclusions
\begin{equation}
\label{8b} \mathcal S\hookrightarrow \mathbf
L_2(\sigma)\hookrightarrow\mathcal S^\prime
\end{equation}
where $\mathcal S$ denotes the Schwartz functions and $\mathcal
S^\prime$ denotes the tempered Schwartz distributions. Let
$\mathscr C$ be the $\mathcal S-\mathcal S^\prime$ cylinder
sigma-algebra of subsets of $\mathcal S^\prime$. We then take
\[
(\Omega,\mathscr C)=(\mathcal S^\prime, \mathscr C),
\]
and we note that the corresponding Gaussian process
$(W^{(\sigma)})$ on $(\mathcal S^\prime,\mathscr C,P)$, determined
by
\begin{equation}
\label{8c} \mathbb
E\left[e^{iW^{(\sigma)}(\varphi)}\right]=e^{-\frac{1}{2}\int_{\mathbb
R} |\widehat{\varphi}(u)|^2d\sigma(u)},\quad \varphi\in\mathcal S,
\end{equation}
satisfies the conditions in Definition \ref{def1} (and of Lemmas
\ref{lem1}-\ref{lemma3}).
\end{Ex}
\begin{Dn}
\label{def6} Let $\gamma_1$ denote the standard $\mathcal N(0,1)$ Gaussian density,
\[
d\gamma_1(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\quad
\mbox{on $\mathbb R$},
\]
and set
\begin{equation}
\label{10} \Omega_\gamma=\times_{\mathbb N}\mathbb R=\mathbb
R\times \mathbb R\times\cdots,
\end{equation}
the infinite Cartesian product, with product measure
\begin{equation}
\label{11} Q\stackrel{\rm def}{=}\times_{\mathbb
N}\gamma_1=\gamma_1\times\gamma_1\times \cdots
\end{equation}
(see \cite{Hida_BM,Ka61}), defined on the cylinder sigma-algebra
$\mathscr C_\gamma$ in $\Omega_\gamma\stackrel{\rm
def}{=}\times_{\mathbb N}\mathbb R$.
\end{Dn}

Specifically, for $f_n$ a measurable
function on $\mathbb R^n$, then on $\Omega_\gamma$ set
$x=(x_1,x_2,\ldots)$ with $x_k\in\mathbb R$ ($k\in\mathbb N$),
\begin{equation}
\label{11b} F(x)=f_n(x_1,\ldots, x_n),
\end{equation}
Then, with this function $f_n$ we set:
\begin{equation}
\label{12} \begin{split}\int_{\Omega_\gamma}FdQ&=\int_{\mathbb
R^n}f_n(x_1,\ldots, x_n)d\gamma_1(x_1)\cdots d\gamma_1(x_n)\\
&=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb R^n}f_n(x_1,\ldots,
x_n)e^{-\frac{1}{2}\sum_{k=1}^nx_k^2}dx_1\cdots dx_n.
\end{split}
\end{equation}

\begin{Rk}{\rm The functions $F$ in \eqref{11b} are called {\it
cylinder functions}. If the conditions in \eqref{12} hold for all
$n$, then $Q$ is uniquely determined.}
\end{Rk}

\begin{Tm}
\label{thm7} \mbox{}\\
$(i)$ Every associated Gaussian process
$(W_A^{(\sigma)})_{A\in\mathscr F}$ (see Definition \ref{def1}), corresponding to a fixed
sigma-finite measure space $(M,\mathscr F,\sigma)$ may be
realized on $(\Omega_\gamma,\mathscr C_\gamma,Q)$, i.e. in the
canonical infinite Cartesian product measure space of Definition
\ref{def6}.\\
$(ii)$ Given $(M,\mathscr F,\sigma)$ and $W^{(\sigma)}$,
$(\Omega,\mathscr C,P)$ as in Definition \ref{def1}, the
realization may be "simulated" by a system of independent,
identically distributed (i.i.d.) $\mathcal N(0,1)$ random variables
$Z_1,Z_2,\ldots$. If $(\varphi_k)_{k\in\mathbb N}$ is an
orthonormal basis of real-valued functions in $\mathbf
L_2(M,\mathscr F,\sigma)$ then a realization of $W^{(\sigma)}$ in
$(\Omega_\gamma,\mathscr C_\gamma,Q)$ may be written as
\begin{equation}
\label{13}
W_A^{(\sigma)}(\cdot)=\sum_{k=1}^\infty\left(\int_A\varphi_k(x)d\sigma(x)\right)
Z_k(\cdot),
\end{equation}
for all $A\in\mathscr F$, where $Z_k((x_i)_{i\in\mathbb
N})\stackrel{\rm def}{=}x_k$ and $(x_i)_{i\in\mathbb
N}\in\times_{\mathbb N}\mathbb R$.
\end{Tm}
\begin{proof} We first note that equation \eqref{13} is a
generalized Karhunen-Lo\`eve formula. We proceed to prove that
the expression in \eqref{13} satisfies the condition in
Definition \ref{def1}. To make sense of \eqref{13} we shall use
the Parseval formula in $\mathbf L_2(M,\mathscr F,\sigma)$. In
particular for $A\in\mathscr F$ we have
\begin{equation}
\label{14} \sigma(A)=\int_M\chi_A(x)d\sigma(x)=\sum_{k\in\mathbb
N}|\langle\chi_A,\varphi_k\rangle_\sigma|^2=\sum_{k\in\mathbb
N}|\int_A\varphi_k(x)d\sigma(x)|^2.
\end{equation}
It follows that the sum in \eqref{13} is convergent in
$\mathbf L_2(\Omega_\gamma,\mathscr C_\gamma,Q)$, and in fact is
convergent pointwise a.e. with respect to $Q$. With the use of
the dominated convergence theorem we obtain:
\[
\mathbb E_Q\left[|\sum_{k\in\mathbb
N}\int_A\varphi_k(x)d\sigma(x)Z_k(\cdot)|^2\right]=
\underbrace{\sum_{k\in\mathbb
N}|\int_A\varphi_k(x)d\sigma(x)|^2}_{\text{(by
\eqref{14})}}=\sigma(A),
\]
where we have used
\begin{equation}
\mathbb E_Q[Z_nZ_m]=\delta_{n,m}\,\quad \forall n,m\in\mathbb N.
\label{15}
\end{equation}
Hence the sum representation of $W^{(\sigma)}$ in \eqref{13} is
a well defined random variable belonging to $\mathbf
L_2(\Omega_\gamma,\mathscr C_\gamma,Q)$. For $A,B\in\mathscr F$
we have
\[
\begin{split}
\mathbb
E_Q\left[W_A^{(\sigma)}W_B^{(\sigma)}\right]&=\sum_{k\in\mathbb
N}\left(\int_A\varphi_k(x)d\sigma(x)\right)\left(\int_B\varphi_k(x)d\sigma(x)\right)\\
&=\sum_{k\in\mathbb N}\langle \chi_A,\varphi_k\rangle_\sigma
\langle \chi_B,\varphi_k\rangle_\sigma\\
&=\langle \chi_A,\chi_B\rangle_\sigma\quad (\text{by Parseval's identity})\\
&=\sigma(A\cap B),
\end{split}
\]
as required in part $(2)$ of Definition \ref{def1}. Verifying the
remaining properties in Definition \ref{def1} is immediate.
\end{proof}

\begin{Rk}{\rm
Note in particular that in representation \eqref{13} of the random
variable $W_A^{(\sigma)}$ is independent of the given choice of
an orthonormal basis in $\mathbf L_2(\sigma)$. If $(M,\mathscr
F,\sigma)=(\mathbb R,\mathscr B,\lambda)$, with $\lambda$ being
the Lebesgue measure on $\mathbb R$, then one can take as the underlying
orthonormal basis a standard wavelet, for instance the Haar
wavelets in $\mathbf L_2(\mathbb R)$. }
\end{Rk}

\begin{La}
\label{lemma8} Let $(M,\mathscr F,\sigma)$ be a sigma-finite
measure space, and let $W^{(\sigma)}$ and $(\Omega,\mathscr C,P)$
be an associated Gaussian process, see Definition \ref{def1} above.
Then for every orthonormal basis $(\varphi_k)_{k\in\mathbb N}$ in
$\mathbf L_2(M,\mathscr F,\sigma)$ the random variables
\begin{equation}
\label{16} X_k^{(\sigma)}=W^{(\sigma)}(\varphi_k),\quad
k\in\mathbb N
\end{equation}
are a collection of i.i.d. $\mathcal N(0,1)$ random variables and
\begin{equation}
\label{17}
W_A^{(\sigma)}(\cdot)=\sum_{k=1}^\infty\left(\int_A\varphi_k(x)
d\sigma(x)\right)X_k^{(\sigma)}(\cdot).
\end{equation}
\end{La}
\begin{proof} Note that the right side of \eqref{16} is the Ito
integral (see \eqref{7} in Notation \ref{n4}). Since
$W^{(\sigma)}$ is an associated Gaussian process, it follows that
$X_k^{(\sigma)}$ in \eqref{16} is Gaussian for all $k\in\mathbb
N$, with $\mathbb E_P[X_k^{(\sigma)}]=0$. Moreover, by \eqref{4} in
Lemma \ref{lemma3} we have
\begin{equation}
\label{18}
\begin{split}
\mathbb
E_P\left[X_k^{(\sigma)}X_n^{(\sigma)}\right]=\int_M\varphi_k(x)\varphi_n(x)d\sigma(x)
&=\langle \varphi_k,\varphi_n\rangle_\sigma =\delta_{m.n}
\end{split}
\end{equation}
since $(\varphi_k)_{k\in\mathbb N}$ is an orthonormal basis in
$\mathbf L_2(\sigma)$. We have already proved that the
$(X_k^{(\sigma)})_{k\in\mathbb N}$ are Gaussian, so it follows
from \eqref{18} that it is an $\mathcal N(0,1)$ i.i.d.
collection on $(\Omega,\mathscr
C,P)$.\smallskip

Moreover, by the argument in the proof of Theorem \ref{thm7}, we may
conclude that representation \eqref{17} holds, with
convergence in $\mathbf L_2(\Omega,\mathscr C,P)$ as well as
pointwise $P$ a.e.
\end{proof}
\begin{Tm}
\label{thm9} Let $(M,\mathscr F,\sigma)$ be a sigma-finite measure
space and let $\left(W_A^{(\sigma)}\right)_{A\in\mathscr F}$ be
an associated Gaussian process on some probability
space $(\Omega,\mathscr C,P)$. Let
$\left(X_k^{(\sigma)}\right)_{k\in\mathbb N}$ be the i.i.d.
$\mathcal N(0,1)$ collection from Lemma \ref{lemma8}. Set
\[
\Gamma\,\,:\,\,\Omega\,\longrightarrow\,\,
\Omega_\gamma=\times_{\mathbb N}\mathbb R,
\]
defined by
\begin{equation}
\label{19} \Gamma(\w)=\left(X_k^{(\sigma)}(\w)\right)_{k\in\mathbb
N}\in\times_{\mathbb N}\mathbb R,\quad \w\in\Omega.
\end{equation}
Then $\Gamma$ is measurable and
\begin{equation}
P(\Gamma^{-1}(C))=Q(C) \label{20}
\end{equation}
holds for all $C\in\mathscr C_\gamma$, i.e. the sigma-algebra of
subsets in $\times_{\mathbb N}\mathbb R$ generated by the
cylinder sets.
\end{Tm}

\begin{Rk}{\rm
For \eqref{20}, we will use the notation
\begin{equation}
\label{21} P\circ\Gamma^{-1}=Q,
\end{equation}
with
\[
\Gamma^{-1}(C)=\left\{\w\in\Omega\,|\, \Gamma(\w)\in C\right\}. \]
}
\end{Rk}
\begin{proof}[Proof of Theorem \ref{thm9}]
Fix $m\in\mathbb N$, and let $I_k=(a_k,b_k]$, $1\le k\le m$ be a series of
finite intervals. It follows from \eqref{19} that if
\[
C=C_m=\left\{(x_k)_{k\in\mathbb N}\,|\, x_k\in I_k,\, 1\le k\le
m\right\}
\]
then
\[
\Gamma^{-1}(C)=\left\{\w\in\Omega\,|\, a_k\le
X^{(\sigma)}_k(\w)<b_k\right\}
\]
and so $\Gamma$, defined by \eqref{19}, is measurable. But we know
that $(X^{(\sigma)}_k)_{k\in\mathbb N}$ is an $\mathcal N(0,1)$ i.i.d.
collection in $(\Omega,\mathscr C,P)$, and $(Z_k)_{k\in\mathbb N}$ is
an $\mathcal N(0,1)$ i.i.d. collection in $(\Omega_\gamma,\mathscr C_\gamma,
Q)$. Hence
\begin{equation}
\label{22} P(\Gamma^{-1}(C))=Q(C_m)=\prod_{k=1}^m\gamma_1(I_k).
\end{equation}
Thus the desired conclusion \eqref{20} holds for cylinders in the
respective probability spaces. On the other hand, the subset of
all sets $C\subset\Omega_\gamma$ for which \eqref{20} holds is a
sigma-algebra. It follows that \eqref{20} is valid on the
respective sigma-algebras generated by the cylinder sets.
\end{proof}
\subsection{Generalized Fourier transform}
\begin{Cy}
\label{cor10} Let $(M,\mathscr F,\sigma)$ be a sigma-finite
measure space, and let $\left(W_A^{(\sigma)}\right)_{A\in\mathscr
F}$ be an associated Gaussian process on the probability space
$(\Omega,\mathscr B, P)$. Then the closed linear span in
$\mathbf L_2(\Omega,\mathscr B, P)$ of the functions
\[
\left\{e^{iW_A^{(\sigma)}(\cdot)}\, |\, A\in\mathscr F\right\}
\]
is $\mathbf L_2(\Omega,\mathscr B, P)$.
\end{Cy}

\begin{proof}
By Theorem \ref{thm9} we may assume that
$\left(W_A^{(\sigma)}\right)_{A\in\mathscr F}$ is realized in
$\mathbf L_2(\times_{\mathbb N}\mathbb R,\mathscr C_\gamma, Q)$,
with $Q=\times_{\mathbb N}\gamma_1$ being the infinite dimensional
measure on $\Omega_\gamma=\times_{\mathbb N}\mathbb R$; see also
Theorem \ref{thm7}. But in this infinite-product space the
conclusion is clear. If $F\in\mathbf L_2(\times_{\mathbb
N}\mathbb R,\mathscr C_\gamma, Q)$ satisfies
\[
\int_{\times_{\mathbb N}\mathbb R} F e^{iW_A}dQ=0,\quad\forall
A\in\mathscr F,
\]
then a direct computation, using \eqref{13}, shows that $F=0$
a.e. $Q$.
\end{proof}

\begin{Dn}
\label{def11} Let $(M,\mathscr F,\sigma)$ be a sigma-finite
measure space, and consider the positive definite function
\begin{equation}
\label{23}
K^{(\sigma)}(A,B)=e^{-\frac{\sigma(A)+\sigma(B)}{2}}e^{\sigma(A\cap
B)}=e^{-\frac{1}{2}\|\chi_A-\chi_B\|^2_\sigma},\quad
A,B\in\mathscr F.
\end{equation}
We shall denote the corresponding reproducing kernel Hilbert
space by $\mathcal H^{(\sigma)}$.
\end{Dn}
\begin{Cy}
\label{cor12} (The generalized Fourier transform.) Let
$\left(W_A^{(\sigma)}\right)_{A\in\mathscr F}$ be a Gaussian
process associated with the sigma-finite measure space  $(M,\mathscr
F,\sigma)$, and let $(\Omega,\mathscr B, P)$ be the underlying
probability space. Let $\mathcal H^{(\sigma)}$ be the reproducing
kernel Hilbert space as in Definition \ref{def11}. For
$F\in\mathbf L_2(\Omega,\mathscr B, P)$, set
\begin{equation}
\label{24} \widehat{F}(A)=\mathbb E_P\left[Fe^{iW_A}\right],\quad
A\in\mathscr F.
\end{equation}
Then the map $F\mapsto \widehat{F}$ is unitary from $\mathbf
L_2(\Omega,\mathscr B, P)$ onto $\mathcal H^{(\sigma)}$.
\end{Cy}
\begin{proof}
Using Corollary \ref{cor10}, it is enough to prove that for all
$A,B\in\mathscr F$ we have
\begin{equation}
\label{25} \mathbb
E\left[e^{-iW_B}e^{iW_A}\right]=K^{(\sigma)}(A,B)=e^{-\left\{\frac{\sigma(A)+\sigma(B)}{2}
+\sigma(A\cap B)\right\}}
\end{equation}
(see \eqref{23}). A direct computation yields
\[
\begin{split}
E\left[e^{-iW_B}e^{iW_A}\right]&=E\left[e^{i(W_A-W_B)}\right]\\
&=e^{-\frac{1}{2}\|\chi_A-\chi_B\|^2_\sigma}\quad (\text{by
Theorem \ref{thm9}})\\
&=K^{(\sigma)}(A,B)\quad\,\,\,\,\,(\text{by \eqref{23}}),
\end{split}
\]
for all $A,B\in\mathscr F$.
\end{proof}

\begin{Cy}
\label{cor13} Let $\left(W_A^{(\sigma)}\right)_{A\in\mathscr
F}$ be a collection of Gaussian random variables associated with a sigma-finite
measure space $(M,\mathscr F,\sigma)$. Let
$(\varphi_k)_{k\in\mathbb N}$ be an orthonormal basis of $\mathbf
L_2(M,\mathscr F,\sigma)$ (we denote this space by $\mathbf
L_2(\sigma)$), and let
$
\Gamma\,\,:\,\,\Omega\,\,\longrightarrow\,\,\times_{\mathbb
N}\mathbb R\stackrel{\rm def.}{=}\Omega_\gamma$, with
\begin{equation}
\label{26} \Gamma(
\w)=(X_k^{(\sigma)}(\w))_{k\in\mathbb N}.
\end{equation}
being the corresponding coordinate system, i.e.
\[
X_k^{(\sigma)}\stackrel{\rm
def.}{=}W^{(\sigma)}(\varphi_k)=\int_M\varphi_k(x)dW^{(\sigma)}(x)
\]
being the Ito-integral representation of \eqref{7}. Then
\begin{equation}
\label{27} \Gamma\left(\mathbf
L_2(\sigma)\right)\subset\ell^2\subset\times_{\mathbb N}\mathbb R
\end{equation}
\end{Cy}

\begin{proof} Recall that we consider real-valued functions. Let
$\langle\cdot, \cdot\rangle_\sigma$ denote the standard inner
product in $\mathbf L_2(\sigma)$. Then for all $f\in \mathbf
L_2(\sigma)$ we have
\[
W^{(\sigma)}(f)\in\mathbf L_2(\Omega, P),
\]
and, by Lemma \ref{lemma3},
\[
\Gamma(f)=\left(\langle
f,\varphi_k\rangle_\sigma\right)_{k\in\mathbb
N}\in\ell_2\subset\times_{\mathbb N}\mathbb R,
\]
and
\begin{equation}
\label{28} \mathbb E_P\left[\left(W^{(\sigma)}(f)\right)^2\right]=\|f\|_\sigma^2=
\sum_{k\in\mathbb N}|\langle f,\varphi_k\rangle|^2,
\end{equation}
and
\begin{equation}
\label{29} W^{(\sigma)}(f)=\sum_{k\in\mathbb N}\langle
f,\varphi_k\rangle_kX_k^{(\sigma)}
\end{equation}
is well defined, Gaussian, and satisfies \eqref{28}.
\end{proof}
\begin{Tm}
\label{thm14} Let $(M,\mathscr F,\sigma)$,
$\left(W_A^{(\sigma)}\right)_{A\in\mathscr F}$ and
$(\Omega,\mathscr B, P)$ be as in Corollary \ref{cor13}. Then the
measure $P$ is quasi-invariant with respect to the $\mathbf
L_2(\sigma)$ translations in $\Omega$. More precisely, we have
\begin{equation}
\label{30}
\int_\Omega  F(\cdot+f)dP(\cdot)=\int_\Omega F(\cdot)e^{-\frac{1}{2}\|f\|_\sigma^2+W^{(\sigma)}(f)(\cdot)}dP(\cdot)
\end{equation}
for all $F\in\mathbf L_2(\Omega, P)$ and all $f\in\mathbf L_2(\sigma)$.
\end{Tm}

\begin{proof}
Note that in the formulation of \eqref{30} we make use of the coordinate system
\[
\Gamma\,:\, \Omega\,\,\longrightarrow\,\, \times_{\mathbb N}\mathbb R
\]
from Theorem \ref{thm9} and Corollary \ref{cor13}.  Hence the term
$e^{-\frac{1}{2}\|f\|_\sigma^2+W^{(\sigma)}(f)(\cdot)}$
is the associated Radon-Nikodym derivative. Using Theorem \ref{thm9}, and passing to cylinder functions, we note that
\eqref{30} follows from a computation  of $\mathbb R^m$-integrals
for all $m\in\mathbb N$. We check that if $F_m$ is $\mathbf L_2$ with respect to the standard $\mathbb R^m$-Gaussian joint density
\[
\gamma_m=\underbrace{\gamma_1\times \cdots\times\gamma_1}_{\text{$m$ times}},
\]
then
\begin{equation}
\label{31}
\begin{split}
\iint_{\mathbb R^m}F_m(x_1+\langle f,\varphi_1\rangle_\sigma,\ldots,
x_m+\langle f,\varphi_m\rangle_\sigma)
d\gamma_m(x_1,\ldots, x_m)&=\\
&\hspace{-6.5cm}= \iint_{\mathbb
R^m}F_m(x_1,\ldots, x_m)e^{-\frac{1}{2}\sum_{k=1}^m\langle
f,\varphi_k\rangle^2_\sigma}e^{\sum_{k=1}^mx_k\langle f, \varphi_k
\rangle_\sigma}d\gamma_m(x_1,\ldots, x_m).
\end{split}
\end{equation}
From Theorem \ref{thm9} and Corollary \ref{cor13} we have that
\begin{equation}
\label{32}
W^{(\sigma)}(f)(\cdot)=\sum_{k=1}^\infty \langle f,\varphi_k\rangle_\sigma X_k^{(\sigma)}(\cdot)
\end{equation}
and
\begin{equation}
\label{33}
\mathbb E_\sigma\left[\left(W^{(\sigma)}(f)\right)^2\right]
=\sum_{k=1}^m\langle f,\varphi_k\rangle_\sigma^2=\|f\|^2_{\mathbf L_2(\sigma)}.
\end{equation}
Using again Theorem \ref{thm9} and Kolmogorov's induction limit construction, we finally note that the desired formula \eqref{30}
follows from \eqref{31}-\eqref{33} above.
\end{proof}

Our quadratic variation result in the first part of the present
section  is motivated by, and is a generalization of a
classical theorem for the Brownian motion, often called L\'evy's
theorem, see e.g. \cite{Buh62}. Similarly, our decomposition
theorem (Theorem \ref{thm7}) is motivated by, and extends a
classical result often called  a {\it Kahrunen-Lo\`eve
decomposition}, see e.g. \cite{Dur98}. In the above,
we have made use of the theory of Ito-integration and its generalizations; see e.g. \cite{MR0388538,MR3051797,MR2508233,MR2454534,MR2444857,MR2397796}
\section{Gaussian stochastic calculus}
\label{sec4}
In this section, a preparation for Section \ref{sec6},  we return to the Gaussian process of Section \ref{sec2},
applying results of Section \ref{sec3}.

\setcounter{equation}{0}
\subsection{Gaussian processes and tempered measures: The Minlos theorem.}

Let $\sigma$ be a tempered measure. The map
\begin{equation}
\label{nnew5}
\varphi\,\mapsto\,e^{-\frac{1}{2}\int_{\mathbb R}|\widehat{\varphi}(u)|^2d\sigma(u)}
\end{equation}
is continuous and positive definite on $\mathcal S$ (see \cite[Proposition 3.3, p. 714]{ajnfao}).
An application of Minlos' theorem (see \cite{MR0435834}) yields:
\begin{Tm}
For every tempered measure $\sigma$ there exists a uniquely defined probability measure $P_\sigma$ on $\mathcal S^\prime$, defined on the
sigma-algebra $\mathscr C=\mathscr C(\mathcal S^\prime)$ generated by the cylinder sets (see \eqref{eqdefcyl}), and determined
by the following condition:
\begin{equation}
\label{new6}
\mathbb E_{P_\sigma}\left[e^{iX_\varphi^{(\sigma)}}\right]=
\int_{\mathcal S^\prime}e^{i\langle \varphi,\xi\rangle}dP_\sigma(\xi)=e^{-\frac{1}{2}\int_{\mathbb R}|\widehat{\varphi}(u)|^2d\sigma(u)}.
\end{equation}
\end{Tm}
Using the corresponding expectation
\[
\mathbb E_\sigma[U]=\int_{\mathcal S^\prime} U(\w)P_\sigma(\w),
\]
\eqref{new6} takes the equivalent form
\[
\mathbb E_\sigma\left[e^{i\langle \cdot,\varphi\rangle}\right]=e^{-\frac{1}{2}\|\widehat{\varphi}\|^2_{\sigma}}.
\]
In the above we make use of reproducing kernel Hilbert spaces and their association with the study of Gaussian processes;
see e.g.
\cite{MR2952544,MR2805533,MR2788708,MR2599216,MR2459226}.
\subsection{A Gaussian process realized on $\mathcal S^\prime$.}
\begin{La}
The pairing \eqref{pairing} between $\mathcal S$ and $\mathcal S^\prime$ extends to a pairing between
$\mathbf L_2(\mathbb R)$ and $\mathcal S^\prime$. The stochastic process
\[
X^{(\sigma)}_\varphi(\xi)=\langle\xi,\varphi\rangle,\quad \varphi\in\mathcal S,
\]
may be extended to a process
\[
X^{(\sigma)}_f(\xi)=\langle\xi,f\rangle,\quad f\in\mathbf L_2(\mathbb R,dx).
\]
\label{la41}
\end{La}

\begin{proof}
Set
\begin{equation}
\label{old7}
X_t^{(\sigma)}(\xi)=\langle\xi,\chi_{[0,t]}\rangle.
\end{equation}
$X_t^{(\sigma)}$ is a zero mean Gaussian process with variance
\begin{equation}
\label{realform}
\int_{\mathbb R}|\widehat{\chi_{[0,t]}}(u)|^2d\sigma(u)=2\int_{\mathbb R}\frac{1-\cos(ut)}{u^2}d\sigma(u)=4
\int_{\mathbb R}\frac{\sin^2(\frac{ut}{2})}{u^2}d\sigma(u).
\end{equation}
\end{proof}

In the following lemma we compute the covariance and related quantities. In formulas \eqref{f1} and \eqref{f2}
below we set
\begin{equation}
\label{f0}
r_\sigma(t)=\mathbb E_\sigma\left[|X_t^{(\sigma)}|^2\right].
\end{equation}

\begin{La} We set $X_0^{(\sigma)}=0$.
The following formulas hold:\\
\begin{eqnarray}
\mathbb E_\sigma\left[|X_t^{(\sigma)}|^2\right]&=&4\int_{\mathbb R}\frac{\sin^2\left(\frac{ut}{2}\right)}{u^2}d\sigma(u),\\
\label{f1}
\mathbb E_\sigma\left[|X_t^{(\sigma)}-X_s^{(\sigma)}|^2\right]&=&r_\sigma(s-t),\\
\label{f2}
\mathbb E_\sigma\left[X_t^{(\sigma)}X_s^{(\sigma)}\right]&=&\frac{r_\sigma(t)+r_\sigma(s)-r_\sigma(s-t)}{2}.
\end{eqnarray}
\label{la42}
\end{La}
The computations are standard and will be omitted.\\

Thus $X^{(\sigma)}$ is a stationary-increment Gaussian process. Recall that these are (not necessarily real)
Gaussian processes with covariance function of the form
\begin{equation}
\label{sincre}
2\mathbb E_\sigma\left[X_t^{(\sigma)}X_s^{(\sigma)}\right]=r(t)+\overline{r(s)}-r(t-s),\quad t,s\in\mathbb R,
\end{equation}
Functions $r$ for which the kernel \eqref{sincre} is positive definite on the real line have been investigated by Schoenberg, von Neumann
and Krein; see \cite{MR0004644,MR0012176}. They are of the form
\[
r(t)=r_0+ict-\int_{\mathbb R}\left\{e^{itv}-1-\frac{itv}{v^2+1}\right\}d\sigma(v)
\]
where $r_0\ge 0$, $c\in\mathbb R$ and $\sigma$ is a tempered measure satisfying \eqref{8} for $p=1$.
In the case where $\sigma$ is even, one recovers formula \eqref{realform}.\smallskip

We note that such processes admit derivatives which are generalized stationary processes
with covariance $\widehat{\sigma}(t-s)$, with the Fourier transform being computed in the sense of distributions. See
\cite{aal2,MR2793121,ajnfao} for more information.

\section{A Gaussian process realized on $C(\mathbb R)$.}
\setcounter{equation}{0}
\label{sec6}
We are now ready to introduce the two realizations of probability spaces for the family of Gaussian processes considered above.
\begin{Tm}
\label{tm61}
There is a unique measure $Q_\sigma$ defined on the cylinder sigma-algebra of $\Omega=C(\mathbb R)$ such that
\begin{equation}
\label{new44}
X_t^{(\sigma)}(\w)=\w(t),\quad t\in\mathbb R,\quad \w\in\Omega=C(\mathbb R),
\end{equation}
and $X^{(\sigma)}_0=0$ is a Gaussian process with zero mean and covariance
\begin{equation}
\mathbb E_{Q_\sigma}\left[X_t^{(\sigma)}X_s^{(\sigma)}\right]=\frac{r_\sigma(t)+r_\sigma(s)-r_\sigma(s-t)}{2}
\end{equation}
(see \eqref{f2}), with $r_\sigma$ as in \eqref{f0}.
\end{Tm}
\begin{proof}
For a finite set $t_1,\ldots, t_n\in\mathbb R$ of sample points and a measurable function $f_n$ on $\mathbb R^n$ we set
\begin{equation}
F(\w)=f_n(\w(t_1),\ldots, \w(t_n)),\quad \w\in\mathbb C(R).
\label{new17}
\end{equation}
For  $a>0$ define
\begin{equation}
\label{ga}
g_a(x)=\frac{1}{a\sqrt{2\pi}}e^{-\frac{x^2}{2a^2}},\quad x\in\mathbb R.
\end{equation}
and for $0<t_1<t_2<\ldots<t_n$ let
\begin{equation}
\begin{split}
\mathcal L_n(F)&=\int_{\mathbb R^n}f_n(x_1,\ldots,
x_n)g_{r_\sigma(t_1)}(x_1)g_{r_\sigma(t_2-t_1)}(x_2-x_1)\cdots
g_{r_\sigma(t_n-t_{n-1})}(x_n-x_{n-1})dx_1dx_2\cdots dx_n\\
&=\int_{\mathbb R^n}f_n(x_1,x_2+x_1,\ldots, x_n+x_{n-1}+\cdots+x_1)
g_{r_\sigma(t_1)}(x_1)g_{r_\sigma(t_2-t_1)}(x_2)\cdots
g_{r_\sigma(t_n-t_{n-1})}(x_n)dx_1dx_2\cdots dx_n,
\end{split}
\end{equation}
where $f_n$ is the function in $\mathbb R^n$ introduced in
\eqref{new17}.\smallskip

Using Kolmogorov's theorem we see that there is a unique measure
$Q_\sigma$ with the property stated in the theorem such that
for a cylinder function $F$ as in \eqref{new17} we have
\begin{equation}
\label{new19}
\int_\Omega FdQ_\sigma=\mathcal L_n(F).
\end{equation}
The other claims are easily verified. We omit the details.
\end{proof}
In this section we make use of the two general procedures for constructing Gaussian processes, and the corresponding
probability spaces. To simplify matters, we divide them into $(i)$ the inductive limit construction first proposed by
Kolmogorov, and $(ii)$ the
alternative approach based on Gelfand triples. For references to the first, see \cite{MR0032961,MR1805833},
for the second, see e.g., \cite{MR1810995}.\\

We further mention the following result from \cite{ajnfao} and
\cite{MR2793121} (we refer the reader to these references for the proof).

\begin{Tm}
\label{place-voltaire}
Let $\sigma$ be a tempered measure (see Definition \ref{def21}) and let $\left(X_\varphi^{(\sigma)}\right)_{\varphi\in\mathcal S}$
be the corresponding Gaussian process indexed by the real Schwartz space $\mathcal S$. Further, let $P_W$ denote the standard Wiener-measure. For $f\in\mathbf L_2(\mathbb R)$, let $\widetilde{f}$ denote the corresponding Gaussian process with zero mean and covariance
\[
\mathbb E_{P_W}(|\widetilde{f}|^2)=\|f\|_2^2.
\]
Then, there exists a continuous linear operator $Q^{(\sigma)}$ from $\mathcal S$ into $\mathbf L_2(\mathbb R)$ such that
\[
\mathbb E_{P_W}\left[e^{i\widetilde{Q^{(\sigma)}(\varphi)}}\right]=\mathbb E_{P_\sigma}\left[e^{iX_\varphi^{(\sigma)}}\right]
=e^{-\frac{1}{2}\int_{\mathbb R}|\widehat{\varphi}(u)|^2d\sigma(u)}.
\]
\end{Tm}

\section{Equivalence of measures}
\label{sec7}
In this section we provide details (Theorem \ref{thm82} below) associated with
the equivalence of the two realizations of probability
spaces for the family of Gaussian processes considered above.
\setcounter{equation}{0}
\subsection{Two probability spaces}
In the approach based on Gelfand triples
we realize the family of stationary-increment processes on the
probability space $(\mathcal S^\prime, \mathscr C(\mathcal
S^\prime),P_\sigma)$, while in the approach based instead on
Kolmogorov consistency theorem, a probability space is
$(C(\mathbb R), \mathscr C,Q_\sigma)$, where $\mathscr C$ is the
corresponding cylinder sigma-algebra. In both cases the Gaussian
process is indexed by a fixed tempered measure $\sigma$.\smallskip

The setting in the present section is as follows. Fix a tempered
measure $\sigma$ on $\mathbb R$ and consider the above two
variations of the $\sigma$-Gaussian process, For $\w\in C(\mathbb
R)$ let $\w^\prime$ be the corresponding Schwartz tempered
distribution.\smallskip

We shall write $X_{\rm Kolm}^{(\sigma)}(t)$ for the process in the
Kolmogorov realization, and $X_{\rm Gel}^{(\sigma)}$ for the
Gelfand-triple realization, i.e.
\begin{eqnarray}
\label{new20} X_{\rm Kolm}^{(\sigma)}(t)(\w)&=&\w(t),\quad\forall
\w\in
C(\mathbb R),\, \forall t\in\mathbb R,\\
X_{\rm
Gel}^{(\sigma)}(t)(\xi)&=&\langle\xi,\chi_{[0,t]}\rangle,\quad\forall
\xi\in\mathcal S^\prime,\, \forall t\in\mathbb R. \label{new21}
\end{eqnarray}
\subsection{Equivalence}
In the following statement recall that
cylinder sets have been defined in \eqref{eqdefcyl}.

\begin{La}
\label{lanew} Let $\sigma$ be a tempered measure as introduced in
Section \ref{sec4} (see Lemmas \ref{la41} and \ref{la42}). Then
the Gaussian process $(X_\varphi^{(\sigma)})_{\varphi\in\mathcal
S}$ determined by
\[
\mathbb E\left[X_\varphi^{(\sigma)}\right]=0\quad\text{and}\quad \mathbb
E\left[e^{iX_\varphi^{(\sigma)}}\right]=e^{-\frac{1}{2}\int_{\mathbb
R} |\widehat{\varphi}(u)|^2d\sigma(u)} \\
\]
extends from $\mathcal S$ to $\mathbf L_2(\mathbb R)$, i.e. the
$\mathcal S$-$\mathcal S^\prime$ pairing
$X_\varphi^{(\sigma)}(\xi)=\langle \xi,\varphi\rangle$ extends
uniquely to $X_f^{(\sigma)}$.
By abuse of notation we shall write
\[
X_f^{(\sigma)}(\xi)=\langle \xi,f\rangle.
\]
\end{La}
\begin{proof}
For $f\in\mathbf L_2(\mathbb R)$, interpret $X_f^{(\sigma)}(\xi)$
as the Ito-integral
\begin{equation}
\label{new456} X_f^{(\sigma)}(\xi)=\int_{\mathbb
R}f(x)dX^{(\sigma)}(x),
\end{equation}
where the right hand side of \eqref{new456} is computed with the
use of Lemma \ref{lemma3}.
\end{proof}

Fix $k\in\mathbb N$, and consider a Gaussian measure $Q$ on $C(\mathbb R^k)$ constructed from Gaussian transition probabilities, see e.g.
\eqref{ga}-\eqref{new17} above. We denote by
\[
C_{\rm temp}(\mathbb R^k)=C(\mathbb R^k)\cap \mathcal S^\prime_{\mathbb R^k}
\]
the set of tempered continuous functions on $\mathbb R^k$ (see \cite{Treves67}). Furthermore, let
$\stackrel{\bullet}{\mathbb R^k}=\left(\mathbb R^k\cup\left\{\infty\right\}\right)^\sim$ denote the one-point
compactification of $\mathbb R^k$. Then we have:

\begin{La}
\label{Lanew}
It holds that
\begin{equation}
\label{eq300814}
C(\mathbb R^k)\cap\prod_{\mathbb R^k}\stackrel{\bullet}{\mathbb R^k}\subset C_{\rm temp}(\mathbb R^k).
\end{equation}
\end{La}
\begin{proof}
The functions $\w$ in the intersection on the left hand side of \eqref{eq300814} have well defined limits at $\infty$. There are in
particular bounded and hence tempered.
\end{proof}

\begin{La}
The Gaussian measure $Q_\sigma$ in Theorem \ref{tm61} is supported in $C(\mathbb R^k)\cap\prod_{\mathbb R^k}\stackrel{\bullet}{\mathbb R^k}$
\end{La}

\begin{proof}
The fact that $Q_\sigma(C(\mathbb R^k)\cap\prod_{\mathbb R^k}\stackrel{\bullet}{\mathbb R^k})=1$ is contained in
\cite[Appendix A]{MR0161189}.
\end{proof}
\begin{Tm}
\label{thm82}
For a fixed tempered measure $\sigma$ let $X^{(\sigma)}_{\rm
Kolm}$ and $X^{(\sigma)}_{\rm Gel}$ be the associated Gaussian
processes arising from Kolmogorov and Gelfand constructions
respectively. Let
\[
T\,:\, C_{\rm temp}(\mathbb R)\,\longrightarrow\, \mathcal S^\prime
\]
be defined by
\begin{equation}
\label{new22} T(\w)=\w^\prime,\quad \w\in C_{\rm temp}(\mathbb R);
\end{equation}
(the prime in $\w^\prime$ denotes the derivative in the sense of
distributions). Then,
\begin{equation}
\label{new222}
X^{(\sigma)}_{\rm Gel}(\w^\prime)=X^{(\sigma)}_{\rm Kolm}(\w)=\w(t),
\end{equation}
and
\[
Q_\sigma\circ T^{-1}=P_\sigma,
\]
meaning that for all cylinder sets $\Delta\subset\mathcal S^\prime$, with
\[T^{-1}(\Delta)=\left\{\w\in C_{\rm temp}(\mathbb R)\,\,\text{such that}\,\, \w^\prime\in\Delta\right\},
\]
then
\begin{equation}
Q_\sigma\left(T^{-1}(\Delta)\right)=P_\sigma(\Delta).
\label{new23}
\end{equation}
\end{Tm}
\begin{proof}
To prove \eqref{new222} let $\w\in C_{\rm temp}(\mathbb R)$ and $t\in\mathbb R$. Since the process is assumed to be zero
at $t=0$ we can write
\[
\begin{split}
X^{(\sigma)}_{\rm Gel}(\w^\prime)&=\langle\w^\prime, \chi_{[0,t]}\rangle\\
&=-\langle \chi_{[0,t]}^\prime, \w\rangle\\
&=\w(t)\\
&=X^{(\sigma)}_{\rm Kolm}(t)(\w).
\end{split}
\]

We now turn to the proof of \eqref{new23}. Note that this yields an explicit formula for the measure $P_\sigma$
on $\mathcal S^\prime$ which was previously obtained indirectly as an application of Minlos' theorem to the
right hand side of \eqref{6}, i.e. the Minlos-Gelfand approach does not yield a construction of $P_\sigma$, only existence.
Let $A\subset \mathbb R^n$ be a Borel set. Consider a cylinder set of the form \eqref{eqdefcyl}, with
\[
\varphi_k(u)=\chi_{[0,t_k]}(u),\quad k=1,\ldots, n.
\]
Then,
\[
\begin{split}
P_\sigma\left\{\w^\prime\in\mathcal S^\prime\,|\, (\langle\w^\prime,\chi_{[0,t_1]}\rangle,\ldots,
\langle\w^\prime,\chi_{[0,t_n]}\rangle)\in A\right\}=Q_\sigma\left\{\w\in C_{\rm temp}(\mathbb R)\,|\,(\w(t_1),\ldots, \w(t_n))\in
A\right\},
\end{split}
\]
where $P_\sigma$ and $Q_\sigma$ refer to the functional measures on $\mathcal S^\prime$ and $C(\mathbb R)$
respectively.\smallskip

\end{proof}

{\bf Acknowledgment:} It is a pleasure to thank the referee for his comments on the first version of the paper.
\bibliographystyle{plain}
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  \setbox7\hbox{\accent"5E#1}\penalty 10000\relax\fi\raise 1\ht7
  \hbox{\lower1.05ex\hbox to 1\wd7{\hss\accent"12\hss}}\penalty 10000
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  \def\cprime{$'$}

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Вакханки, або менади ( — «шалені») — німфи, які супроводжували Діоніса в його мандрах і походах. Іншими словами «вакханка» — жінка, нестримна у проявах своєї пристрасті.

Ще одна назва менад — фіяди, також тіяди, — за ім'ям Фії, дочки річкового бога Цефісса, німфи, першої служниці Діоніса, чия священна територія розташовувалась поблизу стародавніх Дельф. Вона ж першою влаштувала оргії на честь Діоніса.

Мистецтво звичайно зображувало менад у розпущених шатах, оповитих виноградною лозою, з бубном або флейтою в руках. Менади намагалися втягнути всіх зустрічних до оргіастичних обрядів на честь Вакха, а того, хто не хотів підкорятися, роздирали на шматки.

Література

 Словник античної міфології. — : Наукова думка, 1985. — 236 сторінок.

Посилання

Німфи

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Follow this link for custom items for special events related to your department or campus organization. Please contact us for price quotes and art approvals.
One of our top priorities is to be mindful of affordability of and access to course materials for all students. Below are ways we have worked diligently to lower costs and provide a variety of formats so students have many options when securing course materials.
The price comparison tool on our website makes it easy to compare our prices with the online marketplace. Prices from our store along with a variety of other online vendors appear on one convenient page. Students may then choose the purchase option that best meets their needs.
We have a partnership with a third party vendor who provides digital content downloadable on any device, in addition to our used and new print books. Students also are often able to rent books in the store, offering significant savings.
We have year round buyback for students. Once exams are over, if students no longer need a print textbook, they may bring it back to the bookstore for cash. We let students know how much cash we can offer them. We also use an online site that shows ten other vendors and the cash offers they have for that particular book.


A counseling student will be using a curriculum that is structured towards helping people in using the social, psychological and behavioral aspects of the person to help them. The value in a counseling career is determined by the success that is visible from the clients you have helped. To be taken seriously in the industry as a practicing professional you will need to have a master's degree or a doctorate in the field. A good accredited institution, the undergraduate will have all the facilities needed to successfully complete an undergraduate program.
Before proceeding to a graduate program in substance abuse counseling, the student will need to narrow down on what they want to major in. Having identified what your major will be, you will have an easy time looking at the institutions that you feel are best to do your major in. Most careers in counseling and other social sciences need to be looked at as a calling to serve the rest of humanity rather than being money oriented. From children, young adults, adults and senior citizens, drug addiction does not spare.
When working with children, a counselor needs to ensure that they approach the client the proper way as all clients will differ by the age factor. Drug abuse may be linked to other things in life which may need to be looked at so as to effectively help a client. Its important for the client to arrive at the decision to aid their lives themselves rather than the professional to doing it for them. Through the journey of studying, the student will explore the family unit. Drug problems may be traced to some dysfunctional family relations. Unhealthy family relations may influence substance use by children.


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The Bulls Are Blessed With Options
As a result of many injuries, the Chicago Bulls have been able to develop depth that few teams can contend with.
Bogar Alonso
Bogar Alonso is a dedicated student of the hardwood, soccer pitch, boxing ring, and tennis court. He is a regular NBA contributor to XN Sports. His work, involving more than just sports, has appeared on The Creators Project, A&E Networks, XXL Magazine, and others. Follow Bogar on Twitter @blacktiles
Latest posts by Bogar Alonso (see all)
The Bulls Are Blessed With Options - Mar 30, 2015
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Monta Ellis: Man of Misery - Mar 25, 2015
Though certainly not blessed with the best luck in terms of health, the Chicago Bulls are lucky to have the type of deep, playoff-ready squad that nagging injuries has only made deeper.
We all know the Derrick Rose storyline, but even his absence has served as a blessing in disguise for this 2014-15 Bulls posse. Sans Rose, Aaron Brooks has stepped up in a respectable manner, averaging 15.5-4.7-2.5 on 42.6/32.4/86.7 shooting. Not mind-boggling numbers but better in the course of the last 10 games (16.5-4.8-2.9 on 47.3 shooting), and certainly the type of stats that one would expect from a third-string guard.
Most importantly, the nagging injuries has matured the play of second stringers Tony Snell and Nikola Mirotic. Tony has gone cold since a surprising February when he filled in admirably for then-injured Jimmy Butler, bothering and outplaying LeBron in a Bulls win against the Cavs two days before Valentine's. He's since pulled together some nice games, albeit far in between, and remains a nice defender who can space the floor (when on).
Mirotic has been the gaudier storyline. He's been playing like a man—let alone a rookie—possessed. In the month of March, he's averaging 20.8-7.6-1.7-1.2.-0.8 on 44.1/26.3/83.5. Outside of the three-point percentage, those are pristine numbers for a 6-foot-10 power forward who is a rookie that's just recently been getting considerable playing time. With Threekola on the floor, the Bulls are posting a better point differential than the East-leading Hawks. He's posted some of the best bench numbers for the month, passed Marcus Smart for rookie with the most three-pointers, and was leading Westbrook and James in fourth-quarter scoring at some point.
His play has been so superb that he's making fans of Andrew Wiggins question their ROY-loyalty. What's more, Nikola's play has been so eye-opening—he's been draining unreal stepback threes, taking point guards off the dribble, running the pick-and-roll, hitting turnaround fadeaways with ease, taking the ball to the hoop, improving on defense—that some are already concerned as to how the Bulls might weave him into an already crowded frontline. Not a bad problem to have but still a problem and the main catalyst for this article.
The Bulls, now with a healthy Butler and Gibson in tow, are blessed with untold options. Finding the right one, while appeasing its key contributors, will be tough. Mirotic can handle playing the three-spot fine, but will get beaten by guys like James, DeMarre Carroll and even Terrence Ross. At the four is where he reigns which also happens to be where Gasol and Gibson put down work.
Teams across the 50 states (and Toronto) would do immoral things to nab a Gibson or a Gasol, yet, here are the Bulls with them and Noah, and Mirotic, to plug in in some conceivable and playoff-friendly manner.
Once Rose returns, which he plans on doing, it'll be another option with which to contend with. It's a fun time to be a Bulls fan, less so for coach Tom Thibodeau who is trying to cook a meal no one has the recipe to. Still, the best thing about being in a fortunate position is that more options than not are usually at hand. And so, the Bulls are blessed to be in the very position they hoped to be in since before the season started.
Related TopicsChicago BullsfNikola Mirotic
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redpajama_set_name: RedPajamaCommonCrawl


Assistant State's Attorney Mary Fillipitch received her bachelor's degree from the University of Iowa and her law degree from St. Louis University. She started working for the Will County State's Attorney's Office in 1993. She served three years in the Complaint Office where she presented cases to the Grand Jury. She also served for 15 years as a felony prosecutor, seven of those years as a felony courtroom supervisor.


The resurrection of Sliger's Market began last September when Hometown Market owners Bertha and Cayla Furry announced their plans to sell the property to a business developer with Bertha looking to retire and Cayla looking to focus her efforts on the blossoming Early real estate market. Sliger managed the store since 2011 and after hearing the Furrys' decision found a property on Coggin Avenue, then contacted the city for assistance in keeping his family's tradition alive.
"Brownwood went above and beyond to get this going. Any business that wants to come to Brownwood needs to know the city will work with them any way they can," Sliger said.
In preparation for Saturday's soft opening, Sliger and his wife Kim spent much of the day Friday, and into the early hours of Saturday morning, getting the property ready. Two hours into the soft opening customers grabbed lemons trees to purchase Harris Nursery employees set them down after delivering them from their location in east Texas.


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Metallosticha pamphaes is a species of snout moth in the genus Metallosticha. It was described by Turner in 1904, and is known from Australia.

References

Moths described in 1904
Phycitini


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This exclusive Vampire Weekend interview was conducted by The Future Tense as part of the ON/OFF project.
The most special concert was my second ever concert when I was 10 and my father took me to see Tom Petty at Madison Square Garden. Besides just being a memorable performance there's the whole experience around live music of going to see artists with people you are close with and sharing that common passion.
I have a couple of places I really like playing, but a recent memorable one was headlining Super Bock Super Rock in Portugal. The amount of energy from the crowd that day was insane. It was very dry and there was a huge amount of dust kicking up from people jumping up and down, and it just reminded me of being 16 and going to Punk festivals. I think with vampire weekend our performances are generally pretty reliable and consistent, and for me the audience engagement makes a huge difference to me as a performer.
Well you're obviously very reliant on your gear on stage, and there have been many times that bits of kit has gone wrong and left us standing on stage looking like idiots.
I'd like to play Eastern Europe. I studied Russian in college and spent some time in Saint Petersburg, but I would love to go there as a musician and play there as an artist.
I do a lot of ankle roles and stretch my legs out. I think there are a lot of commonalities with being a professional musician and an athlete – it's important to treat your body right when you're performing.


The Broward office of Vernis & Bowling is seeking experienced attorneys in the areas of insurance defense, with a general liability and/or first party property insurance background. Must be a self-starter and team player. This is a great opportunity for a motivated attorney to work independently within a collegial environment.
We offer a friendly atmosphere and competitive compensation package including excellent benefits, which include a 401k and health insurance. Please submit your Resume along with salary requirements in Word format to Hiring Attorney.
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Home / Recipes / Canadian Thanksgiving
The Canadian Thanksgiving makes an interesting counterpoint to the holiday celebrated by its southern neighbor. As mentioned earlier, the first North American thanksgiving event occurred in Newfoundland in 1578. In the 1600s, Samuel de Champlain and the French Settlers who came with him established an "Order of Good Cheer." This group would hold huge celebrations marking the harvests and other events, sharing their food with Native American neighbors.
The First Canadian Thanksgiving
The first Canadian Thanksgiving was celebrated on April 15, 1872 in thanks for the recovery of the future King Edward VII from a serious illness. The next Thanksgiving didn"t occur until 1879 when it was celebrated on a Thursday in November.
Canada Thanksgiving Date
Much like the United States, Canada seemed to have a difficult time deciding when a day of Thanksgiving should occur. From 1879 to 1898 it was celebrated on a Thursday in November; from 1899 to 1907 on a Thursday in October (except in 1901 and 1904 when it was celebrated on a Thursday in November); from 1908 to 1921 on a Monday in October; and between 1922 and 1930 the Armistice Day Act declared that Thanksgiving would be celebrated on Armistice Day, the Monday of November 11. In 1931 the Act was amended and the old practice of Parliament declaring a day of Thanksgiving each year was resumed.
On January 31, 1957 Parliament issued a proclamation to fix permanently the second Monday in October as "a day of general Thanksgiving to Almighty God for the bountiful harvest with which Canada has been blessed."
Much like the United States" Thanksgiving Day, the Canadian celebration includes parades and festive meals, often including turkey and all the "fixins." Yet, again, at the heart of the celebration is the idea of giving thanks for the goodness of the season past.
History of Thanksgiving Day
Hosting a Thanksgiving Dinner
The First Thanksgiving Dinner


IAAPA, the International Association of Amusement Parks and Attractions founded in 1918, is the largest international trade association for permanent situation amusement facilities worldwide, and is dedicated to the preservation and prosperity of the amusement industry. The organization represents more than 4,000 facility supplier and individuals from over 100 countries. Waterparks are a very important constituency.


There are places we each find to be a safe haven, a "sanctuary" if you will.
Mine has always been just that; a sanctuary.
I have, throughout my life, been able to go into such a place and leave most of my troubles on the other side of the door.
This particular Wednesday night was no different.
For on that night, I found my sanctuary during the Wednesday night Bible study at my church.
With our Senior Pastor teaching a series on the book of John.
I settled into the back row.
And not just the back row, mind you.
But also the corner row with only three seats in it.
You cannot get any further back and still be in the sanctuary.
A few minutes into the teaching, our Executive Pastor came in and sat in front of me.
Not only was he in front of me, but, his full head of hair strategically blocked the visual sight line between me and Senior Pastor fully engaged in teaching.
It had been a tiring day.
I was sitting back there, unemployed, with few… well… really, no job prospects.
I will save the details of other circumstances going on in my life.
In short, I was emotionally and physically drained.
So I did what others do in churches across America, across the globe.
I closed my eyes and began to meditate on the goodness of God for a few moments. (In case you are unfamiliar with this phrase, it means I dozed off).
And I did…for a few moments.
"Foot washing", that was the word that woke me up.
He said it again, "foot washing".
As a young child I had grown up in a church that practiced this from time to time, as it symbolized serving others before ourselves.
And then, I was awake, and listening to a passage of scripture I had read and heard many times: the washing of the disciples' feet by, of all people, Jesus!
Our pastor referenced the concerns that the disciples had brought into the room that night.
All of them were looking for a position…a title.
In fact, many times we will take the more powerful sounding title, even if it brings with it a more modest salary.
We want the title printed on our card, so that we have a way of making an indelible mark upon those we present the card to.
What made the teaching on this fall night specific to me?
What was it about this message that had caught my attention?
Inspiration had filled my tired body and troubled mind.
As he closed in prayer I began to send him a text from the back row thanking him for the teaching.
If only I could find a way to get this simple message into the hands of corporate leadership, head hunters, young aggressive executives, leaders in politics and dare I say it, leaders in the churches of all denominations.
We could accomplish so much more in our professional, personal and spiritual lives if we could follow this obvious, but at times, invisible approach to leadership.


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Home > NBA Players > Why Are NBA Players Paid So Much?
There's no doubt that NBA players are paid mountains of money.
But why is it so?
1 The perks of being in the NBA
1.1 Is it the fans or just company luck?
1.2 Could it be the magnetic sport or just the charm at hand?
1.3 Seasons seasons seasons or are the tickets good enough?
1.4 Lavish lifestyle is a must or heath matters more?
2 Then it all comes down to this…
The perks of being in the NBA
Playing in the world's biggest basketball league comes with a wide range of benefits obviously. And getting a paycheck that with a lot of zeros on it is one of these perks. If you did not know, NBA players get paid in millions and billions of dollars.
As eye-catching as it may look, it is something to really ponder upon here.
Yes, they have an excellent set of skills, incredible agility, and the perfect attitude for this particular sport. But is that why they are paid so much? I think not, so let's dive deep into this world of balls(literally).
Is it the fans or just company luck?
Well, there is no hiding when it comes to the fanbase of NBA players and we surely approve of such high spirits.
The NBA system works in such a way that even being second to the NFL it has a fanbase that reaches outside the US and people from every ethnicity, gender and age enjoy the game too greatly.
Having loyal fans does make you a big deal out there but it takes a lot to please everyone too. NBA players work day and night and their company takes special care for them since they are aware of such a large population waiting patiently for the games to start.
The more seats that are filled, the more profit the company makes. Fans have always been the backbone to any celebrity or renowned sport and the feeling is totally different when you have someone cheering your name from the stands and shouting "Dunk! Dunk! Dunk!". So it is safe to say that NBA players do make a lot of profit from their dedicated fanbase! (and of course respect at the most)
Could it be the magnetic sport or just the charm at hand?
Over the years we have come to notice the increasing popularity of basketball and how slowly and sneakily it has started to take over football too!
Many people might disagree with this notion *ehem* football fans *ehem*. Yet, there is no doubt about the thrill the game brings to the spectators and how it makes their blood rush, making the hair rise in table-turning moments.
Imagine standing in the middle of the court while looking at the chairs being filled up slowly and slowly as spectators get in with whispers of excitement and hope to see their dream NBA star play.
When others start getting wind of the popularity it is sure to bring in sponsors.
Where there is profit there are sponsors and the field needs consistent support from any source to keep moving forward and bring remarkable seasons for the NBA fans.
The sport in itself is a pulling factor for all the spectators out there. Don't you feel the need to jump out of your chair when LeBron James gets ready to take his shot and you simultaneously start chanting " LeBrons gonna trade you"( ya that still burns) or the time you are at the edge of your seat hitting your nails due to the severe suspense that hangs in the air(more like flies into the air with the ball)?
This is one of the many reasons the companies need to look after their star players and pay them in huge sums so they take absolute care of themselves otherwise who's gonna 'airball' in their games.
Seasons seasons seasons or are the tickets good enough?
Might sound weird at first and of course, it should but it is what it is. NBA is one of the games with the most seasons at hand and they happen to have 82 games each year! (damn that's a lot of jumping around)
With the whole structure consisting of thirty teams, it does sound ridiculous why would anyone pay so much "cough" millions "cough" but on the other hand not everything is as fishy as it seems or smells.
Where paying such a huge sum sounds absurd and shocking, I for once can understand the whole game behind the real actual game. Do you get my point? No? Let me clarify it for you.
Having thirty seasons can bring up about 21 million people across the UK and even outside then ladies and gentlemen there shall be no doubt on the money they receive.
You can't possibly watch a game without paying for it, I mean it would be great if you could but alas! Not everything can be granted. I feel like half of the world just wakes up in the morning, pack their bags and just rush to the arenas without thinking about the super expensive tickets! Like how? I think twice before buying a ten cent chocolate( don't know if those still exist).
Anyways, the tickets are dripping with money, its value is as high as an antique painting so it is truly very clever to host 82 games per season!! Now I suddenly wish I was an NBA player.
Lavish lifestyle is a must or heath matters more?
Where there is money, there ought to be a lavish lifestyle but it is not necessary for everyone.
Players like Kevin Durant and Stephen Curry need to keep up with their image as they are a few of the top players of the NBA league.
If a player is outstanding then their worth should also reflect their lifestyle.
Maybe it's a scheme to keep the fans content that their favorite players are being treated with the utmost respect and have all the things they deserve.
But nonetheless, it is true that the league takes absolute care of their players and this is why they pay them excessively so that they have access to the best of things be it fitness related like a gym with all the latest equipment or provide them with the adequate sneakers so they always stand tall on their feet ready to make a basket and woo the whole crowd!
Then it all comes down to this…
After all that I have said and ranted out to you all out there, I must say once again that if a genie pops out of nowhere here, my first wish would be, "please turn me into James Harden".
Who wouldn't want to be paid so much but after all, they earned this from working hard and kept striving for the best!
NBA players are paid so much because the league needs them to be the best of the best and maybe this is a way in which they pay their thanks to the dutiful players.
It's never easy to become a person who holds such importance and respect in about half the population on earth. We could all go and say oh they are being paid more than they should and maybe you are right but what matters is that they give their best performance for the promising people out there!
Being on top of the world should feel great but with great power comes great responsibilities so no matter how much a game ticket costs we should all know that a whole team is working behind these outstanding players and managing such tremendous pressure to portray the fairest game ever!!
Moreover, if a team is representing your station then they ought to be paid fairly cause then maybe the matter of pride comes in the way. (only an assumption but everything must be laid out)
The next time you attend a Lakers or a Warriors game, remember to scream at the top of your lungs for your favorite star so he gives his best on that court!
Why Do NBA Players Wear Masks?
Oscar Schmidt Story: Greatest Player Ever?


2017 truly gave us everything when it came to eyebrow trends. There were swiggle Brows, studded brows, feather brows, braided brows... we could go on. But when the year ended, we couldn't help but wonder... what would a wild eyebrow trend look like in 2018? Would there be more glitter? Will there be studs? What will Instagram deliver?
Turns out, the answer to that burning question is actually a 'Halo Brow.' What's a bloody Halo Brow when it's at home? Well, it's exactly what you think it is... it's both of your eyebrows, joined together at the tail to form one mega Halo Brow.
Honestly, what is going on in here on this day?
In an interview with POPSUGAR, Hannah — aka @hannahdoesmakeupp — revealed that she took inspiration from one of this year's brow trends on Instagram. "I was having a conversation with a friend trying to come up with a new idea for a look, and all of a sudden it came to me that I should connect my brow tails, this look was influenced by fishtail brows; seeing the way my brow flicked upwards inspired the idea of just carrying the brow on until it met in the middle."
While we can't say we're a) convinced or b) stop laughing, we've gotta give Hannah props for that texturing on the halo. It looks like real eyebrow hair. LOOK AT IT!
10 out 10 for effort. Have fun coming up with a trend that beats this one, Instagram. See you on the runway.


Often called "nature's cavity fighter," fluoride is an essential part of dental care. The naturally-occurring mineral helps prevent cavities in both children and adults by making enamel (the outer surface of your teeth) stronger and more resistant to acids that cause tooth decay. Fluoride is available in different forms, including toothpastes and at-home rinses.
Choosing the right dentist for yourself or your family is just as important as choosing your doctor. There are certain credentials that qualified and experienced dentists must hold. To help you with your decision-making process, Dr. Robert Milner offers a few guidelines to keep in mind.


Y510 8600gt DDR2 or DDR3?
There are plenty of configs worldwide for the Ideapad Y510.
And even the local Lenovo agent can't tell me for sure if the Nvidia 8600gt found in the Y510 is DDR2 or DDR3, or are both types found in this laptop?
Anybody here can help please?
Your help is appreciated!! Thanks!!
Re: Y510 8600gt DDR2 or DDR3?
I think is should be DDR3 from most of the review I read so far, well mainly from mainland china but should be the same.
To be sure use EVEREST ULTIMATE or RivaTuner (ver 2.06 last as far as I know).
They'll tell you what's under the hood.


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http://www.canada24news.com/opinion/christopher-hume-why-downsview-parks-new-bioswale-is-money-in-the-bank/136938-news
Christopher Hume: Why Downsview Park's new bioswale is money in the bank
By Christopher HumeStar Columnist
At a time when once-a-century storms occur every few years, keeping the city from sinking city isn't as easy as it once was. Subways and basements flood regularly now, sewers overflow and roads disappear beneath the deluge. During the most recent such event, two weeks ago, a month's worth of rain fell in two hours. And who can forget the storm in July, 2013, that left Toronto submerged.
For the most part, the city's response has been all wet. Mayor John Tory proved himself unequal to the task of keeping up with climate and civic change when he led the charge to reject a plan to modernize the city's stormwater management system. The problem was that it included fees. That's a scary prospect for a mayor as timid as Tory, who doesn't believe that Torontonians should have to pay the costs of living in a big city.
In truth the cost of these downpours is far greater than the price of readiness. The 2013 storm alone resulted in more than $1 billion in insurance claims.
Fortunately, not all branches of government are as inept as the political; the proof is a large bioswale commissioned by the Canada Lands Corp. at Downsview Park. If you haven't heard that word, bioswale, don't worry, you will. Essentially a landform feature that drains storm run-off before it can do damage, it goes a step further than the stormwater ponds with which we are now familiar.
"It combines the art of landscape architecture and the performance of engineering," explains Ian Gray, senior landscape architect at WSO Global. "It's designed to take the load off storm sewers, which don't have the capacity to deal with the sort of storms that now occur more often."
And, Gray makes clear, that's money in the bank.
But the bioswale represents more than a simple mechanical response to infrastructural need. It's based on the assumption that the city is a system as much as a place, a human construction as well as a natural phenomenon. Walking past the still unfinished feature, you see what appears to be a linear garden running along Downsview Park Blvd. across the road from a new townhouse subdivision. Filled with native plants — Joe Pye weed, cone flowers, echinacea — along with indigenous grasses and a row of sweet gum trees, it looks more decorative than utilitarian.
Because the dirty work happens underground, the process remains invisible. A series of street-level channels directs water from the road, which means the usual sewer grates are conspicuously absent. The bioswale occupies a trench divided by weirs into sections and filled with a bottom layer of gravel and soil chosen for extra porosity. The soil absorbs and filters run-off as it sinks into the ground and eventually ends up in the water table. In the case of a truly biblical flood, a pipe has been added to convey excess water to the sewer.
Mimicking natural processes gives the city a fighting chance to cope with the realities of climate change. If extreme weather weren't enough, Toronto has also become a place of hard surfaces. Paved in concrete and asphalt, the ground no longer functions as an absorbent. Water has nowhere to go but subways, sewers, basements and similar low points.
One of the most notorious is the underpass on Lower Simcoe St. south of Front. Though completed in 2009, long after climate change was acknowledged as the greatest issue of our time, it floods with every storm. The most famous image of the 2013 deluge showed a Ferrari abandoned to the water in the underpass. It is emblematic of Toronto's failure to understand that 21st-century infrastructure must transcend the sort of traditional engineering that creates as many problems as it solves.
The holistic thinking implicit in the Downsview bioswale anticipates the future. Its starting point is the totality of the context not the application of narrow professional perspectives. Because infrastructure must address unprecedented levels of complexity, only a multidisciplinary approach is up to the challenge.
In addition to stormwater, the bioswale deals with social, economic, environmental and even esthetic issues while enhancing biodiversity. It doesn't exist in physical and intellectual isolation but is fully engaged in the urban landscape.
The implications are significant. Every element of the city must take on critical new duties, especially streets, which form the largest part of the public realm. "We have to start thinking of them as parks," says Gray.
Christopher Hume is a former Star reporter who is a current freelance columnist based in Toronto. Follow him on Twitter: @HumeChristopher
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