1 Introduction
We construct a criterion for testing the hypothesis that the covariance function of measurable real-valued stationary Gaussian stochastic process $X(t)$ equals $\rho (\tau )$. We shall use the correlogram
\[\hat{\rho }(\tau )=\frac{1}{T}\underset{0}{\overset{T}{\int }}X(t+\tau )X(t)dt,\hspace{1em}0\le \tau \le T,\]
as an estimator of the function $\rho (\tau )$.A lot of papers so far have been dedicated to estimation of covariance function with given accuracy in the uniform metric, in particular, the papers [2, 4, 6, 11, 12] and the book [13]. We also note that similar estimates of Gaussian stochastic processes were obtained in books [7] and [1]. The main properties of the correlograms of stationary Gaussian stochastic processes were studied by Buldygin and Kozachenko [3].
The definition of a square Gaussian random vector was introduced by Kozachenko and Moklyachuk [10]. Applications of the theory of square Gaussian random variables and stochastic processes in mathematical statistics were considered in the paper [9] and in the book [3]. In the papers [5] and [8], Kozachenko and Fedoryanich constructed a criterion for testing hypotheses about the covariance function of a Gaussian stationary process with given accuracy and reliability in $L_{2}(\mathbb{T})$.
Our goal is to estimate the covariance function $\rho (\tau )$ of a Gaussian stochastic process with given accuracy and reliability in $L_{p}(\mathbb{T})$, $p\ge 1$. Also, we obtain the estimate for the norm of square Gaussian stochastic processes in the space $L_{p}(\mathbb{T})$. We use this estimate for constructing a criterion for testing hypotheses about the covariance function of a Gaussian stochastic process.
The article is organized as follows. In Section 2, we give necessary information about the square Gaussian random variables. In Section 3, we obtain an estimate for the norm of square Gaussian stochastic processes in the space $L_{p}(\mathbb{T})$. In Section 4, we propose a criterion for testing a hypothesis about the covariance function of a stationary Gaussian stochastic process based on the estimate obtained in Section 3.
2 Some information about the square Gaussian random variables and processes
Definition 1 ([3]).
Let $\mathbb{T}$ be a parametric set, and let $\varXi =\{\xi _{t}:t\in \mathbf{T}\}$ be a family of Gaussian random variables such that $\mathbf{E}\xi _{t}=0$. The space $\mathit{SG}_{\varXi }(\varOmega )$ is called a space of square Gaussian random variables if any $\zeta \in \mathit{SG}_{\varXi }(\varOmega )$ can be represented as
where $\bar{\xi }={(\xi _{1},\dots ,\xi _{N})}^{T}$ with $\xi _{k}\in \varXi $, $k=1,\dots ,n$, and A is an arbitrary matrix with real-valued entries, or if $\zeta \in \mathit{SG}_{\varXi }(\varOmega )$ has the representation
Theorem 1 ([3]).
Assume that $\zeta \in \mathit{SG}_{\varXi }(\varOmega )$ and $\operatorname{Var}\zeta >0$. Then the following inequality holds for $|s|<1$:
Definition 2 ([3]).
A stochastic process Y is called a square Gaussian stochastic process if for each $t\in \mathbb{T}$, the random variable $Y(t)$ belongs to the space $\mathit{SG}_{\varXi }(\varOmega )$.
3 An estimate for the $L_{p}(\mathbb{T})$ norm of a square Gaussian stochastic process
In the following theorem, we obtain an estimate for the norm of square Gaussian stochastic processes in the space $L_{p}(\mathbb{T})$. We shall use this result for constructing a criterion for testing hypotheses about the covariance function of a Gaussian stochastic process.
Theorem 2.
Let $\{\mathbb{T},\mathfrak{A},\mu \}$be a measurable space, where $\mathbb{T}$ is a parametric set, and let $Y=\{Y(t),t\in \mathbb{T}\}$ be a square Gaussian stochastic process. Suppose that Y is a measurable process. Further, let the Lebesgue integral $\int _{\mathbb{T}}{(\mathbf{E}{Y}^{2}(t))}^{\frac{p}{2}}d\mu (t)$ be well defined for $p\ge 1$. Then the integral $\int _{\mathbb{T}}{(Y(t))}^{p}d\mu (t)$ exists with probability 1, and
for all $\varepsilon \ge {(\frac{p}{\sqrt{2}}+\sqrt{(\frac{p}{2}+1)p})}^{p}C_{p}$, where $C_{p}=\int _{\mathbb{T}}{(\mathbf{E}{Y}^{2}(t))}^{\frac{p}{2}}d\mu (t)$.
(2)
\[P\bigg\{\underset{\mathbb{T}}{\int }{\big|Y(t)\big|}^{p}d\mu (t)>\varepsilon \bigg\}\le 2\sqrt{1+\frac{{\varepsilon }^{1/p}\sqrt{2}}{{C_{p}^{\frac{1}{p}}}}}\exp \bigg\{-\frac{{\varepsilon }^{\frac{1}{p}}}{\sqrt{2}{C_{p}^{\frac{1}{p}}}}\bigg\}\]Proof.
Since $\max _{x>0}{x}^{\alpha }{e}^{-x}={\alpha }^{\alpha }{e}^{-\alpha }$, we have ${x}^{\alpha }{e}^{-x}\le {\alpha }^{\alpha }{e}^{-\alpha }$.
If ζ is a random variable from the space $\mathit{SG}_{\varXi }(\varOmega )$ and $x=\frac{s}{\sqrt{2}}\cdot \frac{|\zeta |}{\sqrt{\mathbf{E}{\zeta }^{2}}}$, where $s>0$, then
\[\mathbf{E}{\bigg(\frac{s}{\sqrt{2}}\frac{|\zeta |}{\sqrt{\mathbf{E}{\zeta }^{2}}}\bigg)}^{\alpha }\le {\alpha }^{\alpha }{e}^{-\alpha }\cdot \mathbf{E}\exp \bigg\{\frac{s}{\sqrt{2}}\frac{|\zeta |}{\sqrt{\mathbf{E}{\zeta }^{2}}}\bigg\}\]
and
\[\mathbf{E}|\zeta {|}^{\alpha }\le {\bigg(\frac{\sqrt{2\mathbf{E}{\zeta }^{2}}}{s}\bigg)}^{\alpha }{\alpha }^{\alpha }{e}^{-\alpha }\mathbf{E}\exp \bigg\{\frac{s}{\sqrt{2}}\frac{|\zeta |}{\sqrt{\mathbf{E}{\zeta }^{2}}}\bigg\}.\]
From inequality (1) for $0<s<1$ we get that
(3)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \mathbf{E}|\zeta {|}^{\alpha }& \displaystyle \le {\bigg(\frac{\sqrt{2\mathbf{E}{\zeta }^{2}}}{s}\bigg)}^{\alpha }{\alpha }^{\alpha }{e}^{-\alpha }\bigg(\mathbf{E}\exp \bigg\{\frac{s}{\sqrt{2}}\frac{\zeta }{\sqrt{\mathbf{E}{\zeta }^{2}}}\bigg\}+\mathbf{E}\exp \bigg\{-\frac{s}{\sqrt{2}}\frac{\zeta }{\sqrt{\mathbf{E}{\zeta }^{2}}}\bigg\}\bigg)\\{} & \displaystyle \le \frac{2}{\sqrt{1-s}}{\bigg(\frac{\sqrt{2\mathbf{E}{\zeta }^{2}}}{s}\bigg)}^{\alpha }{\alpha }^{\alpha }{e}^{-\alpha }\exp \bigg\{-\frac{s}{\sqrt{2}}\bigg\}\\{} & \displaystyle =2L_{0}(s){\bigg(\frac{\sqrt{2\mathbf{E}{\zeta }^{2}}}{s}\bigg)}^{\alpha }{\alpha }^{\alpha }{e}^{-\alpha }.\end{array}\]Let $Y(t)$, $t\in \mathbb{T}$, be a measurable square Gaussian stochastic process. Using the Chebyshev inequality, we derive that, for all $l\ge 1$,
\[P\bigg\{\underset{\mathbb{T}}{\int }{\big|Y(t)\big|}^{p}d\mu (t)>\varepsilon \bigg\}\le \frac{\mathbf{E}{(\int _{\mathbb{T}}|Y(t){|}^{p}d\mu (t))}^{l}}{{\varepsilon }^{l}}.\]
Then from the generalized Minkowski inequality together with inequality (3) for $l>1$ we obtain that
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle {\bigg(\mathbf{E}{\bigg(\underset{\mathbb{T}}{\int }{\big|Y(t)\big|}^{p}d\mu (t)\bigg)}^{l}\bigg)}^{\frac{1}{l}}\le \underset{\mathbb{T}}{\int }{\big(\mathbf{E}{\big|Y(t)\big|}^{pl}\big)}^{\frac{1}{l}}d\mu (t)\\{} & \displaystyle \le \underset{\mathbb{T}}{\int }{\big(2L_{0}(s){\big(2\mathbf{E}{Y}^{2}(t)\big)}^{\frac{pl}{2}}{(pl)}^{pl}{s}^{-pl}\exp \{-pl\}\big)}^{\frac{1}{l}}d\mu (t)\\{} & \displaystyle ={\big(2L_{0}(s)\big)}^{\frac{1}{l}}\underset{\mathbb{T}}{\int }{\big(2\mathbf{E}{Y}^{2}(t)\big)}^{\frac{p}{2}}{s}^{-p}{(pl)}^{p}\exp \{-p\}d\mu (t)\\{} & \displaystyle ={\big(2L_{0}(s)\big)}^{\frac{1}{l}}{2}^{\frac{p}{2}}{s}^{-p}{(pl)}^{p}\exp \{-p\}\underset{\mathbb{T}}{\int }{\big(\mathbf{E}{Y}^{2}(t)\big)}^{\frac{p}{2}}d\mu (t).\end{array}\]
Assuming that $C_{p}=\int _{\mathbb{T}}{(\mathbf{E}{Y}^{2}(t))}^{\frac{p}{2}}d\mu (t)$, we deduce that
\[\mathbf{E}{\bigg(\underset{\mathbb{T}}{\int }{\big|Y(t)\big|}^{p}d\mu (t)\bigg)}^{l}\le 2L_{0}(s){2}^{\frac{pl}{2}}{(lp)}^{pl}\exp \{-pl\}{C_{p}^{l}}{s}^{-pl}.\]
Hence,
\[\begin{array}{r@{\hskip0pt}l}\displaystyle P\bigg\{\underset{\mathbb{T}}{\int }{\big|Y(t)\big|}^{p}d\mu (t)>\varepsilon \bigg\}& \displaystyle \le 2\cdot {\big({2}^{\frac{p}{2}}\big)}^{l}L_{0}(s){\big({p}^{p}\big)}^{l}{\big(\exp \{-p\}\big)}^{l}{C_{p}^{l}}{\big({s}^{-p}\big)}^{l}\cdot \frac{{({l}^{p})}^{l}}{{\varepsilon }^{l}}\\{} & \displaystyle =2L_{0}(s){a}^{l}{\big({l}^{p}\big)}^{l},\end{array}\]
where $a=\frac{{2}^{\frac{p}{2}}{p}^{p}C_{p}}{{e}^{p}{s}^{p}\varepsilon }$, that is, ${a}^{\frac{1}{p}}=\frac{{2}^{\frac{1}{2}}p{C_{p}^{\frac{1}{p}}}}{es{\varepsilon }^{\frac{1}{p}}}$. Let us find the minimum of the function $\psi (l)={a}^{l}{({l}^{p})}^{l}$. We can easily check that it reaches its minimum at the point ${l}^{\ast }=\frac{1}{e{a}^{\frac{1}{p}}}$.Then
\[\begin{array}{r@{\hskip0pt}l}\displaystyle 2L_{0}(s)\psi \big({l}^{\ast }\big)& \displaystyle =2L_{0}(s){a}^{\frac{1}{e{a}^{\frac{1}{p}}}}\cdot {\bigg(\frac{1}{e{a}^{\frac{1}{p}}}\bigg)}^{p\cdot \frac{1}{e{a}^{\frac{1}{p}}}}=2L_{0}(s){a}^{\frac{1}{e{a}^{\frac{1}{p}}}}\cdot {a}^{-\frac{1}{e{a}^{\frac{1}{p}}}}\cdot {e}^{-\frac{p}{e{a}^{\frac{1}{p}}}}\\{} & \displaystyle =2L_{0}(s)\exp \bigg\{-\frac{pes{\varepsilon }^{\frac{1}{p}}}{{2}^{\frac{1}{2}}pe{C_{p}^{\frac{1}{p}}}}\bigg\}=2L_{0}(s)\exp \bigg\{-\frac{s{\varepsilon }^{\frac{1}{p}}}{{2}^{\frac{1}{2}}{C_{p}^{\frac{1}{p}}}}\bigg\}\\{} & \displaystyle =\frac{2}{\sqrt{1-s}}\exp \bigg\{-s\bigg(\frac{1}{2}+\frac{{\varepsilon }^{1/p}}{{2}^{\frac{1}{2}}{C_{p}^{\frac{1}{p}}}}\bigg)\bigg\}.\end{array}\]
In turn, minimizing the function $\theta (s)=\frac{2}{\sqrt{1-s}}\exp \{-s(\frac{1}{2}+\frac{{\varepsilon }^{1/p}}{{2}^{\frac{1}{2}}{C_{p}^{\frac{1}{p}}}})\}$ in s, we deduce ${s}^{\ast }=1-\frac{1}{1+\frac{\sqrt{2}{\varepsilon }^{1/p}}{{C_{p}^{1/p}}}}$. Thus,
\[\theta \big({s}^{\ast }\big)=2\sqrt{1+\frac{{\varepsilon }^{1/p}\sqrt{2}}{{C_{p}^{\frac{1}{p}}}}}\exp \bigg\{-\frac{{\varepsilon }^{\frac{1}{p}}}{\sqrt{2}{C_{p}^{\frac{1}{p}}}}\bigg\}.\]
Since ${l}^{\ast }\ge 1$, it follows that inequality (2) holds if $\frac{1}{e{a}^{\frac{1}{p}}}=\frac{s{\varepsilon }^{1/p}}{\sqrt{2}p{C_{p}^{1/p}}}\ge 1$. Substituting the value of ${s}^{\ast }$ into this expression, we obtain the inequality ${\varepsilon }^{2/p}\ge p{C_{p}^{1/p}}({C_{p}^{1/p}}+\sqrt{2}{\varepsilon }^{1/p})$. Solving this inequality with respect to $\varepsilon >0$, we deduce that inequality (2) holds for $\varepsilon \ge {(\frac{p}{\sqrt{2}}+\sqrt{(\frac{p}{2}+1)p})}^{p}C_{p}$. The theorem is proved. □4 The construction of a criterion for testing hypotheses about the covariance function of a stationary Gaussian stochastic process
Consider a measurable stationary Gaussian stochastic process X defined for all $t\in \mathbb{R}$. Without any loss of generality, we can assume that $X=\{X(t)$, $t\in \mathbb{T}=[0,T+A]$, $0<T<\infty $, $0<A<\infty \}$ and $\mathbf{E}X(t)=0$. The covariance function $\rho (\tau )=\mathbf{E}X(t+\tau )X(t)$ of this process is defined for any $\tau \in \mathbb{R}$ and is an even function. Let $\rho (\tau )$ be continuous on $\mathbb{T}$.
Theorem 3.
Let the correlogram
be an estimator of the covariance function $\rho (\tau )$. Then the following inequality holds for all $\varepsilon \ge {(\frac{p}{\sqrt{2}}+\sqrt{(\frac{p}{2}+1)p})}^{p}C_{p}$:
(4)
\[\hat{\rho }(\tau )=\frac{1}{T}\underset{0}{\overset{T}{\int }}X(t+\tau )X(t)dt,\hspace{1em}0\le \tau \le A,\]
\[P\bigg\{\underset{0}{\overset{A}{\int }}{\big(\hat{\rho }(\tau )-\rho (\tau )\big)}^{p}d\tau >\varepsilon \bigg\}\le 2\sqrt{1+\frac{{\varepsilon }^{1/p}\sqrt{2}}{{C_{p}^{\frac{1}{p}}}}}\exp \bigg\{-\frac{{\varepsilon }^{\frac{1}{p}}}{\sqrt{2}{C_{p}^{\frac{1}{p}}}}\bigg\},\]
where $C_{p}={\int _{0}^{A}}{(\frac{2}{{T}^{2}}{\int _{0}^{T}}(T-u)({\rho }^{2}(u)+\rho (u+\tau )\rho (u-\tau ))du)}^{\frac{p}{2}}d\tau $ and $0<A<\hspace{2.5pt}\infty $.
Remark 1.
Since the sample paths of the process $X(t)$ are continuous with probability one on the set $\mathbb{T}$, $\hat{\rho }(\tau )$ is a Riemann integral.
Proof.
Consider
Since $\hat{\rho }(\tau )-\rho (\tau )$ is a square Gaussian stochastic process (see Lemma 3.1, Chapter 6 in [3]), it follows from Theorem 2 that
\[\mathbf{E}{\big(\hat{\rho }(\tau )-\rho (\tau )\big)}^{2}=\mathbf{E}{\big(\hat{\rho }(\tau )\big)}^{2}-{\rho }^{2}(\tau ).\]
From the Isserlis equality for jointly Gaussian random variables it follows that
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle \mathbf{E}{\big(\hat{\rho }(\tau )\big)}^{2}-{\rho }^{2}(\tau )=\mathbf{E}\bigg(\frac{1}{{T}^{2}}\underset{0}{\overset{T}{\int }}\underset{0}{\overset{T}{\int }}X(t+\tau )X(t)X(s+\tau )X(s)dtds\bigg)-{\rho }^{2}(\tau )\\{} & \displaystyle \hspace{1em}=\frac{1}{{T}^{2}}\underset{0}{\overset{T}{\int }}\underset{0}{\overset{T}{\int }}\big(\mathbf{E}X(t+\tau )X(t)\mathbf{E}X(s+\tau )X(s)+\mathbf{E}X(t+\tau )X(s+\tau )\\{} & \displaystyle \hspace{2em}\times \mathbf{E}X(t)X(s)+\mathbf{E}X(t+\tau )X(s)\mathbf{E}X(s+\tau )X(t)\big)dtds-{\rho }^{2}(\tau )\\{} & \displaystyle \hspace{1em}=\frac{1}{{T}^{2}}\underset{0}{\overset{T}{\int }}\underset{0}{\overset{T}{\int }}\big({\rho }^{2}(\tau )+{\rho }^{2}(t-s)+\rho (t-s+\tau )\rho (t-s-\tau )\big)dtds-{\rho }^{2}(\tau )\\{} & \displaystyle \hspace{1em}=\frac{1}{{T}^{2}}\underset{0}{\overset{T}{\int }}\underset{0}{\overset{T}{\int }}\big({\rho }^{2}(t-s)+\rho (t-s+\tau )\rho (t-s-\tau )\big)dtds\\{} & \displaystyle \hspace{1em}=\frac{2}{{T}^{2}}\underset{0}{\overset{T}{\int }}(T-u)\big({\rho }^{2}(u)+\rho (u+\tau )\rho (u-\tau )\big)du.\end{array}\]
We have obtained that
(5)
\[\mathbf{E}{\big(\hat{\rho }(\tau )-\rho (\tau )\big)}^{2}=\frac{2}{{T}^{2}}\underset{0}{\overset{T}{\int }}(T-u)\big({\rho }^{2}(u)+\rho (u+\tau )\rho (u-\tau )\big)du.\]
\[P\bigg\{\underset{0}{\overset{A}{\int }}{\big(\hat{\rho }(\tau )-\rho (\tau )\big)}^{p}d\tau >\varepsilon \bigg\}\le 2\sqrt{1+\frac{{\varepsilon }^{1/p}\sqrt{2}}{{C_{p}^{\frac{1}{p}}}}}\exp \bigg\{-\frac{{\varepsilon }^{\frac{1}{p}}}{\sqrt{2}{C_{p}^{\frac{1}{p}}}}\bigg\}.\]
Applying Eq. (5), we get
The theorem is proved. □Denote
\[g(\varepsilon )=2\sqrt{1+\frac{{\varepsilon }^{1/p}\sqrt{2}}{{C_{p}^{\frac{1}{p}}}}}\exp \bigg\{-\frac{{\varepsilon }^{\frac{1}{p}}}{\sqrt{2}{C_{p}^{\frac{1}{p}}}}\bigg\}.\]
From Theorem 3 it follows that if $\varepsilon \ge z_{p}=C_{p}{(\frac{p}{\sqrt{2}}+\sqrt{(\frac{p}{2}+1)p})}^{p}$, then
\[P\bigg\{\underset{0}{\overset{A}{\int }}{\big(\hat{\rho }(\tau )-\rho (\tau )\big)}^{p}d\tau >\varepsilon \bigg\}\le g(\varepsilon ).\]
Let $\varepsilon _{\delta }$ be a solution of the equation $g(\varepsilon )=\delta $, $0<\delta <1$. Put $S_{\delta }=\max \{\varepsilon _{\delta },z_{p}\}$. It is obvious that $g(S_{\delta })\le \delta $ and
Let $\mathbb{H}$ be the hypothesis that the covariance function of a measurable real-valued stationary Gaussian stochastic process $X(t)$ equals $\rho (\tau )$ for $0\le \tau \le A$. From Theorem 3 and (6) it follows that to test the hypothesis $\mathbb{H}$, we can use the following criterion.
Remark 2.
The equation $g(\varepsilon )=\delta $ has a solution for any $\delta >0$ since $g(\varepsilon )$ is a decreasing function. We can find the solution of the equation using numerical methods.
The next theorem contain assumptions under which $C_{p}\to 0$ as $T\to \infty $.
Theorem 4.
Let $\rho (\tau )$ be the covariance function of a centered stationary random process. Let $\rho (\tau )$ be a continuous function. If $\rho (T)\to 0$ as $T\to \infty $, then $C_{p}\to 0$ as $T\to \infty $, where $C_{p}={\int _{0}^{A}}{(\psi (T,\tau ))}^{p/2}dt$ and
Proof.
We have $\psi (T,\tau )\le \frac{2}{T}{\int _{0}^{T}}({\rho }^{2}(u)+\rho (u+\tau )\rho (u-\tau ))du\le 4{\rho }^{2}(0)$. Now it is suffices to prove that $\psi (T,\tau )\to 0$ as $T\to \infty $. From the L’Hopital’s rule it follows that
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \underset{T\to \infty }{\lim }\psi (T,\tau )& \displaystyle =\underset{T\to \infty }{\lim }\frac{2}{T}\underset{0}{\overset{T}{\int }}\big({\rho }^{2}(u)+\rho (u+\tau )\rho (u-\tau )\big)du\\{} & \displaystyle =\underset{T\to \infty }{\lim }\big({\rho }^{2}(T)+\rho (T+\tau )\rho (T-\tau )\big)=0.\end{array}\]
Application of Lebesgue’s dominated convergence theorem completes the proof. □
Here are examples in which we find the estimates for $C_{p}$.
Example 1.
Let $\mathbb{H}$ be the hypothesis that the covariance function of a centered measurable stationary Gaussian stochastic process equals $\rho (\tau )=B\exp \{-a|\tau |\}$, where $B>0$ and $a>0$.
To test the hypothesis $\mathbb{H}$, we can use Criterion 1 by selecting $\hat{\rho }_{T}(\tau )$ that is defined in (4) as an estimator of the function $\rho (\tau )$. Let $0<A<\infty $. We shall find the value of the expression
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I& \displaystyle =\underset{0}{\overset{T}{\int }}(T-u)\big({e}^{-2au}+{e}^{-a|u+\tau |}{e}^{-a|u-\tau |}\big)du\\{} & \displaystyle =\underset{0}{\overset{T}{\int }}T{e}^{-2au}du+T\underset{0}{\overset{T}{\int }}{e}^{-a|u+\tau |}{e}^{-a|u-\tau |}du-\underset{0}{\overset{T}{\int }}u{e}^{-2au}du\\{} & \displaystyle \hspace{1em}-\underset{0}{\overset{T}{\int }}u{e}^{-a|u+\tau |}{e}^{-a|u-\tau |}du\\{} & \displaystyle =I_{1}+I_{2}+I_{3}+I_{4}.\end{array}\]
Now lets us calculate the summands:
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{1}& \displaystyle =T\underset{0}{\overset{T}{\int }}{e}^{-2au}du=\frac{T}{2a}\big(1-{e}^{-2aT}\big),\\{} \displaystyle I_{2}& \displaystyle =T\underset{0}{\overset{T}{\int }}{e}^{-a|u+\tau |}{e}^{-a|u-\tau |}du\\{} & \displaystyle =T\bigg(\underset{0}{\overset{\tau }{\int }}{e}^{-a(u+\tau )}{e}^{a(u-\tau )}du+\underset{\tau }{\overset{T}{\int }}{e}^{-a(u+\tau )}{e}^{-a(u-\tau )}du\bigg)\\{} & \displaystyle =T\bigg(\underset{0}{\overset{\tau }{\int }}{e}^{-2a\tau }du+\underset{\tau }{\overset{T}{\int }}{e}^{-2au}du\bigg)\\{} & \displaystyle =T\bigg(\tau {e}^{-2a\tau }-\frac{1}{2a}{e}^{-2aT}+\frac{1}{2a}{e}^{-2a\tau }\bigg),\\{} \displaystyle I_{3}& \displaystyle =\underset{0}{\overset{T}{\int }}u{e}^{-2au}du=-\frac{T}{2a}{e}^{-2aT}+\frac{1}{2a}\underset{0}{\overset{T}{\int }}{e}^{-2au}du\\{} & \displaystyle =-\frac{T}{2a}{e}^{-2aT}-\frac{1}{4{a}^{2}}{e}^{-2aT}+\frac{1}{4{a}^{2}},\\{} \displaystyle I_{4}& \displaystyle =\underset{0}{\overset{T}{\int }}u{e}^{-a|u+\tau |}{e}^{-a|u-\tau |}du\\{} & \displaystyle =\underset{0}{\overset{\tau }{\int }}u{e}^{-a(u+\tau )}{e}^{a(u-\tau )}du+\underset{\tau }{\overset{T}{\int }}u{e}^{-a(u+\tau )}{e}^{-a(u-\tau )}du\\{} & \displaystyle =\underset{0}{\overset{\tau }{\int }}u{e}^{-2a\tau }du+\underset{\tau }{\overset{T}{\int }}u{e}^{-2au}du\\{} & \displaystyle =\frac{{\tau }^{2}}{2}{e}^{-2a\tau }-\frac{T}{2a}{e}^{-2aT}+\frac{\tau }{2a}{e}^{-2a\tau }-\frac{1}{4{a}^{2}}{e}^{-2aT}+\frac{1}{4{a}^{2}}{e}^{-2a\tau }.\end{array}\]
Therefore,
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I& \displaystyle =\bigg(T\tau +\frac{T}{2a}-\frac{{\tau }^{2}}{2}-\frac{\tau }{2a}-\frac{1}{4{a}^{2}}\bigg){e}^{-2a\tau }+\frac{1}{2{a}^{2}}{e}^{-2aT}+\frac{T}{2a}-\frac{1}{4{a}^{2}}\\{} & \displaystyle \le \bigg(T\tau +\frac{T}{2a}\bigg){e}^{-2a\tau }+\frac{T}{2a}+\frac{1}{2{a}^{2}}{e}^{-2aT}.\end{array}\]
Thus, we obtain
\[\begin{array}{r@{\hskip0pt}l}\displaystyle C_{p}& \displaystyle \le {\bigg(\frac{2B}{{T}^{2}}\bigg)}^{\frac{p}{2}}\underset{0}{\overset{A}{\int }}{\bigg(\bigg(T\tau +\frac{T}{2a}\bigg){e}^{-2a\tau }+\frac{T}{2a}+\frac{1}{2{a}^{2}}{e}^{-2aT}\bigg)}^{p/2}d\tau \\{} & \displaystyle ={(2B)}^{\frac{p}{2}}\frac{{T}^{p/2}}{{T}^{p}}I_{5}={(2B)}^{\frac{p}{2}}\frac{1}{{T}^{p/2}}I_{5},\end{array}\]
where $I_{5}={\int _{0}^{A}}{((\tau +\frac{1}{2a}){e}^{-2a\tau }+\frac{1}{2a}+\frac{1}{2{a}^{2}}{e}^{-2aT})}^{p/2}d\tau $.Example 2.
Let $\mathbb{H}$ be the hypothesis that the covariance function of a centered measurable stationary Gaussian stochastic process equals $\rho (\tau )=B\exp \{-a|\tau {|}^{2}\}$, where $B>0$ and $a>0$.
Similarly as in the previous example, to test the hypothesis $\mathbb{H}$, we can use Criterion 1 by selecting $\hat{\rho }_{T}(\tau )$ defined in (4) as the estimator of the function $\rho (\tau )$. Let $0<A<\infty $. Let us find the value of the expression
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I& \displaystyle =\underset{0}{\overset{T}{\int }}(T-u)\big({e}^{-2a{u}^{2}}+{e}^{-a|u+\tau {|}^{2}}{e}^{-a|u-\tau {|}^{2}}\big)du\\{} & \displaystyle =\underset{0}{\overset{T}{\int }}T{e}^{-2a{u}^{2}}du+T\underset{0}{\overset{T}{\int }}{e}^{-a|u+\tau {|}^{2}}{e}^{-a|u-\tau {|}^{2}}du-\underset{0}{\overset{T}{\int }}u{e}^{-2a{u}^{2}}du\\{} & \displaystyle \hspace{1em}-\underset{0}{\overset{T}{\int }}u{e}^{-a|u+\tau {|}^{2}}{e}^{-a|u-\tau {|}^{2}}du\\{} & \displaystyle =I_{1}+I_{2}+I_{3}+I_{4}.\end{array}\]
Now let us calculate the summands:
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{1}& \displaystyle =T\underset{0}{\overset{T}{\int }}{e}^{-2a{u}^{2}}du\le T\underset{0}{\overset{\infty }{\int }}{e}^{-2a{u}^{2}}du=\frac{\sqrt{\pi }T}{2\sqrt{2a}},\\{} \displaystyle I_{2}& \displaystyle =T\underset{0}{\overset{T}{\int }}{e}^{-a|u+\tau {|}^{2}}{e}^{-a|u-\tau {|}^{2}}du=T{e}^{-2a{\tau }^{2}}\underset{0}{\overset{T}{\int }}{e}^{-2a{u}^{2}}du\le \frac{\sqrt{\pi }T}{2\sqrt{2a}}{e}^{-2a{\tau }^{2}},\\{} \displaystyle I_{3}& \displaystyle =\underset{0}{\overset{T}{\int }}u{e}^{-2a{u}^{2}}du=-\frac{1}{4a}\underset{0}{\overset{T}{\int }}{e}^{-2a{u}^{2}}d\big(-2a{u}^{2}\big)=-\frac{1}{4a}\big({e}^{-2a{T}^{2}}-1\big),\\{} \displaystyle I_{4}& \displaystyle =\underset{0}{\overset{T}{\int }}u{e}^{-2a({u}^{2}+{\tau }^{2})}du={e}^{-2a{\tau }^{2}}\underset{0}{\overset{T}{\int }}u{e}^{-2a{u}^{2}}du=-\frac{1}{4a}{e}^{-2a{\tau }^{2}}\big({e}^{-2a{T}^{2}}-1\big).\end{array}\]
Hence,
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I& \displaystyle \le \frac{\sqrt{\pi }T}{2\sqrt{2a}}+\frac{\sqrt{\pi }T}{2\sqrt{2a}}{e}^{-2a{\tau }^{2}}+\frac{1}{4a}\big({e}^{-2a{T}^{2}}-1\big)+\frac{1}{4a}{e}^{-2a{\tau }^{2}}\big({e}^{-2a{T}^{2}}-1\big)\\{} & \displaystyle \le T\bigg(\frac{\sqrt{\pi }}{2\sqrt{2a}}+\frac{\sqrt{\pi }}{2\sqrt{2a}}{e}^{-2a{\tau }^{2}}\bigg).\end{array}\]
Thus, we obtain
\[C_{p}\le {\bigg(\frac{2B}{{T}^{2}}\bigg)}^{\frac{p}{2}}\underset{0}{\overset{A}{\int }}{\bigg(T\bigg(\frac{\sqrt{\pi }}{2\sqrt{2a}}+\frac{\sqrt{\pi }}{2\sqrt{2a}}{e}^{-2a{\tau }^{2}}\bigg)\bigg)}^{p/2}d\tau ={(2B)}^{\frac{p}{2}}\frac{1}{{T}^{p/2}}I_{6},\]
where $I_{6}={\int _{0}^{A}}{(\frac{\sqrt{\pi }}{2\sqrt{2a}}+\frac{\sqrt{\pi }}{2\sqrt{2a}}{e}^{-2a{\tau }^{2}})}^{p/2}d\tau $.
