Let T be a parametric set, and let Ξ={ξt:t∈T} be a family of Gaussian random variables such that Eξt=0 . The space SGΞ(Ω) is called a space of square Gaussian random variables if any ζ∈SGΞ(Ω) can be represented as ζ=ξ¯TAξ¯−Eξ¯TAξ¯, where ξ¯=(ξ1,…,ξN)T with ξk∈Ξ , k=1,…,n , and A is an arbitrary matrix with real-valued entries, or if ζ∈SGΞ(Ω) has the representation ζ=limn→∞(ξ¯nTAξ¯n−Eξ¯nTAξ¯n).

Assume that ζ∈SGΞ(Ω) and Varζ>0 0$]]>. Then the following inequality holds for |s|<1 : (1) Eexp{s2(ζVarζ)}≤11−|s|exp{−|s|2}=L0(s).

A stochastic process Y is called a square Gaussian stochastic process if for each t∈T , the random variable Y(t) belongs to the space SGΞ(Ω) .

Since maxx>0xαe−x=ααe−α 0}{x}^{\alpha }{e}^{-x}={\alpha }^{\alpha }{e}^{-\alpha }$]]>, we have xαe−x≤ααe−α .

If ζ is a random variable from the space SGΞ(Ω) and x=s2·|ζ|Eζ2 , where s>0 0$]]>, then E(s2|ζ|Eζ2)α≤ααe−α·Eexp{s2|ζ|Eζ2} and E|ζ|α≤(2Eζ2s)αααe−αEexp{s2|ζ|Eζ2}. From inequality (1) for 0<s<1 we get that (3) E|ζ|α≤(2Eζ2s)αααe−α(Eexp{s2ζEζ2}+Eexp{−s2ζEζ2})≤21−s(2Eζ2s)αααe−αexp{−s2}=2L0(s)(2Eζ2s)αααe−α.

Let Y(t) , t∈T , be a measurable square Gaussian stochastic process. Using the Chebyshev inequality, we derive that, for all l≥1 , P{∫T|Y(t)|pdμ(t)>ε}≤E(∫T|Y(t)|pdμ(t))lεl. \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 1$]]> we obtain that (E(∫T|Y(t)|pdμ(t))l)1l≤∫T(E|Y(t)|pl)1ldμ(t)≤∫T(2L0(s)(2EY2(t))pl2(pl)pls−plexp{−pl})1ldμ(t)=(2L0(s))1l∫T(2EY2(t))p2s−p(pl)pexp{−p}dμ(t)=(2L0(s))1l2p2s−p(pl)pexp{−p}∫T(EY2(t))p2dμ(t).

Assuming that Cp=∫T(EY2(t))p2dμ(t) , we deduce that E(∫T|Y(t)|pdμ(t))l≤2L0(s)2pl2(lp)plexp{−pl}Cpls−pl. Hence, P{∫T|Y(t)|pdμ(t)>ε}≤2·(2p2)lL0(s)(pp)l(exp{−p})lCpl(s−p)l·(lp)lεl=2L0(s)al(lp)l, \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=2p2ppCpepspε , that is, a1p=212pCp1pesε1p . Let us find the minimum of the function ψ(l)=al(lp)l . We can easily check that it reaches its minimum at the point l∗=1ea1p .

Then 2L0(s)ψ(l∗)=2L0(s)a1ea1p·(1ea1p)p·1ea1p=2L0(s)a1ea1p·a−1ea1p·e−pea1p=2L0(s)exp{−pesε1p212peCp1p}=2L0(s)exp{−sε1p212Cp1p}=21−sexp{−s(12+ε1/p212Cp1p)}.

In turn, minimizing the function θ(s)=21−sexp{−s(12+ε1/p212Cp1p)} in s, we deduce s∗=1−11+2ε1/pCp1/p . Thus, θ(s∗)=21+ε1/p2Cp1pexp{−ε1p2Cp1p}. Since l∗≥1 , it follows that inequality (2) holds if 1ea1p=sε1/p2pCp1/p≥1 . Substituting the value of s∗ into this expression, we obtain the inequality ε2/p≥pCp1/p(Cp1/p+2ε1/p) . Solving this inequality with respect to ε>0 0$]]>, we deduce that inequality (2) holds for ε≥(p2+(p2+1)p)pCp . The theorem is proved.  □

Let the correlogram (4) ρˆ(τ)=1T ∫0TX(t+τ)X(t)dt,0≤τ≤A, be an estimator of the covariance function ρ(τ) . Then the following inequality holds for all ε≥(p2+(p2+1)p)pCp : P{ ∫0A(ρˆ(τ)−ρ(τ))pdτ>ε}≤21+ε1/p2Cp1pexp{−ε1p2Cp1p}, \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 Cp=∫0A(2T2∫0T(T−u)(ρ2(u)+ρ(u+τ)ρ(u−τ))du)p2dτ and 0<A<∞ .

Since the sample paths of the process X(t) are continuous with probability one on the set T , ρˆ(τ) is a Riemann integral.

Consider E(ρˆ(τ)−ρ(τ))2=E(ρˆ(τ))2−ρ2(τ). From the Isserlis equality for jointly Gaussian random variables it follows that E(ρˆ(τ))2−ρ2(τ)=E(1T2 ∫0T ∫0TX(t+τ)X(t)X(s+τ)X(s)dtds)−ρ2(τ)=1T2 ∫0T ∫0T(EX(t+τ)X(t)EX(s+τ)X(s)+EX(t+τ)X(s+τ)×EX(t)X(s)+EX(t+τ)X(s)EX(s+τ)X(t))dtds−ρ2(τ)=1T2 ∫0T ∫0T(ρ2(τ)+ρ2(t−s)+ρ(t−s+τ)ρ(t−s−τ))dtds−ρ2(τ)=1T2 ∫0T ∫0T(ρ2(t−s)+ρ(t−s+τ)ρ(t−s−τ))dtds=2T2 ∫0T(T−u)(ρ2(u)+ρ(u+τ)ρ(u−τ))du. We have obtained that (5) E(ρˆ(τ)−ρ(τ))2=2T2 ∫0T(T−u)(ρ2(u)+ρ(u+τ)ρ(u−τ))du. Since ρˆ(τ)−ρ(τ) is a square Gaussian stochastic process (see Lemma 3.1, Chapter 6 in [3]), it follows from Theorem 2 that P{ ∫0A(ρˆ(τ)−ρ(τ))pdτ>ε}≤21+ε1/p2Cp1pexp{−ε1p2Cp1p}. \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 Cp= ∫0A(2T2 ∫0T(T−u)(ρ2(u)+ρ(u+τ)ρ(u−τ))du)p2dτ. The theorem is proved.  □

For a given level of confidence δ the hypothesis H is accepted if ∫0A(ρˆ(τ)−ρ(τ))pdμ(τ)<Sδ; otherwise, the hypothesis is rejected.

The equation g(ε)=δ has a solution for any δ>0 0$]]> since g(ε) is a decreasing function. We can find the solution of the equation using numerical methods.

We can easily see that Criterion 1 can be used if Cp→0 as T→∞ .

Let ρ(τ) be the covariance function of a centered stationary random process. Let ρ(τ) be a continuous function. If ρ(T)→0 as T→∞ , then Cp→0 as T→∞ , where Cp=∫0A(ψ(T,τ))p/2dt and ψ(T,τ)=2T2 ∫0T(T−u)(ρ2(u)+ρ(u+τ)ρ(u−τ))du,A>0,T>0. 0,\hspace{2.5pt}T>0.\]]]>

We have ψ(T,τ)≤2T∫0T(ρ2(u)+ρ(u+τ)ρ(u−τ))du≤4ρ2(0) . Now it is suffices to prove that ψ(T,τ)→0 as T→∞ . From the L’Hopital’s rule it follows that limT→∞ψ(T,τ)=limT→∞2T ∫0T(ρ2(u)+ρ(u+τ)ρ(u−τ))du=limT→∞(ρ2(T)+ρ(T+τ)ρ(T−τ))=0.

Application of Lebesgue’s dominated convergence theorem completes the proof.  □

Let H be the hypothesis that the covariance function of a centered measurable stationary Gaussian stochastic process equals ρ(τ)=Bexp{−a|τ|} , where B>0 0$]]> and a>0 0$]]>.

To test the hypothesis H , we can use Criterion 1 by selecting ρˆT(τ) that is defined in (4) as an estimator of the function ρ(τ) . Let 0<A<∞ . We shall find the value of the expression I= ∫0T(T−u)(e−2au+e−a|u+τ|e−a|u−τ|)du= ∫0TTe−2audu+T ∫0Te−a|u+τ|e−a|u−τ|du− ∫0Tue−2audu− ∫0Tue−a|u+τ|e−a|u−τ|du=I1+I2+I3+I4. Now lets us calculate the summands: I1=T ∫0Te−2audu=T2a(1−e−2aT),I2=T ∫0Te−a|u+τ|e−a|u−τ|du=T( ∫0τe−a(u+τ)ea(u−τ)du+ ∫τTe−a(u+τ)e−a(u−τ)du)=T( ∫0τe−2aτdu+ ∫τTe−2audu)=T(τe−2aτ−12ae−2aT+12ae−2aτ),I3= ∫0Tue−2audu=−T2ae−2aT+12a ∫0Te−2audu=−T2ae−2aT−14a2e−2aT+14a2,I4= ∫0Tue−a|u+τ|e−a|u−τ|du= ∫0τue−a(u+τ)ea(u−τ)du+ ∫τTue−a(u+τ)e−a(u−τ)du= ∫0τue−2aτdu+ ∫τTue−2audu=τ22e−2aτ−T2ae−2aT+τ2ae−2aτ−14a2e−2aT+14a2e−2aτ. Therefore, I=(Tτ+T2a−τ22−τ2a−14a2)e−2aτ+12a2e−2aT+T2a−14a2≤(Tτ+T2a)e−2aτ+T2a+12a2e−2aT.

Thus, we obtain Cp≤(2BT2)p2 ∫0A((Tτ+T2a)e−2aτ+T2a+12a2e−2aT)p/2dτ=(2B)p2Tp/2TpI5=(2B)p21Tp/2I5, where I5=∫0A((τ+12a)e−2aτ+12a+12a2e−2aT)p/2dτ .

Let H be the hypothesis that the covariance function of a centered measurable stationary Gaussian stochastic process equals ρ(τ)=Bexp{−a|τ|2} , where B>0 0$]]> and a>0 0$]]>.

Similarly as in the previous example, to test the hypothesis H , we can use Criterion 1 by selecting ρˆT(τ) defined in (4) as the estimator of the function  ρ(τ) . Let 0<A<∞ . Let us find the value of the expression I= ∫0T(T−u)(e−2au2+e−a|u+τ|2e−a|u−τ|2)du= ∫0TTe−2au2du+T ∫0Te−a|u+τ|2e−a|u−τ|2du− ∫0Tue−2au2du− ∫0Tue−a|u+τ|2e−a|u−τ|2du=I1+I2+I3+I4. Now let us calculate the summands: I1=T ∫0Te−2au2du≤T ∫0∞e−2au2du=πT22a,I2=T ∫0Te−a|u+τ|2e−a|u−τ|2du=Te−2aτ2 ∫0Te−2au2du≤πT22ae−2aτ2,I3= ∫0Tue−2au2du=−14a ∫0Te−2au2d(−2au2)=−14a(e−2aT2−1),I4= ∫0Tue−2a(u2+τ2)du=e−2aτ2 ∫0Tue−2au2du=−14ae−2aτ2(e−2aT2−1). Hence, I≤πT22a+πT22ae−2aτ2+14a(e−2aT2−1)+14ae−2aτ2(e−2aT2−1)≤T(π22a+π22ae−2aτ2). Thus, we obtain Cp≤(2BT2)p2 ∫0A(T(π22a+π22ae−2aτ2))p/2dτ=(2B)p21Tp/2I6, where I6=∫0A(π22a+π22ae−2aτ2)p/2dτ .