1 Introduction
The convergence of the scaled cumulant generating functions of a sequence of random variables implies a large deviation principle; this is known as the Gärtner–Ellis condition [6, p. 43]. Our main result is that condition for the square of an asymptotically stationary Gaussian process. Reasons for studying squared Gaussian processes come from different fields: large deviation theory [19, 5], time series analysis [10], or ancestry-dependent branching processes [16]. Since only nonnegative real-valued random variables are considered here, we shall use logarithms of Laplace transforms instead of cumulant generating functions.
Theorem 1.
Let $(X_{t})_{t\in \mathbb{N}}$ be a Gaussian process with mean $m=(m(t))$ and covariance kernel $K=(K(t,s))$: for all $t,s\in \mathbb{Z}$,
Assume that there exist a constant $m_{\infty }$ and a positive definite symmetric function k such that:
Denote by f the spectral density of k:
For $t\geqslant 0$, consider the following Laplace transform:
Then for all $\alpha \geqslant 0$,
with
and
\[ \mathbb{E}[X_{t}]=m(t)\hspace{1em}\textit{and}\hspace{1em}\mathbb{E}\big[\big(X_{t}-m(t)\big)\big(X_{s}-m(s)\big)\big]=K(t,s).\]
Assume:
(H2)
\[ \underset{t\geqslant 1}{\sup }\hspace{0.1667em}{\underset{s=0}{\overset{t-1}{\max }}}{\sum \limits_{r=0}^{t-1}}\big|K(s,r)\big|<+\infty .\](H4)
\[ \underset{t\to +\infty }{\lim }\frac{1}{t}{\sum \limits_{s=0}^{t-1}}\big|m(s)-m_{\infty }\big|=0;\](H5)
\[ \underset{t\to +\infty }{\lim }\frac{1}{t}{\sum \limits_{s,r=0}^{t-1}}\big|K(s,r)-k(r-s)\big|=0.\](2)
\[ L_{t}(\alpha )=\mathbb{E}\Bigg[\exp \Bigg(-\alpha {\sum \limits_{s=0}^{t-1}}{X_{s}^{2}}\Bigg)\Bigg].\](3)
\[ \underset{t\to +\infty }{\lim }\frac{1}{t}\log \big(L_{t}(\alpha )\big)=-\ell (\alpha )=-\ell _{0}(\alpha )-\ell _{1}(\alpha )\hspace{0.2778em}\]Theorem 1 yields as a particular case the following result of weak multiplicative ergodicity.
Proposition 1.
Under the hypotheses of Theorem 1 and assuming $K(0,0)$ positive, consider
where $\mathbb{E}_{x}$ denotes the conditional expectation given $X_{0}=x$.
(6)
\[ L_{x,t}(\alpha )=\mathbb{E}_{x}\Bigg[\exp \Bigg(-\alpha {\sum \limits_{s=0}^{t-1}}{X_{s}^{2}}\Bigg)\Bigg],\]The analogue for finite-state Markov chains has long been known [6, p. 72]. It was extended to strong multiplicative ergodicity of exponentially converging Markov chains by Meyn and his coworkers; see [14]. In [13], the square of a Gauss–Markov process was studied, strong multiplicative ergodicity was proved, and the limit was explicitly computed. This motivated the present generalization.
The particular case of a centered stationary process ($m(t)=0$, $K(t,s)=k(t-s)$) can be considered as classical: in that case, the limit (4) follows from Szegő’s theorem on Toeplitz matrices: see [9, 4] as a general reference on Toeplitz matrices and [2] for a review of probabilistic applications of Szegő’s theory. The extension to the centered asymptotically stationary case follows from the notion of asymptotically equivalent matrices in the ${L}^{2}$ sense; see Section 7.4, p. 104, of [9], and [8]. The noncentered stationary case ($m(t)=m_{\infty }$ and $K(s,t)=k(s-t)$) has received much less attention. In Proposition 2.2 of [5], the large deviation principle is obtained for a squared noncentered stationary Gaussian process. There, the centered case is deduced from Szegő’s theorem, whereas the noncentered case follows from the contraction principle. A similar approach to the general case can be found in [1].
We propose here a different method. Instead of the spectral decomposition and Szegő’s theorem, a Wiener–Hopf factorization is used. The limits (4) and (5) are both deduced from the asymptotics of that factorization. The technique is close to those developed in [12] and used in [13]. One advantage is that the coefficients of the Wiener–Hopf factorization can be given a probabilistic interpretation in terms of a regression problem. This approach will be detailed in Section 2.
To go from the stationary to the asymptotically stationary case, the asymptotic equivalence of matrices is needed. But the classical ${L}^{2}$ definition of [8, Sect. 2.3] does not suffice for the noncentered case. A stronger notion, linked to the ${L}^{1}$ norm of vectors instead of the ${L}^{2}$ norm, will be developed in Section 3.
Joining the stationary case to asymptotic equivalence, we get the conclusion of Theorem 1, but only for small enough values of α. To deduce that the convergence holds for all $\alpha \geqslant 0$, an extension of Lévy’s continuity theorem will be used: if both ${(L_{t}(\alpha ))}^{1/t}$ and ${\mathrm{e}}^{-\ell (\alpha )}$ are the Laplace transforms of probability distributions on ${\mathbb{R}}^{+}$, then the convergence over an interval implies the weak convergence of measures and hence the convergence of Laplace transforms for all $\alpha \geqslant 0$. In fact, ${(L_{t}(\alpha ))}^{1/t}$ and ${\mathrm{e}}^{-\ell (\alpha )}$ both are the Laplace transforms of infinitely divisible distributions, more precisely, convolutions of gamma distributions with Poisson compounds of exponentials. Details will be given in Section 4, together with the particular case of a Gauss–Markov process.
2 The stationary case
This section treats the stationary case: $m(t)=m_{\infty }$ and $K(s,t)=k(t-s)$. We shall denote by $c_{t}=(m_{\infty })_{s=0,\dots ,t-1}$ the constant vector with coordinates all equal to $m_{\infty }$ and by $H_{t}$ the Toeplitz matrix with symbol k: $H_{t}=(k(s-r))_{s,r=0,\dots ,t-1}$. The main result of this section is a particular case of Theorem 1. It entails Proposition 2.2 of Bryc and Dembo [5].
Proposition 2.
Assume that k is a positive definite symmetric function such that
and denote by f the corresponding spectral density:
Let $Z=(Z_{t})_{t\in \mathbb{Z}}$ be a centered stationary process with covariance function k. Let $m_{\infty }$ be a real. For all α such that $0\leqslant \alpha <1/(2M)$,
\[ \underset{t\to +\infty }{\lim }\frac{1}{t}\log \Bigg(\mathbb{E}\Bigg[\exp \Bigg(-\alpha {\sum \limits_{s=0}^{t-1}}{(Z_{s}+m_{\infty })}^{2}\Bigg)\Bigg]\Bigg)=-\ell _{0}(\alpha )-\ell _{1}(\alpha ),\]
where $\ell _{0}(\alpha )$ and $\ell _{1}(\alpha )$ are defined by (4) and (5).
Denote by $m_{t}$ and $K_{t}$ the mean and covariance matrix of the vector $(X_{s})_{s=0,\dots ,t-1}$. The Laplace transform of the squared norm of a Gaussian vector has a well-known explicit expression; see, for instance, [19, p. 6]. The identity matrix indexed by $0,\dots ,t-1$ is denoted by $I_{t}$, and the transpose of a vector m is denoted by ${m}^{\ast }$. Then
In the stationary case, $m_{t}=c_{t}$ and $K_{t}=H_{t}$. From (7) we must prove that the following two limits hold:
and
Here, $I_{t}+2\alpha H_{t}$ will be interpreted as the covariance matrix of the random vector $(Y_{s})_{s=0,\dots ,t-1}$ from the process
where $\varepsilon =(\varepsilon _{t})_{t\in \mathbb{Z}}$ is a sequence of i.i.d. standard normal random variables, independent from Z. The limits (8) and (9) will be deduced from a Cholesky decomposition of $I_{t}+2\alpha H_{t}$. We begin with an arbitrary positive definite matrix A. The Cholesky decomposition writes it as the product of a lower triangular matrix by its transpose. Thus, ${A}^{-1}$ is the product of an upper triangular matrix by its transpose. Write it as ${A}^{-1}={T}^{\ast }DT$, where T is a unit lower triangular matrix (the diagonal coefficients equal to 1), and D is a diagonal matrix with positive coefficients. Denote by G the lower triangular matrix $DT$. Then $GA={({T}^{\ast })}^{-1}$ is a unit upper triangular matrix. Hence, the coefficients $G(s,r)$ of G are uniquely determined by the following system of linear equations: for $0\leqslant s\leqslant t$,
where $\delta _{t,s}$ denotes the Kronecker symbol equal to 1 if $t=s$ and 0 else. Notice that ${A}^{-1}={G}^{\ast }{D}^{-1}G$, and $TA{T}^{\ast }={D}^{-1}$, where D is the diagonal matrix with diagonal entries $G(s,s)$. In particular,
and for any vector $m=(m(r))$,
Here is the probabilistic interpretation of the coefficients $G(t,s)$. Consider a centered Gaussian vector Y with covariance matrix A. For $t=0,\dots ,n$, denote by $\mathcal{Y}_{⟦0,t⟧}$ the σ-algebra generated by $Y_{0},\dots ,Y_{t}$, and by $\nu _{t}$ the partial innovation
with the convention $\nu _{0}=Y_{0}$. Using elementary properties of Gaussian vectors, it is easy to check that
Moreover, the $\nu _{t}$ are independent, and the variance of $\nu _{t}$ is $1/G(t,t)$.
(7)
\[ L_{t}(\alpha )={\big(\mathrm{det}(I_{t}+2\alpha K_{t})\big)}^{-1/2}\exp \big(-\alpha {m_{t}^{\ast }}{(I_{t}+2\alpha K_{t})}^{-1}m_{t}\big),\](8)
\[ \underset{t\to +\infty }{\lim }\frac{1}{2t}\log \big(\mathrm{det}(I_{t}+2\alpha H_{t})\big)=\ell _{0}(\alpha )=\frac{1}{4\pi }{\int _{0}^{2\pi }}\log \big(1+2\alpha f(\lambda )\big)\hspace{0.1667em}\mathrm{d}\lambda \](9)
\[ \underset{t\to +\infty }{\lim }\frac{\alpha }{t}{c_{t}^{\ast }}{(I_{t}+2\alpha H_{t})}^{-1}c_{t}=\ell _{1}(\alpha )={m_{\infty }^{2}}\alpha {\big(1+2\alpha f(0)\big)}^{-1}.\](13)
\[ {m}^{\ast }{A}^{-1}m=\sum \limits_{s}\frac{1}{G(s,s)}{\Bigg({\sum \limits_{r=0}^{s}}G(s,r)m(r)\Bigg)}^{2}.\]When this is applied to $A=I_{t}+2\alpha H_{t}$, another interesting interpretation arises. For $t=0,\dots ,n$, $(G(t,s))_{s=0,\dots ,t}$ is the unique solution to the system
Observe that Eqs. (15) are the normal equations of the regression of the $\varepsilon _{t}$ over the $Y_{t}$ in the model (10). Actually, since $\mathbb{E}[Y_{r}Y_{s}]=\delta _{s,r}+2\alpha k(r-s)$ and $\mathbb{E}[\varepsilon _{t}Y_{s}]=\delta _{t,s}$, setting
Eq. (15) says that for $s=0,\dots ,t$,
This means that
(16)
\[ \mu _{t}=G(t,t)\hspace{0.1667em}\nu _{t}={\sum \limits_{r=0}^{t}}G(t,r)\hspace{0.1667em}Y_{r},\]
\[ \mu _{t}=\mathbb{E}[\varepsilon _{t}\hspace{0.1667em}|\hspace{0.1667em}\mathcal{Y}_{⟦0,t⟧}\hspace{0.1667em}].\]
Obviously, the $\mu _{t}$ are independent, the variance of $\mu _{t}$ is $G(t,t)$, and the filtering error is
In particular, it follows that $0<G(t,t)<1$.The asymptotics of $G(t,s)$ will now be related to the spectral density f. Denote $g_{t}(s)=G(t,t-s)$. A change of index in (15) shows that $(g_{t}(s))_{s=0,\dots ,t}$ is the unique solution to the system
Proposition 3.
Assume that k is a positive definite symmetric function such that
and denote by f the corresponding spectral density:
For all α such that $0\leqslant \alpha <1/(2M)$, the following equation has a unique solution in ${L}^{1}(\mathbb{Z})$:
We have:
and
Moreover, if $g_{t}(s)$ is defined for all $0\leqslant s\leqslant t$ by (17), then for all $s\geqslant 0$,
and
(19)
\[ g(0)=\exp \Bigg(-\frac{1}{2\pi }{\int _{0}^{2\pi }}\log \big(1+2\alpha f(\lambda )\big)\hspace{0.1667em}\mathrm{d}\lambda \Bigg)\](20)
\[ {\sum \limits_{s=0}^{+\infty }}g(s)=\exp \Bigg(-\frac{1}{2}\log \big(1+2\alpha f(0)\big)-\frac{1}{4\pi }{\int _{0}^{2\pi }}\log \big(1+2\alpha f(\lambda )\big)\hspace{0.1667em}\mathrm{d}\lambda \Bigg).\]The proof is equivalent to writing the Wiener–Hopf factorization of the operator $I+2\alpha H$: compare with Section 1.5 of [4], in particular, with the proof of Theorem 1.14 on p. 17. The main idea is to reduce Eq. (18) to the problem of finding a sectionally holomorphic function satisfying a boundary condition on the unit circle. This idea is originally due to Krein [15].
Proof.
Conditions of invertibility for Toeplitz operators are well known. They are treated in Sections 2.3 and 7.2 of [4]. Here, the ${L}^{1}$ norm of the Toeplitz operator H with symbol k is M, and the condition $0\leqslant \alpha <1/(2M)$ permits to write the inverse as
This property implies the existence and uniqueness of the solution to Eq. (18). The convergence of the truncated inverse ${(I_{t}+2\alpha H_{t})}^{-1}$ to ${(I+2\alpha H)}^{-1}$ is deduced for the ${L}^{2}$ case from [4, p. 42]. The convergence of entries follows, and hence (21). To obtain (22), consider $\Delta _{t}(s)=g(s)-g_{t}(s)$. From (17) and (18) we have
\[ \Delta _{t}(s)=-2\alpha {\sum \limits_{r=0}^{t}}k(r-s)\Delta _{t}(r)-2\alpha {\sum \limits_{r=t+1}^{+\infty }}g(r)k(r-s).\]
Hence,
\[\begin{array}{r@{\hskip0pt}l}\displaystyle {\sum \limits_{s=0}^{t}}\big|\Delta _{t}(s)\big|& \displaystyle \leqslant 2\alpha \Bigg({\sum \limits_{r=0}^{t}}\big|\Delta _{t}(r)\big|\Bigg)\Bigg({\sum \limits_{s=-\infty }^{+\infty }}\big|k(s)\big|\Bigg)\\{} & \displaystyle \hspace{1em}+2\alpha \Bigg({\sum \limits_{s=-\infty }^{+\infty }}\big|k(s)\big|\Bigg)\Bigg({\sum \limits_{r=t+1}^{+\infty }}\big|g(r)\big|\Bigg).\end{array}\]
Thus, we obtain the following bound:
\[ {\sum \limits_{s=0}^{t}}\big|\Delta _{t}(s)\big|\leqslant \frac{2\alpha M}{1-2\alpha M}{\sum \limits_{r=t+1}^{+\infty }}\big|g(r)\big|,\]
which yields (22).Now we prove identities (19) and (20). The generating function of $(g(s))_{s\geqslant 0}$ will be first related to the spectral density f. Define for all $s\in \mathbb{Z}$,
where $\widetilde{f}(\zeta )$ denotes the value of $f(\lambda )$ for $\zeta ={\mathrm{e}}^{\mathrm{i}\lambda }$. Problem (24) is a well-known homogeneous Riemann problem. Since by construction φ is bounded near infinity and for $|\zeta |=1$, $1+2\alpha \widetilde{f}(\zeta )>0$, the solution of (24) can be written explicitly [18, §35]. Assuming for a moment that $\widetilde{f}$ satisfies the Hölder condition on the unit circle, we have that for all $\zeta _{0}$,
Observe that the choice of a branch for the logarithm does not change the result. From now on, the principal branch will be taken.
\[ {g}^{+}(s)=\bigg\{\begin{array}{l@{\hskip10.0pt}l}g(s)& \text{if}\hspace{2.5pt}s\geqslant 0,\\{} 0& \text{else,}\end{array}\hspace{1em}\text{and}\hspace{1em}{g}^{-}(s)=\bigg\{\begin{array}{l@{\hskip10.0pt}l}g(s)& \text{if}\hspace{2.5pt}s<0,\\{} 0& \text{else.}\end{array}\]
Denote by ${F}^{+}$ and ${F}^{-}$ the Fourier transforms of ${g}^{+}$ and ${g}^{-}$:
\[ {F}^{\pm }(\lambda )=\sum \limits_{s\in \mathbb{Z}}{\mathrm{e}}^{\mathrm{i}s\lambda }{g}^{\pm }(s).\]
Take the Fourier transforms in both members of (18):
or
Let us define the sectionally holomorphic function φ as follows (see [18]):
\[ \varphi (\zeta )=\left\{\begin{array}{l@{\hskip10.0pt}l}{\varphi }^{+}(\zeta )=\sum _{s\geqslant 0}{\zeta }^{s}{g}^{+}(s)\hspace{1em}& \text{if}\hspace{2.5pt}|\zeta |<1,\\{} {\varphi }^{-}(\zeta )=1-\sum _{s<0}{\zeta }^{s}{g}^{-}(s)\hspace{1em}& \text{if}\hspace{2.5pt}|\zeta |>1.\end{array}\right.\]
Then:
\[\begin{array}{r@{\hskip0pt}l}\displaystyle {F}^{+}(\lambda )& \displaystyle =\underset{\begin{array}{c}\zeta \to {\mathrm{e}}^{\mathrm{i}\lambda }\\{} |\zeta |<1\end{array}}{\lim }\varphi (\zeta )\\{} \displaystyle {F}^{-}(\lambda )& \displaystyle =\underset{\begin{array}{c}\displaystyle \zeta \to {\mathrm{e}}^{\mathrm{i}\lambda }\\{} |\zeta |>1\end{array}}{\lim }1-\varphi (\zeta ),\end{array}\]
and Eq. (23) expresses the boundary condition
(24)
\[ {\varphi }^{+}(\zeta )=\frac{1}{1+2\alpha \widetilde{f}(\zeta )}{\varphi }^{-}(\zeta ),\hspace{1em}|\zeta |=1,\]Equation (25) for $\zeta _{0}=0$ implies immediately that
\[\begin{array}{r@{\hskip0pt}l}\displaystyle {g}^{+}(0)& \displaystyle =\exp \bigg(-\frac{1}{2\pi \mathrm{i}}\oint _{|\zeta |=1}\frac{\log (1+2\alpha \widetilde{f}(\zeta ))}{\zeta }\hspace{0.1667em}\mathrm{d}\zeta \hspace{0.1667em}\bigg)\\{} & \displaystyle =\exp \Bigg(-\frac{1}{2\pi }{\int _{0}^{2\pi }}\log \big(1+2\alpha f(\lambda )\big)\hspace{0.1667em}\mathrm{d}\lambda \Bigg),\end{array}\]
which is (19).To prove (20), we will calculate
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle \underset{\begin{array}{c}\zeta _{0}\to 1\\{} |\zeta _{0}|<1\end{array}}{\lim }-\frac{1}{2\pi \mathrm{i}}\oint _{|\zeta |=1}\frac{\log (1+2\alpha \widetilde{f}(\zeta ))}{\zeta -\zeta _{0}}\hspace{0.1667em}\mathrm{d}\zeta \\{} & \displaystyle \hspace{1em}=\underset{\begin{array}{c}\zeta _{0}\to 1\\{}|\zeta _{0}|<1\end{array}}{\lim }-\frac{1}{2\pi \mathrm{i}}\oint _{|\zeta |=1}\frac{\log (1+2\alpha \widetilde{f}(1))}{\zeta -\zeta _{0}}\hspace{0.1667em}\mathrm{d}\zeta \\{} & \displaystyle \hspace{2em}-\underset{\begin{array}{c}\zeta _{0}\to 1\\{} |\zeta _{0}|<1\end{array}}{\lim }\frac{1}{2\pi \mathrm{i}}\oint _{|\zeta |=1}\frac{\log (1+2\alpha \widetilde{f}(\zeta ))-\log (1+2\alpha \widetilde{f}(1))}{\zeta -\zeta _{0}}\hspace{0.1667em}\mathrm{d}\zeta .\end{array}\]
The first integral does not depend on $\zeta _{0}$: it is equal to
Still assuming that $\widetilde{f}$ satisfies a Hölder condition on the unit circle, the second limit exists and is equal to Cauchy’s principal value integral [18]:
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle \frac{1}{2\pi \mathrm{i}}\oint _{|\zeta |=1}\frac{\log (1+2\alpha \widetilde{f}(\zeta ))-\log (1+2\alpha \widetilde{f}(1))}{\zeta -1}\hspace{0.1667em}\mathrm{d}\zeta \\{} & \displaystyle \hspace{1em}=\underset{\epsilon \to 0}{\lim }\frac{1}{2\pi \mathrm{i}}\oint _{\begin{array}{c} |\zeta |=1\\{}|\arg (\zeta )|>\epsilon \end{array}}\frac{\log (1+2\alpha \widetilde{f}(\zeta ))-\log (1+2\alpha \widetilde{f}(1))}{\zeta -1}\hspace{0.1667em}\mathrm{d}\zeta \\{} & \displaystyle \hspace{1em}=\underset{\epsilon \to 0}{\lim }\frac{1}{2\pi }\int _{\begin{array}{c} [-\pi ,\pi ]\\{} |\lambda |>\epsilon \end{array}}\frac{\log (1+2\alpha f(\lambda ))-\log (1+2\alpha f(0))}{1-{\mathrm{e}}^{\mathrm{i}\lambda }}\hspace{0.1667em}\mathrm{d}\lambda .\end{array}\]
Now for $\epsilon <|\lambda |<\pi $,
\[ \frac{1}{1-{\mathrm{e}}^{-\mathrm{i}\lambda }}=\frac{1}{2}+\mathrm{i}\frac{\sin (\lambda )}{2(1-\cos (\lambda ))}.\]
The imaginary part is an odd function of λ, which is multiplied by an even function inside the integral. Hence, the imaginary part in the last integral vanishes. Therefore,
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle \frac{1}{2\pi \mathrm{i}}\oint _{|\zeta |=1}\frac{\log (1+2\alpha \widetilde{f}(\zeta ))-\log (1+2\alpha \widetilde{f}(1))}{\zeta -1}\hspace{0.1667em}\mathrm{d}\zeta \\{} & \displaystyle \hspace{1em}=-\frac{1}{2}\log \big(1+2\alpha f(0)\big)+\frac{1}{4\pi }{\int _{-\pi }^{\pi }}\log \big(1+2\alpha f(\lambda )\big)\hspace{0.1667em}\mathrm{d}\lambda .\end{array}\]
Substracting the last equation from (26) and taking exponential, we get
\[\begin{array}{r@{\hskip0pt}l}\displaystyle {\varphi }^{+}(1)& \displaystyle =\underset{\begin{array}{c} \zeta \to 1\\{} |\zeta |<1\end{array}}{\lim }\varphi (\zeta )\\{} & \displaystyle =\exp \Bigg(-\frac{1}{2}\log \big(1+2\alpha f(0)\big)-\frac{1}{4\pi }{\int _{0}^{2\pi }}\log \big(1+2\alpha f(\lambda )\big)\hspace{0.1667em}\mathrm{d}\lambda \Bigg),\end{array}\]
which is (20).To finish the proof, we must explain how the extra Hölder condition on $\widetilde{f}$ can be removed. It must be emphasized here that the problem is not to obtain the solution of a Riemann problem without Hölder condition on the boundary, but only the values of $\varphi (0)$ and ${\varphi }^{+}(1)$. For this, a truncation argument can be used. From the covariance function k on $\mathbb{Z}$, define
\[ k_{N}(s)=\left\{\begin{array}{l@{\hskip10.0pt}l}k(s)\hspace{1em}& \text{if}\hspace{2.5pt}|s|\leqslant N,\\{} 0\hspace{1em}& \text{else.}\end{array}\right.\]
Replace k by $k_{N}$ in (18) and denote the solution by $g_{N}$. The spectral density $f_{N}$, which is the Fourier transform of $k_{N}$, is smooth. Therefore, the Hölder condition on the unit circle is satisfied for $\widetilde{f}_{N}$. The previous proof shows that Eqs. (19) and (20) hold for $g_{N}$ and $f_{N}$. But $g_{N}$ converges to g in ${L}^{1}(\mathbb{Z})$, and $f_{N}$ converges uniformly to f. Taking the limit in N yields the desired result. □Here is the probabilistic interpretation. Consider a centered stationary process $(Y_{t})_{t\in \mathbb{Z}}$ with covariance function $A(t,s)=a(t-s)$. For $s\leqslant t$, denote by $\mathcal{Y}_{⟦s,t⟧}$ the σ-algebra generated by $(Y_{r})_{r=s,\dots ,t}$. Consider again the partial innovation $\nu _{t}=Y_{t}-\mathbb{E}[Y_{t}\hspace{0.1667em}|\hspace{0.1667em}\mathcal{Y}_{⟦0,t-1⟧}]$. From (14) and using stationarity, $\nu _{t}$ has the same distribution as
which is
As t tends to infinity, $\eta _{t}$ converges almost surely to
\[ \eta _{\infty }=Y_{0}-\mathbb{E}[Y_{0}\hspace{0.1667em}|\hspace{0.1667em}\mathcal{Y}_{⟦-\infty ,-1⟧}].\]
Observe by stationarity that for all r,
\[ \eta _{\infty }\stackrel{\mathcal{D}}{=}Y_{r}-\mathbb{E}[Y_{r}\hspace{0.1667em}|\hspace{0.1667em}\mathcal{Y}_{⟦-\infty ,r-1⟧}],\]
which is the innovation process associated to Y. Now the variance of $\nu _{t}$, $1/G(t,t)$, tends to the variance of $\eta _{\infty }$. By the Szegő–Kolmogorov formula (see, e.g., Theorem 3 on p. 137 of [10]) that variance is
\[ \exp \Bigg(\frac{1}{2\pi }{\int _{0}^{2\pi }}\log \big(\phi (\lambda )\big)\hspace{0.1667em}\mathrm{d}\lambda \Bigg),\]
where $\phi (\lambda )$ is the spectral density of Y. Let X be a centered stationary process with covariance function k, ε be a standard Gaussian noise, and $Y=\varepsilon +\sqrt{2\alpha }X$. The spectral densities ϕ of Y and f of X are related by $\phi (\lambda )=1+2\alpha f(\lambda )$. Hence,
\[ \underset{t\to +\infty }{\lim }\mathrm{var}(\nu _{t})=\underset{t\to +\infty }{\lim }\frac{1}{G(t,t)}=\exp \Bigg(\frac{1}{2\pi }{\int _{0}^{2\pi }}\log \big(1+2\alpha f(\lambda )\big)\hspace{0.1667em}\mathrm{d}\lambda \Bigg),\]
which is equivalent to (19).Alternatively, observe that, due to stationarity, $\mu _{t}$ defined by (16) has the same distribution as
which is
As t tends to infinity, $\xi _{t}$ converges a.s. to
\[ \xi _{\infty }=\mathbb{E}[\varepsilon _{0}\hspace{0.1667em}|\hspace{0.1667em}\mathcal{Y}_{⟦-\infty ,0⟧}].\]
Of course, since $\mathbb{E}[\varepsilon _{-s}Y_{-r}]=\delta _{s,r}$ for all $s=0,\dots ,t$,
Hence, the limiting property (21) says that
\[ \mathbb{E}[\xi _{\infty }\hspace{0.1667em}\varepsilon _{-s}]=\underset{t\to +\infty }{\lim }G(t,t-s)=g(s).\]
In fact, $\xi _{\infty }$ admits the representation
Similarly, for all t,
\[ \mathbb{E}[\varepsilon _{t}\hspace{0.1667em}|\hspace{0.1667em}\mathcal{Y}_{⟦-\infty ,t⟧}]={\sum \limits_{s=0}^{+\infty }}g(s)\hspace{0.1667em}Y_{t-s},\]
which means that $(g(s))$ realizes the optimal causal Wiener filter of $\varepsilon _{t}$ from the $Y_{t-s}$.Now, Proposition 2 is a straightforward consequence of Proposition 3.
Proof.
Let the coefficients $g_{\tau }(s)$ be defined by (17). Applying (12) to $A=I_{t}+2\alpha H_{t}$, we get
\[ {\big(\mathrm{det}(I_{t}+2\alpha H_{t})\big)}^{-1/2}={\Bigg({\prod \limits_{\tau =0}^{t-1}}g_{\tau }(0)\Bigg)}^{1/2}.\]
Therefore,
\[ \frac{1}{t}\log \big({\big(\mathrm{det}(I_{t}+2\alpha H_{t})\big)}^{-1/2}\big)=\frac{1}{2t}{\sum \limits_{\tau =0}^{t-1}}\log \big(g_{\tau }(0)\big).\]
From Proposition 3 we have
\[ \underset{\tau \to +\infty }{\lim }g_{\tau }(0)=g(0)=\exp \Bigg(-\frac{1}{2\pi }{\int _{0}^{2\pi }}\log \big(1+2\alpha f(\lambda )\big)\hspace{0.1667em}\mathrm{d}\lambda \Bigg).\]
Hence,
\[ \underset{t\to +\infty }{\lim }\frac{1}{t}\log \big({\big(\mathrm{det}(I_{t}+2\alpha H_{t})\big)}^{-1/2}\big)=-\ell _{0}(\alpha ).\]
Applying now (13) to $A=I_{t}+2\alpha H_{t}$, we get
\[ {c_{t}^{\ast }}{G_{t}^{\ast }}{D_{t}^{-1}}G_{t}c_{t}={m_{\infty }^{2}}{\sum \limits_{\tau =0}^{t-1}}\frac{1}{g_{\tau }(0)}{\Bigg({\sum \limits_{s=0}^{\tau }}g_{\tau }(s)\Bigg)}^{2}.\]
From Proposition 3 we have
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \underset{\tau \to +\infty }{\lim }\frac{1}{g_{\tau }(0)}{\Bigg({\sum \limits_{s=0}^{\tau }}g_{\tau }(s)\Bigg)}^{2}& \displaystyle =\frac{1}{g(0)}{\Bigg({\sum \limits_{s=0}^{+\infty }}g(s)\Bigg)}^{2}\\{} & \displaystyle ={\big(1+2\alpha f(0)\big)}^{-1}.\end{array}\]
Hence,
□3 Asymptotic equivalence
Proposition 2 only treats the stationary case. To extend the result under the hypotheses of Theorem 1, a notion of asymptotic equivalence of matrices and vectors is needed. It is developed in this section.
From (7), we must prove that, under the hypotheses of Theorem 1,
and
If $K_{t}=H_{t}$ (centered stationary case), then (27) is (8). It can also be obtained by a straightforward application of Szegő’s theorem; see [4, 2]. Relation (27) (centered asymptotically stationary case) is a consequence of the theory of asymptotically Toeplitz matrices; see Section 7.4 on p. 104 of [9] and also [8, Theorem 4 on p. 178]. Asymptotic equivalence of matrices in Szegő’s theory is taken in the ${L}^{2}$ sense, which is weaker than that considered here. In other words, (27) holds under weaker hypotheses than (H1–H5). In order to prove (28), we shall develop asymptotic equivalence of matrices and vectors along the same lines as [8, Sect. 2.3], but in a stronger sense, replacing ${L}^{2}$ by ${L}^{\infty }$ and ${L}^{1}$, for boundedness and convergence. The norms used here for a vector $v=(v(s))_{s=0,\dots ,t-1}$ are
(27)
\[ \underset{t\to +\infty }{\lim }\frac{1}{2t}\log \big(\mathrm{det}(I_{t}+2\alpha K_{t})\big)=\ell _{0}(\alpha )=\frac{1}{4\pi }{\int _{0}^{2\pi }}\log \big(1+2\alpha f(\lambda )\big)\hspace{0.1667em}\mathrm{d}\lambda \](28)
\[ \underset{t\to +\infty }{\lim }\frac{\alpha }{t}{m_{t}^{\ast }}{(I_{t}+2\alpha K_{t})}^{-1}m_{t}=\ell _{1}(\alpha )={m_{\infty }^{2}}\alpha {\big(1+2\alpha f(0)\big)}^{-1}.\]
\[ \| v\| _{\infty }={\underset{s=0}{\overset{t-1}{\max }}}\big|v(s)\big|\hspace{1em}\text{and}\hspace{1em}\| v\| _{1}={\sum \limits_{s=0}^{t-1}}\big|v(s)\big|.\]
For symmetric matrices, the norm subordinate to $\| \hspace{0.1667em}\cdot \hspace{0.1667em}\| _{\infty }$ is equal to the norm subordinate to $\| \hspace{0.1667em}\cdot \hspace{0.1667em}\| _{1}$. It will be denoted by $\| \hspace{0.1667em}\cdot \hspace{0.1667em}\| $ and referred to as the strong norm. For $A=(A(s,r))_{s,r=0,\dots ,t-1}$ such that ${A}^{\ast }=A$,
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \| A\| & \displaystyle ={\underset{s=0}{\overset{t-1}{\max }}}{\sum \limits_{r=0}^{t-1}}\big|A(s,r)\big|=\underset{\| v\| _{\infty }=1}{\max }\| Av\| _{\infty }\\{} & \displaystyle ={\underset{r=0}{\overset{t-1}{\max }}}{\sum \limits_{s=0}^{t-1}}\big|A(s,r)\big|=\underset{\| v\| _{1}=1}{\max }\| Av\| _{1}.\end{array}\]
The following weak norm will be denoted by $|A|$:
Clearly, $|A|\leqslant \| A\| $. Moreover, the following bounds hold.Proof.
$|AB|$ is the arithmetic mean of the ${L}^{1}$ norms of column vectors of $AB$. If b is any column vector of B, then
because the strong norm is subordinate to the ${L}^{1}$ norm of vectors. Hence, the first result. For the second result, replace columns by rows. □
Here is a definition of asymptotic equivalence for vectors.
Definition 1.
Let $(v_{t})_{t\geqslant 0}$ and $(w_{t})_{t\geqslant 0}$ be two sequences of vectors such that for all $t\geqslant 0$, $v_{t}=(v_{t}(s))_{s=0,\dots ,t-1}$ and $w_{t}=(w_{t}(s))_{s=0,\dots ,t-1}$. They are said to be asymptotically equivalent if:
The asymptotic equivalence of $(v_{t})$ and $(w_{t})$ will be denoted by $v_{t}\sim w_{t}$.
Asymptotic equivalence for matrices is defined as follows (compare with [8, p. 172]).
Definition 2.
Let $(A_{t})_{t\geqslant 0}$ and $(B_{t})_{t\geqslant 0}$ be two sequences of symmetric matrices, where for all $t\geqslant 0$, $A_{t}=(A_{t}(s,r))_{s,t=0,\dots ,t-1}$ and $B_{t}=(B_{t}(s,r))_{s,t=0,\dots ,t-1}$. They are said to be asymptotically equivalent if:
The asymptotic equivalence of $(A_{t})$ and $(B_{t})$ will still be denoted by $A_{t}\sim B_{t}$.
Here are some elementary results, analogous to those stated in Theorem 1 on p. 172 of [8].
Lemma 2.
Let $(A_{t}),(B_{t}),(C_{t}),(D_{t})$ be four sequences of symmetric matrices.
-
1. If $A_{t}\sim B_{t}$ and $B_{t}\sim C_{t}$, then $A_{t}\sim C_{t}$.
-
2. If $A_{t}\sim B_{t}$ and $C_{t}\sim D_{t}$, then $A_{t}+C_{t}\sim B_{t}+D_{t}$.
-
3. If $A_{t}\sim B_{t}$ and $C_{t}\sim D_{t}$, then $A_{t}C_{t}\sim B_{t}D_{t}$.
-
4. If $A_{t}\sim B_{t}$ and F is an analytic function with radius R such that $R>\max \| A_{t}\| ,\max \| B_{t}\| $, then $F(A_{t})\sim F(B_{t})$.
Proof.
Points 1 and 2 follow from the triangle inequality for the weak norm. For point 3, because $\| \hspace{0.1667em}\cdot \hspace{0.1667em}\| $ is a norm of matrices, $\| A_{t}C_{t}\| \leqslant \| A_{t}\| \hspace{0.1667em}\| C_{t}\| $, and $\| B_{t}D_{t}\| \leqslant \| B_{t}\| \hspace{0.1667em}\| D_{t}\| $ are uniformly bounded. Moreover by Lemma 1,
\[\begin{array}{r@{\hskip0pt}l}\displaystyle |A_{t}C_{t}-B_{t}D_{t}|& \displaystyle \leqslant \big|(A_{t}-B_{t})C_{t}\big|+\big|B_{t}(C_{t}-D_{t})\big|\\{} & \displaystyle \leqslant |A_{t}-B_{t}|\hspace{0.1667em}\| C_{t}\| +\| B_{t}\| \hspace{0.1667em}|C_{t}-D_{t}|.\end{array}\]
Since $\| C_{t}\| $ and $\| B_{t}\| $ are uniformly bounded and
the result follows. For point 4, let F be analytic with radius of convergence R. For $|z|<R$, let
and
The matrices $F(A_{t})$, $F(B_{t})$ are defined as the limits of $F_{n}(A_{t})$, $F_{n}(B_{t})$; from the hypothesis it follows that the convergence is uniform in t. Because $\| \hspace{0.1667em}\cdot \hspace{0.1667em}\| $ is a matrix norm, $\| F(A_{t})\| \leqslant F(\| A_{t}\| ),$ and the same holds for $B_{t}$: $\| F(A_{t})\| $ and $\| F(B_{t})\| $ are uniformly bounded. Let ϵ be a positive real. Fix n such that for all t,
\[ \big\| F(A_{t})-F_{n}(A_{t})\big\| <\frac{\epsilon }{3}\hspace{1em}\text{and}\hspace{1em}\big\| F(B_{t})-F_{n}(B_{t})\big\| <\frac{\epsilon }{3}.\]
By induction on n using points 2 and 3, $F_{n}(A_{t})\sim F_{n}(B_{t})$. There exists $t_{0}$ such that for all $t>t_{0}$,
Thus, for all $t>t_{0}$,
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \big|F(A_{t})-F(B_{t})\big|& \displaystyle \leqslant \big|F(A_{t})-F_{n}(A_{t})\big|+\big|F_{n}(A_{t})-F_{n}(B_{t})\big|+\big|F_{n}(B_{t})-F(B_{t})\big|\\{} & \displaystyle \leqslant \big\| F(A_{t})-F_{n}(A_{t})\big\| +\big|F_{n}(A_{t})-F_{n}(B_{t})\big|+\big\| F_{n}(B_{t})-F(B_{t})\big\| \\{} & \displaystyle <\epsilon .\end{array}\]
Hence the result. □Hypothesis (H3) implies that $\| H_{t}\| $ is uniformly bounded, (H2) and (H5) that $K_{t}\sim H_{t}$. Point 4 will be applied to $F(z)={(1+2\alpha z)}^{-1}$, which has the radius of convergence $R=1/2\alpha $. Let M be defined as
\[ M=\max \bigg\{\underset{t\geqslant 1}{\max }\| K_{t}\| ,\sum \limits_{t\in \mathbb{Z}}\big|k(t)\big|\bigg\}.\]
For all $\alpha <\alpha _{0}=1/(2M)$,
Here is the relation between asymptotic equivalence of vectors and matrices.
Proof.
The norms $\| A_{t}v_{t}\| _{\infty }$, $\| B_{t}v_{t}\| _{\infty }$, $\| A_{t}w_{t}\| _{\infty }$ are uniformly bounded because of the fact that $\| \hspace{0.1667em}\cdot \hspace{0.1667em}\| $ is subordinate to $\| \hspace{0.1667em}\cdot \hspace{0.1667em}\| _{\infty }$. Next, for point 1,
For point 2,
□
The relation between asymptotic equivalence of vectors and our goal is the following.
Proof.
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \frac{1}{t}\big|{v_{t}^{\ast }}u_{t}-{w_{t}^{\ast }}z_{t}\big|& \displaystyle \leqslant \frac{1}{t}\big(\big|{v_{t}^{\ast }}(u_{t}-z_{t})\big|+\big|\big({v_{t}^{\ast }}-{w_{t}^{\ast }}\big)z_{t}\big|\big)\\{} & \displaystyle \leqslant \frac{1}{t}\big(\| v_{t}\| _{\infty }\hspace{0.1667em}\| u_{t}-z_{t}\| _{1}+\| z_{t}\| _{\infty }\| v_{t}-w_{t}\| _{1}\big).\end{array}\]
Hence the result. □
Using asymptotic equivalence, (27) and (28) can easily be deduced from (8) and (9) for $0<\alpha <1/(2M)$. We shall not detail the passage from (8) to (27); see Theorem 4 on p. 178 of [8]. Here is the passage from (9) to (28). For all $\alpha <1/(2M)$, it follows from (29) by point 1 of Lemma 3 that
\[ {(I_{t}+2\alpha K_{t})}^{-1}c_{t}\hspace{0.1667em}\sim \hspace{0.1667em}{(I_{t}+2\alpha H_{t})}^{-1}c_{t}.\]
By point 2 of Lemma 3, we have
\[ {(I_{t}+2\alpha K_{t})}^{-1}m_{t}\hspace{0.1667em}\sim \hspace{0.1667em}{(I_{t}+2\alpha H_{t})}^{-1}c_{t}.\]
Lemma 4 implies
\[ \underset{t\to +\infty }{\lim }\frac{1}{t}{m_{t}^{\ast }}(I_{t}+2\alpha K_{t})m_{t}=\underset{t\to +\infty }{\lim }\frac{1}{t}{c_{t}^{\ast }}{(1+2\alpha H_{t})}^{-1}c_{t}.\]
Hence (28).Still using asymptotic equivalence, it will now be shown that Proposition 1 is just a particular case of Theorem 1. Indeed, consider the Gaussian process ${X}^{x}$ with mean
and covariance function
The distribution of $({X_{t}^{x}})_{t\in \mathbb{N}}$ and the conditional distribution of $(X_{t})_{t\in \mathbb{N}}$ given $X_{0}=x$ are the same. Denote by $m_{x,t}$ and ${K_{t}^{\bullet }}$ the mean and covariance matrix of $({X_{s}^{x}})_{s=0,\dots ,t-1}$. Theorem 1 applies to ${X}^{x}$, provided that it is proved that $m_{x,t}\sim c_{t}$ and ${K_{t}^{\bullet }}\sim H_{t}$. By (H1) and (H2), $\| m_{x,t}\| _{\infty }$ is uniformly bounded. Moreover, by (30),
(31)
\[ {K}^{\bullet }(t,s)=\mathbb{E}\big[\big({X_{t}^{x}}-m_{x}(t)\big)\big({X_{s}^{x}}-m_{x}(s)\big)\big]=K(t,s)-\frac{K(t,0)K(s,0)}{K(0,0)}.\]
\[ \frac{1}{t}\| m_{x,t}-m_{t}\| _{1}\leqslant \frac{|x|+\| m_{t}\| _{\infty }}{tK(0,0)}{\sum \limits_{s=0}^{t-1}}\big|K(0,s)\big|\leqslant \frac{|x|+\| m_{t}\| _{\infty }}{tK(0,0)}\| K_{t}\| ;\]
thus, $m_{x,t}\sim m_{t}$, and hence $m_{x,t}\sim c_{t}$ by transitivity. Now from (31) we have
\[ \| {K_{t}^{\bullet }}\| \leqslant \| K_{t}\| +{\underset{r=0}{\overset{t-1}{\max }}}\frac{\big|K(0,r)\big|}{K(0,0)}{\sum \limits_{s=0}^{t-1}}K(0,s)\leqslant \| K_{t}\| +\frac{\| K_{t}{\| }^{2}}{K(0,0)}.\]
Moreover,
\[ \big|{K_{t}^{\bullet }}-K_{t}\big|\leqslant \frac{1}{tK(0,0)}{\Bigg({\sum \limits_{s=0}^{t-1}}\big|K(0,s)\big|\Bigg)}^{2}\leqslant \frac{\| K_{t}{\| }^{2}}{tK(0,0)};\]
thus, ${K_{t}^{\bullet }}\sim K_{t}$, and hence ${K_{t}^{\bullet }}\sim H_{t}$ by transitivity (point 1 of Lemma 2).4 Asymptotic distributions
The results of the two previous sections establish that the conclusion of Theorem 1 holds for a small enough α. To finish the proof, the convergence must be extended to all $\alpha \geqslant 0$. The following variant of Lévy’s continuity theorem applies (see Chapter 4 of [11] and, in particular, Exercise 9 on p. 78).
Lemma 5.
Let $\pi ,\pi _{1},\pi _{2},\dots \hspace{0.1667em}$, be probability measures on ${\mathbb{R}}^{+}$. Assume that for some $\alpha _{0}>0$ and all $\alpha \in [0,\alpha _{0}[$,
\[ \underset{n\to \infty }{\lim }{\int _{0}^{+\infty }}{\mathrm{e}}^{-\alpha x}\hspace{0.1667em}\mathrm{d}\pi _{n}(x)={\int _{0}^{+\infty }}{\mathrm{e}}^{-\alpha x}\hspace{0.1667em}\mathrm{d}\pi (x).\]
Then $(\pi _{n})$ converges weakly to π, and the convergence holds for all $\alpha \geqslant 0$.
To apply this lemma, we have to check that ${(L_{t}(\alpha ))}^{1/t}$ and ${\mathrm{e}}^{-\ell (\alpha )}$ are the Laplace transforms of probability distributions on ${\mathbb{R}}^{+}$. It turns out that in our case, the function $L_{t}(\alpha )$ defined by (2) is the Laplace transform of an infinitely divisible distribution, and thus so are ${(L_{t}(\alpha ))}^{1/t}$ and its limit. We give here the probabilistic interpretation of ${\mathrm{e}}^{-\ell _{0}(\alpha )}$ and ${\mathrm{e}}^{-\ell _{1}(\alpha )}$ as the Laplace transforms of two infinitely divisible distributions. Next, the particular case of a Gauss–Markov process will be considered.
Through an orthogonal transformation diagonalizing its covariance matrix, the squared norm of any Gaussian vector can be written as the sum of independent random variables, each being the square of a Gaussian variable and thus having noncentral chi-squared distribution. If Z is Gaussian with mean μ and variance v, then the Laplace transform of ${Z}^{2}$ is
The first factor is the Laplace transform of the gamma distribution with shape parameter $1/2$ and scale parameter $2v$. Assuming μ and v nonnull, rewrite the second factor as
This is the Laplace transform of a Poisson compound of the exponential with expectation $2v$ by the Poisson distribution with rate $\frac{{\mu }^{2}}{2v}$. Therefore, the squared norm of a Gaussian vector has an infinitely divisible distribution, which is a convolution of gamma distributions with Poisson compounds of exponentials. Squared Gaussian vectors have received a lot of attention since even in dimension 2, the mean and covariance matrix must satisfy certain conditions for the distribution of the vector to be infinitely divisible [17]. Yet the sum of coordinates of such a vector always has an infinitely divisible distribution.
For all t, the distribution with Laplace transform ${(L_{t}(\alpha ))}^{1/t}$ is the convolution of gamma distributions with Poisson compounds of exponentials. As t tends to infinity, ${(L_{t}(\alpha ))}^{1/t}$ tends to ${\mathrm{e}}^{-\ell _{0}(\alpha )}\hspace{0.1667em}{\mathrm{e}}^{-\ell _{1}(\alpha )}$. The first factor ${\mathrm{e}}^{-\ell _{0}(\alpha )}$ is the Laplace transform of a limit of convolutions of gamma distributions, which belongs to the Thorin class $T({\mathbb{R}}^{+})$ (see [3] as a general reference). Consider now ${\mathrm{e}}^{-\ell _{1}(\alpha )}$. Rewrite $\ell _{1}(\alpha )$ as
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \ell _{1}(\alpha )& \displaystyle ={m_{\infty }^{2}}\alpha {\big(1+2\alpha f(0)\big)}^{-1}\\{} & \displaystyle =\frac{{m_{\infty }^{2}}}{2f(0)}\big(1-{\big(1+2\alpha f(0)\big)}^{-1}\big).\end{array}\]
Thus, ${\mathrm{e}}^{-\ell _{1}(\alpha )}$ is the Laplace transform of a Poisson compound of the exponential distribution with expectation $2f(0)$ by the Poisson distribution with parameter $\frac{{m_{\infty }^{2}}}{2f(0)}$.As an illustrating example, consider the Gauss–Markov process defined as follows. Let θ be a real such that $-1<\theta <1$. Let $(\varepsilon _{t})_{t\geqslant 1}$ be a sequence of i.i.d. standard Gaussian random variables. Let $Y_{0}$, independent from the sequence $(\varepsilon _{t})_{t\geqslant 1}$, follow the normal $\mathcal{N}(0,{(1-{\theta }^{2})}^{-1})$ distribution. For all $t\geqslant 1$, let
Thus, $(Y_{t})_{t\in \mathbb{N}}$ is a stationary centered autoregressive process. Consider the noncentered process $(X_{t})_{t\in \mathbb{N}}$ with $X_{t}=Y_{t}+m_{\infty }$. This is the case considered in [13], where a stronger result was proved. Formula (10) on p. 72 of that reference matches (4) and (5) here. Indeed, the spectral density is
Write $\ell _{0}(\alpha )$ as a contour integral over the unit circle:
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \ell _{0}(\alpha )& \displaystyle =\frac{1}{4\pi }{\int _{0}^{2\pi }}\log \big(1+2\alpha f(\lambda )\big)\hspace{0.1667em}\mathrm{d}\lambda \\{} & \displaystyle =\frac{1}{4\pi \mathrm{i}}\oint _{|\zeta |=1}\frac{1}{\zeta }\log \bigg(1+\frac{2\alpha }{1+{\theta }^{2}-\theta (\frac{1}{\zeta }+\zeta )}\bigg)\hspace{0.1667em}\mathrm{d}\zeta .\end{array}\]
Now we have
\[ 1+\frac{2\alpha }{1+{\theta }^{2}-\theta (\frac{1}{\zeta }+\zeta )}=\frac{{\zeta }^{2}-(\theta +\frac{1}{\theta }+\frac{2\alpha }{\theta })\zeta +1}{{\zeta }^{2}-(\theta +\frac{1}{\theta })\zeta +1}.\]
Observe that the two roots of the numerator have the same sign as θ, and their product is 1. Denote them by ${\zeta }^{-}$ and ${\zeta }^{+}$, so that $0<|{\zeta }^{-}|<1<|{\zeta }^{+}|$. The two roots of the denominator are θ and $\frac{1}{\theta }$. The function to be integrated has five poles, among which $0,\theta ,{\zeta }^{-}$ are inside the unit disk, and $\frac{1}{\theta },{\zeta }^{+}$ are outside. Rewrite $\ell _{0}$ as
\[ \ell _{0}(\alpha )=\frac{1}{4\pi \mathrm{i}}\oint _{|\zeta |=1}\frac{1}{\zeta }\log \bigg(\frac{\zeta -{\zeta }^{-}}{\zeta -\theta }\bigg)\hspace{0.1667em}\mathrm{d}\zeta +\frac{1}{4\pi \mathrm{i}}\oint _{|\zeta |=1}\frac{1}{\zeta }\log \bigg(\frac{\zeta -{\zeta }^{+}}{\zeta -\frac{1}{\theta }}\bigg)\hspace{0.1667em}\mathrm{d}\zeta .\]
The first integral is null since
\[ \oint _{|\zeta |=1}\frac{1}{\zeta }\log \big(\zeta -{\zeta }^{-}\big)\hspace{0.1667em}\mathrm{d}\zeta =\oint _{|\zeta |=1}\frac{1}{\zeta }\log (\zeta -\theta )\hspace{0.1667em}\mathrm{d}\zeta ,\]
the two functions having the same residues inside the unit disk. The second integral is
\[ \frac{1}{4\pi \mathrm{i}}\oint _{|\zeta |=1}\frac{1}{\zeta }\log \bigg(\frac{\zeta -{\zeta }^{+}}{\zeta -\frac{1}{\theta }}\bigg)\hspace{0.1667em}\mathrm{d}\zeta =\frac{1}{2}\log \big(\theta {\zeta }^{+}\big).\]
Therefore,
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \ell _{0}(\alpha )& \displaystyle =\frac{1}{2}\log \big(\theta {\zeta }^{+}\big)\\{} & \displaystyle =\frac{1}{2}\log \bigg(\frac{1}{2}\big({\theta }^{2}+1+2\alpha +\sqrt{\big({(\theta +1)}^{2}+2\alpha \big)\big({(\theta -1)}^{2}+2\alpha \big)}\big)\bigg).\end{array}\]
The expression of $\ell _{1}$ is
\[ \ell _{1}(\alpha )=\frac{{m_{\infty }^{2}}\alpha {(1-\theta )}^{2}}{{(1-\theta )}^{2}+2\alpha }.\]
It turns out that the probability distribution with Laplace transform
\[ {\mathrm{e}}^{-\ell _{0}(\alpha )}={\bigg(\frac{1}{2}\big({\theta }^{2}+1+2\alpha +\sqrt{\big({(\theta +1)}^{2}+2\alpha \big)\big({(\theta -1)}^{2}+2\alpha \big)}\big)\bigg)}^{-1/2}\]
has an explicit density $f_{0}(x)$ defined on $(0,+\infty )$, which is related to the modified Bessel function of the first kind with order $1/2$ (compare with formula (3.10) on p. 437 in [7]):
\[\begin{array}{r@{\hskip0pt}l}\displaystyle f_{0}(x)& \displaystyle ={\mathrm{e}}^{-\frac{1+{\theta }^{2}}{2}x}\big({2}^{-1}|\theta {|}^{-1/2}{x}^{-1}I_{1/2}(|\theta |x)\big)\\{} & \displaystyle ={\mathrm{e}}^{-\frac{1+{\theta }^{2}}{2}x}\big({(2\pi )}^{-1/2}|\theta {|}^{-1}{x}^{-3/2}\sinh (|\theta |x)\big).\end{array}\]