Let $\mathbf{T}$ be the translation operator/matrix in discrete-time domain which can be written as $\mathbf{T} = \mathbf{\Phi} \mathbf{P} \mathbf{\Phi}^*$ where $\mathbf{P} = \exp(-i Diag([w_0, w_1, \ldots, w_M]))$ is a diagonal matrix where $w_k= 2\pi(k-1)/M$. We know that the stochastic signal $\mathbf{x}$ is called wide-sense stationary (WSS) w.r.t. the translation operator on in discrete-time domain $\mathbf{T}$ if and only if for all $r$:

${\mathbf R}_{\mathbf{x}} = \mathbb{E}[\mathbf{x}\mathbf{x}^{*}] = \mathbb{E}\big[(\mathbf{T}^{r} \mathbf{x})\big(\mathbf{T}^{r} \mathbf{x}\big)^{*}\big].$

We need to show that: a process in second-order moment is WSS if and only if \begin{align} {\mathbf R}_{{\mathbf{x}}} = \mathbf{\Phi} {\mathbf S}_{{\mathbf{x}}} \mathbf{\Phi}^{*}, \end{align} where $\mathbf{S}_{\mathbf x}$ is a diagonal matrix with non-negative entries on its main diagonal and $\mathbf{\Phi}^{*}$ is the DFT matrix. How can we prove this result?

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    $\begingroup$ I am certain that the definition of P is incorrect here. The translation operator T should be unitary, i.e. $\mathbf{T}^{-1} = \mathbf{T}^{\dagger}$, where $\dagger$ indicates complex conjugate transpose (* in your notation). T is unitary if you change the definition of P to $\textrm{diag}(1,e^{2\pi i/M},\ldots,e^{2\pi i (M-1)/M})$, where $i = \sqrt{-1}$. With that definition, $\mathbf{P}\mathbf{P}^{\dagger} = \mathbf{I}$. $\endgroup$
    – Joe Mack
    Jun 12, 2020 at 18:34
  • $\begingroup$ @JoeMac: Sorry. That was typo. Corrected. $\endgroup$
    – Amin
    Jun 12, 2020 at 22:56

1 Answer 1


Given the definition of the correlation matrix $\mathbf{R}_{\mathbf{x}}$ here, I am assuming that $\mathsf{E}[\mathbf{x}] = \mathbf{0}$. I do this because the correlation matrix is usually defined as $\mathsf{E}[(\mathbf{x} - \mathsf{E}[\mathbf{x}])(\mathbf{x} - \mathsf{E}[\mathbf{x}])^{\dagger}]$, where $\dagger$ indicates complex conjugate tranpose.

Note that since $\mathbf{R}_{\mathbf{x}}$ is a correlation matrix, it is Hermitian positive semi-definite, which means that it is unitarily diagonalizable and has all non-negative eigenvalues: \begin{equation} \mathbf{R}_{\mathbf{x}} = \mathbf{U}\mathbf{\Lambda}\mathbf{U}^{\dagger}, \end{equation} where $\mathbf{U}$ is a unitary matrix and $\mathbf{\Lambda}$ is a diagonal matrix with non-negative numbers on its diagonal.
Let's consider the translation operator $\mathbf{T}$: \begin{equation} \mathbf{T} = \mathbf{\Phi}\mathbf{P}\mathbf{\Phi}^{\dagger}, \end{equation} where $\mathbf{\Phi}^{\dagger}$ is the DFT matrix and $\mathbf{P}$ is diagonal. As I mentioned in a comment, $\mathbf{T}$ should be unitary, and that requires that the diagonal matrix $\mathbf{P}$ have complex exponentials on its diagonal. In fact, to make $\mathbf{\Phi}\mathbf{P}\mathbf{\Phi}^{\dagger}$ a translation operator, $\mathbf{P}$ should be \begin{equation} \mathbf{P} = \textrm{diag}(1,e^{2\pi i/M},\ldots,e^{2\pi i(M-1)/M}). \end{equation}

I do not think this will have an impact on the proof, but it is important for applications.
Now let's assume that $\mathbf{R}_{\mathbf{T}\mathbf{x}} = \mathbf{R}_{\mathbf{x}}$: \begin{eqnarray} \mathsf{E}[(\mathbf{T}\mathbf{x})(\mathbf{T}\mathbf{x})^{\dagger}] &=& \mathbf{R}_{\mathbf{x}}\\ \mathbf{T}\mathsf{E}[\mathbf{x}\mathbf{x}^{\dagger}]\mathbf{T}^{\dagger} &=& \mathbf{R}_{\mathbf{x}}\\ \mathbf{T}\mathbf{R}_{\mathbf{x}}\mathbf{T}^{\dagger} &=& \mathbf{R}_{\mathbf{x}}\\ \mathbf{T}\mathbf{R}_{\mathbf{x}} &=& \mathbf{R}_{\mathbf{x}}\mathbf{T} \end{eqnarray} Since

  • $\mathbf{R}_{\mathbf{x}}$ and $\mathbf{T}$ are both diagonalizable
  • and they commute with each other,

they are simultaneously diagonalizable. This means that there is a single matrix that diagonalizes both.

We have been told from the beginning that $\mathbf{\Phi}$ diagonalizes $\mathbf{T}$, so now we know that $\mathbf{\Phi}$ diagonalizes $\mathbf{R}_{\mathbf{x}}$, too. This means that the unitary matrix $\mathbf{U}$ that diagonalizes $\mathbf{R}_{\mathbf{x}}$ must be $\mathbf{\Phi}$: \begin{equation} \mathbf{R}_{\mathbf{x}} = \mathbf{\Phi}\mathbf{\Lambda}\mathbf{\Phi}^{\dagger}. \end{equation} We have already established that $\mathbf{\Lambda}$ is diagonal with non-negative real numbers on its diagonal. The $\mathbf{S}_{\mathbf{x}}$ that we have sought, is $\mathbf{\Lambda}$.

  • $\begingroup$ Thanks @JoeMac. My confusion is resolved. $\endgroup$
    – Amin
    Jun 12, 2020 at 23:08

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