strong textI have been reading "The Stationary Wavelet Transform and some Statistical Applications" by Nason and Silverman, and there is a claim in the their paper of which I cannot convince myself.


  • $\mathcal{H}$: A low-pass filter.
    • The action of the low pass filter on a doubly infinite sequence $\{..., x_{-1}, x_0, x_1, x2, ... \}$ is defined $(\mathcal{H}x)_k = \sum_n h_{n-k} x_n$
    • The filter satisfied internal orthogonality, i.e. $\sum_n h_{n} h_{n + 2j} = 0$
  • $\mathcal{G}$: A high-pass filter, defined by the filter $g_n = (-1)^n h_{1-n}$
    • Clearly $\mathcal{G}$ satisfied the same internal orthogonality, and is mutually orthogonal to $\mathcal{H}$, i.e. $\sum_n h_{n} g_{n + 2j} = 0$ $\forall j$
  • $\mathcal{S}$: A shift operator defined by $(\mathcal{S} x)_j = x_{j+1}$
  • $\mathcal{D}_0$: A binary decimation operator that chooses every even member of a sequence.
    • $(\mathcal{D}_0 x)_j = x_{2j}$
    • $\mathcal{R}_0$ the inverse operation.
  • $\mathcal{D}_1$: A binary decimation operator that chooses every odd member of a sequence, similar to $\mathcal{D}_0$ above.
  • It is not even necessary for the same choice of "odd" or "even" to be used throughout.
    • Suppose $\epsilon_{J-1}, \epsilon_{J-2}, ..., \epsilon_{0}$ are a sequence of 0's and 1's.
    • One can use the binary operator $\mathcal{D}_{\epsilon_j}$ at level $j$, and the original signal can be recovered by applying the corresponding sequence $\mathcal{R}_{\epsilon_j}$

On page 4, the paper claims,

"If $x$ is a finite sequence, define the shift periodically at the boundary. It then immediate from the definitions that $\mathcal{D}_1$ = $\mathcal{D}_0 \mathcal{S}$ and hence that $\mathcal{R}_1$ = $\mathcal{S}^{-1} \mathcal{R}_0$.

NOW comes the portion I can't seem to grasp...

It is also easy to see that $\mathcal{S} \mathcal{D}_0 = \mathcal{D}_0 \mathcal{S}^2$. and that the operator $\mathcal{S}$ commutes with $\mathcal{H}$ and $\mathcal{G}$.

How is this obvious? I have tried recreating these properties in R, but I cannot convince myself that $\mathcal{S} \mathcal{D}_0 = \mathcal{D}_0 \mathcal{S}^2$, or that S and the filters commute. Can anyone help?


I am suspicious due to the claim that $\mathcal{R}_0$ is the inverse of $\mathcal{D}_0$. Decimation is not invertible. Once samples/entries are deleted, those values are forgotten. They cannot be reconstructed from what remains.

Let's say the input finite sequences are in $\mathbb{R}^{2N}$, numbered as $x = (x[0],x[1],\ldots,x[2N-1])$. Then $\mathcal{D}_0:\mathbb{R}^{2N}\to\mathbb{R}^N$, and $\mathcal{S}$ is overloaded, so that it maps from $\mathbb{R}^{2N}$ to $\mathbb{R}^{2N}$ and from $\mathbb{R}^{N}$ to $\mathbb{R}^{N}$: \begin{eqnarray} (\mathcal{S}x)[n] &=& x[n+1~\textrm{mod}~2N]~~\textrm{for}~~x\in\mathbb{R}^{2N},\\ (\mathcal{S}y)[n] &=& y[n+1~\textrm{mod}~N]~~\textrm{for}~~y\in\mathbb{R}^{N}. \end{eqnarray}

Let $u = \mathcal{D}_0x\in\mathbb{R}^{N}$:

\begin{equation} u[n] ~=~ (\mathcal{D}_0x)[n] ~=~ x[2n], \end{equation} where $n$ runs from 0 to $N-1$.

Then $\mathcal{S}u = \mathcal{S}\mathcal{D}_0x\in\mathbb{R}^{N}$:

\begin{equation} \begin{split} (\mathcal{S}u)[n] &=~ u[n+1 ~\textrm{mod}~N]\\ &=~ x[2\times((n+1)~\textrm{mod}~N)]\\ &=~ x[2n+2~\textrm{mod}~2N] \end{split} \end{equation}

Now let $v = \mathcal{S}^2x\in\mathbb{R}^{2N}$:

\begin{equation} v[n] ~=~ (\mathcal{S}^2x)[n] ~=~ x[n+2~\textrm{mod}~2N] \end{equation}

Then $\mathcal{D}_0v = \mathcal{D}_0\mathcal{S}^2x\in\mathbb{R}^{N}$:

\begin{equation} \begin{split} (\mathcal{D}_0v)[n] &=~ v[2n]\\ &=~ x[(2n)+2~\textrm{mod}~2N]\\ &=~ x[2n+2~\textrm{mod}~2N] \end{split} \end{equation}

Hence \begin{equation} (\mathcal{D}_0\mathcal{S}^2x)[n] ~=~ x[2n+2~\textrm{mod}~2N] ~=~ (\mathcal{S}\mathcal{D}_0x)[n] \end{equation} for $0\leq n < N$ (because all results are in $\mathbb{R}^{N}$), so $\mathcal{D}_0\mathcal{S}^2$ and $\mathcal{S}\mathcal{D}_0$ are the same.


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