# What exactly is meant by "translation invariant dictionaries/wavelets"?

I'm trying to wrap my head around the notion of translation invariance in terms of dictionaries/wavelets. For example in Lecture Notes, Page 41 its written that one starts with a family of atoms/wavelets $$(\psi_j)_{j\in \mathbb{Z}}$$ and adds all translates $$t\in \mathbb{R}$$ via defining $$\begin{equation*} \psi_{j,t}(x) := \psi_j (x - t). \end{equation*}$$ Now the calculation of the coefficients for a singal $$f$$ becomes $$$$\label{calc coeff} \Phi f (j,t) = \left\langle f, \psi_{j,t}\right\rangle = \int_\mathbb{R} f(x) \psi_j^\ast (t - x) \mathrm{d}x = f \ast \psi_j^\ast (t),$$$$ where $$\psi_j^\ast(x) := \psi_j(-x)$$. To me, translation invariance means something like $$f(x) = f(x-t)$$, but I don't understand/see where this equations holds for the above "argumentation"?

Furthermore, reading some Wikipedia on Stationary Wavelet Transform, it says in the first sentence "to overcome the lack of translation-invariance of the discrete wavelet transform", i.e. when the translation $$t\in \mathbb{R}$$ becomes $$t\in \mathbb{Z}$$. Why is the discrete wavelet transform not translation invariant? The calculation for the coefficients $$\left\langle f, \psi_{j,t} \right\rangle$$ is the same, but only discrete convolution instead of continuous?

It's rather translation equivariant:

$$\text{CWT}_{s, t}x(t - t_0) = \text{CWT}_{s, t - t_0}x(t) \tag{1}$$

and

$$\langle x(t−t_0),\psi(t) \rangle= \langle x(t), \psi(t+t_0)\rangle \tag{2}$$

That is, when a signal is shifted, its representation is also shifted but not modified (like LTI). This makes its derived features, such as energy and norm, or most manipulations of coefficients, invariant - hence the terminology (though agreeably misleading).

### DWT

The DWT is likewise equivariant at source - i.e. before subsampling. Suppose subsampling by x2:

$$[0, 1, 2, 3, 4, 5, 6, 0, 0] \rightarrow [0, 2, 4, 6, 0]$$ now (circular-)shift by 1: $$[0, 0, 1, 2, 3, 4, 5, 6, 0] \rightarrow [0, 1, 3, 5, 0]$$

If it were equivariant, we'd get $$[0, 0, 2, 4, 6]$$ - but now, our manipulations (e.g. $$\sum |\text{coef}|^2$$) will no longer yield the same results. Nonetheless, "at source" equivariance is much more than no equivariance at all, and some manipulations can take advantage of it.

### Equivariant dictionaries

One can notice that $$(2)$$ is an identity - i.e. it's always true. The idea is that "shift" for wavelets is defined this way to begin with, i.e. "shift by $$t_0$$" means $$\psi(t - t_0)$$, which isn't the case for all functions (e.g. its Fourier transform, shifted as a function of frequency, also changes its width, breaking equivariance). To define a shift this way means to make it a convolutional operator, and the derived representation LTI.

### Invariance

There is an important sense in which complex wavelets are invariant:

$$\text{CWT}(x(t)) = \text{CWT}(x(t - t_0))$$

While the equality never strictly holds with wavelets alone (except at infinite scale), the distance between shifted coefficients can be made as low as the application demands with further steps; with wavelets alone, the greater the scale, the lesser the distance. Time-shift invariance is a bedrock of the scattering transform.

See related post.

but only discrete convolution instead of continuous

The distinction is unimportant: they're both implemented as discrete convolutions. The difference is in subsampling scheme and choice of wavelets.

• Thanks that clarified already a lot! I would have a follow-up question: For the last point on distinction between discrete/continuous, could you give examples of wavelets which can be used both, for CWT and DWT and/or which can be used for one but not the other? (I'm mostly familiar the analytical definition of Shannon-, Haar- or Meyer-Wavelet) Aug 20, 2021 at 10:51
• @stish Any DWT wavelet can be used for CWT, but not other way around; most CWT wavelets are ill-suited (or invalid) for DWT. E.g. Morlet, Morse, Mexican Hat, Bump - anything that cannot effectively tile the frequency domain without overlaps. Aug 20, 2021 at 11:16
• This is what I'm looking for. Just a clarification, this means that if I have two overlapping scalograms (by CWT) produced by two overlapping windows, the overlapped parts would show the same? Jun 17, 2022 at 8:32
• @EddyPiedad Unsure what you mean. Two separate scalograms cannot be easily overlapped as they're multi-resolution, also likely computed with different paddings. If we the signal is split in two, then the in-between region, outside the influence of boundary effects, will be identical yes. Jun 17, 2022 at 14:11
• If this is about whether we can do overlap-add/overlap-save for CWT, it's a yes. Jun 17, 2022 at 14:32