0
$\begingroup$

Referencing this article here https://arxiv.org/pdf/1203.1513.pdf

It states "A wavelet transform commutes with translations, and is therefore not translation invariant". Now I understand why it is a problem that the result is not translation invariant, however, I'm confused as to why it is.

What does it mean for a transform to commute with translation and why does the Wavelet transform commute with translations (i.e. why is the Wavelet transformation shift invariant)?.

|improve this question
$\endgroup$
  • $\begingroup$ Doesn't shift = translation? $\endgroup$ – Cherny Apr 15 '19 at 10:52
  • $\begingroup$ @Cherny Correct, one way I could pose the question is "What does it mean for the wavelet transform to commute with shifts?" $\endgroup$ – Izzo Apr 15 '19 at 13:42
1
$\begingroup$

We can start from what is "shift invarient":

Transform G is shift invariant if - $$\forall x:\sigma^nG(x) = G(x)$$ $\sigma^n$ being shift by n. Examples for transforms that are invarient to shifts are histogram and the amplitude of Fourier transform.

Commuting with shift is - $$\forall x:\sigma^nG(x) = G(\sigma^nx)$$ So it can't be shift invariant (unless G(x) is constant).

Note: I'm actually not certain that wavelet transform commutes with shift, but it's most certainly is not shift inveriant.

|improve this answer
$\endgroup$
  • $\begingroup$ Alright, so you've technically answered my question as to what it means for a transform to commute with translations. However, I'm having a difficult time understanding why commuting with translations determines that the transformation is shift invariant. $\endgroup$ – Izzo Apr 18 '19 at 2:17
  • $\begingroup$ I'm sorry if I answered the wrong question! but I'll try to clarify and give another perspective. Shift-invariant as a property can be useful for multiple reasons, mainly as features that are the same if you shift the signal around. Now imagine you have a situation where your problem is independent of shift, and you use commuting with translation transform - you just added a full dimension over an invariant transform. Hope it's clearer, I'd be happy to try again if you'd like $\endgroup$ – Cherny Apr 18 '19 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.