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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)?.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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