I am reading a book titled "Two Dimensional Wavelets and their relatives" by Antoine et al. and it talks about vanishing moments. I have trouble understanding the exact significance of it. Can anybody give an idea on vanishing moments?

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    $\begingroup$ Maybe you could tell which of the hundreds of papers on wavelets you are reading? and in what context the phrase "vanishing moment" is used? $\endgroup$ Commented Oct 26, 2012 at 21:27
  • $\begingroup$ I am reading a book titled "Two Dimensional Wavelets and their relatives" by Antoine et al. I have a pic of the exact place where I am referring to. Please find it here dl.dropbox.com/u/28068989/IMAG0746.jpg $\endgroup$
    – mkuse
    Commented Oct 26, 2012 at 21:35
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    $\begingroup$ In brief, if a wavelet has $n$ vanishing moments, the output of filtering a $n$-th degree polynomial with this wavelet will be 0. $\endgroup$
    – Phonon
    Commented Oct 27, 2012 at 0:04
  • $\begingroup$ This is an intuitive explanation "for dummies" I don't know about continuous wavelets, but in discrete wavelets, a wavelet with $n$ vanishing moments, produce low coefficients on the parts of the data which can be approached by a polynomial with degree $n$. It facilitates the identification of parts of the data as "polynomial of order $n$" This should be a comment, and not an answer, but I'm not allowed to comment. $\endgroup$
    – zexot
    Commented Apr 24, 2018 at 15:54

2 Answers 2


A moment is a generalization of the notion in physics of moment of a (point) mass about an axis being the product of the mass and the distance from the axis.

For a continuous random variable $X$ with probability density function $f(x)$, the $n$-th moment is $$m_n = \int_{-\infty}^\infty x^n f(x)\,\mathrm dx.$$ The zero-th moment is $1$ (the area under the density is $1$), the first moment is called the mean or expected value of the random variable and the second moment the mean square value. Note that since $f(x) \geq 0$, the second moment cannot be zero.

Even more generally, the $n$-th moment of an arbitrary function $f(x)$ can be defined as $$m_n = \int_{-\infty}^\infty x^n f(x)\,\mathrm dx.$$ Now the restriction of zero-th moment being $1$ and second moment being positive is not applicable any more, and the "vanishing moment" is merely a fancy way of saying that $f(x)$ must be such that $m_0 = m_1 = m_2 = \cdots m_N = 0$. In particular, $m_0$ is the DC value of the wavelet and the authors are insisting that the DC value be $0$.


One of the applications of the (continuous!) wavelet transform is the detection and characterization of fractal signals. For that in particular the nature of the underlying singularities become important. Singularities are characterized by their Höldner exponent. In that context the number of vanishing moments of the analysis wavelet becomes important. It needs to have at least as many vanishing moments as the order of Höldner exponent to be discovered by it.


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