# Vanishing Moments

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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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? –  Dilip Sarwate Oct 26 '12 at 21:27
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 –  mkuse Oct 26 '12 at 21:35
In brief, if a wavelet has $n$ vanishing moments, the output of filtering a $n$-th degree polynomial with this wavelet will be 0. –  Phonon Oct 27 '12 at 0:04

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 $0$.

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