Wavelet bicoherence was given by Van Milligen1995, which used to analyze turbulence. And the normalized squared wavelet bicoherence (usually called wavelet bicoherence) is shown below. $$ WBC(a_1,a_2)=\frac{|\int_TW_f(a_1,\tau)W_f(a_2,\tau)W_g^*(a,\tau)\mathrm{d}\tau|^2}{\int_T|W_f(a_1,\tau)W_f(a_2,\tau)|^2\mathrm{d}\tau\int_T|W_g(a,\tau)|^2\mathrm{d}\tau} $$

Van Milligen1995 proposed a concept called statistical noise level which puzzled me. The content of statistical noise is as follows

However, the wavelet coefficients are not all statistically independent, since the chosen wavelet family is not orthogonal. Each coefficient is calculated by evaluating Eq. (4), integrating over the range $- \infty < t< +\infty$. Due to the periodicity $a$ of the wavelets of scale $a$, two statistically independent estimates of the wavelet coefficients are separated by a time $a/2$, or by a number of points $M(a)=a\omega_{samp}/(4\pi)$ ($w_{samp}$ being the sampling frequency). Thus, the summation done in the evaluation of the bicoherence $WBC(a_1,a_2)$ is not really carried out over $N$ points, but only over $N/max[M(a)]$, where the maximum is taken over the values of a that come into play for the evaluation of a specific value of the squared bicoherence. An estimate for the statistical noise level in $WBC(a_1,a_2)$ is, therefore $$ \epsilon(WBC)\approx \Big[ \frac{\omega_{samp}}{2\cdot \mathrm{Min}(|w_1|,|w_2|,|w_1+w_2|)}\frac{1}{N}\Big] $$ where $w=2\pi/a$.

My questions are

  1. We all know that the wavelet functions are well-localized. So why did Milligen say that "the periodicity $a$ of the wavelets of scale $a$"? And why the wavelet coefficients separated by a time $a/2$ are statistically independent?
  2. Why take the maximum value of $M(a)$ in $N/max[M(a)]$?
  3. Milligen didn't show the formal derivation of the statistical noise level, i.e. $\epsilon(WBC)$. So I want to know how to relate the formula of $\epsilon(WBC)$ with $N/max[M(a)]$.

1 Answer 1


I'll be answering 1 and 2, and partly 3 & 4.

1. $a$-periodicity

The usage of "periodic" here differs from typical but is appropriate for analysis of independence; the author refers to figure 1 (red markings mine):

In the time interval $\Delta t = a = 1$, the wavelet completes one period at its center frequency.

2. $a/2$-independence

A complex sinusoid from $0$ to $T/2$ will correlate to zero with itself from $T/2$ to $T$, where $T$ is its period:

enter image description here

The correlation is sum(prod), and we have a full period complex sinusoid which is naturally zero-sum.

3, 4

This is e.g. $\max(M(a_1), M(a_2))$ ("values that come into play"), i.e. we take the larger of two periods, i.e. lesser of two frequencies $\omega_1, \omega_2$. I don't know what this is about. Note that sum of functions of periods $T_1$ and $T_2$ will have a periodicity of least common multiple of $T_1, T_2$, e.g. $6, 10 \rightarrow 30$, so if we seek to evaluate only at independent points, that'd be $N/30$, which isn't $N/10$. Also the lower the frequency the more subsample-able a waveform is (the fewer points we need to represent it, hence $N/\max{}$).

However, I don't see how any of this allows fft(f1) * fft(f2) * fft(f1 + f2) to be uniquely represented by N/max(T1, T2) points, especially when instead of pure tones we have wavelets that take more points to represent in either domain. Hopefully this gives you clues.

  • $\begingroup$ Thank you for your answer. By the way, what is the periodicity of the Mexican hat wavelet ? $\endgroup$
    – Wang Yun
    Jun 29, 2021 at 6:52
  • $\begingroup$ @WangYun That's a separate question, but briefly, you can find the periodicity and independence interval of any wavelet by sampling it in the frequency domain (ensure entire shape is captured until decay), taking IFFT, and taking sum-product of it with its own conjugate at different shifts (autocorrelation with conjugate, sum(p*conj(p))); the first shift to produce a zero is the period, or for some wavelets, half-period. I don't know the exact relation for cmhat. $\endgroup$ Jun 29, 2021 at 15:24
  • $\begingroup$ I didn't figure out the meaning of the bottom figure of the second part of your answer. What do the solid blue line and orange line refer to, respectively? $\endgroup$
    – Wang Yun
    Jul 4, 2021 at 0:38
  • $\begingroup$ @WangYun It's top left multiplied by top right. Suppose c is a complex sinusoid of one period and N samples; then left is c[:N//2], right is c[N//2:], bottom is c[:N//2]*c[N//2:]. The sum of this product (over all N//2 samples) is zero. $\endgroup$ Jul 4, 2021 at 0:46
  • $\begingroup$ Could you express your content in continuous form, i.e. integral form? $\endgroup$
    – Wang Yun
    Aug 7, 2021 at 12:15

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