Hierarchical Wiener filtering

In the above process, we do not use the information between the wavelet coefficients at different scales. We modify the previous algorithm by introducing a prediction w_h of the wavelet coefficient from the upper scale. This prediction could be determined from the regression [2] between the two scales but better results are obtained when we only set w_h to W_i+1. Between the expectation coefficient W_i and the prediction, a dispersion exists where we assume that it is a Gaussian distribution:

$$P(W_i/w_h) = \frac{1}{\sqrt{2\pi}T_i}e^{-\frac{(W_i- w_h)^2}{2T_i^2}}$$ (14.84)

The relation which gives the coefficient W_i knowing w_i and w_h is:

$$P(W_i/w_i \mbox{ and } w_h) = \frac{1}{\sqrt{2\pi}\beta_i} e^{-\f... ...i w_i)^2}{2\beta_i^2}} \frac{1}{\sqrt{2\pi}T_i} e^{-\frac{(W_i-w_h)^2}{2T_i^2}}$$ (14.85)

with:

$$\beta_i^2 = \frac{S_i^2B_i^2}{S^2 + B_i^2}$$ (14.86)

and:

$$\alpha_i = \frac{S_i^2}{S_i^2+B_i^2}$$ (14.87)

This follows a Gaussian distribution with a mathematical expectation:

$$W_i = \frac{T_i^2}{B_i^2+T_i^2+Q_i^2} w_i + \frac{B_i^2}{B_i^2+T_i^2+Q_i^2} w_h$$ (14.88)

with:

$$Q_i^2 = \frac{T_i^2B_i^2}{S_i^2}$$ (14.89)

W_i is the barycentre of the three values w_i, w_h, 0 with the weights T_i², B_i², Q_i². The particular cases are:

• If the noise is large ( $S_i \ll B_i$) and even if the correlation between the two scales is good (T_i is low), we get $W_i \rightarrow 0$.
• if $B_i \ll S_i \ll T$ then $W_i \rightarrow w_i$.
• if $B_i \ll T_i \ll S$ then $W_i \rightarrow w_i$.
• if $T_i \ll B_i \ll S$ then $W_i \rightarrow w_h$.

At each scale, by changing all the wavelet coefficients w_i of the plane by the estimate value W_i, we get a Hierarchical Wiener Filter. The algorithm is:

1.

Compute the wavelet transform of the data. We get w_i.

2.

Estimate the standard deviation of the noise B₀ of the first plane from the histogram of w₀.

3.

Set i to the index associated with the last plane: i = n

4.

Estimate the standard deviation of the noise B_i from B₀.

5.

S_i² = s_i² - B_i² where s_i² is the variance of w_i

6.

Set w_h to W_i+1 and compute the standard deviation T_iof w_i - w_h.

7.

$W_i = \frac{T_i^2}{B_i^2+T_i^2+Q_i^2} w_i + \frac{B_i^2}{B_i^2+T_i^2+Q_i^2} w_h$

8.

i = i - 1. If i > 0 go to 4

9.

Reconstruct the picture