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Image denoising in undecimated dual-tree complex wavelet domain using multivariate t-distribution

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Abstract

Denoising of natural images is a basic problem in image processing. The present paper proposes a new algorithm for image denoising based on the maximum a-posteriori (MAP) estimator in undecimated dual-tree complex wavelet transform. The undecimated dual-tree complex wavelet transform (UDT-CWT), along with the directional selectivity of the dual-tree complex wavelet transform (DT-CWT), offers exact translational invariance property through removing the down-sampling of filter outputs together with the up-sampling of the complex filter pairs of DT-CWT. These properties are very important in image denoising. The performance of the MAP estimator depends strongly on the probability of noise-free wavelet coefficients. In our proposed denoising method, multivariate t-distribution is applied as the prior probability of noise-free coefficients. The t-distribution can accurately model the statistics of wavelet coefficients, which have peaky and heavy-tailed characteristics. On the other hand, the multivariate model makes it possible to take into account the dependencies of wavelet coefficients and their neighbors. Also, in our work, the necessary parameters of the multivariate distribution will be estimated in a locally-adaptive way to improve the denoising results via using the correlations among the amplitudes of neighbor coefficients. Simulation results delineate that the proposed algorithm outperforms state-of-the-art denoising algorithms in the literature.

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Author information

Authors and Affiliations

  1. Intelligent Data Processing Laboratory (IDPL), Department of Electrical Engineering, Shahid Bahonar University of Kerman, Kerman, Iran

    Mansoore Saeedzarandi, Hossein Nezamabadi-pour & Saeid Saryazdi

  2. Department of Statistics, Faculty of Mathematics & Computer, Shahid Bahonar University of Kerman, Kerman, Iran

    Ahad Jamalizadeh

Authors
  1. Mansoore Saeedzarandi
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  2. Hossein Nezamabadi-pour
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  3. Saeid Saryazdi
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  4. Ahad Jamalizadeh
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Correspondence to Mansoore Saeedzarandi.

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Appendix 1

Appendix 1

From eq. (15), we have:

$$ {\displaystyle \begin{array}{c}\hat{x}=\underset{x}{\arg\;\max}\kern0.24em lnp\left(x,u\left|y\right.\right)=\underset{x}{\arg\;\max}\kern0.24em \mathit{\ln}\frac{p\left(y\left|x\right.,u\right){p}_X\left(x,u\right)}{p_Y(y)}\\ {}=\underset{x}{\arg\;\max}\kern0.24em \mathit{\ln}\frac{p\left(y\left|x\right.,u\right){p}_X\left(x|u\right){p}_U(u)}{p_Y(y)}\\ {}=\underset{x}{\arg\;\max}\kern0.24em \left( lnp\left(y\left|x,u\right.\right)+\mathit{\ln}{p}_X\left(x|u\right)+\mathit{\ln}{p}_U(u)-\mathit{\ln}{p}_Y(y)\right),\end{array}} $$
(34)

The probability density functions of y and u do not depend on x thus, it can be concluded that:

$$ \hat{\boldsymbol{x}}\propto \underset{\boldsymbol{x}}{\arg\;\max}\kern0.24em \left(\mathit{\ln}\ p\left(\boldsymbol{y}\left|\boldsymbol{x}\right.,u\right)+\mathit{\ln}\ {p}_X\left(\boldsymbol{x}|u\right)\right). $$
(35)

From eq. (10), we can easily conclude that

$$ p\left(\boldsymbol{y}|\boldsymbol{x},u\right)=p\left(\boldsymbol{y}|\boldsymbol{x}\right)={\varphi}_p\left(\boldsymbol{y};\boldsymbol{x},{\boldsymbol{\varSigma}}_{\boldsymbol{n}}\right)=\frac{1}{{\left(2\pi \right)}^{p/2}{\left|{\boldsymbol{\varSigma}}_{\boldsymbol{n}}\right|}^{1/2}}\mathit{\exp}\left(\frac{-{\left(\boldsymbol{y}-\boldsymbol{x}\right)}^T{{\boldsymbol{\varSigma}}_n}^{-1}\left(\boldsymbol{y}-\boldsymbol{x}\right)}{2}\right). $$
(36)

Therefore, we obtain:

$$ \hat{\boldsymbol{x}}\propto \underset{\boldsymbol{x}}{\arg\;\max}\kern0.24em \left(\ln\ {\varphi}_p\left(\boldsymbol{y};\boldsymbol{x},{\boldsymbol{\varSigma}}_{\boldsymbol{n}}\right)+\mathit{\ln}\ {p}_X\left(\boldsymbol{x}|u\right)\right). $$
(37)

On the other hand, from eq. (3), we can conclude that:

$$ p\left(\boldsymbol{x}|u\right)={\varphi}_p\left(\boldsymbol{x};\mathbf{0},\boldsymbol{\varSigma} {u}^{-1}\right)=\frac{1}{{\left(2\pi \right)}^{p/2}{\left|\boldsymbol{\varSigma} {u}^{-1}\right|}^{1/2}}\mathit{\exp}\left(\frac{-{\boldsymbol{x}}^Tu{\boldsymbol{\varSigma}}^{-1}\boldsymbol{x}}{2}\right). $$
(38)

From (37) and (38), we obtain:

$$ \hat{\boldsymbol{x}}\propto \underset{\boldsymbol{x}}{\arg\;\max}\kern0.49em \left(\mathit{\ln}\ {\varphi}_p\left(\boldsymbol{y};\boldsymbol{x},{\boldsymbol{\varSigma}}_{\boldsymbol{n}}\right)+\mathit{\ln}\ {\varphi}_p\left(\boldsymbol{x};\mathbf{0},\boldsymbol{\varSigma} {u}^{-1}\right)\right) $$
(39)

and from eqs. (36) and (38) we concluded that:

$$ {\varphi}_p\left(\boldsymbol{y};\boldsymbol{x},{\boldsymbol{\varSigma}}_{\boldsymbol{n}}\right)=\frac{1}{{\left(2\pi \right)}^{p/2}{\left|{\boldsymbol{\varSigma}}_{\boldsymbol{n}}\right|}^{1/2}}\mathit{\exp}\left(\frac{-{\left(\boldsymbol{y}-\boldsymbol{x}\right)}^T{{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}\left(\boldsymbol{y}-\boldsymbol{x}\right)}{2}\right), $$
(40)
$$ {\varphi}_p\left(\boldsymbol{x};\mathbf{0},\boldsymbol{\varSigma} {u}^{-1}\right)=\frac{1}{{\left(2\pi \right)}^{p/2}{\left|\boldsymbol{\varSigma} {u}^{-1}\right|}^{1/2}}\mathit{\exp}\left(\frac{-{\boldsymbol{x}}^Tu{\boldsymbol{\varSigma}}^{-1}\boldsymbol{x}}{2}\right). $$
(41)

By substituting eqs. (40) and (41) into eq. (39) we have:

$$ \hat{\boldsymbol{x}}\propto \underset{\boldsymbol{x}}{\arg\;\max}\kern0.24em \left(\mathit{\ln}\ \left(\frac{1}{{\left(2\pi \right)}^{\frac{p}{2}}{\left|{\boldsymbol{\varSigma}}_{\boldsymbol{n}}\right|}^{\frac{1}{2}}}\right)+\left(\frac{-{\left(\boldsymbol{y}-\boldsymbol{x}\right)}^T{{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}\left(\boldsymbol{y}-\boldsymbol{x}\right)}{2}\right)+\mathit{\ln}\ \left(\frac{1}{{\left(2\pi \right)}^{p/2}{\left|\boldsymbol{\varSigma} {u}^{-1}\right|}^{1/2}}\right)+\left(\frac{-{\boldsymbol{x}}^Tu{\boldsymbol{\varSigma}}^{-1}\boldsymbol{x}}{2}\right)\right), $$
(42)

because the first and third terms do not depend on x thus we can conclude that:

$$ \hat{\boldsymbol{x}}\propto \underset{\boldsymbol{x}}{\arg\;\max}\kern0.24em \left(\left(\frac{-{\left(\boldsymbol{y}-\boldsymbol{x}\right)}^T{{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}\left(\boldsymbol{y}-\boldsymbol{x}\right)}{2}\right)+\left(\frac{-{\boldsymbol{x}}^Tu{\boldsymbol{\varSigma}}^{-1}\boldsymbol{x}}{2}\right)\right) $$
$$ \hat{\boldsymbol{x}}\propto \underset{\boldsymbol{x}}{\arg\;\min}\kern0.24em \left(\left(\frac{{\left(\boldsymbol{y}-\boldsymbol{x}\right)}^T{{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}\left(\boldsymbol{y}-\boldsymbol{x}\right)}{2}\right)+\left(\frac{{\boldsymbol{x}}^Tu{\boldsymbol{\varSigma}}^{-1}\boldsymbol{x}}{2}\right)\right) $$
$$ \hat{\boldsymbol{x}}\propto \underset{\boldsymbol{x}}{\arg\;\min}\kern0.24em \left(\left({\left(\boldsymbol{y}-\boldsymbol{x}\right)}^T{{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}\left(\boldsymbol{y}-\boldsymbol{x}\right)\right)+\left({\boldsymbol{x}}^Tu{\boldsymbol{\varSigma}}^{-1}\boldsymbol{x}\right)\right) $$
$$ \hat{\boldsymbol{x}}\propto \underset{\boldsymbol{x}}{\arg\;\min}\kern0.24em \left({\boldsymbol{y}}^T{\Sigma}_{\boldsymbol{n}}^{-1}\boldsymbol{y}-{\boldsymbol{y}}^T{\Sigma}_{\boldsymbol{n}}^{-1}\boldsymbol{x}-{\boldsymbol{x}}^T{\Sigma}_{\boldsymbol{n}}^{-1}\boldsymbol{y}+{\boldsymbol{x}}^T{\Sigma}_{\boldsymbol{n}}^{-1}\boldsymbol{x}+{\boldsymbol{x}}^T{\Sigma}_{\boldsymbol{n}}^{-1}\boldsymbol{x}\right) $$
$$ \hat{\boldsymbol{x}}\propto \underset{\boldsymbol{x}}{\arg\;\min}\kern0.24em \left({\boldsymbol{y}}^T{{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}\boldsymbol{y}-2{\boldsymbol{y}}^T{{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}\boldsymbol{x}+{\boldsymbol{x}}^T{{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}\boldsymbol{x}+{\boldsymbol{x}}^Tu{\boldsymbol{\varSigma}}^{-1}\boldsymbol{x}\right) $$
(43)

Now by differentiating of (43) with respect to x and using the following derivative rules:

$$ \frac{\partial {b}^T\theta }{\partial \theta }=b $$
(44)
$$ \frac{\partial {\theta}^T B\theta}{\partial \theta }=2 B\theta $$
(45)

where b and θ are p × 1 real vectors and B is a p × p symmetric matrix [6], we have:

$$ -2{{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}\boldsymbol{y}+2{{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}\hat{\boldsymbol{x}}+2u{\boldsymbol{\varSigma}}^{-1}\hat{\boldsymbol{x}}=0, $$
(46)

thus, we can compute \( \hat{\boldsymbol{x}} \) as follows:

$$ {\displaystyle \begin{array}{l}\hat{\boldsymbol{x}}={\left({{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}+u{\boldsymbol{\varSigma}}^{-1}\right)}^{-1}{{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}\boldsymbol{y}={\left({{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}\left(\mathbf{I}+u{\boldsymbol{\varSigma}}_{\boldsymbol{n}}{\boldsymbol{\varSigma}}^{-1}\right)\right)}^{-1}{{\boldsymbol{\varSigma}}_{\boldsymbol{n}}}^{-1}\boldsymbol{y}={\left(\boldsymbol{I}+u{\boldsymbol{\varSigma}}_{\boldsymbol{n}}{\boldsymbol{\varSigma}}^{-1}\right)}^{-1}\boldsymbol{y}\\ {}\kern1em ={\left({\boldsymbol{\varSigma} \boldsymbol{\varSigma}}^{-1}+u{\boldsymbol{\varSigma}}_{\boldsymbol{n}}{\boldsymbol{\varSigma}}^{-1}\right)}^{-1}\boldsymbol{y}={\left(\left(\boldsymbol{\varSigma} +u{\boldsymbol{\varSigma}}_{\boldsymbol{n}}\right){\boldsymbol{\varSigma}}^{-1}\right)}^{-1}\boldsymbol{y}=\boldsymbol{\varSigma} {\left(\boldsymbol{\varSigma} +u{\boldsymbol{\varSigma}}_{\boldsymbol{n}}\right)}^{-1}\boldsymbol{y}\end{array}} $$
(47)

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Saeedzarandi, M., Nezamabadi-pour, H., Saryazdi, S. et al. Image denoising in undecimated dual-tree complex wavelet domain using multivariate t-distribution. Multimed Tools Appl 79, 22447–22471 (2020). https://doi.org/10.1007/s11042-020-08954-y

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