Dual-Tree Complex Wavelet Coefficient Magnitude Modeling Using Scale Mixtures of Rayleigh Distribution for Image Denoising

Abstract

Denoising an image, while retaining the important features of the image, has been a fundamental problem in image processing. Dual-tree complex wavelet transform is a recently created transform that offers both near shift invariance and improved directional selectivity properties. This transform has been used in many techniques, including denoising. However, these techniques have used the real and imaginary components of the complex-valued sub-band coefficients separately. This paper proposes the use of coefficient magnitudes to provide an improvement in image denoising. Our proposed algorithm is based on the maximum a posteriori estimator, wherein the heavy-tailed scale mixtures of bivariate Rayleigh distribution are considered as the noise-free wavelet coefficient magnitudes’ prior distribution. Also, in our work, the necessary parameters of the bivariate distributions are estimated in a locally adaptive way to improve the denoising results via using the correlation between the amplitudes of neighbor coefficients. Simulation results delineate the performance of the proposed algorithm in both MSSIM and PSNR metrics.

Appendices

Appendix 1

1.1 Univariate Rayleigh–Student’s t Distribution

Using Eq. (12) and by assuming gamma distribution for $u$ as follows:

$$ h(u;\upsilon ) = \frac{{(\upsilon /2)^{\upsilon /2} }}{\varGamma (\upsilon /2)}u^{\upsilon /2 - 1} \exp ( - \upsilon u/2), $$

(52)

we obtain

$$ \begin{aligned} f_{{rt_{\text{SMN}} }} (r,\sigma ,\upsilon ) & = \int_{0}^{\infty } {\frac{ru}{{\sigma^{2} }}\exp \left( {\frac{{ - r^{2} u}}{{2\sigma^{2} }}} \right)h(u;\upsilon ){\text{d}}u} \\ & = \int_{0}^{\infty } {\frac{ru}{{\sigma^{2} }}\exp \left( {\frac{{ - r^{2} u}}{{2\sigma^{2} }}} \right)\frac{{(\upsilon /2)^{\upsilon /2} }}{\varGamma (\upsilon /2)}u^{\upsilon /2 - 1} \exp ( - \upsilon u/2)} \\ & = \frac{r}{{\sigma^{2} }}\frac{{(\upsilon /2)^{\upsilon /2} }}{\varGamma (\upsilon /2)}\int_{0}^{\infty } {u^{{\left( {\frac{\upsilon + 2}{2}} \right) - 1}} } \exp \left( { - \left( {\frac{\upsilon + 2}{2}} \right)u\left( {1 - \left( {\frac{{2\sigma^{2} - r^{2} }}{{\sigma^{2} (\upsilon + 2)}}} \right)} \right)} \right){\text{d}}u. \\ \end{aligned} $$

(53)

Let

$$ 1 - \left( {\frac{{2\sigma^{2} - r^{2} }}{{\sigma^{2} (\upsilon + 2)}}} \right) = A\quad {\text{and}}\quad uA = u^{\prime } . $$

(54)

Also, we have

$$ \int_{0}^{\infty } {h(u;\upsilon )\,{\text{d}}u} = \int_{0}^{\infty } {\frac{{(\upsilon /2)^{\upsilon /2} }}{\varGamma (\upsilon /2)}u^{\upsilon /2 - 1} \exp ( - \,\upsilon u/2) = 1} . $$

(55)

Therefore,

$$ \int_{0}^{\infty } {h(u;\upsilon ){\text{d}}u} = \int_{0}^{\infty } {\frac{{\left( {\frac{\upsilon + 2}{2}} \right)^{{\frac{\upsilon + 2}{2}}} }}{{\varGamma \left( {\frac{\upsilon + 2}{2}} \right)}}u^{{\left( {\frac{\upsilon + 2}{2}} \right) - 1}} \exp \left( {\frac{ - (\upsilon + 2)u}{2}} \right) = 1} . $$

(56)

Using Eqs. (53), (54), and (56), we have

$$ \begin{aligned} f_{{rt_{\text{SMN}} }} (r;\sigma ,\upsilon ) & = \int_{0}^{\infty } {f(r|u)f(u){\text{d}}u} = \frac{r}{{\sigma^{2} }}\frac{{(\upsilon /2)^{\upsilon /2} }}{\varGamma (\upsilon /2)}\frac{1}{{A^{{\left( {\frac{\upsilon + 2}{2}} \right)}} }}\int_{0}^{\infty } {u^{{\prime \left( {\frac{\upsilon + 2}{2}} \right) - 1}} } \exp \left( { - \left( {\frac{\upsilon + 2}{2}} \right)u^{\prime } } \right){\text{d}}u^{\prime } \\ & = \frac{r}{{\sigma^{2} }}\frac{{(\upsilon /2)^{\upsilon /2} }}{\varGamma (\upsilon /2)}\frac{1}{{A^{{\left( {\frac{\upsilon + 2}{2}} \right)}} }}\frac{{\varGamma \left( {\frac{\upsilon + 2}{2}} \right)}}{{\left( {\frac{\upsilon + 2}{2}} \right)^{{\frac{\upsilon + 2}{2}}} }}. \\ \end{aligned} $$

(57)

By substituting $A$ into Eq. (57) and after some algebra, we have

$$ f_{{rt_{\text{SMN}} }} (r;\sigma ,\upsilon ) = \upsilon^{{\frac{\upsilon + 2}{2}}} \frac{{r\sigma^{\upsilon } }}{{(\upsilon \sigma^{2} + r^{2} )^{{\frac{\upsilon + 2}{2}}} }}. $$

(58)

1.2 Univariate Rayleigh–Laplace Distribution

Using Eq. (12) and by assuming $u \sim {\text{IG}}\left( {1,\frac{1}{2}} \right)$ as follows:

$$ h(u) = \frac{1}{2}u^{ - 2} \exp ( - 1/2u), $$

(59)

we will have:

$$ \begin{aligned} f_{{rL_{\text{SMN}} }} (r;\sigma ) & = \int_{0}^{\infty } {\frac{ru}{{\sigma^{2} }}\exp \left( {\frac{{ - r^{2} u}}{{2\sigma^{2} }}} \right)h(u){\text{d}}u} = \int_{0}^{\infty } {\frac{ru}{{\sigma^{2} }}\exp \left( {\frac{{ - r^{2} u}}{{2\sigma^{2} }}} \right)\frac{1}{2}u^{ - 2} \exp ( - 1/2u)} \\ & = \frac{r}{{\sigma^{2} }}\frac{1}{2}\int_{0}^{\infty } {u^{ - 1} } \exp \left( {\frac{ - 1}{2}\left( {\frac{{r^{2} u}}{{\sigma^{2} }} + \frac{1}{u}} \right)} \right){\text{d}}u. \\ \end{aligned} $$

(60)

Also, we have [17]

$$ \int_{0}^{\infty } {x^{\lambda - 1} } \exp \left( {\frac{ - 1}{2}(\chi x^{ - 1} + \psi x)} \right) = \frac{{2K_{\lambda } (\sqrt {\chi \psi } )}}{{\chi^{ - 1} (\sqrt {\chi \psi } )^{\lambda } }}. $$

(61)

Using Eqs. (60) and (61), we derive

$$ f_{{rL_{\text{SMN}} }} (r;\sigma ) = \frac{r}{{\sigma^{2} }}K_{0} \left( {\frac{r}{\sigma }} \right). $$

(62)

Appendix 2

Using Eq. (19) and by the assuming gamma distribution for $u,$ we have

$$ \begin{aligned} f_{BRt} (r_{1} ,r_{2} ;\sigma ,\upsilon ) & = \int_{0}^{\infty } {\frac{{r_{1} r_{2} u^{2} }}{{\sigma^{4} }}\exp \left( {\frac{{ - r_{1}^{2} u}}{{2\sigma^{2} }}} \right)\exp \left( {\frac{{ - r_{2}^{2} u}}{{2\sigma^{2} }}} \right)h(u;\upsilon ){\text{d}}u} \\ & = \int_{0}^{\infty } {\frac{{r_{1} r_{2} u^{2} }}{{\sigma^{4} }}\exp \left( {\frac{{ - r_{1}^{2} u}}{{2\sigma^{2} }}} \right)\exp \left( {\frac{{ - r_{2}^{2} u}}{{2\sigma^{2} }}} \right)} \frac{{(\upsilon /2)^{\upsilon /2} }}{\varGamma (\upsilon /2)}u^{\upsilon /2 - 1} \exp ( - \upsilon u/2) \\ & = \frac{{(\upsilon /2)^{\upsilon /2} }}{\varGamma (\upsilon /2)}\frac{{r_{1} r_{2} }}{{\sigma^{4} }}\int_{0}^{\infty } {u^{{\left( {\frac{\upsilon + 4}{2}} \right) - 1}} \exp \left( { - \left( {\frac{\upsilon + 4}{2}} \right)u\left( {1 - \left( {\frac{{4\sigma^{2} - (r_{1}^{2} + r_{2}^{2} )}}{{\sigma^{2} (\upsilon + 4)}}} \right)} \right)} \right){\text{d}}u} . \\ \end{aligned} $$

(63)

Let

$$ 1 - \left( {\frac{{4\sigma^{2} - (r_{1}^{2} + r_{2}^{2} )}}{{\sigma^{2} (\upsilon + 4)}}} \right) = A\quad {\text{and}}\quad uA = u^{\prime } , $$

(64)

therefore

$$ \begin{aligned} f_{BRt} (r_{1} ,r_{2} ;\sigma ,\upsilon ) & = \frac{{r_{1} r_{2} }}{{\sigma^{4} }}\frac{{(\upsilon /2)^{\upsilon /2} }}{\varGamma (\upsilon /2)}\frac{1}{{A^{{\left( {\frac{\upsilon + 4}{2}} \right)}} }}\int_{0}^{\infty } {u^{{\prime \left( {\frac{\upsilon + 4}{2}} \right) - 1}} } \exp \left( { - \left( {\frac{\upsilon + 4}{2}} \right)u^{\prime } } \right){\text{d}}u^{\prime } \\ & = \frac{{r_{1} r_{2} }}{{\sigma^{4} }}\frac{{(\upsilon /2)^{\upsilon /2} }}{\varGamma (\upsilon /2)}\frac{1}{{A^{{\left( {\frac{\upsilon + 4}{2}} \right)}} }}\frac{{\varGamma \left( {\frac{\upsilon + 4}{2}} \right)}}{{\left( {\frac{\upsilon + 4}{2}} \right)^{{\frac{\upsilon + 4}{2}}} }}. \\ \end{aligned} $$

(65)

By substituting $A$ into Eq. (65), we have

$$ f_{BRt} \left( {r_{1} ,r_{2} ;\sigma ,\upsilon } \right) = \frac{{\left( {\upsilon + 2} \right)r_{1} r_{2} }}{{\upsilon \left( \sigma \right)^{4} }}\left( {1 + \frac{{r_{1}^{2} + r_{2}^{2} }}{{\sigma^{2} \upsilon }}} \right)^{{ - \left( {\frac{\upsilon + 4}{2}} \right)}} \text{, }\quad r_{1} > 0,\,\,\,r_{2} > 0. $$

(66)

2.1 Bivariate Rayleigh–Laplace Distribution

Using Eq. (19) and by assuming $u \sim {\text{IG}}\left( {1,\frac{1}{2}} \right),$ we derive

$$ \begin{aligned} f_{\text{BRL}} (r_{1} ,r_{2} ;\sigma ) & = \int_{0}^{\infty } {\frac{{r_{1} r_{2} u^{2} }}{{\sigma^{4} }}\exp \left( {\frac{{ - r_{1}^{2} u}}{{2\sigma^{2} }}} \right)\exp \left( {\frac{{ - r_{2}^{2} u}}{{2\sigma^{2} }}} \right)h(u){\text{d}}u} \\ & = \int_{0}^{\infty } {\frac{{r_{1} r_{2} u^{2} }}{{\sigma^{4} }}\exp \left( {\frac{{ - r_{1}^{2} u}}{{2\sigma^{2} }}} \right)\exp \left( {\frac{{ - r_{2}^{2} u}}{{2\sigma^{2} }}} \right)} \frac{1}{2}u^{ - 2} \exp ( - 1/2u) \\ & = \frac{1}{2}\frac{{r_{1} r_{2} }}{{\sigma^{4} }}\int_{0}^{\infty } {\exp \left( { - \frac{1}{2}\left( {\frac{{r_{1}^{2} + r_{2}^{2} }}{{\sigma^{2} }} + \frac{1}{2u}} \right)} \right){\text{d}}u} . \\ \end{aligned} $$

(67)

Using Eqs. (61) and (67), we derive

$$ f_{\text{BRL}} \left( {r_{1} ,r_{2} ;\sigma } \right) = \frac{{r_{1} r_{2} }}{{\left( \sigma \right)^{4} }}\frac{{K_{1} \left( {\sqrt {\frac{{r_{1}^{2} + r_{2}^{2} }}{{\sigma^{2} }}} } \right)}}{{\left( {\frac{{r_{1}^{2} + r_{2}^{2} }}{{\sigma^{2} }}} \right)^{1/2} }}\text{,}\quad r_{1} > 0,\,\,\,r_{2} > 0. $$

(68)

Appendix 3

In probability theory and statistics, kurtosis is a measure of the tailedness of the probability distribution of a random variable and variance is the expectation of the squared deviation of a random variable from its mean and is defined as:

$$ \text{var}\left( x \right) = E\left( {x - \mu } \right)^{2} = M_{2} , $$

(69)

$$ \text{kurtosis}\left( x \right) = \frac{{E\left[ {\left( {x - \mu } \right)^{4} } \right]}}{{E\left[ {\left( {x - \mu } \right)^{2} } \right]^{2} }} = \frac{{M_{4} }}{{M_{2}^{2} }}, $$

(70)

where $M_{4}$ and $M_{2}$ are fourth and second central moments, respectively.

The cumulants $k_{n}$ of a random variable $x$ are defined via the cumulant generating function $(K(t))$, which is the natural logarithm of the moment generation function:

$$ K(t) = \log E\left[ {e^{tx} } \right]. $$

(71)

The fourth and second cumulants are related to the central moments by the following equations [11, 44]:

$$ M_{2} = k_{2} , $$

(72)

$$ M_{4} = k_{4} + 3k_{2}^{2} . $$

(73)

Kurtosis and variance in terms of cumulants are:

$$ \text{var} \left( x \right) = M_{2} = k_{2} , $$

(74)

$$ \text{Kurtosis}\left( x \right) = \frac{{M_{4} }}{{M_{2}^{2} }} = \frac{{k_{4} + 3k_{2}^{2} }}{{k_{2}^{2} }} = \frac{{k_{4} }}{{k_{2}^{2} }} + 3. $$

(75)