Communication, Signal Processing & Information Technology

Solving (22), we get the update of πi,j,

${\pi }_{i,j}^{\left(k+1\right)}=w_{i,j}^{\left(k\right)},$

and the mean and the central moments are updated as follows,

$\begin{array}{l}{\mu }_j^{\left(k+1\right)}=\frac{{\displaystyle \sum _{i=1}^{N_u}{u}_i{\pi }_{i,j}^{\left(k+1\right)}}}{{\displaystyle \sum _{i=1}^{N_u}{\pi }_{i,j}^{\left(k+1\right)}}}=\frac{{\displaystyle \sum _{i=1}^{N_u}{u}_i{w}_{i,j}^{\left(k\right)}}}{{\displaystyle \sum _{i=1}^{N_u}{w}_{i,j}^{\left(k\right)}}}\\ {\zeta }_{n,j}^{\left(k+1\right)}=\frac{{{\displaystyle \sum _{i=1}^N\left(u_i-{\mu }_{i,j}^{\left(k+1\right)}\right)}}^n{\pi }_{i,j}^{\left(k+1\right)}}{{\displaystyle \sum _{i=1}^N{\pi }_{i,j}^{\left(k+1\right)}}}=\frac{{{\displaystyle \sum _{i=1}^N\left(u_i-{\mu }_j^{\left(k+1\right)}\right)}}^n{w}_{i,j}^{\left(k\right)}}{{\displaystyle \sum _{i=1}^N{w}_{i,j}^{\left(k\right)}}}\end{array}$

for n = 2,3,4, and where μj and ζn,j are the mean value and the n-th central moment of the pixels belonging to class j, respectively. With the results in (24) and (25), s1