# Fast Minimum Error Entropy for Linear Regression
## 3 Methodology
\begin{split}&\int\varphi(\xi)(\frac{\xi-\mu}{\sigma})^{2}H_{4} (\frac{\xi-\mu}{\sigma})\mathrm{d}\xi\\ &=\int\frac{1}{\sqrt{2\pi}\sigma}\exp\big{[}-\frac{(\xi-\mu)^{ 2}}{2\sigma^{2}}\big{]}*\\ &\quad\big{[}(\frac{\xi-\mu}{\sigma})^{6}-6(\frac{\xi-\mu}{\sigma })^{4}+3(\frac{\xi-\mu}{\sigma})^{2}\big{]}\mathrm{d}\xi\\ &=\int\frac{1}{\sqrt{2\pi}}\exp(-\frac{t^{2}}{2})(t^{6}-6t^{4}+ 3t^{2})\mathrm{d}t\\ &=3\int\frac{1}{\sqrt{2\pi}}\exp(-\frac{t^{2}}{2})t^{2}\mathrm{ d}t-6\int\frac{1}{\sqrt{2\pi}}\exp(-\frac{t^{2}}{2})t^{4}\mathrm{d}t\\ &\quad+\int\frac{1}{\sqrt{2\pi}}\exp(-\frac{t^{2}}{2})t^{6} \mathrm{d}t\end{split}  \tag{19}\]The first term in Equation (19) can be calculated as Equation (20),  
\[ \begin{split}\int\frac{1}{\sqrt{2\pi}}&\exp(-\frac{t^{2 }}{2})t^{2}\mathrm{d}t\\ &=\int\frac{1}{\sqrt{2\pi}}(-t)\mathrm{d}\big{(}\exp(-\frac{t^{2} }{2})\big{)}\\ &=\frac{1}{\sqrt{2\pi}}(-t)\big{(}\exp(-\frac{t^{2}}{2})\big{)} \bigg{|}_{-\infty}^{+\infty}+\\ &\int\frac{1}{\sqrt{2\pi}}\exp(-\frac{t^{2}}{2})\mathrm{d}t\\ &=1\end{split}  \tag{20}\]  
The second term in Equation (19) can be calculated as Equation (21),  
\[ \begin{split}\int\frac{1}{\sqrt{2\pi}}&\exp(-\frac{t ^{2}}{2})t^{4}\mathrm{d}t\\ &=\int\frac{1}{\sqrt{2\pi}}(-t^{3})\mathrm{d}\big{(}\exp(-\frac{t ^{2}}{2})\big{)}\\ &=\frac{1}{\sqrt{2\pi}}(-t^{3})\big{(}\exp(-\frac{t^{2}}{2})\big{)} \bigg{|}_{-\infty}^{+\infty}+\\ & 3\int\frac{1}{\sqrt{2\pi}}\exp(-\frac{t^{2}}{2})t^{2}\mathrm{d} t\\ &=3\end{split}  \tag{21}\]  
The last term in Equation (19) can be calculated as Equation (22),  
\[ \begin{split}\int\frac{1}{\sqrt{2\pi}}&\exp(-\frac{t ^{2}}{2})t^{6}\mathrm{d}t\\ &=\int\frac{1}{\sqrt{2\pi}}(-t^{5})\mathrm{d}\big{(}\exp(-\frac{t ^{2}}{2})\big{)}\\ &=\frac{1}{\sqrt{2\pi}}(-t^{5})\big{(}\exp(-\frac{t^{2}}{2}) \big{)}\bigg{|}_{-\infty}^{+\infty}+\\ & 5\int\frac{1}{\sqrt{2\pi}}\exp(-\frac{t^{2}}{2})t^{4}\mathrm{d} t\\ &=15\end{split}  \tag{22}\]  
According to Equations (20)-(22),  
\[\int\varphi(\xi)(\frac{\xi-\mu}{\sigma})^{2}H_{4}(\frac{\xi-\mu}{\sigma}) \mathrm{d}\xi=15-6*3+3=0 \tag{23}\]  
and Equation (16) holds. Then, for Equation (13),  
\[ \begin{split} H(x)&=-\int\varphi(\xi)\log\varphi(\xi) \mathrm{d}\xi-\\ &\int\varphi(\xi)\bigg{[}\frac{\kappa_{3}(x)}{3!\sigma^{3}}H_{3}( \frac{\xi-\mu}{\sigma})+\frac{\kappa_{4}(x)}{4!\sigma^{4}}H_{4}(\frac{\xi-\mu} {\sigma})\bigg{]}\mathrm{d}\xi+\\ &\int\frac{1}{2}\varphi(\xi)\bigg{[}\frac{\kappa_{3}(x)}{3! \sigma^{3}}H_{3}(\frac{\xi-\mu}{\sigma})+\frac{\kappa_{4}(x)}{4!\sigma^{4}}H_ {4}(\frac{\xi-\mu}{\sigma})\bigg{]}^{2}\mathrm{d}\xi-\\ &\int\varphi(\xi)\log\varphi(\xi)\bigg{[}\frac{\kappa_{3}(x)}{3! \sigma^{3}}H_{3}(\frac{\xi-\mu}{\sigma})+\\ &\frac{\kappa_{4}(x)}{4!\sigma^{4}}H_{4}(\frac{\xi-\mu}{\sigma}) \bigg{]}\mathrm{d}\xi\\ &-\int\varphi(\xi)\bigg{[}\frac{\kappa_{3}(x)}{3!\sigma^{3}}H_{3} (\frac{\xi-\mu}{\sigma})+\\ &\frac{\kappa_{4}(x)}{4!\sigma^{4}}H_{4}(\frac{\xi-\mu}{\sigma}) \bigg{]}^{2}\mathrm{d}\xi+\\ &\int\frac{1}{2}\varphi(\xi)\bigg{[}\frac{\kappa_{3}(x)}{3! \sigma^{3}}H_{3}(\frac{\xi-\mu}{\sigma})+\frac{\kappa_{4}(x)}{4!\sigma^{4}}H_ {4}(\frac{\xi-\mu}{\sigma})\bigg{]}^{3}\mathrm{d}\xi\end{split}  \tag{24}\]  
Notice that \(H_{0}=1\), then the second term of Equation (24) is 0. As it is assumed that \(x\) is close to a normal random variable, \(\kappa_{3}(x)\) and \(\kappa_{4}(x)\) are very small. Then, the last term of Equation (24) can be omitted as 0, as it involves third-order monomials, which are infinitely smaller than the terms only containing second-order monomials. Then, for Equation (24), take Equations (15) and (16) into consideration, and notice again that \(H_{0}=1\),  
\[