# Fault Estimation for Semi-Markov Jump Neural Networks Based on the Extended State Method
## 3 Main Results
\]  
Let us define the forward difference:  
\[\mathrm{E}\{\Delta(\mathbb{V}(\mathcal{V}(g),\sigma(g),p)\}=\mathrm{E}\Bigg{\{} \sum\limits_{j=3}^{3}\left[\mathbb{V}_{j}(\mathcal{V}(g+1),\sigma(g+1),p)- \mathbb{V}_{j}(\mathcal{V}(g),\sigma(g),p)\right]\Bigg{\}} \tag{30}\]  
\[\mathrm{P}_{p,\sigma(g)}= \begin{bmatrix}\mathrm{P}_{1p,\sigma(g)}&0\\ 0&\mathrm{P}_{1p,\sigma(g)}\end{bmatrix} ,\quad\widetilde{\mathrm{P}}_{p,\sigma (g+1)}= \begin{bmatrix}\widetilde{\mathrm{P}}_{1p,\sigma(g+1)}&0\\ 0&\widetilde{\mathrm{P}}_{1p,\sigma(g+1)}\end{bmatrix} ,\quad\widetilde{ \mathrm{P}}_{p,\sigma(g+1)}=\sum\limits_{q=1}^{Q}\pi_{pq}(\zeta)\mathrm{P}_{1q,\sigma(g+1)}\]  
Based on Assumption 1, we have \(\widetilde{\mathrm{P}}_{1p,\sigma(g+1)}=\mathrm{P}_{1q,\sigma(g+1)}=\mathrm{ P}_{1p,\sigma(g+1)}\). Then, \(\widetilde{\mathrm{P}}_{p,\sigma(g+1)}=\mathrm{P}_{p,\sigma(g+1)}\). Now, from Equation (30), we can get Equation (31) as follows:  
\[ \begin{array}{l}\mathrm{E}\{\Delta\mathbb{V}_{1}(\mathcal{V}(g),\sigma(g),p )\}\\ =\mathrm{E}\{\mathbb{V}_{1}(\mathcal{V}(g+1),\sigma(g+1),p)-\mathbb{V}_{1}( \mathcal{V}(g),\sigma(g),p)\}\\ =\mathrm{E}\Big{\{}\mathcal{V}^{\mathrm{T}}(g)\,\mathrm{P}_{p,\sigma(g)} \mathcal{V}(g)-\rho\mathcal{V}^{\mathrm{T}}(g)\,\mathrm{P}_{p,\sigma(g)} \mathcal{V}(g)+(\rho-1)\mathbb{V}_{1}(\mathcal{V}(g),\sigma(g),p)\Big{\}}\\ \end{array}  \tag{31}\]  
\[ \begin{array}{l}\mathrm{E}\{\Delta\mathbb{V}_{2}(\mathcal{V}(g),\sigma(g),p )\}\\ =\mathrm{E}\{\mathbb{V}_{2}(\mathcal{V}(g+1),\sigma(g+1),p)-\mathbb{V}_{2}( \mathcal{V}(g),\sigma(g),p)\}\\ =\mathrm{E}\Big{\{}\rho^{d_{1}-1}\mathcal{V}^{\mathrm{T}}(g)\mathrm{X}_{1} \mathcal{V}(g)+\rho^{d(g+1)-1}\mathcal{V}^{\mathrm{T}}(g)\mathrm{X}_{2} \mathcal{V}(g)+\rho^{d_{2}-1}\mathcal{V}^{\mathrm{T}}(g)\mathrm{X}_{1} \mathcal{V}(g)\\ -\sum\limits_{s=g-d_{1}}^{g-1}\rho^{g-s-1}\mathcal{V}^{\mathrm{T}}(g-d_{1}) \mathrm{X}_{1}\mathcal{V}(g-d_{1})+\sum\limits_{s=g-d(g)}^{g-1}\rho^{d(g+1)-1} \mathcal{V}^{\mathrm{T}}(g-d_{1})\mathrm{X}_{2}\mathcal{V}(g-d_{1})\\ +\sum\limits_{s=g-d_{2}}^{g-1}\rho^{d_{2}-1}\mathcal{V}^{\mathrm{T}}(g-d_{1}) \mathrm{X}_{3}\mathcal{V}(g-d_{1})\}\end{array}  \tag{32}\]  
Through the conditions \(0<\rho<1\), we obtain  
\[ \begin{array}{l}\mathrm{E}\{\Delta\mathbb{V}_{2}(\mathcal{V}(g),\sigma(g),p )\}\\ =\mathrm{E}\Big{\{}\mathcal{V}^{\mathrm{T}}(g)\Big{[}\sum\limits_{a=1}^{3} \mathrm{X}_{a}\Big{]}\mathcal{V}(g)\,-\rho^{d_{1}}\mathcal{V}^{\mathrm{T}}(g-d (g))\mathrm{X}_{1}\mathcal{V}(g-d(g))-\rho^{d_{2}}\mathcal{V}^{\mathrm{T}}(g-d _{1})\mathrm{X}_{2}\mathcal{V}(g-d_{1})\\ -\rho^{d_{2}}\mathcal{V}^{\mathrm{T}}(g-d_{2})\mathrm{X}_{3}\mathcal{V}(g-d_{2} )+\sum\limits_{s=g+1-d_{2}}^{g-1}\rho^{g-s}\mathcal{V}^{\mathrm{T}}(s)\mathrm{ X}_{2}\mathcal{V}(s)+(\rho-1)\mathbb{V}_{2}(\mathcal{V}(g),\sigma(g),p) \Bigg{\}}\end{array}  \tag{33}\]\[ \begin{split}&\mathrm{E}\{\Delta\mathbb{V}_{3}(\mathcal{V}(g),\sigma(g),p) \}\\ &=\mathrm{E}\{\mathbb{V}_{3}(\mathcal{V}(g+1),\sigma(g+1),p)- \mathbb{V}_{3}(\mathcal{V}(g),\sigma(g),p)\}\\ &=\mathrm{E}\{\sum\limits_{k=-d_{2}+1}^{-d_{1}}\sum\limits_{s= g+1+h}^{g+1-1}\rho^{g+1-s-1}\mathcal{V}^{\mathrm{T}}(s)X_{2}\mathcal{V}(s)+(d_{2}-d_{1}) \sum\limits_{h=-d_{2}}^{-d_{1}-1}\sum\limits_{s=g+1+h}^{g+1-s-1}\rho^{g+1-s-1 }\mathcal{V}^{\mathrm{T}}(s)X_{2}\mathcal{V}(s)\\ &-\left(\sum\limits_{h=-d_{2}+1}^{-d_{1}}\sum\limits_{s=g+h}^{g- 1}\rho^{g-s-1}\mathcal{V}^{\mathrm{T}}(s)X_{2}\mathcal{V}(s)+(d_{2}-d_{1}) \sum\limits_{h=-d_{2}}^{-d_{1}-1}\sum\limits_{s=g+h}^{g-1}\rho^{g-s-1}\mathcal{ V}^{\mathrm{T}}(s)X_{2}\mathcal{V}(s)\right)\}\end{split}  \tag{34}\]  
By dislocation subtraction, we finally obtain  
\[