# 马尔科夫链删除节点的性质

$\displaystyle{ \sum_{j=1}^{N}m_{ij}\lt 1, i\in [0,N] }$

$\displaystyle{ U=I+M+M^2+\cdot\cdot\cdot=(I-M)^{-1} }$

$\displaystyle{ M=M_{-d}+\Delta M }$ 模板:Align

$\displaystyle{ \Delta M=\left\{\begin{array}{ll} m_{i,j} & \mbox {if } i=d, \\ 0 & \mbox {if } i \neq d\end{array}\right. }$

$\displaystyle{ U_{-d}=I+M_{-d}+M^2_{-d}+\cdot\cdot\cdot=(I-M_{-d})^{-1} }$

### 性质1

$\displaystyle{ MU=U-I }$

$\displaystyle{ I=(I-M)U=U-MU }$

$\displaystyle{ \sum_k m_{ik}u_{kj}=u_{ij}-\delta_{ij} }$

$\displaystyle{ \delta_{ij}=\left\{\begin{array}{ll} 1 & \mbox {if } i=j, \\ 0 & \mbox {if } i \neq j\end{array}\right. }$

### 性质2

$\displaystyle{ (I-M)U=(I-M_{-d})U_{-d}=I }$

### 定理1

$\displaystyle{ (U_{-d})_{ij}=u_{ij}-(U_{-d})_{id}u_{dj}=u_{ij}-\frac{u_{id}}{u_{dd}}u_{dj}-\frac{u_{id}}{u_{dd}}\delta_{dj} }$

$\displaystyle{ U-MU=U-(M_{-d}+\Delta M)U=U_{-d}-M_{-d}U_{-d} }$

$\displaystyle{ U-U_{-d}=U_{-d}\Delta M U }$

$\displaystyle{ U_{-d}(\Delta M U)_{ij}=(U_{-d})_{id}\sum_k m_{dk}u_{kj}=(U_{-d})_{id}(u_{dj}-\delta_{dj}) }$

$\displaystyle{ u_{ij}-(U_{-d})_{ij}=(U_{-d})_{id}(u_{dj}-\delta_{dj}) }$ 模板:Align

$\displaystyle{ u_{id}-(U_{-d})_{id}=(U_{-d})_{id}(u_{dd}-1) }$

$\displaystyle{ (U_{-d})_{id}=\frac{u_{id}}{u_{dd}} }$

$\displaystyle{ u_{ij}-(U_{-d})_{ij}=\frac{u_{id}}{u_{dd}}(u_{dj}-\delta_{dj}) }$

$\displaystyle{ (U_{-d})_{ij}=u_{ij}-\frac{u_{id}}{u_{dd}}u_{dj}-\frac{u_{id}}{u_{dd}}\delta_{dj} }$

#### 推论1

$\displaystyle{ (U_{-j}^2)_{ij}=\frac{(U^2)_{ij}}{u_{jj}}-\frac{u_{ij}}{u_{jj}^2}(U^2)_{jj}+\frac{u_{ij}}{u_{jj}} }$

#### 推论2

$\displaystyle{ \frac{1}{u_{ij}}\left[(MU^2)_{ij}-u_{jj}(M_{-j}U^2_{-j})_{ij}\right]=\frac{(MU^2)_{jj}}{u_{jj}} }$