668
个编辑
更改
→马尔科夫链的EI
<math>
<math>
\begin{aligned}
\begin{aligned}
EI &= I(X_t,X_{t+1}|do(X_t)\sim U(\mathcal{X}))=I(\tilde{X}_t,\tilde{X}_{t+1}) \\
EI &= I(X_t,X_{t+1}|do(X_t\sim U(\mathcal{X})))=I(\tilde{X}_t,\tilde{X}_{t+1}) \\
&= \sum^N_{i=1}\sum^N_{j=1}Pr(\tilde{X}_t=i,\tilde{X}_{t+1}=j)\log \frac{Pr(\tilde{X}_t=i,\tilde{X}_{t+1}=j)}{Pr(\tilde{X}_t=i)Pr(\tilde{X}_{t+1}=j)}\\
&= \sum^N_{i=1}\sum^N_{j=1}Pr(\tilde{X}_t=i,\tilde{X}_{t+1}=j)\log \frac{Pr(\tilde{X}_t=i,\tilde{X}_{t+1}=j)}{Pr(\tilde{X}_t=i)Pr(\tilde{X}_{t+1}=j)}\\
&= \sum^N_{i=1}Pr(\tilde{X}_t=i)\sum^N_{j=1}Pr(\tilde{X}_{t+1}=j|\tilde{X}_t=i)\log \frac{Pr(\tilde{X}_{t+1}=j|\tilde{X}_t=i)}{Pr(\tilde{X}_{t+1}=j)}\\
&= \sum^N_{i=1}Pr(\tilde{X}_t=i)\sum^N_{j=1}Pr(\tilde{X}_{t+1}=j|\tilde{X}_t=i)\log \frac{Pr(\tilde{X}_{t+1}=j|\tilde{X}_t=i)}{Pr(\tilde{X}_{t+1}=j)}\\