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大小无更改 、 2024年9月20日 (星期五)
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<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)}\\
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