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→马尔科夫链的EI
<math>
<math>
EI = I(X_t,X_{t+1}|do(X_t)\sim U(\mathcal{X}))=I(\tilde{X}_t,\tilde{X}_{t+1}) \\
\begin{aligned}
= \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)}\\
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}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}\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)}\\
= \frac{1}{N}\sum^N_{i=1}\sum^N_{j=1}p_{ij}\log\frac{N\cdot p_{ij}}{\sum_{k=1}^N p_{kj}}\\
&= \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)}\\
&= \frac{1}{N}\sum^N_{i=1}\sum^N_{j=1}p_{ij}\log\frac{N\cdot p_{ij}}{\sum_{k=1}^N p_{kj}}
\end{aligned}
</math>
</math>