请证明$H(\Theta \given \Theta^t)$满足以下性质: \begin{align*} \Theta^t = \operatorname{\arg \max}_{\Theta} H(\Theta \given \Theta^t). \end{align*}

根据条件熵的定义，我们有：
\begin{align*}
H(\Theta \given \Theta^t) = -\sum_{\theta} \sum_{\theta^t} p(\theta, \theta^t) \log \frac{p(\theta \given \theta^t)}{p(\theta)}
\end{align*}
其中，$p(\theta)$是先验概率分布，$p(\theta \given \theta^t)$是后验概率分布。我们可以对式子做一些变形：
\begin{align*}
H(\Theta \given \Theta^t) &= -\sum_{\theta} \sum_{\theta^t} p(\theta, \theta^t) \log \frac{p(\theta \given \theta^t)}{p(\theta)} \\
&= -\sum_{\theta} \sum_{\theta^t} p(\theta, \theta^t) \log p(\theta \given \theta^t) + \sum_{\theta} \sum_{\theta^t} p(\theta, \theta^t) \log p(\theta) \\
&= -\sum_{\theta} \sum_{\theta^t} p(\theta, \theta^t) \log p(\theta \given \theta^t) + \sum_{\theta} p(\theta) \log p(\theta) \sum_{\theta^t} p(\theta^t) \\
&= -\sum_{\theta} p(\theta) \log p(\theta) \sum_{\theta^t} p(\theta^t) + \sum_{\theta} \sum_{\theta^t} p(\theta, \theta^t) \log p(\theta^t \given \theta) \\
&= -\sum_{\theta} p(\theta) \log p(\theta) + \sum_{\theta} \sum_{\theta^t} p(\theta^t \given \theta) p(\theta) \log p(\theta^t \given \theta) \\
&= -\sum_{\theta} p(\theta) \log p(\theta) - \sum_{\theta} p(\theta) \sum_{\theta^t} p(\theta^t \given \theta) \log \frac{1}{p(\theta^t \given \theta)} \\
&= H(\Theta) - \sum_{\theta} p(\theta) H(\Theta^t \given \Theta = \theta)
\end{align*}
其中，$H(\Theta)$是先验分布的熵，$H(\Theta^t \given \Theta = \theta)$是给定先验分布$\Theta$时后验分布$\Theta^t$的条件熵。因此，我们可以得到：
\begin{align*}
\operatorname{\arg \max}_{\Theta} H(\Theta \given \Theta^t) &= \operatorname{\arg \max}_{\Theta} \left\{ H(\Theta) - \sum_{\theta} p(\theta) H(\Theta^t \given \Theta = \theta) \right\} \\
&= \operatorname{\arg \min}_{\Theta} \sum_{\theta} p(\theta) H(\Theta^t \given \Theta = \theta) \\
&= \Theta^t
\end{align*}
因此，$H(\Theta \given \Theta^t)$满足要求的性质。