This gives a mixed source where all type classes are equiprobable, as shown now.

Let the realizations of $\{X_i\}_{i=1}^n$ be expressed as $z \in \{0,\ldots,n\}$, the number of ones. $$\Pr[\{X_i\}_{i=1}^n = z] = \int_{0}^{1} q_{\Theta}(\theta) \Pr[\{X_i\}_{i=1}^n = z \mid \Theta = \theta] \, d\theta \mbox{,}$$ where $q_{\Theta}(\theta)$ simply equals one over the range of integration, and $\Pr[\{X_i\}_{i=1}^n = z \mid \Theta = \theta]$ is given by a binomial distribution: $$\Pr[\{X_i\}_{i=1}^n = z \mid \Theta = \theta] = \binom{n}{z}\theta^z (1-\theta)^{n-z} \mbox{.}$$ So, $$\Pr[\{X_i\}_{i=1}^n = z] = \int_{0}^{1} \binom{n}{z}\theta^z (1-\theta)^{n-z} \, d\theta = \binom{n}{z} B(z+1,n-z+1) = \frac{1}{1+n} \mbox{,}$$ where the beta function $B(\cdot,\cdot)$ has been used. Thus, the result is that the type classes are equiprobable.

Since the types are equiprobable, the entropy $H(\{X_i\}_{i=1}^n)$ is just the logarithm of the number of the types: $$H(\{X_i\}_{i=1}^n) = \log \binom{n+|\mathcal{X}|-1}{|\mathcal{X}|-1} = \log(n+1) \mbox{.}$$ The entropy conditioned on $\Theta = \theta$ is simply the entropy of a binomial random variable [@JacquetS1999] and so the conditional entropy is $$\begin{aligned}
H(\{X_i\}_{i=1}^n \mid \Theta)
 &= \int H(\{X_i\}_{i=1}^n \mid \theta) q_{\Theta}(\theta) \, d\theta
  = \int \left[\tfrac{1}{2}\log\left(2\pi e n \theta(1-\theta)\right) + {\ts\sum_{k\ge 1}a_k n^{-k}}\right]q_{\Theta}(\theta) \, d\theta \\
 &= \tfrac{1}{2}\log n + \sum_{k\ge 1}n^{-k} \int a_k q_{\Theta}(\theta) \, d\theta  + \int 