we can show that $H(\Delta | \{\Psi_i\})$ is $o(n)$, then it will follow that the entropy rate of the multiset, $H(\mathfrak{X})$, is equal to the entropy rate of the pattern of the multiset $$H(\mbox{\Neptune}) = \lim_{n\to\infty}\tfrac{1}{n} H(\{\Psi_i\}_{i=1}^n) \mbox{.}$$

Noting the fact that the dictionary $\Delta$ is independent of the order of the pattern, $J(\Psi)$, we establish that $$\begin{aligned}
\lim_{n\to\infty}\tfrac{1}{n} H(\Delta|(\Psi_i)_{i=1}^n) &= \lim_{n\to\infty}\tfrac{1}{n} H(\Delta | J((\Psi_i)_{i=1}^n),\{\Psi_i\}_{i=1}^n) \\                            
&= \lim_{n\to\infty}\tfrac{1}{n} H(\Delta \mid \{\Psi_i\}_{i=1}^n) \mbox{,}\end{aligned}$$ where the first step follows since the order and type of pattern determine the pattern, and the second step is due to independence. The result [@OrlitskySVZ2006 Theorem 9] shows that for all finite-entropy, discrete stationary processes, the asymptotic per-letter values of $H((\Psi_i)))$ and $H((X_i))$ are equal. Hence, $$\lim_{n\to\infty}\tfrac{1}{n} H(\Delta \mid (\Psi_i)_{i=1}^n) = 0 \mbox{,}$$ and so $$\lim_{n\to\infty}\tfrac{1}{n} H(\Delta \mid \{\Psi_i\}_{i=1}^n) = 0 \mbox{.}$$ This implies that $H(\Delta | \{\Psi_i\})$ is $o(n)$ and yields the following theorem.

\[thm:coincide\] The entropy rate of a multiset generated by a discrete finite-entropy stationary process and the entropy rate of its pattern coincide: $$H(\mathfrak{X}) = H(\mbox{\Neptune}) \mbox{.}$$

Computing the entropy rate of the multiset or equivalently of the pattern of the multiset can be difficult. See [@Shamir2007] and references therein for a discussion 