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--- technical:finite-entropy-estimate [2022/11/19 04:47] – chunchung
+++ technical:finite-entropy-estimate [2022/11/27 05:36] (current) – [Distribution] chunchung
@@ Line 19: / Line 19: @@
 while the length of the sequence is by
 \begin{equation} N = \sum_n n k_n . \end{equation}
+Here, the value of $k_0$ is not directly known from the observed sequence and is linked to the total number of states $\Gamma$. Since each element of the sequence is drawn from the states, we have the total number of possible sequences
+\begin{equation} N_\mathrm{seq} = \Gamma^N. \end{equation}
+The number of of the sequences among $N_\mathrm{seq}$ that will give rise to a given combination of $\{k_n\}$ can be calculated from the product of two factors. First is the number of ways to pick the states with $n$ repeats for all $n$ values,
+\begin{equation} \frac{\Gamma !}{\prod_n k_n!}. \end{equation}
+Second is the number of ways for assigning the orders in the sequence to the occurrences of the states,
+\begin{equation} \frac{N!}{\prod_n (n!)^{k_n}}. \end{equation}
+Overall, the likelihood of getting the combination $\{k_n| n \in \mathbb{N} \}$ ($\mathbb{N}$ is the set of positive integers) is
+\begin{equation} \frac{\Gamma! N!}{\Gamma^N \prod_n k_n! (n!)^{k_n}}. \end{equation}
+In the expression, the number of unobserved states is $k_0 = \Gamma - \sum_{n\in\mathbb{N}}k_n$.
+=====Distribution=====
+Consider a given state, the probability that it gets hit by $n$ points is
+\begin{align}
+P_n & = C^N_n \Gamma^{-n}\left(1-\Gamma^{-1}\right)^{N-n} \\
+& = \frac{N!}{n!(N-n)!} \Gamma^{-n}\left(1-\Gamma^{-1}\right)^{N-n} \\
+& = \frac{1}{n!}\frac{N(N-1)\ldots(N-n+1)}{\Gamma^n}\left(1-\Gamma^{-1}\right)^{N-n} \\
+& \approx \frac{\lambda^n}{n!}(1-\frac{\lambda}{N})^N \\
+& \approx \frac{\lambda^n}{n!}e^{-\lambda}
+\end{align}
+approaching the Poisson distribution with $\lambda \equiv N/\Gamma$ at the large $N$, $\Gamma$ limit.