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| technical:finite-entropy-estimate [2022/11/27 04:29] – [Number of states with given number of repeats] chunchung | technical:finite-entropy-estimate [2022/11/27 05:36] (current) – [Distribution] chunchung | ||
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| Second is the number of ways for assigning the orders in the sequence to the occurrences of the states, | 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} | \begin{equation} \frac{N!}{\prod_n (n!)^{k_n}}. \end{equation} | ||
| - | Overall, the likelihood of getting the combination $\{k_n| n \in \mathbb{N}_0 \}$ ($\mathbb{N}_0$ is the set of non-negative | + | Overall, the likelihood of getting the combination $\{k_n| n \in \mathbb{N} \}$ ($\mathbb{N}$ is the set of positive |
| \begin{equation} \frac{\Gamma! N!}{\Gamma^N \prod_n k_n! (n!)^{k_n}}. \end{equation} | \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 $k_0 = \Gamma - \sum_{n\in\mathbb{N}}k_n$. | In the expression, the number of unobserved states is $k_0 = \Gamma - \sum_{n\in\mathbb{N}}k_n$. | ||
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| & \approx \frac{\lambda^n}{n!}e^{-\lambda} | & \approx \frac{\lambda^n}{n!}e^{-\lambda} | ||
| \end{align} | \end{align} | ||
| - | approaching the Poisson distribution with $\lambda \equiv N/\Gamma$. | + | approaching the Poisson distribution with $\lambda \equiv N/ |