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| technical:finite-entropy-estimate [2022/11/19 04:24] – chunchung | technical:finite-entropy-estimate [2022/11/27 05:36] (current) – [Distribution] chunchung | ||
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| The objective is to estimate $\Gamma$ by sampling the space $\mathbb{X}$. The only information we have is whether a sampled point is the same as the other: $\delta_{x_i, | The objective is to estimate $\Gamma$ by sampling the space $\mathbb{X}$. The only information we have is whether a sampled point is the same as the other: $\delta_{x_i, | ||
| - | The consideration of Ma is that the number of repeats or the number of coincident pairs in the sequence of length $N$ is dependent on the space size $\Gamma$. If the sequence has zero correlation time and the element are drawn independently from the space, the chance of finding a pair to be coincident ($x_i=x_j$) is estimated by $P_\mathrm{c} = \Gamma^{-1}$. Since there are $N(N-1)/2$ pairs in a sequence of length $N$, the expected number of coincident pairs is estimated | + | The consideration of Ma is that the number of repeats or the number of coincident pairs in the sequence of length $N$ is dependent on the space size $\Gamma$. If the sequence has zero correlation time and the element are drawn independently from the space, the chance of finding a pair to be coincident ($x_i=x_j$) is estimated by $P_\mathrm{c} = \Gamma^{-1}$. Since there are $N(N-1)/2$ pairs in a sequence of length $N$, the expected number of coincident pairs is given by |
| \begin{equation} | \begin{equation} | ||
| - | N_\mathrm{c} = \frac{N(N-1)}{2\Gamma} | + | N_\mathrm{c} = \frac{N(N-1)}{2\Gamma}. |
| \end{equation} | \end{equation} | ||
| + | =====Number of states with given number of repeats===== | ||
| + | From the perspective of states, we assume that there are $k_n$ states out of $\Gamma$ that occur $n$ times in the sequence of size $N$. The total number of states is given by | ||
| + | \begin{equation} \Gamma = \sum_n k_n \end{equation} | ||
| + | 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. | ||