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<math>\Pr(P_i|X,{\mathbf r}) \propto \Pr({\mathbf r}|P_i,X)\Pr(P_i|X)</math>
We express <math>\Pr({\mathbf r}|P_i,X)</math> as <math>p^{d_i} (1-p)^{n_ib-d_i} 2^{-(N-n_i)b}</math> (by using the binary symmetric channel likelihood for the first <math>n_ib</math> bits followed by a uniform prior over the remaining bits). We express <math>\Pr(P_i|X)</math> in terms of the number of branch choices one has made, <math>n_i</math>, and the number of branches from each node, <math>2^{Rb}</math>. Therefore:
<math>\Pr(P_i|X,{\mathbf r}) \propto p^{d_i} (1-p)^{n_ib-d_i} 2^{-(N-n_i)b} 2^{-n_iRb}\propto p^{d_i} (1-p)^{n_ib-d_i} 2^{n_ib} 2^{-n_iRb}</math>
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