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Bayes Error Rate For the general case of K different classes, the probability of classifing x instance correctly is: \begin{equation} \label{eq1} \begin{split} P(correct) & = \sum_{i=1}^{K} p(x \in H_i, C_i) \\ & = \sum_{i=1}^{K} \int_{x \in H_i} p(x,C_i) \, dx\\ & = \sum_{i=1}^{K} \int_{x \in H_i} P(C_i|x)p(x)\,dx \\ \end{split} \end{equation} ...


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Your probability hasn't been normalized! In this case, you are computing the probability of being good, given that the other features have a fixed value. To obtain the correct probability, you need to normalize (divide) the value from your calculation by the probability that the features have taken on those fixed values. You can calculate this as follows: ...


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Position Detection In a traditional data acquisition and control scenario, with some assumptions, the relation between sensors signals $s_i$, emitters drive $\epsilon_j$, distances $x_{ij}$, and calibration factors is modelled as follows. $$ \forall \, (i, j) \text{,} \quad \frac {s_i} {v_i} = \frac {\epsilon_j} {v_j \, x_{ij}^2} $$ The assumptions include ...


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Model input: 1 mean scaled input for each emitter 1 distance value for each distance Multiple input You mentioned there is noise. If the noise is constant, ie you test it in place A and the values returned are always the same, then it means training in different places. If you place it in a place and the first reading is different from the second reading....


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Bernoulli naïve Bayes $P(x|c_k) = \prod^{n}_{i=1} p^{x_i}_{ki} (1-p_{ki})^{(1-x_i)}$ Let's examine the example of document classification. Let K different text classes and n different terms that our vocabulary contains. $x_i$ are boolean variables (0, 1) expressing if the $i^{th}$ term exists in document x. x is a vector of dimension n. $P(x|c_k)$ ...


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This is a bit of a puzzle but you can compute a reasonable narrow limit even without knowing whether or not $P(S,A) = P(S) P(A)$. Start with the contingency table relating $P(S, A)$, $P(S,\neg A)$, $P(\neg S, A)$, $P(\neg S,\neg A)$ to $P(S)$ and $P(A)$ : $$\begin{array}{cc|c} P( S,A)& P(\neg S,A) & P(A) \\ P(S,\neg A)& P(\neg S,\neg A) & P(...


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