vault backup: 2025-03-16 18:59:42
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@@ -55,13 +55,17 @@
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# Modified Probability Estimates
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- Consider attribute *outlook* for class *yes*
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# $\frac{2+\frac{1}{3}\mu}{9+\mu}$
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Sunny
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# $\frac{4+\frac{1}{3}\mu}{9+\mu}$
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Overcast
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# $\frac{3+\frac{1}{3}\mu}{9+\mu}$
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Rainy
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- Each value treated the same way
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@@ -73,13 +77,16 @@ Rainy
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## Fully Bayesian Formulation
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# $\frac{2+\frac{1}{3}\mu p_1}{9+\mu}$
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Sunny
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# $\frac{4+\frac{1}{3}\mu p_2}{9+\mu}$
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Overcast
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# $\frac{3+\frac{1}{3}\mu p_3}{9+\mu}$
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Rainy
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- Where $p_1 + p_2 + p_3 = 1$
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- $p_1, p_2, p_3$ are prior probabilities of outlook being sunny, overcast or rainy before seeing the training set. However, in practice it is not clear how these prior probabilities should be assigned.
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- Where $p_1 + p_2 + p_3 = 1$
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- $p_1, p_2, p_3$ are prior probabilities of outlook being sunny, overcast or rainy before seeing the training set. However, in practice it is not clear how these prior probabilities should be assigned.
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@@ -27,20 +27,25 @@
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| High | 2/5 | 1/9 | Red | 3/5 | 2/9 | | | | | | | | |
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# Problem 1
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# $Pr[Diagnosis=N|E] = \frac{2}{5} \times \frac{2}{5} \times \frac{4}{5} \times \frac{3}{5} \times \frac{5}{14} = 0.027428571$
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# $Pr[Diagnosis = B|E] = \frac{3}{9} \times \frac{4}{9} \times \frac{3}{9} \times \frac{3}{9} \times \frac{9}{14} = 0.010582011$
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# $p(B) = \frac{0.0106}{0.0106+0.0274} = 0.2789$
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# $p(N) = \frac{0.0274}{0.0106+0.0274} = 0.7211$
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# $p(N) = \frac{0.0274}{0.0106+0.0274} = 0.7211$
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Diagnosis N is much more likely than Diagnosis B
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# Problem 2
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# $Pr[Diagnosis = N|E] = \frac{2}{5} \times \frac{1}{5} \times \frac{3}{5} \times \frac{5}{14} = 0.0171$
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# $Pr[Diagnosis = B|E] = \frac{3}{9} \times \frac{6}{9} \times \frac{3}{9} \times \frac{9}{14} = 0.0476$
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# $p(N) = \frac{0.0171}{0.0171+0.0476} = 0.2643$
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# $p(B) = \frac{0.0474}{0.0476+0.0171} = 0.7357$
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Diagnosis B is much more likely than Diagnosis N
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@@ -48,4 +53,5 @@ Diagnosis B is much more likely than Diagnosis N
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# Problem 3
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# $Pr[Diagnosis = N|E] = \frac{0}{5} \times \frac{2}{5} \times \frac{4}{5} \times \frac{3}{5} \times \frac{5}{14} = 0$
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# $Pr[Diagnosis = B|E] = \frac{5}{9} \times \frac{4}{9} \times \frac{3}{9} \times \frac{3}{9} \times \frac{9}{14} = 0.018$
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@@ -274,4 +274,4 @@ Root mean squared error 0.3223
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Relative absolute error 70.1487 %
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Root relative squared error 68.0965 %
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Total Number of Instances 3
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```
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```
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