Talk:AdaBoost: Difference between revisions
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I'm not sure with the way the weight α is computed. In Viola and Jones paper, they define α as 0.5*log(1-ε/ε) and not ln. If this is ln, α is negative, which is not possible. |
I'm not sure with the way the weight α is computed. In Viola and Jones paper, they define α as 0.5*log(1-ε/ε) and not ln. If this is ln, α is negative, which is not possible. |
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:First, ''log'' is not necessarily logarithm with base 10, often it is the same as ''ln''. Second, if it is, then log10(x) = ln(x)/ln(10), so log10(x) and ln(x) are either both positive or both negative. Since it is postulated that ε<0.5, then α is always positive. Finally, (1-ε)/ε ≠ 1-ε/ε. [[User:Victordk13|Victordk13]] ([[User talk:Victordk13|talk]]) 10:19, 7 July 2008 (UTC) |
:First, ''log'' is not necessarily logarithm with base 10, often it is the same as ''ln''. Second, if it is, then log10(x) = ln(x)/ln(10), so log10(x) and ln(x) are either both positive or both negative. Since it is postulated that ε<0.5, then α is always positive. Finally, (1-ε)/ε ≠ 1-ε/ε. [[User:Victordk13|Victordk13]] ([[User talk:Victordk13|talk]]) 10:19, 7 July 2008 (UTC) |
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What is a weak classifier? |
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Revision as of 06:38, 27 March 2009
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I'm not sure with the way the weight α is computed. In Viola and Jones paper, they define α as 0.5*log(1-ε/ε) and not ln. If this is ln, α is negative, which is not possible.
- First, log is not necessarily logarithm with base 10, often it is the same as ln. Second, if it is, then log10(x) = ln(x)/ln(10), so log10(x) and ln(x) are either both positive or both negative. Since it is postulated that ε<0.5, then α is always positive. Finally, (1-ε)/ε ≠ 1-ε/ε. Victordk13 (talk) 10:19, 7 July 2008 (UTC)
What is a weak classifier?