- Get link
- Google+
- Other Apps

Competitions are scored using a variety of functions, and the most common for binary classification tasks with confidence is something called log-loss, which is essentially \(\sum_{i=1}^{n} p_i\cdot\log(p_i)\), where \(p_i\) is your model's claimed confidence for test data point \(i\)'s correct label. Why does Kaggle use this scoring function? Here I'll follow Terry Tao's argument.

Ideally what we'd like is a scoring function \(f(x)\) that yields the maximum expected score precisely when the claimed confidence \(x_i\) in the correct label for \(i\) is actually what the submitter believes is the true probability (or frequency) of that outcome. This means that we want \[L(x)=p\cdot f(x) + (1-p)\cdot f(1-x)\] for fixed \(p\) to be maximized when \(x=p\). Differentiating, this means \[L'(x) = p\cdot f'(x) - (1-p)\cdot f'(1-x) = 0\] when \(x=p\), hence \(p\cdot f'(p) = (1-p)\cdot f'(1-p)\) for all \(p\). This will be satisfied by any admissible \(f(x)\) with \(x\cdot f'(x)\) symmetric around \(x=\frac{1}{2}\), but if we extend our analysis to multinomial outcomes we get the stronger conclusion that in fact \(x\cdot f'(x) = c_0\) for some constant \(c_0\). This in turn implies \(f(x)=c_0\cdot \log(x)+c_1\). If we want \(f(1/2)=0\) and \(f(1)=1\), we end up with \(f(x)={\log}_2(2x)\) and the expected score is \[L(x)=x\cdot {\log}_2(2x) + (1-x)\cdot {\log}_2(2(1-x)).\]

Mantab mas/mbak artikelnya (y)

ReplyDeletejangan lupa kunjungi: http://pdaagar.com/

terimakasih..

E-commerce Solutions:

ReplyDelete- Graphic Design Services

- Shopping Cart Software

- Secure eCommerce Hosting

- Expert Consultation

- Custom Solutions

ReplyDeleteشركة كشف تسربات المياه بالاحساء

شركة عزل اسطح بالاحساء

شركة كشف تسربات المياه بالاحساء

شركة كشف تسربات المياه بالقطيف

شركة عزل اسطح بالقطيف

شركة كشف تسربات المياه بالقطيف