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Machine Learning: Difference between revisions
→Cross Entropy
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We show that the Hessian is positive semi definite.<br> | We show that the Hessian is positive semi definite.<br> | ||
<math> | <math> | ||
\nabla_\theta J(\theta) | \nabla_\theta J(\theta) = -\nabla_\theta \sum [(y^{(i)})\log(g(\theta^t x^{(i)})) + (1-y^{(i)})\log(1-g(\theta^t x^{(i)}))] | ||
= -\sum [(y^{(i)})\frac{g(\theta^t x^{(i)})(1-g(\theta^t x^{(i)}))}{g(\theta^t x^{(i)})}x^{(i)} + (1-y^{(i)})\frac{-g(\theta^t x^{(i)})(1-g(\theta^t x^{(i)}))}{1-g(\theta^t x^{(i)})}x^{(i)}] | = -\sum [(y^{(i)})\frac{g(\theta^t x^{(i)})(1-g(\theta^t x^{(i)}))}{g(\theta^t x^{(i)})}x^{(i)} + (1-y^{(i)})\frac{-g(\theta^t x^{(i)})(1-g(\theta^t x^{(i)}))}{1-g(\theta^t x^{(i)})}x^{(i)}] | ||
= -\sum [(y^{(i)})(1-g(\theta^t x^{(i)}))x^{(i)} - (1-y^{(i)})g(\theta^t x^{(i)})x^{(i)}] | = -\sum [(y^{(i)})(1-g(\theta^t x^{(i)}))x^{(i)} - (1-y^{(i)})g(\theta^t x^{(i)})x^{(i)}] | ||