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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 \sum [(y^{(i)})\log(g(\theta^t x^{(i)})) + (1-y^{(i)})\log(1-g(\theta^t x^{(i)}))] | |||
\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)})(1-g(\theta^t x^{(i)}))x^{(i)} - (1-y^{(i)})g(\theta^t x^{(i)})x^{(i)}] | |||
= -\sum [(y^{(i)})x^{(i)} -(y^{(i)}) g(\theta^t x^{(i)}))x^{(i)} - g(\theta^t x^{(i)})x^{(i)} + y^{(i)}g(\theta^t x^{(i)})x^{(i)}] | |||
= -\sum [(y^{(i)})x^{(i)} - g(\theta^t x^{(i)})x^{(i)}] | |||
</math><br> | |||
\implies \nabla^2_\theta J(\theta) | <math> | ||
\implies \nabla^2_\theta J(\theta) = \nabla_\theta -\sum [(y^{(i)})x^{(i)} - g(\theta^t x^{(i)})x^{(i)}] | |||
= \sum_i g(\theta^t x^{(i)})(1-g(\theta^t x^{(i)})) x^{(i)} (x^{(i)})^T | |||
</math><br> | </math><br> | ||
which is a PSD matrix | which is a PSD matrix |