Deep Learning3: Softmax Regression

Softmax就是依概率预测分类标签

训练集\{ (x^{(1)}, y^{(1)}), \ldots, (x^{(m)}, y^{(m)}) \} 标签 y^{(i)} \in \{0,1\}

\begin{align} h_\theta(x) = \frac{1}{1+\exp(-\theta^Tx)}, \end{align}

此时，cost function为

\begin{align} J(\theta) = -\frac{1}{m} \left[ \sum_{i=1}^m y^{(i)} \log h_\theta(x^{(i)}) + (1-y^{(i)}) \log (1-h_\theta(x^{(i)})) \right] \end{align}

若标签y^{(i)} \in \{1, 2, \ldots, k\}设定

\begin{align} h_\theta(x^{(i)}) = \begin{bmatrix} p(y^{(i)} = 1 | x^{(i)}; \theta) \\ p(y^{(i)} = 2 | x^{(i)}; \theta) \\ \vdots \\ p(y^{(i)} = k | x^{(i)}; \theta) \end{bmatrix} = \frac{1}{ \sum_{j=1}^{k}{e^{ \theta_j^T x^{(i)} }} } \begin{bmatrix} e^{ \theta_1^T x^{(i)} } \\ e^{ \theta_2^T x^{(i)} } \\ \vdots \\ e^{ \theta_k^T x^{(i)} } \\ \end{bmatrix} \end{align}

\theta_1, \theta_2, \ldots, \theta_k \in \Re^{n+1}

令

\theta = \begin{bmatrix} \mbox{---} \theta_1^T \mbox{---} \\ \mbox{---} \theta_2^T \mbox{---} \\ \vdots \\ \mbox{---} \theta_k^T \mbox{---} \\ \end{bmatrix}

1\{\cdot\} is the indicator function, so that 1{a true statement} = 1, and 1{a false statement} = 0

此时，cost function为

 

\begin{align} J(\theta) = - \frac{1}{m} \left[ \sum_{i=1}^{m} \sum_{j=1}^{k} 1\left\{y^{(i)} = j\right\} \log \frac{e^{\theta_j^T x^{(i)}}}{\sum_{l=1}^k e^{ \theta_l^T x^{(i)} }}\right] \end{align}

\begin{align} J(\theta) &= -\frac{1}{m} \left[ \sum_{i=1}^m (1-y^{(i)}) \log (1-h_\theta(x^{(i)})) + y^{(i)} \log h_\theta(x^{(i)}) \right] \\ &= - \frac{1}{m} \left[ \sum_{i=1}^{m} \sum_{j=0}^{1} 1\left\{y^{(i)} = j\right\} \log p(y^{(i)} = j | x^{(i)} ; \theta) \right] \end{align}

对于softmax来说，设定indicator function为

p(y^{(i)} = j | x^{(i)} ; \theta) = \frac{e^{\theta_j^T x^{(i)}}}{\sum_{l=1}^k e^{ \theta_l^T x^{(i)}} }

梯度如下

\begin{align} \nabla_{\theta_j} J(\theta) = - \frac{1}{m} \sum_{i=1}^{m}{ \left[ x^{(i)} \left( 1\{ y^{(i)} = j\} - p(y^{(i)} = j | x^{(i)}; \theta) \right) \right] } \end{align}

迭代如下

\theta_j := \theta_j - \alpha \nabla_{\theta_j} J(\theta)

cost function上加入weight decay term得

\begin{align} J(\theta) = - \frac{1}{m} \left[ \sum_{i=1}^{m} \sum_{j=1}^{k} 1\left\{y^{(i)} = j\right\} \log \frac{e^{\theta_j^T x^{(i)}}}{\sum_{l=1}^k e^{ \theta_l^T x^{(i)} }} \right] + \frac{\lambda}{2} \sum_{i=1}^k \sum_{j=0}^n \theta_{ij}^2 \end{align}

此时梯度为

\begin{align} \nabla_{\theta_j} J(\theta) = - \frac{1}{m} \sum_{i=1}^{m}{ \left[ x^{(i)} ( 1\{ y^{(i)} = j\} - p(y^{(i)} = j | x^{(i)}; \theta) ) \right] } + \lambda \theta_j \end{align}

 

posted on 2016-10-13 17:11  Beginnerpatienceless  阅读(157)  评论(0)    收藏  举报