Maximum Entropy Inverse Reinforcementの行間埋め

$$\begin{array}{r}L(p_i,\theta ,\lambda )=-\sum _{i=1}^M{p}_i\log p_i+{\theta }^T(\sum _{i=1}^M{p}_i{\mathbf{f}}_i-\mathbb{E}[\mathbf{f}])+{\mu }^T(\sum _{i=1}^M{p}_i-1)\end{array}$$

${\displaystyle \frac{\partial L}{\partial p_i}}=0$より、

$-\log p_i-1+{\theta }^T{\mathbf{f}}_i+\mu =0$

よって、

$p_i=\exp (-1+{\theta }^T{\mathbf{f}}_i+\mu )$

${\displaystyle \frac{\partial L}{\partial \mu }}=0$より、

$$\begin{array}{r}\sum _{i=1}^M{p}_i=1\end{array}$$

よって、

$$\begin{array}{rcl}& & \sum _{i=1}^M\exp (-1+{\theta }^T{\mathbf{f}}_i+\mu )=1\\ & & \exp (-1+\mu )\sum _{i=1}^M\exp \left({\theta }^T{\mathbf{f}}_i\right)=1\\ & & \exp (-1+\mu )={\displaystyle \frac{1}{\sum _{i=1}^M\exp \left({\theta }^T{\mathbf{f}}_i\right)}}\end{array}$$
よって、

$$\begin{array}{r}{p}_i={\displaystyle \frac{\exp \left({\theta }^T{\mathbf{f}}_i\right)}{\sum _{i=1}^M\exp \left({\theta }^T{\mathbf{f}}_i\right)}}\end{array}$$