update notations

content/CSE510/CSE510_L14.md

@@ -27,7 +27,7 @@ $\theta_{new}=\theta_{old}+d$
 First order Taylor expansion for the loss and second order for the KL:
 
 $$
-\approx \arg\max_{d} J(\theta_{old})+\nabla_\theta J(\theta)\mid_{\theta=\theta_{old}}d-\frac{1}{2}\lambda(d^T\nabla_\theta^2 D_{KL}\left[\pi_{\theta_{old}}||\pi_{\theta}\right]\mid_{\theta=\theta_{old}}d)+\lambda \delta
+\approx \arg\max_{d} J(\theta_{old})+\nabla_\theta J(\theta)\mid_{\theta=\theta_{old}}d-\frac{1}{2}\lambda(d^\top\nabla_\theta^2 D_{KL}\left[\pi_{\theta_{old}}||\pi_{\theta}\right]\mid_{\theta=\theta_{old}}d)+\lambda \delta
 $$
 
 If you are really interested, try to fill the solving the KL Constrained Problem section.
@@ -38,7 +38,7 @@ Setting the gradient to zero:
 
 $$
 \begin{aligned}
-0&=\frac{\partial}{\partial d}\left(-\nabla_\theta J(\theta)\mid_{\theta=\theta_{old}}d+\frac{1}{2}\lambda(d^T F(\theta_{old})d\right)\\
+0&=\frac{\partial}{\partial d}\left(-\nabla_\theta J(\theta)\mid_{\theta=\theta_{old}}d+\frac{1}{2}\lambda(d^\top F(\theta_{old})d\right)\\
 &=-\nabla_\theta J(\theta)\mid_{\theta=\theta_{old}}+\frac{1}{2}\lambda F(\theta_{old})d
 \end{aligned}
 $$
@@ -58,15 +58,15 @@ $$
 $$
 
 $$
-D_{KL}(\pi_{\theta_{old}}||\pi_{\theta})\approx \frac{1}{2}(\theta-\theta_{old})^T F(\theta_{old})(\theta-\theta_{old})
+D_{KL}(\pi_{\theta_{old}}||\pi_{\theta})\approx \frac{1}{2}(\theta-\theta_{old})^\top F(\theta_{old})(\theta-\theta_{old})
 $$
 
 $$
-\frac{1}{2}(\alpha g_N)^T F(\alpha g_N)=\delta
+\frac{1}{2}(\alpha g_N)^\top F(\alpha g_N)=\delta
 $$
 
 $$
-\alpha=\sqrt{\frac{2\delta}{g_N^T F g_N}}
+\alpha=\sqrt{\frac{2\delta}{g_N^\top F g_N}}
 $$
 
 However, due to the quadratic approximation, the KL constrains may be violated.