提交 4dea6021 编写于 作者: X xiaowei_xing

test

上级 3d438f47
......@@ -148,5 +148,9 @@ $$
最后的等式中,所有轨迹的概率和为 1。在倒数第二个等式中,由于 $b$ 为常值,我们可以将提出至积分外面(例如,$b$ 可以为平均回报,$b = \frac{1}{N}\sum_{i=1}^{N}r(\tau)$)。即使 $b$ 是状态 $s$ 的函数,这一项也是无偏的:
$$
\mathbb{E}_ {\tau\sim\pi_{\theta}(\tau)} [\nabla_{\theta}\log \pi_{\theta}(\tau) b(s_t)] = \mathbb{E}_ {s_{0:t},a_{0:(t-1)}} [\mathbb{E}_ {s_{(t+1):T},a_{t:(T-1)}} [\nabla_{\theta}\log \pi_{\theta}(a_t|s_t)b(s_t)]]
\mathbb{E}_ {\tau\sim\pi_{\theta}(\tau)} [\nabla_{\theta}\log \pi_{\theta}(\tau) b(s_t)]
$$
$$
= \mathbb{E}_ {s_{0:t},a_{0:(t-1)}} [\mathbb{E}_ {s_{(t+1):T},a_{t:(T-1)}} [\nabla_{\theta}\log \pi_{\theta}(a_t|s_t)b(s_t)]]
$$
\ No newline at end of file
Markdown is supported
0% .
You are about to add 0 people to the discussion. Proceed with caution.
先完成此消息的编辑!
想要评论请 注册