提交 193d3d93 编写于 作者: X xiaowei_xing

test

上级 16f9377e
......@@ -251,4 +251,19 @@ $$
\nabla_{\theta} \mathbb{E}_ {\pi_{\theta}}[r_{t'}] = \mathbb{E}_ {\pi_{\theta}}[r_{t'}\sum_{t=0}^{t'} \nabla_{\theta} \log \pi_{\theta}(a_t|s_t)]。
$$
由于 $\sum_{t'=t}^{T-1}r_{t'}^{(i)}$ 就是回报 $G_t^{(i)}$,
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由于 $\sum_{t'=t}^{T-1}r_{t'}^{(i)}$ 就是回报 $G_t^{(i)}$,我们可以将其随时间步累加,从而得到:
$$
\nabla_{\theta} V(\theta) = \nabla_{\theta} \mathbb{E}_ {\tau \sim \pi_{\theta}}[R(\tau)] = \mathbb{E}_ {\pi_{\theta}}[\sum_{t'=0}^{T-1}r_{t'}\sum_{t=0}^{t'} \nabla_{\theta} \log \pi_{\theta}(a_t|s_t)]
\tag{12}
$$
$$
= \mathbb{E}_ {\pi_{\theta}}[\sum_{t'=0}^{T-1} \nabla_{\theta} \log \pi_{\theta}(a_t|s_t) \sum_{t'=t}^{T-1}r_{t'}]
\tag{13}
$$
$$
= \mathbb{E}_ {\pi_{\theta}}[\sum_{t'=0}^{T-1} G_t \nabla_{\theta} \log \pi_{\theta}(a_t|s_t)]。
\tag{14}
$$
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