提交 8ab0c195 编写于 作者: X xiaowei_xing

test

上级 5dd190af
......@@ -270,7 +270,35 @@ $$
**引理 4.1**
$$
J(\pi')-J(\pi) = \mathbb{E}_ {\tau\sim\pi'}[\sum_{t=0}^{\inf}\gamma^{t} A^{\pi}(s_t,a_t)]。
J(\pi')-J(\pi) = \mathbb{E}_ {\tau\sim\pi'}[\sum_{t=0}^{\infty}\gamma^{t} A^{\pi}(s_t,a_t)]。
$$
证明:
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证明:
$$
\mathbb{E}_ {\tau\sim\pi'} [\sum_{t=0}^{\infty}\gamma^{t} A^{\pi}(s_t,a_t)] = \mathbb{E}_ {\tau\sim\pi'} [\sum_{t=0}^{\infty}\gamma^{t} [r(s_t,a_t)+\gamma V^{\pi}(s_{t+1}) - V^{\pi}(s_t)]]
$$
$$
= \mathbb{E}_ {\tau\sim\pi'} [\sum_{t=0}^{\infty} \gamma^{t}r(s_t,a_t)] + \mathbb{E}_ {\tau\sim\pi'}[\sum_{t=0}^{\infty}\gamma^{t}[\gamma V^{\pi}(s_{t+1}) - V^{\pi}(s_t)]]
$$
$$
= J(\pi') - \mathbb{E}_ {\tau\sim\pi'}[V^{\pi}(s_0)]
$$
$$
J(\pi')-J(\pi)。
$$
证明完毕。$\diamondsuit$
因此,我们有:
$$
\mathop{\max}_ {\pi'} J(\pi') = \mathop{\max}_{\pi'} J(\pi')-J(\pi)
$$
$$
= \mathop{\max}_ {\pi'} \mathbb{E}_ {\tau\sim\pi'}[\sum_{t=0}^{\infty}\gamma^{t} A^{\pi}(s_t,a_t)]。
$$
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