From 3acf10274ae3391a29c9d9095998559f27037749 Mon Sep 17 00:00:00 2001
From: xiaowei_xing <997427575@qq.com>
Date: Fri, 8 Nov 2019 14:22:36 +0900
Subject: [PATCH] test

---
 docs/10.md | 6 +++++-
 1 file changed, 5 insertions(+), 1 deletion(-)

diff --git a/docs/10.md b/docs/10.md
index 875dd37..b6247b1 100644
--- a/docs/10.md
+++ b/docs/10.md
@@ -21,9 +21,13 @@ $$
 我们将目标函数记为 $J(\theta)$，可以用蒙特卡洛方法估计 $J(\theta)$。我们用 $r(\tau)$ 来代表轨迹 $\tau$ 的衰减奖励总和。
 
 $$
-J(\theta) = \mathbb{E}_{\tau\sim \pi _{\theta}(\tau)}[\sum_t \gamma^t r (s_t,a_t)] = \int \pi_\theta(\tau)r(\tau)\text{d}\tau \approx \frac{1}{N}\sum_{i=1}^N \sum_{t=1}^T \gamma^t r(s_{i,t},a_{i,t})
+J(\theta) = \mathbb{E}_{\tau\sim \pi _{\theta}(\tau)}[\sum_t \gamma^t r (s_t,a_t)] = \int \pi_\theta(\tau)r(\tau)\text{d}\tau
 $$
 
+$$
+ \approx \frac{1}{N}\sum_{i=1}^N \sum_{t=1}^T \gamma^t r(s_{i,t},a_{i,t})
+ $$
+
 $$
 \theta^* = \mathop{\arg\max}_\theta J(\theta)
 $$
\ No newline at end of file
-- 
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