我继续在学习《ML Lecture 23-1: Deep Reinforcement Learning by Hung-yi Lee》中的视频教程https://youtu.be/W8XF3ME8G2I?si=zEQ3qj_iXzZZ-n85，其中提到：
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Gradient Ascent θ new ← θ old + η ∇ R ˉ θ old = ∑ t = 1 T ∇ log p ( a t ∣ s t , θ ) ∇ R ˉ θ ≈ 1 N ∑ n = 1 N R ( τ n ) ∇ log P ( τ n ∣ θ ) = 1 N ∑ n = 1 N R ( τ n ) ∑ t = 1 T n ∇ log p ( a t n ∣ s t n , θ ) = 1 N ∑ n = 1 N ∑ t = 1 T n R ( τ o n ) ∇ log ‾ p ( a t n ∣ s t n , θ ) \begin{aligned} & \begin{array}{l} \text { Gradient Ascent } \\ \theta^{\text {new }} \leftarrow \theta^{\text {old }}+\eta \nabla \bar{R}_{\theta^{\text {old }}} \end{array} \quad=\sum_{t=1}^T \nabla \log p\left(a_t \mid s_t, \theta\right) \\ & \nabla \bar{R}_\theta \approx \frac{1}{N} \sum_{n=1}^N R\left(\tau^n\right) \nabla \log P\left(\tau^n \mid \theta\right)=\frac{1}{N} \sum_{n=1}^N R\left(\tau^n\right) \sum_{t=1}^{T_n} \nabla \log p\left(a_t^n \mid s_t^n, \theta\right) \\ & =\frac{1}{N} \sum_{n=1}^N \sum_{t=1}^{T_n} R\left(\tau_o^n\right) \nabla \underline{\log } p\left(a_t^n \mid s_t^n, \theta\right) \\ & \end{aligned} Gradient Ascent θnew ←θold +η∇Rˉθold =t=1∑T∇logp(at∣st,θ)∇Rˉθ≈N1n=1∑NR(τn)∇logP(τn∣θ)=N1n=1∑NR(τn)t=1∑Tn∇logp(atn∣stn,θ)=N1n=1∑Nt=1∑TnR(τon)∇logp(atn∣stn,θ)
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“这里的 Gradient Ascent 的微分是很符合人类直觉的， R ( τ n ) R\left(\tau^n\right) R(τn)为正则会提升获得此次胜利的过程中采取的每一次动作的概率；而 R ( τ n ) R\left(\tau^n\right) R(τn)为负，则会降低这些动作出现的概率”，请问，这种说法正确吗