16.2
$$ Q_{n}(k)=\frac{1}{n}\left((n-1)\times Q_{n-1}(k)+v_{n}\right) $$
[推导]： $$ \begin{aligned} Q_{n}(k)&=\frac{1}{n}\sum_{i=1}^{n}v_{i}\\ &=\frac{1}{n}\left(\sum_{i=1}^{n-1}v_{i}+v_{n}\right)\\ &=\frac{1}{n}\left((n-1)\times Q_{n-1}(k)+v_{n}\right)\\ &=Q_{n-1}(k)+\frac{1}{n}\left(v_n-Q_{n-1}(k)\right) \end{aligned} $$
16.3
$$ \begin{aligned} &Q_{n}(k)=\frac{1}{n}\left((n-1) \times Q_{n-1}(k)+v_{n}\right)\\ &=Q_{n-1}(k)+\frac{1}{n}\left(v_{n}-Q_{n-1}(k)\right) \end{aligned} $$
[推导]：参见 16.2
16.4
$$ P(k)=\frac{e^{\frac{Q(k)}{\tau }}}{\sum_{i=1}^{K}e^{\frac{Q(i)}{\tau}}} $$
[解析]： $$ P(k)=\frac{e^{\frac{Q(k)}{\tau }}}{\sum_{i=1}^{K}e^{\frac{Q(i)}{\tau}}}\propto e^{\frac{Q(k)}{\tau }}\propto\frac{Q(k)}{\tau }\propto\frac{1}{\tau} $$
16.7
$$ \begin{aligned} V_{T}^{\pi}(x)&=\mathbb{E}{\pi}[\frac{1}{T}\sum{t=1}^{T}r_{t}\mid x_{0}=x]\\ &=\mathbb{E}{\pi}[\frac{1}{T}r{1}+\frac{T-1}{T}\frac{1}{T-1}\sum_{t=2}^{T}r_{t}\mid x_{0}=x]\\ &=\sum_{a\in A}\pi(x,a)\sum_{x{}’\in X}P_{x\rightarrow x{}’}^{a}(\frac{1}{T}R_{x\rightarrow x{}’}^{a}+\frac{T-1}{T}\mathbb{E}{\pi}[\frac{1}{T-1}\sum{t=1}^{T-1}r_{t}\mid x_{0}=x{}’])\\ &=\sum_{a\in A}\pi(x,a)\sum_{x{}’\in X}P_{x\rightarrow x{}’}^{a}(\frac{1}{T}R_{x\rightarrow x{}’}^{a}+\frac{T-1}{T}V_{T-1}^{\pi}(x{}’)]) \end{aligned} $$
[解析]：
因为 $$ \pi(x,a)=P(action=a|state=x) $$ 表示在状态$x$下选择动作$a$的概率，又因为动作事件之间两两互斥且和为动作空间，由全概率展开公式 $$ P(A)=\sum_{i=1}^{\infty}P(B_{i})P(A\mid B_{i}) $$ 可得 $$ \begin{aligned} &\mathbb{E}{\pi}[\frac{1}{T}r{1}+\frac{T-1}{T}\frac{1}{T-1}\sum_{t=2}^{T}r_{t}\mid x_{0}=x]\\ &=\sum_{a\in A}\pi(x,a)\sum_{x{}’\in X}P_{x\rightarrow x{}’}^{a}(\frac{1}{T}R_{x\rightarrow x{}’}^{a}+\frac{T-1}{T}\mathbb{E}{\pi}[\frac{1}{T-1}\sum{t=1}^{T-1}r_{t}\mid x_{0}=x{}’]) \end{aligned} $$ 其中 $$ r_{1}=\pi(x,a)P_{x\rightarrow x{}’}^{a}R_{x\rightarrow x{}’}^{a} $$ 最后一个等式用到了递归形式。
16.8
$$ V_{\gamma }^{\pi}(x)=\sum {a\in A}\pi(x,a)\sum{x{}’\in X}P_{x\rightarrow x{}’}^{a}(R_{x\rightarrow x{}’}^{a}+\gamma V_{\gamma }^{\pi}(x{}’)) $$
[推导]： $$ \begin{aligned} V_{\gamma }^{\pi}(x)&=\mathbb{E}{\pi}[\sum{t=0}^{\infty }\gamma^{t}r_{t+1}\mid x_{0}=x]\\ &=\mathbb{E}{\pi}[r{1}+\sum_{t=1}^{\infty}\gamma^{t}r_{t+1}\mid x_{0}=x]\\ &=\mathbb{E}{\pi}[r{1}+\gamma\sum_{t=1}^{\infty}\gamma^{t-1}r_{t+1}\mid x_{0}=x]\\ &=\sum {a\in A}\pi(x,a)\sum{x{}’\in X}P_{x\rightarrow x{}’}^{a}(R_{x\rightarrow x{}’}^{a}+\gamma \mathbb{E}{\pi}[\sum{t=0}^{\infty }\gamma^{t}r_{t+1}\mid x_{0}=x{}’])\\ &=\sum {a\in A}\pi(x,a)\sum{x{}’\in X}P_{x\rightarrow x{}’}^{a}(R_{x\rightarrow x{}’}^{a}+\gamma V_{\gamma }^{\pi}(x{}’)) \end{aligned} $$
16.10
$$ \left\{\begin{array}{l} Q_{T}^{\pi}(x, a)=\sum_{x^{\prime} \in X} P_{x \rightarrow x^{\prime}}^{a}\left(\frac{1}{T} R_{x \rightarrow x^{\prime}}^{a}+\frac{T-1}{T} V_{T-1}^{\pi}\left(x^{\prime}\right)\right) \\ Q_{\gamma}^{\pi}(x, a)=\sum_{x^{\prime} \in X} P_{x \rightarrow x^{\prime}}^{a}\left(R_{x \rightarrow x^{\prime}}^{a}+\gamma V_{\gamma}^{\pi}\left(x^{\prime}\right)\right) \end{array}\right. $$
[推导]：参见 16.7, 16.8
16.14
$$ V^{}(x)=\max _{a \in A} Q^{\pi^{}}(x, a) $$
[解析]：为了获得最优的状态值函数$V$，这里取了两层最优，分别是采用最优策略$\pi^{*}$和选取使得状态动作值函数$Q$最大的状态$\max_{a\in A}$。
16.16
$$ V^{\pi}(x)\leqslant V^{\pi{}’}(x) $$ [推导]： $$ \begin{aligned} V^{\pi}(x) & \leqslant Q^{\pi}\left(x, \pi^{\prime}(x)\right) \\ &=\sum_{x^{\prime} \in X} P_{x \rightarrow x^{\prime}}^{\pi^{\prime}(x)}\left(R_{x \rightarrow x^{\prime}}^{\pi^{\prime}(x)}+\gamma V^{\pi}\left(x^{\prime}\right)\right) \\ & \leqslant \sum_{x^{\prime} \in X} P_{x \rightarrow x^{\prime}}^{\pi^{\prime}(x)}\left(R_{x \rightarrow x^{\prime}}^{\pi^{\prime}(x)}+\gamma Q^{\pi}\left(x^{\prime}, \pi^{\prime}\left(x^{\prime}\right)\right)\right) \\ &= \sum_{x^{\prime} \in X} P_{x \rightarrow x^{\prime}}^{\pi^{\prime}(x)}\left(R_{x \rightarrow x^{\prime}}^{\pi^{\prime}(x)}+ \sum_{x’^{\prime} \in X} P_{x’ \rightarrow x^{’’}}^{\pi^{\prime}(x’)}\left(\gamma R_{x’ \rightarrow x^{\prime \prime}}^{\pi^{\prime}(x’)}+ \gamma^2 V^{\pi}\left(x^{\prime \prime}\right)\right)\right)\\ & \leqslant \sum_{x^{\prime} \in X} P_{x \rightarrow x^{\prime}}^{\pi^{\prime}(x)}\left(R_{x \rightarrow x^{\prime}}^{\pi^{\prime}(x)}+ \sum_{x’^{\prime} \in X} P_{x’ \rightarrow x^{’’}}^{\pi^{\prime}(x’)} \left( \gamma R_{x’ \rightarrow x^{\prime \prime}}^{\pi^{\prime}(x’)} + \gamma^2 Q^{\pi}\left(x^{\prime \prime}, \pi^{\prime }\left(x^{\prime \prime}\right)\right)\right)\right) \\ &\leqslant \cdots \\ &\leqslant \sum_{x^{\prime} \in X} P_{x \rightarrow x^{\prime}}^{\pi^{\prime}(x)}\left(R_{x \rightarrow x^{\prime}}^{\pi^{\prime}(x)}+\sum_{x’^{\prime} \in X} P_{x’ \rightarrow x^{’’}}^{\pi^{\prime}(x’)}\left(\gamma R_{x’ \rightarrow x^{\prime \prime}}^{\pi^{\prime}(x’)}+\sum_{x’^{\prime} \in X} P_{x’’ \rightarrow x^{’’’}}^{\pi^{\prime}(x’’)} \left(\gamma^2 R_{x’’ \rightarrow x^{\prime \prime \prime}}^{\pi^{\prime}(x’’)}+\cdots \right)\right)\right) \\ &= V^{\pi’}(x) \end{aligned} $$ 其中，使用了动作改变条件 $$ Q^{\pi}(x,\pi{}’(x))\geqslant V^{\pi}(x) $$ 以及状态-动作值函数 $$ Q^{\pi}(x{}’,\pi{}’(x{}’))=\sum_{x{}’\in X}P_{x{}’\rightarrow x{}’}^{\pi{}’(x{}’)}(R_{x{}’\rightarrow x{}’}^{\pi{}’(x{}’)}+\gamma V^{\pi}(x{}’)) $$ 于是，当前状态的最优值函数为 $$ V^{\ast}(x)=V^{\pi{}’}(x)\geqslant V^{\pi}(x) $$
16.31
$$ Q_{t+1}^{\pi}(x,a)=Q_{t}^{\pi}(x,a)+\alpha (R_{x\rightarrow x{}’}^{a}+\gamma Q_{t}^{\pi}(x{}’,a{}’)-Q_{t}^{\pi}(x,a)) $$
[推导]：根据累计折扣奖励的定义（P373）可知：
$$Q_{t+1}^{\pi}(x, a)=\gamma Q_{t}^{\pi}(x’, a’)+R_{x\to x’}^{a}$$
将上式进行类似于公式(16.29)的形式改写，可以得到：
$$Q_{t+1}^{\pi}(x, a)= Q_{t}^{\pi}(x, a) + (R_{x\to x’}^{a} + \gamma Q_{t}^{\pi}(x’, a’) - Q_{t}^{\pi}(x, a) )$$
括号中的部分即为累计折扣奖励下的需要学习的增量，然后乘以学习率$\alpha$，即可得到公式16.31.