# Writing action-value function in terms of state-value function for a Markov decision process

I am working through Sutton and Barto's text on reinforcement learning (2nd edition), and am stuck on exercise 3.13 (p. 58). The exercise is to write $$q_\pi$$ in terms of $$v_\pi$$ and $$p(s',r\mid s,a)$$, where $$q_\pi$$ is the action-value function defined by $$q_\pi(s,a) := \mathbb E_\pi [G_t \mid S_t=s, A_t=a];$$ $$v_\pi$$ is the state-value function defined by $$v_\pi(s) := \mathbb E_\pi[G_t \mid S_t=s];$$ and $$G_t$$ is the discounted return $$G_t := \sum_{k=0}^\infty \gamma^k R_{t+k+1}.$$

My attempt so far is to write $$G_t = R_{t+1} + \gamma G_{t+1}$$ so that $$q_\pi(s,a) = \mathbb E_\pi[R_{t+1} \mid S_t=s,A_t=a] + \gamma \mathbb E_\pi[G_{t+1} \mid S_t=s,A_t=a].$$

Now $$\mathbb E_\pi[G_{t+1} \mid S_t=s,A_t=a]$$ becomes:

• $$\sum_g g \Pr(G_{t+1}=g\mid S_t=s,A_t=a)$$ (definition of expected value)
• $$\sum_g g \sum_{s'} \Pr(G_{t+1}=g, S_{t+1}=s'\mid S_t=s,A_t=a)$$ (law of total probability)
• $$\sum_g g \sum_{s'} \Pr(G_{t+1}=g \mid S_t=s,A_t=a,S_{t+1}=s') \Pr(S_{t+1}=s' \mid S_t=s,A_t=a)$$
• (definition of conditional probability; the expression was too long so I've placed this parenthetical on the line below)
• $$\sum_{s'} \Pr(S_{t+1}=s'\mid S_t=s,A_t=a) \sum_g g \Pr(G_{t+1}=g\mid S_t=s,A_t=a,S_{t+1}=s')$$
• (rearranging sums)
• $$\sum_{s'}\Pr(S_{t+1}=s'\mid S_t=s,A_t=a) \mathbb E_\pi[G_{t+1} \mid S_t=s,A_t=a,S_{t+1}=s']$$
• (definition of expected value)

In turn, $$\Pr(S_{t+1}=s'\mid S_t=s,A_t=a)$$ can be written as $$\sum_r p(s',r\mid s,a)$$ (this is equation 3.5 in the book).

Similarly $$\mathbb E_\pi[R_{t+1} \mid S_t=s,A_t=a]$$ can be written as $$\sum_{s'} \Pr(S_{t+1}=s'\mid S_t=s,A_t=a) \mathbb E_\pi[R_{t+1} \mid S_t=s,A_t=a,S_{t+1}=s']$$ where $$\mathbb E_\pi[R_{t+1} \mid S_t=s,A_t=a,S_{t+1}=s']$$ can be written in terms of $$p(s',r\mid s,a)$$ (using equation 3.6 in the book).

So the only problem left is the expression $$\mathbb E_\pi[G_{t+1} \mid S_t=s,A_t=a,S_{t+1}=s']$$. If I squint my eyes, it looks like $$v_\pi(s')$$, and indeed, looking around, I see formulas indicating that it should be $$v_\pi(s')$$ (e.g. PDF page 35 of these slides, page 13 of these slides, and the equation given in this section). But I don't see how to convince myself that this is the case. Sutton and Barto write that "the probability of each possible value for $$S_t$$ and $$R_t$$ depends only on the immediately preceding state and action, $$S_{t-1}$$ and $$A_{t-1}$$, and, given them, not at all on earlier states and actions" (p. 49). But here, we only have $$S_{t+1}$$ and not $$A_{t+1}$$, so we can't just ignore the "$$S_t=s,A_t=a$$" part. Given the policy $$\pi$$, the action $$A_{t+1}$$ is completely determined by $$S_{t+1}$$, so my attempt is to write:

• $$\mathbb E_\pi[G_{t+1} \mid S_t=s,A_t=a,S_{t+1}=s']$$
• $$\sum_g g \Pr(G_{t+1}=g \mid S_t=s,A_t=a,S_{t+1}=s')$$ (definition of expected value)
• $$\sum_g g \sum_{a'} \Pr(G_{t+1}=g,A_{t+1}=a' \mid S_t=s,A_t=a,S_{t+1}=s')$$ (law of total probability)
• $$\sum_g g \sum_{a'} \Pr(G_{t+1}=g \mid S_t=s,A_t=a,S_{t+1}=s',A_{t+1}=a')\pi(a'\mid s')$$
• (definition of conditional probability)
• $$\sum_g g \sum_{a'} \Pr(G_{t+1}=g \mid S_{t+1}=s',A_{t+1}=a')\pi(a'\mid s')$$ (Markov property)
• $$\sum_g g \sum_{a'} \Pr(G_{t+1}=g,A_{t+1}=a' \mid S_{t+1}=s')$$ (definition of conditional probability)
• $$\sum_g g \Pr(G_{t+1}=g\mid S_{t+1}=s')$$ (law of total probability)
• $$\mathbb E_\pi[G_{t+1} \mid S_{t+1}=s']$$ (definition of expected value)

Is this reasoning valid?

My other confusion is that, supposing $$\mathbb E_\pi[G_{t+1} \mid S_t=s,A_t=a,S_{t+1}=s'] = \mathbb E_\pi[G_{t+1} \mid S_{t+1}=s']$$, it seems like we could also say that $$\mathbb E_\pi[R_t\mid S_{t-1}=s,A_{t-1}=a,S_t=s'] = \mathbb E_\pi[R_t \mid S_t=s']$$. But if that's the case, I'm confused about why Sutton and Barto derive an expression for $$r(s,a,s'):=\mathbb E_\pi[R_t\mid S_{t-1}=s,A_{t-1}=a,S_t=s']$$ rather than one for $$r(s')$$ (which can be defined to equal $$\mathbb E_\pi[R_t\mid S_t=s']$$). In other words my question is something like "if the $$s$$ and $$a$$ in $$r(s,a,s')$$ don't matter, why include them in the notation?"

Is this reasoning valid?

I think this is correct.

My other confusion is that, supposing $$\mathbb E_\pi[G_{t+1} \mid S_t=s,A_t=a,S_{t+1}=s'] = \mathbb E_\pi[G_{t+1} \mid S_{t+1}=s']$$, it seems like we could also say that $$\mathbb E_\pi[R_t\mid S_{t-1}=s,A_{t-1}=a,S_t=s'] = \mathbb E_\pi[R_t \mid S_t=s']$$.

I think my confusion was that I was forgetting that $$G_{t+1}$$ actually starts at $$R_{t+2}$$, i.e. $$G_{t+1} = R_{t+2} + \gamma R_{t+3} + \cdots$$. So I think we could say $$\mathbb E_\pi[R_{t+2} \mid S_t=s,A_t=a,S_{t+1}=s'] = \mathbb E_\pi[R_{t+2} \mid S_{t+1}=s']$$, which, shifting the time index down by one, is the same as $$\mathbb E_\pi[R_{t+1} \mid S_{t-1}=s,A_{t-1}=a,S_t=s'] = \mathbb E_\pi[R_{t+1} \mid S_t=s']$$. But we can't say $$\mathbb E_\pi[R_t\mid S_{t-1}=s,A_{t-1}=a,S_t=s'] = \mathbb E_\pi[R_t \mid S_t=s']$$ which is why Sutton and Barto want to derive the expression for $$r(s,a,s')$$.

If I was doing the exercise today, I would do it like this:

$$q_\pi(s,a) := \mathbb E_\pi[G_t \mid S_t=s, A_t=a]$$ becomes:

• $$\sum_g g \Pr(G_t = g\mid S_t=s, A_t=a)$$ (definition of expectation)
• $$\sum_g g \sum_{s'} \Pr(G_t=g, S_{t+1}=s' \mid S_t=s, A_t=a)$$ (law of total probability)
• $$\sum_{s'} \sum_g g \Pr(G_t=g, S_{t+1}=s' \mid S_t=s, A_t=a)$$ (reorder sums)
• $$\sum_{s'} \sum_g g \Pr(S_{t+1}=s'\mid S_t=s, A_t=a) \Pr(G_t=g \mid S_t=s, A_t=a, S_{t+1}=s')$$ (definition of conditional probability)
• $$\sum_{s'} \Pr(S_{t+1}=s'\mid S_t=s, A_t=a) \sum_g g \Pr(G_t=g \mid S_t=s, A_t=a, S_{t+1}=s')$$ (factor)
• $$\sum_{s'} \Pr(S_{t+1}=s'\mid S_t=s, A_t=a) \mathbb E_\pi[G_t \mid S_t=s, A_t=a, S_{t+1}=s']$$ (definition of expectation)
• $$\sum_{s'} \Pr(S_{t+1}=s'\mid S_t=s, A_t=a) \mathbb E_\pi[R_{t+1} + \gamma G_{t+1} \mid S_t=s, A_t=a, S_{t+1}=s']$$ (definition of return)
• $$\sum_{s'} \Pr(S_{t+1}=s'\mid S_t=s, A_t=a) (\mathbb E_\pi[R_{t+1}\mid S_t=s, A_t=a, S_{t+1}=s'] + \gamma\mathbb E_\pi[ G_{t+1} \mid S_t=s, A_t=a, S_{t+1}=s'])$$
• (linearity of expectation)
• $$\sum_{s'} p(s'\mid s,a) (r(s,a,s') + \gamma\mathbb E_\pi[ G_{t+1} \mid S_t=s, A_t=a, S_{t+1}=s'])$$ (equations 3.4 and 3.6 in the book)
• $$\sum_{s'} p(s'\mid s,a) (r(s,a,s') + \gamma v_\pi(s'))$$ (using reasoning given in the question)

This is probably the "simplest" form of $$q_\pi$$ which uses $$v_\pi$$, but the exercise actually says to use "the four-argument $$p$$", so we can expand $$\sum_{s'} p(s'\mid s,a) (r(s,a,s') + \gamma v_\pi(s'))$$ to $$\sum_{s'} \left(\sum_r p(s',r\mid s,a)\right) \left(\sum_r r \frac{p(s',r\mid s,a)}{\sum_{r'} p(s',r'\mid s,a)} + \gamma v_\pi(s')\right)$$ which can be simplified to $$\sum_{s',r} p(s',r\mid s,a) (r + \gamma v_\pi(s'))$$

• Thanks for the derivation! I am curious about the 2nd to the last expression. Why do you need the denominator? And I think the $r$ expected reward in the 2nd parenthesis is not the $r$ in the 4-parameter $p$. Maybe some confusion there? Jan 18, 2021 at 7:38
• @YoHsiao The way I did that was to first expand $r(s,a,s')$ using equation 3.6 in the book (2nd ed). That gives an expression with denominator $p(s'\mid s,a)$, so I used equation 3.4 to rewrite that in terms of the 4-parameter $p$. To avoid clobbering up the "$r$" in the inner and outer loops, I used a separate indexing variable $r'$. Jan 31, 2021 at 1:27
• @IssaRice What definition of markov property you used? For me $S_t$ is Markov if $p(S_t=s_t|S_{t-1}=s_{t-1}) = p(S_t=s_t|S_{t-1}=s_{t-1},...,S_{1}=s_1)$ Sep 22, 2023 at 0:49
• @piero See equation 3.2 in the book. So something like $\Pr(S_t = s_t, R_t = r_t \mid S_{t-1}=s_{t-1}, A_{t-1}=a_{t-1}) = \Pr(S_t = s_t, R_t = r_t \mid S_{t-1}=s_{t-1}, A_{t-1}=a_{t-1}, \ldots, S_0=s_0, R_1=r_1, A_0=a_0)$. Sep 23, 2023 at 7:12