# Regarding the arbitrariness of states and controls

I am trying to better understand the process of deriving Euler Equations using the first order condition of the problem on the right hand side of a Bellman equation and the Benveniste-Scheinkman formula. In particular, there is a line in Ljungqvist and Sargent's (LS) Recursive Macroeconomic Theory that I don't quite understand in practice. It is this:

When the state and controls can be defined in such a way that only $$u$$ appears in the transition equation, i.e., $$x' = g(u)$$: the derivative of the value function becomes,... $$V'(x)=r_1(x,h(x)).$$

I see why the result follows, granted that states and controls can be defined in the way described. But lets look at a simple optimal growth example.

Let the sequential problem be

$$\max\sum_{t=0}^\infty\beta^t\log(c_t) \\ \text{s.t.} \\ k_{t+1} + c_t = Ak_t^\alpha.$$

The associated Bellman equation is $$V(k)=\max\log(c) + \beta V(k') \\ \text{s.t.} \\ k' + c = Ak^\alpha.$$

Now, in my current understanding $$k'$$ and $$c$$ are both controls and $$k$$ is the state. But, (SL) say "let the state be $$k$$ and the control be $$k'$$ where $$k'$$ denotes next period's value of $$k$$." However, this does not give me the desired form, i.e. writing $$k'=Ak^\alpha-c$$ has both a control and a state on the right hand side. How does this fit into the form $$x'=g(u)$$?

The change from $$c$$ as a decision variable towards $$k'$$ as a decision variable is by a simple change of variables.

In the original setup, $$k$$ is the state and $$c$$ is the control (decision) variable. The Bellman equation is the following. $$v(k) = \max_{c \in [0, Ak^\alpha]} u(c) + \beta v(Ak^\alpha - c).$$ (here the constraint $$k' = Ak^\alpha - c$$ is already substituted into the Bellman equation).

Now, you can define $$k' = Ak^\alpha - c$$. As $$c \in [0, Ak^\alpha]$$, we have that $$k' \in [0, Ak^\alpha]$$. Notice that there is a one-to-one correspondence between the values of $$c$$ and the values of $$k'$$. Substituting $$c$$ by $$Ak^\alpha - k'$$ in the Bellman equation gives: $$v(k) = \max_{k' \in [0, Ak^\alpha]} u(Ak^\alpha - k') + \beta v(k').$$ In this problem $$k$$ is the state and $$k'$$ is the control (and also the state in the next period.

The Envelope condition gives: $$v'(k) = A \alpha k^{\alpha - 1} u'(Ak^\alpha - k')$$ The first order condition gives: $$-u'(Ak^\alpha - k') + \beta v'(k') = 0.$$ These two can be combined to give the desired Euler equation.

General setting

In general, the Bellman equation takes the following form: $$v(x) = \max_{u \in \Gamma(x)} f(x,u) + \beta v(r(x,u)).$$ where $$x$$ is the state variable and $$u$$ is the control. The law-of-motion function $$x' = r(x,u)$$ gives the state tomorrow $$(x')$$ as a function of the state today $$(x)$$ and the control today $$(u)$$.

If $$r$$ is invertible in $$u$$ (as a function of $$u$$) we can sometimes rewrite this as: $$v(x) = \max_{x' \in \Delta(x)} g(x,x') + \beta v(x').$$ where $$g(x,x') = f(x, r^{-1}(x,x'))$$ and $$\Delta(x) = \{x'| \exists u \in \Gamma(x), x' = r(x,u)\}.$$

This new Bellman equation has as state $$(x)$$ and control $$(x')$$ which happen to coincide with the state tomorrow.