Stochastic AK model derivation Consider the following version of the stochastic Ak model written as a Bellman equation:
$$v(A,k)=max\ log(c)+\beta E[v(A',k')|A]$$
$$s.t\ k'+c\leq Ak$$ and non-negatitvities.
$A$ is a stationary first order Markov process. $E[\cdot|A]$ denotes the expectations of · conditional on $A$. We assume that $\beta\in(0, 1)$.
Now we can also show that the value function $v$ can be expressed as $$v(A,k)=\frac{log(k)}{1-\beta}+v(A,1).$$
If we denote $E[v(A',1)|A]$ as $D(A)$, which is a constant (i.e., is only a function of the exogenous parameters of the model and the realization of the stochastic process A, but is not a function of the endogenous variables). Use this and the new equation to substitute into the maximization problem in the Bellman equation. We should get a simple maximization problem in $c$ and $k$.
And what the solution is given as $$v(A,k)=max\ log(c)+\frac{\beta}{1-\beta}log(k')+\beta D(A).$$
Could someone show the steps behind this derivation
I think you may substitute it directly as mentioned?
$$v(A,k) = \max log(c) + \beta E[v(A',k')|A]$$
Applying the value function expression (note period change):
$v(A',k') = \frac{log(k')}{1-\beta} + v(A',1)$
Substituting:
$$v(A,k) = \max log(c) + \beta E\left[\frac{log(k')}{1-\beta} + v(A',1)|A\right]$$
By linearity:
$$v(A,k) = \max log(c) + \beta E\left[\frac{log(k')}{1-\beta}|A\right] + E\left[v(A',1)|A\right]$$
Since $E\left[v(A',1)|A\right] = D(A)$:
$$v(A,k) = \max log(c) + \beta \frac{log(k')}{1-\beta} + \beta D(A)$$