# 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

• Hi I am struggled with the equation $v(A, k)=\frac{\log (k)}{1-\beta}+v(A, 1)$. The answer below takes it as given. Could you show the derivation of it? Thanks. – Alalalalaki Mar 27 at 16:59
• I was looking through my notes and what my professor had derived goes something like this: He has considered $\sigma=1$ so we have $u(c)=log c$. Value function is something like this: $V(\lambda k_0, A_0)=\sum\beta_t\sum P(A^t)log(\lambda c_t (A^t))$ or $V(\lambda k_0, A_0)=\sum\beta_t\sum P(A^t)[log\lambda +log(c_t (A^t))$ where $\sum\beta_t\sum P(A^t)[log\lambda= log\lambda/1-\beta$ and $\sum\beta_t\sum P(A^t)log(c_t (A^t)=V(k_0,A_0)$ thus, $v(k,A)=log k/1-\beta+V(1,A)$. I guess a similar logic @Alalalalaki but someone can add something that may be missing – Maybeline Lee Mar 27 at 17:24
• Thank you very much for your reply. However I am still a little bit confused about the derivation. First, do you know why is the case that $V\left(\lambda k_{0}, A_{0}\right)=\sum \beta_{t} \sum P\left(A^{t}\right) \log \left(\lambda c_{t}\left(A^{t}\right)\right)$, i.e. why can we simply scale the consumption function at each period when we have a scaled initial capital? Second, given your derivation, it seems that we only have $v(k, A)=\log k / 1-\beta+V(1, A)$ when $k_0=1$. Do you assume that $k_0=1$? Can we get the same result if $k_0 \neq1$? – Alalalalaki Mar 27 at 18:11
• I'm not sure if this reply will be helpful but previously (before going on to this special case) we have considered $v(k,A)=max\ c^{1-\sigma}/1-\sigma + \beta k^{1-\sigma}E[v(1,A')|A]$ s.t the constraint. Then we have considered that the utility is homothetic in c,k'. In regard to the second question, I assume that we don't get the same result if $k_0\neq 1$, maybe no closed-form solution or something. – Maybeline Lee Mar 27 at 18:30

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 $$E\left[v(A',1)|A\right] = D(A)$$: $$v(A,k) = \max log(c) + \beta \frac{log(k')}{1-\beta} + \beta D(A)$$

• I see. It was the period change in the value function that threw me off. Thank you :) – Maybeline Lee Mar 27 at 10:55
• Hi I am struggled with the equation $v(A, k)=\frac{\log (k)}{1-\beta}+v(A, 1)$. Could you show also the derivation of it? – Alalalalaki Mar 27 at 12:20