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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

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  • $\begingroup$ 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. $\endgroup$
    – Alalalalaki
    Commented Mar 27, 2021 at 16:59
  • $\begingroup$ 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 $\endgroup$
    – Maybeline Lee
    Commented Mar 27, 2021 at 17:24
  • $\begingroup$ 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 $k_0 \neq1$? $\endgroup$
    – Alalalalaki
    Commented Mar 27, 2021 at 18:11
  • $\begingroup$ 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. $\endgroup$
    – Maybeline Lee
    Commented Mar 27, 2021 at 18:30

1 Answer 1

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

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  • $\begingroup$ I see. It was the period change in the value function that threw me off. Thank you :) $\endgroup$
    – Maybeline Lee
    Commented Mar 27, 2021 at 10:55
  • $\begingroup$ 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? $\endgroup$
    – Alalalalaki
    Commented Mar 27, 2021 at 12:20

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