<p>Your reasoning is valid, although I am not sure you can do this efficiently with continuous time...</p>
<p>Using discrete time to convey the same idea (with continuous time, no, I don't know how to do this, since then you need to linearize the Euler equations, but a linear approximation is not a good idea if you have to compare back and forth the value functions), something like:</p>
<p><span class="math-container">$W(k) = \max \{V_a(k), V_b(k)\} + \beta W(k')$</span></p>
<p>Where</p>
<p><span class="math-container">$V_i(k) = \max_{0\leq k'\leq z_i k^{\alpha} + (1-\delta_i)k} \{U(z_i k^{\alpha} + (1-\delta_i) k - k') \} + \beta V_i(k')$</span></p>
<p>And <span class="math-container">$z_a &gt; z_b$</span>, <span class="math-container">$\delta_a &gt; \delta_b$</span></p>
<p>The control variable (namely, the choice variable, what you optimize) of the problem above is <span class="math-container">$k'$</span>.</p>
<p>So, here I've chosen productivity as a fixed constant, but obviously you can make this way more messy and realistic...</p>

CC BY-SA 4.0
2024-01-13T10:10:55.217Z
2024-01-17T09:31:04.880Z
2024-01-17T09:31:04.880Z
Matteo Bulgarelli
<p>Suppose we have the following problem:</p>
<p><span class="math-container">$\max \int_0^\infty \exp(-\rho t) u(c(t))dt$</span></p>
<p>where <span class="math-container">$c(t)$</span> is consumption at time <span class="math-container">$t$</span>. Subject to:</p>
<p><span class="math-container">$\dot{k}(t)= f(k(t))- c(t) - \delta k(t)$</span>.</p>
<p>where <span class="math-container">$k$</span> is capital, <span class="math-container">$f(.)$</span> production function and <span class="math-container">$\delta$</span> depreciation.</p>
<p>This problem is not hard to solve. However, suppose that we now add a twist. Suppose that agent can choose between two production functions:</p>
<p><span class="math-container">$f_A(k(t))&gt;f_B(k(t))$</span></p>
<p>but if agent chooses <span class="math-container">$f_A(k(t))$</span> there is a downside that depreciation is larger <span class="math-container">$\delta_A &gt;\delta_B$</span>.</p>
<p>I am not sure how to solve this problem. My idea was as follows:</p>
<ol>
<li>I will set up two separate optimization problems; problem A where I only use parameters for A production function and problem B where I only use B production function,</li>
<li>Calculate the optimal <span class="math-container">$c_A$</span> and <span class="math-container">$c_B$</span> and corresponding utilities,</li>
<li>Compare <span class="math-container">$U(c_A)$</span> to <span class="math-container">$U(c_B)$</span>.</li>
</ol>
<p>My reasoning is that if the person has choice between two production technologies they would simply pick which one gives highest utility. Is this valid reasoning?</p>
<p>However, what if the person can switch back and forth? Can we still apply this recipe (if my recipe is valid)?</p>

CC BY-SA 4.0
2023-12-24T16:47:28.627Z
**Incorrect comment see below**:
$$\textrm{}$$

Looks like you would update your law of motion to:
$$k(t+1)-k(t)=\max\{f_A(k(t))-\delta_A k(t), f_B(k(t))-\delta_B k(t)\}-c(t)$$

There would be no switching if both $f_A(k(t))$ and $f_B(k(t))$ are concave.
EconJohn
CC BY-SA 4.0
2023-12-25T21:39:20.687Z
@EconJohn is that really it? If you want to post it as an answer I can accept it
WilliamT
CC BY-SA 4.0
2023-12-26T05:20:21.300Z
EconJohn: That was great but could you or WilliamT explain where the expression for $\dot{k}(t)$ comes from ? Thanks.
mark leeds
CC BY-SA 4.0
2023-12-26T05:20:25.230Z
@WilliamT, I have made a mistake. There can be switching even with concave functions. The law of motion is the same.

to see switching with concave functions: https://www.desmos.com/calculator/vtumhnfhfd

In case the link dies: https://imgur.com/a/kMNAaDF
EconJohn
CC BY-SA 4.0
2023-12-26T05:23:20.647Z
I guess in the discrete case, it represents the change and in the continuous case, the derivative with respect to $t$ but I don't get  the RHS. Thanks again.
mark leeds
CC BY-SA 4.0
2023-12-26T05:29:03.017Z
@markleeds its because the problem is in continuous time
EconJohn
CC BY-SA 4.0
2023-12-26T05:29:47.873Z
I think I see it. For the discrete case,  the change in capital between time t and t-1 is equal to some function of [ capital at time t - consumption of capital between time t and (t-1) - some depreciation multiplier times capital at time t ]. Continuous case follows similarly.  Apologies for noise.
mark leeds
CC BY-SA 4.0
2023-12-26T11:14:28.930Z
I am not sure if an ex ante choice of the production is relevant (or interesting). However, it would be interesting to write a problem where an individual (or a social planner) chooses optimally the technology.
optimal control
CC BY-SA 4.0
2024-01-11T14:00:15.783Z
Question: the choice of the production function, is it a commitment in period 0 forever, or is it flexible, so the planner can switch in every period? In the latter case, you need dynamic programming for sure (linearizing won't produce a good approximation of the true solution), in the former I don't know
Matteo Bulgarelli
CC BY-SA 4.0
2024-01-12T17:25:09.810Z
@MatteoBulgarelli it can change every period, I know I need some sort of dynamic programming (that is needed even without switching) but you can't solve this using regular dynamic programming (e.g. using classic bellman)
WilliamT
<mathematical-economics><economic-growth><dynamic-programming><dynamic-optimization>
How to solve dynamic problem with 2 production functions?
WilliamT
# How to solve dynamic problem with 2 production functions?

(Asked by: username_0 on 2023-12-24)

Suppose we have the following problem:

$\max \int\_0^\infty \exp(-\rho t) u(c(t))dt$

where $c(t)$ is consumption at time $t$. Subject to:

$\dot{k}(t)= f(k(t))- c(t) - \delta k(t)$.

where $k$ is capital, $f(.)$ production function and $\delta$ depreciation.

This problem is not hard to solve. However, suppose that we now add a twist. Suppose that agent can choose between two production functions:

$f\_A(k(t))>f\_B(k(t))$

but if agent chooses $f\_A(k(t))$ there is a downside that depreciation is larger $\delta\_A >\delta\_B$.

I am not sure how to solve this problem. My idea was as follows:

1. I will set up two separate optimization problems; problem A where I only use parameters for A production function and problem B where I only use B production function,
2. Calculate the username_3 $c\_A$ and $c\_B$ and corresponding utilities,
3. Compare $U(c\_A)$ to $U(c\_B)$.

My reasoning is that if the person has choice between two production technologies they would simply pick which one gives highest utility. Is this valid reasoning?

However, what if the person can switch back and forth? Can we still apply this recipe (if my recipe is valid)?

**Tags:** mathematical-economics, economic-growth, dynamic-programming, dynamic-optimization

> **Comments:**
> **username_1**: **Incorrect comment see below**: $$\textrm{}$$  Looks like you would update your law of motion to: $$k(t+1)-k(t)=\max\{f_A(k(t))-\delta_A k(t), f_B(k(t))-\delta_B k(t)\}-c(t)$$  There would be no switching if both $f_A(k(t))$ and $f_B(k(t))$ are concave.  
> **username_0**: @username_1 is that really it? If you want to post it as an answer I can accept it  
> **username_2**: username_1: That was great but could you or username_0 explain where the expression for $\dot{k}(t)$ comes from ? Thanks.  
> **username_1**: @username_0, I have made a mistake. There can be switching even with concave functions. The law of motion is the same.  to see switching with concave functions: https://www.desmos.com/calculator/vtumhnfhfd  In case the link dies: https://imgur.com/a/kMNAaDF  
> **username_2**: I guess in the discrete case, it represents the change and in the continuous case, the derivative with respect to $t$ but I don't get  the RHS. Thanks again.  
> **username_1**: @username_2 its because the problem is in continuous time  
> **username_2**: I think I see it. For the discrete case,  the change in capital between time t and t-1 is equal to some function of [ capital at time t - consumption of capital between time t and (t-1) - some depreciation multiplier times capital at time t ]. Continuous case follows similarly.  Apologies for noise.  
> **username_3**: I am not sure if an ex ante choice of the production is relevant (or interesting). However, it would be interesting to write a problem where an individual (or a social planner) chooses optimally the technology.  
> **username_4**: Question: the choice of the production function, is it a commitment in period 0 forever, or is it flexible, so the planner can switch in every period? In the latter case, you need dynamic programming for sure (linearizing won't produce a good approximation of the true solution), in the former I don't know  
> **username_0**: @username_4 it can change every period, I know I need some sort of dynamic programming (that is needed even without switching) but you can't solve this using regular dynamic programming (e.g. using classic bellman)  

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## 1 answers

### Answer by username_4 on 2024-01-13. Score: 1

Your reasoning is valid, although I am not sure you can do this efficiently with continuous time...

Using discrete time to convey the same idea (with continuous time, no, I don't know how to do this, since then you need to linearize the Euler equations, but a linear approximation is not a good idea if you have to compare back and forth the value functions), something like:

$W(k) = \max \{V\_a(k), V\_b(k)\} + \beta W(k')$

Where

$V\_i(k) = \max\_{0\leq k'\leq z\_i k^{\alpha} + (1-\delta\_i)k} \{U(z\_i k^{\alpha} + (1-\delta\_i) k - k') \} + \beta V\_i(k')$

And $z\_a > z\_b$, $\delta\_a > \delta\_b$

The username_3 variable (namely, the choice variable, what you optimize) of the problem above is $k'$.

So, here I've chosen productivity as a fixed constant, but obviously you can make this way more messy and realistic...


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