<p><span class="math-container">$$V(v, t)=\int_{t}^{t+\Delta t} \exp \left(-\int_{t}^{s} r\left(s^{\prime}\right) d s^{\prime}\right) \pi(v, s) d s + \exp \left(-\int_{t}^{t+\Delta t} r\left(s^{\prime}\right) d s^{\prime}\right) V(v, t+\Delta t)$$</span></p>
<p>Use Taylor expansion, <span class="math-container">$$V(v, t+\Delta t) = V(v, t)+\frac{\partial V(v, t)}{\partial t} \Delta t + o(\Delta t)$$</span>.</p>
<p>Replace it into the first equation: <span class="math-container">$$-\int_{t}^{t+\Delta t} \exp \left(-\int_{t}^{s} r\left(s^{\prime}\right) d s^{\prime}\right) \pi(v, s) d s = \left( \exp \left(-\int_{t}^{t+\Delta t} r\left(s^{\prime}\right) d s^{\prime}\right) - 1\right) V(v, t) \\ + \exp \left(-\int_{t}^{t+\Delta t} r\left(s^{\prime}\right) d s^{\prime}\right) \left(\frac{\partial V(v, t)}{\partial t} \Delta t + o(\Delta t)\right)$$</span>.</p>
<p>Divide this equation  by <span class="math-container">$\Delta t$</span>, and take the limit as <span class="math-container">$\Delta t \to 0$</span> (a.k.a taking derivative), you get the HJB equation.</p>

CC BY-SA 4.0
2022-08-04T18:58:49.027Z
@Mike if you approved the question, please accept it as the best answer
Marcelo Gelati
CC BY-SA 4.0
2022-08-05T13:28:16.400Z
Can you explain the economic meaning of what you have done? To be honest, I did not get the derivation you have done. What is $x(t) ?$
Maximilian
CC BY-SA 4.0
2022-08-05T13:43:50.530Z
Sorry for the typo. I change all $x(t)$ to $v$ now. But note that this derivation works also when $v$ is a function of $t$, i.e. $v=x(t)$.
Alalalalaki
CC BY-SA 4.0
2022-08-05T16:08:57.183Z
To be honest, I do not understand how can you rewrite the first integral in that way. I mean, I'd have been expected to see the sum of two integrals, i.e., one ranging from $[t, t +\Delta t]$ and the other ranging from $[t+\Delta t, \infty]$
Maximilian
CC BY-SA 4.0
2022-08-05T18:25:45.447Z
That's why this is called Bellman equation. You can define the second integration as $V(v, t+\Delta t)$, subject to the discounting at $t+\Delta t$.
Alalalalaki
CC BY-SA 4.0
2022-08-05T18:32:51.243Z
Now, it seems that makes sense. Thanks
Maximilian
CC BY-SA 4.0
2022-08-04T00:38:49.663Z
2022-08-05T13:42:23.493Z
2022-08-05T13:42:23.493Z
Alalalalaki
<p>Let us consider the lab-equipment model developed by Romer with input varieties.</p>
<p>The value of owning the blueprint of a machine of a variety <span class="math-container">$\nu$</span> is given by:</p>
<p><a href="https://i.sstatic.net/4mRGz.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4mRGz.png" alt="enter image description here" /></a></p>
<p>And:</p>
<p><a href="https://i.sstatic.net/s6aKz.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/s6aKz.png" alt="enter image description here" /></a></p>
<p>denotes the profits of the monopolist producing machine <span class="math-container">$\nu$</span> at time <span class="math-container">$t$</span>, <span class="math-container">$x(\nu, t)$</span> and <span class="math-container">$p^x(\nu, t)$</span> are the profit-maximizing choices for the monopolist and <span class="math-container">$r(t)$</span> is the market interest rate at time <span class="math-container">$t$</span>. Finally, <span class="math-container">$\psi$</span> is the marginal cost for producing one unit of that machine. This marginal cost is equal to <span class="math-container">$\psi$</span> units of the final good.</p>
<p>Alternatively, assuming that the value function is differentiable in time, this equation could be
written in the form of a <strong>Hamilton-Jacobi-Bellman equation</strong> as follows:</p>
<p><a href="https://i.sstatic.net/ats3i.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ats3i.png" alt="enter image description here" /></a></p>
<p>Could you show me how to derive the last equation?</p>

<macroeconomics><endogenous-growth>
Hamilton-Jacobi-Bellman equation
Maximilian
# Hamilton-Jacobi-Bellman equation

(Asked by: username_0 on 2022-08-03)

Let us consider the lab-equipment model developed by Romer with input varieties.

The value of owning the blueprint of a machine of a variety $\nu$ is given by:

[![enter image description here](https://i.sstatic.net/4mRGz.png)](https://i.sstatic.net/4mRGz.png)

And:

[![enter image description here](https://i.sstatic.net/s6aKz.png)](https://i.sstatic.net/s6aKz.png)

denotes the profits of the monopolist producing machine $\nu$ at time $t$, $x(\nu, t)$ and $p^x(\nu, t)$ are the profit-maximizing choices for the monopolist and $r(t)$ is the market interest rate at time $t$. Finally, $\psi$ is the marginal cost for producing one unit of that machine. This marginal cost is equal to $\psi$ units of the final good.

Alternatively, assuming that the value function is differentiable in time, this equation could be
written in the form of a **Hamilton-Jacobi-Bellman equation** as follows:

[![enter image description here](https://i.sstatic.net/ats3i.png)](https://i.sstatic.net/ats3i.png)

Could you show me how to derive the last equation?

**Tags:** macroeconomics, endogenous-growth


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## 1 answers

### Answer by username_1 on 2022-08-04. Score: 3 | Accepted answer

$$V(v, t)=\int\_{t}^{t+\Delta t} \exp \left(-\int\_{t}^{s} r\left(s^{\prime}\right) d s^{\prime}\right) \pi(v, s) d s + \exp \left(-\int\_{t}^{t+\Delta t} r\left(s^{\prime}\right) d s^{\prime}\right) V(v, t+\Delta t)$$

Use Taylor expansion, $$V(v, t+\Delta t) = V(v, t)+\frac{\partial V(v, t)}{\partial t} \Delta t + o(\Delta t)$$.

Replace it into the first equation: $$-\int\_{t}^{t+\Delta t} \exp \left(-\int\_{t}^{s} r\left(s^{\prime}\right) d s^{\prime}\right) \pi(v, s) d s = \left( \exp \left(-\int\_{t}^{t+\Delta t} r\left(s^{\prime}\right) d s^{\prime}\right) - 1\right) V(v, t) \\ + \exp \left(-\int\_{t}^{t+\Delta t} r\left(s^{\prime}\right) d s^{\prime}\right) \left(\frac{\partial V(v, t)}{\partial t} \Delta t + o(\Delta t)\right)$$.

Divide this equation by $\Delta t$, and take the limit as $\Delta t \to 0$ (a.k.a taking derivative), you get the HJB equation.

> **Comments:**
> **username_2**: @Mike if you approved the question, please accept it as the best answer  
> **username_0**: Can you explain the economic meaning of what you have done? To be honest, I did not get the derivation you have done. What is $x(t) ?$  
> **username_1**: Sorry for the typo. I change all $x(t)$ to $v$ now. But note that this derivation works also when $v$ is a function of $t$, i.e. $v=x(t)$.  
> **username_0**: To be honest, I do not understand how can you rewrite the first integral in that way. I mean, I'd have been expected to see the sum of two integrals, i.e., one ranging from $[t, t +\Delta t]$ and the other ranging from $[t+\Delta t, \infty]$  
> **username_1**: That's why this is called Bellman equation. You can define the second integration as $V(v, t+\Delta t)$, subject to the discounting at $t+\Delta t$.  
> **username_0**: Now, it seems that makes sense. Thanks  

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