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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:

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And:

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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:

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Could you show me how to derive the last equation?

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

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.

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  • $\begingroup$ @Mike if you approved the question, please accept it as the best answer $\endgroup$ Aug 4 at 18:58
  • $\begingroup$ 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 $x(t) ?$ $\endgroup$
    – Mike
    Aug 5 at 13:28
  • $\begingroup$ 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. $v=x(t)$. $\endgroup$ Aug 5 at 13:43
  • $\begingroup$ 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 $[t+\Delta t, \infty]$ $\endgroup$
    – Mike
    Aug 5 at 16:08
  • $\begingroup$ 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 $t+\Delta t$. $\endgroup$ Aug 5 at 18:25

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