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I've been stuck on an exam question regarding long run growth

Consider the following model of economic growth

enter image description here

A) Derive the equilibrium growth rate of capital, and hence find the steady-state values of capital and output.

If there is no $K_t$ term in (4) how does one find the growth rate? Since the growth is found using $\Delta K/K_t \equiv \frac {Kt+1 - Kt} {Kt}$.

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    $\begingroup$ Are you sure that equation $(4)$ is correct? In absence of depreciation we should have $\Delta K= K_{t+1}-K_t=I_t$. An equation as $(4)$ implies that the capital stock of the preceding period $K_t$ is $0$, so that in each period the existing stock of capital is destroyed. It could be, even if it s unusual. $\endgroup$ Apr 10, 2023 at 17:07
  • $\begingroup$ Moreover, how is the steady state defined here? In the usual Solow model it is the state in which the capital-labor ratio remains constant, but here we haven’t any information about population growth and per capita variables. I suppose $\Delta K=0$ $\endgroup$ Apr 10, 2023 at 17:52
  • $\begingroup$ @BakerStreet Yes I copied it straight from the past paper, it's unusual as no other questions given to me have depreciation and/or Kt missing. The steady state is defined as DeltaKt+1 = 0. $\endgroup$
    – sffffii
    Apr 10, 2023 at 18:04
  • $\begingroup$ Ok, the absence of $K_t$ is not a problem, I give my suggestions in an answer below. $\endgroup$ Apr 10, 2023 at 18:08
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    $\begingroup$ Those equations could perhaps model a primitive agricultural economy with seed retained from one year to the next as the only capital, also output less than proportional to seed input because of a limited supply of land and/or labour. $\endgroup$ Apr 11, 2023 at 18:27

2 Answers 2

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To find the rate of growth of capital just subtract $K_t$ from both sides of $(4)$, to obtain:

$$\Delta K\equiv K_{t+1}-K_t= I_t-K_t \qquad (1)$$

from which, recalling equality of investment and saving

$$\frac {\Delta K}{K_t}= \frac {I_t-K_t}{K_t} =\frac {sA{K_t}^{1/2}- K_t}{K_t}\qquad (2)$$ .

You can find the steady state value of capital and output, defined as a state in which the stock of capital is constant, setting the preceding equation equal to $0$:

$$\frac {\Delta K}{K_t}= \frac {sA{K_t}^{1/2}- K_t}{K_t}=0\qquad (3)$$

From $(3)$ you obtain the steady state value of $K_t$ and then you can derive the steady state value of output plugging this value of $K_t$ into the production function as usual.

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  • $\begingroup$ Thats very helpful, thanks! In the second line, why didn't you divide by the Kt to cancel out Kt/Kt and reduce Kt^1/2 to Kt^-1/2? $\endgroup$
    – sffffii
    Apr 10, 2023 at 18:15
  • $\begingroup$ Yes, you can do it, it is the same. You can do the steps you prefer to find the solution. $\endgroup$ Apr 10, 2023 at 18:17
  • $\begingroup$ Much appreciated, thanks again. $\endgroup$
    – sffffii
    Apr 10, 2023 at 18:18
  • $\begingroup$ You are welcome! $\endgroup$ Apr 10, 2023 at 18:19
  • $\begingroup$ Hey, one more question. How would I go about deriving the saving rate that maximises the steady state consumption? $\endgroup$
    – sffffii
    Apr 10, 2023 at 18:47
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Are you sure that equation (4) is correct? In absence of depreciation we should have ΔK=Kt+1−Kt=It . An equation as (4) implies that the capital stock of the preceding period Kt is 0 , so that in each period the existing stock of capital is destroyed. It could be, even if it s unusual.

It must be that the depreciation rate is 100% indeed.

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  • $\begingroup$ Yes, it is the same, capital 'disappears' for some reason. It is a possible assumption, all models are not realistic to some extent. Just, to answer the question, one must be sure that the text is correct and there aren't typos. $\endgroup$ Apr 10, 2023 at 18:01
  • $\begingroup$ I thought so but I'm not sure if it's right, or how to go about solving it. $\endgroup$
    – sffffii
    Apr 10, 2023 at 18:04
  • $\begingroup$ Looking for a solution is possible even if $(4)$ is correct. $\endgroup$ Apr 10, 2023 at 18:06
  • $\begingroup$ @BakerStreet How would I go about it? $\endgroup$
    – sffffii
    Apr 10, 2023 at 18:08
  • $\begingroup$ I write my suggestions in an answer $\endgroup$ Apr 10, 2023 at 18:09

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