<p>Is that derivation for <span class="math-container">$k(t+1)$</span> correct? Technically, you never reach the steady state, but only asimptotically as <span class="math-container">$t\rightarrow\infty$</span>, but at infinity the <span class="math-container">$A(t)$</span> will also be infinite because it grows exponentially with time. I suspect that <span class="math-container">$k(t+1)$</span> should depend on <span class="math-container">$(1+g)$</span> instead of <span class="math-container">$A(t)$</span>.</p>
<p>Usually these models are expressed in units of effective labor, that is, dividing the capital stock <span class="math-container">$K(t)$</span> and the labor <span class="math-container">$L(t)$</span> (and hence, the production and consumption) by <span class="math-container">$N(t)$</span> and <span class="math-container">$A(t)$</span>.</p>

CC BY-SA 4.0
2021-01-22T05:55:02.293Z
This is much better posed as a comment, you don't answer the problem. Also as you can see this is not one of those problems in which we impose Harrod-neutral technological progress. While your point is valid, that is not what is being used for this specific problem
Brennan
CC BY-SA 4.0
2021-01-22T19:40:02.377Z
SE does not allow me to write comments. My point is that the capital stock, as is expressed in the formula written by OP does not have a steady state, because $A(t)$ grows exponentially. Maybe OP can share how he arrived at that law of motion for capital.
Pekisch
CC BY-SA 4.0
2021-01-22T01:54:22.820Z
2021-01-22T02:42:51.097Z
2021-01-22T02:42:51.097Z
Pekisch
<p>My question is from over lapping generations</p>
<p>Question is as follows</p>
<p><a href="https://i.sstatic.net/hirTi.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/hirTi.jpg" alt="enter image description here" /></a></p>
<p>I found that</p>
<p><span class="math-container">$$k(t+1)= \frac{\beta(1-\alpha)}{(1+\beta)(1+n)}A(t)k(t)^{\alpha}$$</span></p>
<p>How can I deal with A(t) to find the steady state <span class="math-container">$k^*$</span>?
By the way, at steady state,<span class="math-container">$ k(t+1)=k(t)=k^*$</span></p>

<macroeconomics><self-study><steady-state><endogenous-growth>
Question on overlapping generations
studentp
# Question on overlapping generations

(Asked by: username_0 on 2021-01-14)

My question is from over lapping generations

Question is as follows

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

I found that

$$k(t+1)= \frac{\beta(1-\alpha)}{(1+\beta)(1+n)}A(t)k(t)^{\alpha}$$

How can I deal with A(t) to find the steady state $k^\*$?
By the way, at steady state,$ k(t+1)=k(t)=k^\*$

**Tags:** macroeconomics, self-study, steady-state, endogenous-growth


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## 1 answers

### Answer by username_1 on 2021-01-22. Score: 1 | Accepted answer

Is that derivation for $k(t+1)$ correct? Technically, you never reach the steady state, but only asimptotically as $t\rightarrow\infty$, but at infinity the $A(t)$ will also be infinite because it grows exponentially with time. I suspect that $k(t+1)$ should depend on $(1+g)$ instead of $A(t)$.

Usually these models are expressed in units of effective labor, that is, dividing the capital stock $K(t)$ and the labor $L(t)$ (and hence, the production and consumption) by $N(t)$ and $A(t)$.

> **Comments:**
> **username_2**: This is much better posed as a comment, you don't answer the problem. Also as you can see this is not one of those problems in which we impose Harrod-neutral technological progress. While your point is valid, that is not what is being used for this specific problem  
> **username_1**: SE does not allow me to write comments. My point is that the capital stock, as is expressed in the formula written by OP does not have a steady state, because $A(t)$ grows exponentially. Maybe OP can share how he arrived at that law of motion for capital.  

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