Suppose a Solow-Swan model with $Y_t = F(K,AL)$ as the production function. Suppose the growth rate of $A$ and $L$ are $g > 0$ and $n > 0$ respectively. It is also given that $\dot{K} = sY - \delta K$ where $\delta$ is the rate of depreciation and $s$ is the savings rate. If $Z = \frac{K}{AL}$ and $Q_t = \frac{Y_t}{L_t}$, show that $\overline{Q_t} = A_t f(\overline{Z}) e^{gt}$ where $\overline{Q}$ is the balanced growth path value of $Q$, $\overline{Z}$ is the steady state value of $Z$ and $f(Z) := F(Z,1)$.


$Q_t = A_t \frac{F(K_t,A_tL_t)}{A_tL_t} = A_t f(Z_t) = A_t f(Z_t)$. At the steady state, $\frac{\dot{Z}}{Z} = \frac{s f(Z)}{Z} - (n + \delta + g) = 0 \implies f(\overline{Z}) = \frac{n+\delta+g}{s}\overline{Z}$.

Along the balanced growth path, let's assume $\frac{\dot{Y}}{Y} = g_Y$.

How do I proceed from here?

  • $\begingroup$ Are you sure that in the text is $\overline{Q_t} = A_t f(\overline{Z}) e^{gt}$ and not $\overline{Q_t} = A_0 f(\overline{Z}) e^{gt}$ , where $A_0$ is the initial value of $A_t$? $\endgroup$ Mar 27, 2023 at 11:49
  • $\begingroup$ @BakerStreet Exactly what came to my mind. I still don't know how to get there but I realized that $A_t = A_0 e^{gt}$ would probably mean that $\bar{Q_t} = A_0 e^{gt} f(\bar{Z})$. I don't understand the balanced growth path - as in, which all variables are supposed to be constants and which all should be equal to one another. Can you please suggest me a reference and maybe answer this too if time permits? Thank you. $\endgroup$ Mar 27, 2023 at 11:51
  • $\begingroup$ It could be a typo. $A_t$ is strange in this formula. $\endgroup$ Mar 27, 2023 at 11:52
  • $\begingroup$ @BakerStreet Very possibly so. Please read my previous comment; I have edited it. You don't have to solve the problem; you can just add a sketch of how the variables should be like along the balanced growth path. Thanks. $\endgroup$ Mar 27, 2023 at 11:54
  • $\begingroup$ In the steady state the ratio $Y_t/A_tL_t$, that is the output per effective worker is constant. It means that the output per worker $Q_t$ grows at the rate of growth of $A_t$, that is $g$. The exponential form of $\overline {Q} = A_0f(\overline ({Z}) e^{gt}$ means exactly that. $\endgroup$ Mar 27, 2023 at 12:00

1 Answer 1


To answer the question it is necessary to remember the behaviour of the variables in the steady state, so of course it is necessary to have solved previously the model for the steady state.$^1$

In your question we have the Solow model with labour augmenting technical progress, that is the production function is

$Y_t=f(K_t,A_t L_t)$

where $A_t$ represents technical progress and $A_t L_t$ is the so-called effective labour.

Solving the model for the steady state, we obtain that the variables per effective labour are constant, that is $Z_t\equiv \frac {K_t}{A_tL_t}= \overline {Z}$ and $y_t \equiv \frac {Y_t}{A_tL_t}$ are constant, where $\overline {Z}$ is the constant steady state value of $Z_t$.

Therefore we have, in steady state:

$y_t\equiv \frac {Y_t}{A_tL_t}= f(\overline {Z})\;\;\;\;\quad (1)$

from which

$ Q_t\equiv \frac {Y_t}{L_t}= A_tf(\overline {Z})\qquad (2)$

But, by assumption in the model, we have $^2$:

$A_t= A_0 e^{gt}\qquad \;\;\;\;\;\;\;\;\;\;\;(3)$

where $g$ is the rate of growth of technology $A_t$ and $A_0$ is the initial value of $A_t$.

Therefore, substituting in $(2)$, we obtain:

$ Q_t\equiv \frac {Y_t}{L_t}= A_0 f(\overline {Z})e^{gt}.\qquad (4)$ $$\;$$

$^1$ Of course, the solution of the model is too long to be reported here, you can look at a good text of macroeconomics, as Romer, Advanced macroeconomics or at some standard references on economic growth, as the books of Acemoglu or Barro & Sala i Martin.

$^2$ In the model there is the assumption that $A_t$ grows at constant rate $g$. Remember, very important, that the exponential form with $e^{at}$, as in $(3)$, is the mathematical expression to denote a growth at constant rate $a$ of a variable. If you calculate the growth rate in that case you can realize it.

  • $\begingroup$ Thanks a lot for the effort. (The website doesn't allow me to upvote you, so sorry about that, but your answer is extremely helpful.) $\endgroup$ Mar 27, 2023 at 23:58
  • $\begingroup$ You are welcome, thank you! $\endgroup$ Mar 28, 2023 at 0:46

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