Consider the following question:

enter image description here

So, assume the standard function for production:

$$Y_t = A_t K^\alpha_tL^{1-\alpha}$$

where $L$ is fixed.

Then, the growth rate of output is:

$$g_Y \approx g_A + \alpha g_K $$

Case without technological change

In the standard Solow model with constant $A$, $g_A=0$. In the steady state, $sY=dK$ (investment equal depreciation), and $g_Y = g_K = 0$, consistent with the above equation.

Case with exogenous technological change

Here, $g_A=2\%$ (as per the question). But then, there is clearly no steady state, as the marginal product of capital is permanently increasing, thereby expanding capital. In the standard diagram, this means a shift of the function $sY_t$ upwards every period, meaning an ever increasing "steady state", which is never achieved. Hence, to me the question does not make sense.

Finally, here is the official answer:

enter image description here

Now, to me this answer is wrong. How can $Y$, $A$, and $K$ growth at the same rate? The equation of growth written above denies this!

Is the question and/or answer wrong? Am I missing something? Am I wrong?

  • 2
    $\begingroup$ " But then, there is clearly no steady state". How do you define steady state ? In this context I think it refers to balanced growth equilibrium. $\endgroup$
    – user11629
    Feb 5, 2017 at 17:20
  • $\begingroup$ In any case, there is no such equilibrium where $A, K, Y$ grow at 2%. According to the growth formula, and intuitively, that is impossible. What is possible instead is, for instance, $g_Y=g_A=2\%$ and $g_K=4\%$ with $\alpha = 0.5$. But then the answer is wrong. $\endgroup$
    – luchonacho
    Feb 5, 2017 at 17:46
  • $\begingroup$ It would be useful to see you type out the exact contradiction, because I can't seem to find any. More precisely, how do you derive the growth rate of output? $\endgroup$
    – Giskard
    Feb 5, 2017 at 19:23
  • $\begingroup$ @denesp How can $Y, K, A$ all grow at the same rate? If you could prove that, then most of my cues would be gone. $\endgroup$
    – luchonacho
    Feb 6, 2017 at 11:03
  • $\begingroup$ The question might be improved by noting that the original Solow paper (piketty.pse.ens.fr/fichiers/Solow1956.pdf) did consider (p 85) a production function with neutral (rather than labour-augmenting) technical progress. $\endgroup$ Feb 8, 2017 at 11:07

2 Answers 2


To simplistically answer your question, use the following: $Y = K^\alpha (AL)^{1 - \alpha}$

In order to prove all three can be equal: We will assume that technological progress is labor augmenting (Harrod Neutral) or "labor saving", while holding the following true:

  • Returns to capital are roughly constant over time.
  • Capital share of income is roughly constant over time.

The Harrod Neutral is associated with a capital-output ratio that is constant (K/Y) in our "steady state".

Capital Accumulation In the basic Solow model, (1). $$ \frac{\dot{k}}{k} \equiv s\frac{y}{k}-(d+n)$$

For growth rate per capita capital to be constant, y/k = Y/K must also be constant. (Characteristic of the Harrod Neutral)

We Need To Define The Balanced Growth Path With Technological Progress Taking the original production equation: $Y = K^\alpha (AL)^{1 - \alpha}$

and dividing by AL (number of effective labor units) we are given our production function representing effective labor. $$ \tilde{y} \equiv \tilde{k}^\alpha$$

Doing The Same For Capital Accumulation Follow steps below: $$\frac{\tilde{\dot{k}}}{k} = \frac{dln(k/AL)}{dt}$$

$$ \frac{dlnK}{dt} - \frac{dlnA}{dt} - \frac{dlnL}{dt} $$

$$\frac{\dot{K}}{K} - \frac{\dot{A}}{A} - \frac{\dot{L}}{L} = \frac{\dot{K}}{K} - g - n$$

then using this to finding our Capital Accumulation equation below:

$$\frac{\dot{K}}{K} = S \frac{{Y}}{K} - d$$ $$\frac{\tilde{\dot{k}}}{\tilde{k}} = s\frac{Y}{K} - d - g - n$$ And, $$\frac{Y}{K} = \frac{Y}{AL} \frac{AL}{K} = \frac{\tilde{y}}{\tilde{k}}$$

Then, $$\frac{\tilde{\dot{k}}}{\tilde{k}} = s\frac{\tilde{y}}{\tilde{k}} - d - g - n$$

$$\tilde{\dot{k}} = s \tilde{y} - (d+g+n)\tilde{k}$$

Finally, combining our production function with capital accumulation we get an equation (6) that represents the Solow model with tech changes. $$\tilde{\dot{k}} = s \tilde{k^\alpha} - (d+g+n)\tilde{k}$$

So our steady state is when (7) is true. $$\tilde{\dot{k}^*} = 0$$

If we set this to be true for our steady state (denoted by *) levels of capital and output per effective labor. (8 and 9) 8: $$\tilde{k^*} = [\frac{s}{d+n+g}]^\frac{1}{1-\alpha}$$ 9: $$\tilde{y^*} = [\frac{s}{d+n+g}]^\frac{\alpha}{1-\alpha}$$

In the steady state: (10) Capital (and income) per capita grow at the rate of exogenous technological growth.

$$\tilde{k} = K/AL$$ $$k = K/L$$ $$k = A\tilde{k}$$

$$\tilde{k} = \tilde{k}^* = [\frac{s}{d+n+g}]^\frac{1}{1-\alpha}$$ $$k(t) = A(t)\tilde{k}^*$$ $$lnk(t) = lnA(t) + ln\tilde{k}^*$$ $$\frac{\dot{k}}{k} = \frac{\dot{A}}{A}$$

If you graph out $$\tilde{\dot{k}} = s \tilde{k^\alpha} - (d+g+n)\tilde{k}$$

  • Any savings rate changes, s, change the savings curve and therefore change $$\tilde{k^*}$$
  • Population or depreciation changes, n and d respectively, rotates the straight line, (d+g+n)k. Changes in these affect the growth rate of per capita capital and per capita income only outside the steady-state.
  • exo-tech growth changes the graph sort of like n and d by changing the level of $$\tilde{k^*}$$

This exo-tech growth changes the growth rate of per capita capital and per capita income in the steady state (along balanced growth paths).

  1. " But then, there is clearly no steady state". How do you define steady state ? In this context I think it refers to balanced growth equilibrium. See here.

  2. Yes there is a contradiction. Maybe $Y = K^\alpha (AL)^{1 - \alpha}$ ? This would mean $A$ is some measure of the efficiency of labor.

  • 1
    $\begingroup$ I think you did not realise how your answer was helpful. If you use the formula you have, you can achieve a balanced growth path where the three variables grow at the same rate, as in the official answer. Can you write that down in your answer? $\endgroup$
    – luchonacho
    Feb 8, 2017 at 14:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.