# Steady state Solow model with exogenous technological change

Consider the following question: 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:

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?

• " 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. – Ululo Feb 5 '17 at 17:20
• 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. – luchonacho Feb 5 '17 at 17:46
• 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? – Giskard Feb 5 '17 at 19:23
• @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. – luchonacho Feb 6 '17 at 11:03
• 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. – Adam Bailey Feb 8 '17 at 11:07

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.

• 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? – luchonacho Feb 8 '17 at 14:09