K Problem Set 3 - Solution

K.1 The Solow Model with Exogenous Growth

  1. The saving rate is exogenous and equal to \(s\) in the Solow growth model, and the depreciation rate is \(\delta\). Therefore, the law of motion for capital is: \[\Delta K_{t+1}=K_{t+1}-K_t=sY_t-\delta K_t.\] Using the value for \(Y_t\), we get: \[\boxed{K_{t+1}=s A_t K_t^{\alpha} L_t^{1-\alpha} + (1-\delta) K_t}.\] which is a law of motion for \(K_t\): a value for \(K_{t+1}\) as a function of \(K_t\) and the exogenous parameters in the model.

  2. Defining \(k_t\) as: \[k_t\equiv\frac{K_t}{A_t^{1/(1-\alpha)} L_t},\] as is suggested, we divide both the left-hand side and the right-hand side of the equation by \(A_t^{1/(1-\alpha)} L_t\).

This gives: \[\frac{K_{t+1}}{A_t^{1/(1-\alpha)} L_t}=s \frac{A_t K_t^{\alpha} L_t^{1-\alpha}}{A_t^{1/(1-\alpha)} L_t} + (1-\delta) \frac{K_t}{A_t^{1/(1-\alpha)} L_t}\] We may proceed to a simplification of the first term on the right-hand side by putting the \(A_t\) and the \(L_t^{1-\alpha}\) from the numerator to the denominator (using that \(f/g=1/(g/f)\)): \[ \begin{aligned} \frac{A_t K_t^{\alpha} L_t^{1-\alpha}}{A_t^{1/(1-\alpha)} L_t}&=\frac{K_t^\alpha}{A_t^{1/(1-\alpha)-1}L_t^{1-(1-\alpha)}}\\ &=\frac{K_t^\alpha}{A_t^{\alpha/(1-\alpha)}L_t^{\alpha}}\\ \frac{A_t K_t^{\alpha} L_t^{1-\alpha}}{A_t^{1/(1-\alpha)} L_t}&=\left(\frac{K_t}{A_t^{1/(1-\alpha)}L_t}\right)^\alpha. \end{aligned} \] Thus, replacing out the expression for the first term on the right-hand side allows to write: \[\frac{K_{t+1}}{A_t^{1/(1-\alpha)} L_t}=s\left(\frac{K_t}{A_t^{1/(1-\alpha)}L_t}\right)^\alpha+(1-\delta)\frac{K_t}{A_t^{1/(1-\alpha)} L_t}.\] And therefore, the right-hand side is now expressed only a function of \(k_t\): \[\frac{K_{t+1}}{A_t^{1/(1-\alpha)} L_t}=sk_t^\alpha+(1-\delta)k_t.\] The left-hand side of the equation can also be simplified (we want to express it also only as a function of \(k_t\) (or rather, \(k_{t+1}\)): \[ \begin{aligned} \frac{K_{t+1}}{A_t^{1/(1-\alpha)} L_t}&= \frac{A_{t+1}^{1/(1-\alpha)} L_{t+1}}{A_t^{1/(1-\alpha)} L_t} \cdot \frac{K_{t+1}}{A_{t+1}^{1/(1-\alpha)} L_{t+1}} \\ \frac{K_{t+1}}{A_t^{1/(1-\alpha)} L_t}&=(1+g)^{1/(1-\alpha)}(1+n) k_{t+1}. \end{aligned} \] Therefore: \[(1+g)^{1/(1-\alpha)}(1+n) k_{t+1} =sk_t^\alpha+(1-\delta)k_t.\] If \(g\) and \(n\) are small then: \[(1+g)^{1/(1-\alpha)}(1+n)\approx1+\frac{1}{1-\alpha}g+n.\] Thus: \[\left(1+\frac{1}{1-\alpha}g+n\right)k_{t+1}\approx s k_t^\alpha+(1-\delta)k_t.\] A law of motion for \(k_{t+1}\) is thus (we use equal signs now, even though it is really an approximation): \[\boxed{k_{t+1}=\frac{s}{1+g/(1-\alpha)+n}k_t^{\alpha}+\frac{1-\delta}{1+g/(1-\alpha)+n}k_t}.\]

  1. The steady-state is such that: \[\left(1+\frac{1}{1-\alpha}g+n\right)k^{*} = s(k^{*})^\alpha + (1-\delta)k^{*}.\] Therefore: \[\left(\delta+\frac{1}{1-\alpha}g+n\right)k^{*} = s(k^{*})^\alpha.\] Finally, this gives \(k^{*}\): \[\boxed{k^{*}=\left(\frac{s}{\delta+g/(1-\alpha)+n}\right)^{\frac{1}{1-\alpha}}}.\]

  2. In this exercise, we make intensive use of the following rules on growth rates: \[ \begin{aligned} g_{XY}&=g_X+g_Y\\ g_{X/Y}&=g_X-g_Y\\ g_{X^a}&=ag_X. \end{aligned} \] On the balanced growth path: \[\frac{K_t}{A_t^{1/(1-\alpha)} L_t}=k^{*} \quad \Rightarrow \quad K_t = k^{*}A_t^{1/(1-\alpha)} L_t.\] We may apply the rule above on products (\(g_{XY}=g_X+g_Y\)) to see that the growth rate of \(K_t\) is the growth rate of \(A_t^{1/(1-\alpha)}\) plus the growth rate of \(L_t\), since \(k^{*}\) is simply a constant which does not grow. In turn, using the rule on “powers” (that is \(g_{X^a}=ag_X\), with \(a=1/(1-\alpha)\)) we get that the growth rate of \(A_t^{1/(1-\alpha)}\) is the growth rate of \(A_t\) times \(1/(1-\alpha)\). Finally, the growth rate of \(A_t\) is \(g\) and the growth rate of \(L_t\) is \(n\) by assumption. Thus, finally: \[ \begin{aligned} g_K &=g_{A^{1/(1-\alpha)}L}\\ &=g_{A^{1/(1-\alpha)}} + g_L\\ &=\frac{1}{1-\alpha}g_A + g_L\\ g_K &=\frac{1}{1-\alpha}g + n \end{aligned} \] Output is given by: \[Y_t=A_t K_t^{\alpha} L_t^{1-\alpha}\] Therefore, the rate of growth of output is: \[ \begin{aligned} g_Y &= g + \alpha g_K +(1-\alpha)g_L\\ &= g+ \alpha\left(n+\frac{1}{1-\alpha}g\right)+(1-\alpha)n\\ &= g+ \alpha n+\frac{\alpha}{1-\alpha}g+(1-\alpha)n \\ &= \left[\alpha n+(1-\alpha)n\right] + \left[g+ \frac{\alpha}{1-\alpha}g\right] \\ g_Y&=n+\frac{1}{1-\alpha}g. \end{aligned} \] \(C_t\) grows at the same rate as \(Y_t\) since \(C_t=(1-s)Y_t\), thus: \[ \begin{aligned} g_C&=g_Y \\ g_C&=n+\frac{1}{1-\alpha}g. \end{aligned} \] The rate of growth of \(K_t/Y_t\) is zero since \(Y_t\) and \(k_t\) grow at the same rate: \[ \begin{aligned} g_{K/Y}&=g_K-g_Y\\ &=\left(\frac{1}{1-\alpha}g + n\right)-\left(\frac{1}{1-\alpha}g + n\right)\\ g_{K/Y}&=0 \end{aligned} \] The rate of growth of \(K_t/L_t\) is: \[ \begin{aligned} g_{K/L}&=g_K-g_L\\ &= \left(\frac{1}{1-\alpha}g + n\right)-n \\ g_{K/L}&=\frac{1}{1-\alpha}g. \end{aligned} \] The wage is equal to the marginal product of labor from firms’ optimality condition, as in lecture ??: \[ \begin{aligned} w_t&=\frac{\partial Y_t}{\partial L_t}\\ &=(1-\alpha)A_t K_t^\alpha L_t^{-\alpha}\\ w_t&=(1-\alpha)A_t \left(\frac{K_t}{L_t}\right)^{\alpha}. \end{aligned} \] Thus, the rate of growth of \(w_t\) is: \[ \begin{aligned} g_w&=g_A+\alpha g_{K/L}\\ &=g+\frac{\alpha}{1-\alpha}g\\ g_w&=\frac{1}{1-\alpha}g. \end{aligned} \] The rate of growth of \(w_t L_t\) is the sum of that of \(w\) and that of \(L_t\) thus: \[ \begin{aligned} g_{wL}&=g_w + g_L \\ g_{wL}&=\frac{1}{1-\alpha}g+n. \end{aligned} \] The marginal product of capital \(R_t\) is: \[ \begin{aligned} R_t&=\frac{\partial Y_t}{\partial K_t}\\ &=\alpha A_t K_t^{\alpha-1}L_t^{1-\alpha}\\ R_t&= \alpha A_t \left(\frac{K_t}{L_t}\right)^{\alpha-1}. \end{aligned} \] Thus, the rate of growth of the marginal product of capital \(R_t\) is: \[ \begin{aligned} g_R&=g_A+(\alpha-1) g_{K/L}\\ &=g +\frac{\alpha-1}{1-\alpha}g\\ &=g-g\\ g_R&=0 \end{aligned} \] The rate of growth of capital income \(R_t K_t\) is given by: \[ \begin{aligned} g_{R K}&=g_R + g_K \\ g_{R K}&= n + \frac{g}{1-\alpha}. \end{aligned} \] Finally, the growth in the labor share \(w_t L_t\) and that in the capital share \(R_t K_t\) are equal to zero which can be inferred from the fact that they are constant with a Cobb-Douglas production function, or that the growth of \(w_t L_t\) and \(r_t K_t\) are equal to that of output.

  3. Using the expression for \(y_t\) allows to write what is called the intensive form of the production function: \[ \begin{aligned} y_t&=\frac{Y_t}{A_t^{1/(1-\alpha)} L_t}\\ &= \left(\frac{K_t}{A_t^{1/(1-\alpha)}L_t}\right)^\alpha\\ y_t&= k_t^\alpha. \end{aligned} \] This implies that the relationship applies also to the steady state, and allows us to calculate steady-state \(y^{*}\) corresponding to steady-state \(k^{*}\): \[ \begin{aligned} y^*=\left(k^*\right)^\alpha \end{aligned} \] From question 3, we replace out \(k^{*}\) in the equation above which gives directly: \[ \begin{aligned} \boxed{y^{*}=\left(\frac{s}{\delta+g/(1-\alpha)+n}\right)^{\frac{\alpha}{1-\alpha}}}. \end{aligned} \] Finally, using that \(C_t = (1-s)Y_t\) and dividing on both sides by \(A_t^{1/(1-\alpha)} L_t\) gives: \[\frac{C_t}{A_t^{1/(1-\alpha)} L_t}=(1-s)\frac{Y_t}{A_t^{1/(1-\alpha)} L_t}.\] Using the given definitions for \(c_t\) and \(y_t\), this implies that: \[c_t = (1-s)y_t.\] From this we can see that this relationship applies also to steady states so that: \[c^{*} = (1-s)y^{*}.\] Thus, replacing \(y^{*}\) by its expression from previously: \[ \begin{aligned} \boxed{c^{*}=(1-s)\left(\frac{s}{\delta+g/(1-\alpha)+n}\right)^{\frac{\alpha}{1-\alpha}}}. \end{aligned} \]

  4. Just as in the Solow growth model of lecture ??, we see that we have a constant times a function of \(s\), which simplifies the maximization problem a lot: \[c^{*}=\frac{1}{\left(\delta+g/(1-\alpha)+n\right)^{\frac{\alpha}{1-\alpha}}} (1-s)s^{\frac{\alpha}{1-\alpha}}\] We are thus led to maximize only the part which depends on \(s\) (if you are not convinced, you can leave the constant there, your calculations will just be more complicated!): \[ \begin{aligned} \max_s \quad (1-s)s^{\frac{\alpha}{1-\alpha}}. \end{aligned} \] Taking the first-order condition as in lecture ??: \[ \begin{aligned} &-s^{\frac{\alpha}{1-\alpha}}+\frac{\alpha}{1-\alpha}(1-s)s^{\frac{\alpha}{1-\alpha}-1} =0 \quad \Rightarrow\quad\frac{\alpha}{1-\alpha}\frac{1-s}{s}=1\\ &\quad \quad \Rightarrow\quad\alpha-\alpha s=s-\alpha s \quad\Rightarrow\quad\boxed{s=\alpha}. \end{aligned} \]

  5. The marginal product of capital \(R_t\) is then equal to: \[ \begin{aligned} R_t&=\alpha A_t \left(\frac{K_t}{L_t}\right)^{\alpha-1}\\ &=\alpha \left(\frac{K_t}{A_t^{1/(1-\alpha)} L_t}\right)^{\alpha-1}\\ R_t&=\alpha k_t^{\alpha-1}. \end{aligned} \] Therefore, in the steady state, using the above expression for \(k^{*}\) (question 3) we get: \[ \begin{aligned} R^{*}&=\alpha \left[\left(\frac{s}{\delta + g/(1-\alpha)+n}\right)^{\frac{1}{1-\alpha}}\right]^{\alpha-1}\\ &=\alpha \frac{\delta + g/(1-\alpha)+n}{s}\\ &=\delta + g/(1-\alpha)+n\\ R^{*}&=\frac{\alpha}{s}\left(\delta + g_Y\right). \end{aligned} \] where we have used that the rate of growth of output \(g_Y\) is given by \(g_Y = g/(1-\alpha)+n\) which was proved in an earlier question. Using that \(s=\alpha\) at the Golden Rule, we get an expression for the steady-state marginal product of capital: \[s= \alpha \quad \Rightarrow \quad \boxed{R^{*} = \delta + g_Y}.\] Finally, note that the net interest rate at the Golden Rule \(R^{*}-\delta\), which is often denoted by \(r^{*}\) needs to be equal to the rate of growth of output \(g_Y\), to be at the Golden Rule level of capital accumulation: \[\boxed{r^{*}=R^{*}-\delta=g_Y}.\] We shall encounter this condition again in lecture 10 when we study the sustainability of public debt.

K.2 Solow Growth Model if \(\alpha = 1/3\)

  1. Yes, there are constant returns to scale. When one doubles all inputs, one gets double the output. This is true more generally for any \(x\): \[F(xK,xL)=(xK)^\alpha (xL)^{1-\alpha} = x K^\alpha L^{1-\alpha}= x F(K,L).\]

  2. Yes, returns are decreasing with respect to capital. The reason is that the derivative of the production function with respect to the capital stock, which is: \[\frac{\partial F}{\partial K}=\alpha K^{\alpha-1} L^{1-\alpha},\] is decreasing in the amount of capital (indeed, since \(\alpha=1/3\) we have that the exponent on \(K\) is \(-2/3\), so that this is a decreasing function). This implies that the “gross returns to capital”, defined as the additional production allowed by one additional unit of capital, given by the derivative \(\partial F/\partial K\), are decreasing in \(K\). Another way to see this is to note that the production function is concave in the capital stock, as the second derivative is negative: \[\frac{\partial^2 F}{\partial K^2}=\alpha (\alpha-1) K^{\alpha-2} L^{1-\alpha}<0\] which is just another characterization of decreasing returns. This is because \(\alpha-1=-2/3\) which is negative. See lecture ?? for more detail.

  3. Yes, returns are decreasing with respect to labor, for the same reason as returns are decreasing with respect to capital. The reason is that the derivative of the production function with respect to the amount of labor (number of employees, or number of hours), which is: \[\frac{\partial F}{\partial L}=(1-\alpha) K^{\alpha} L^{-\alpha},\] is decreasing in the amount of labor (indeed, since \(\alpha=1/3\) we have that the exponent on \(L\) is \(-1/3\), so that this is a decreasing function). This implies that the “returns to labor”, defined as the additional production allowed by one additional unit of labor, given by the derivative \(\partial F/\partial L\), are decreasing in \(L\). Another way to see this is to note that the production function is concave in the stock of labor, as the second derivative is negative: \[\frac{\partial^2 F}{\partial L^2}=-(1-\alpha)\alpha K^{\alpha} L^{-\alpha-1}<0\] which is just another characterization of decreasing returns. See Lecture 2 for more detail.

  4. Dividing the LHS and the RHS by \(L\): \[\frac{Y}{L}=\frac{K^\alpha L^{1-\alpha}}{L} = \left(\frac{K}{L}\right)^\alpha = F\left(\frac{K}{L}, 1\right).\] Definining the intensive form of the production function by \(f(.)\): \[f(k) \equiv F(k, 1),\] we can then write: \[\frac{Y}{L}=f\left(\frac{K}{L}\right).\]

  5. Again, we write the lax of motion of the capital stock as: \[ \begin{aligned} K_{t+1}&=(1-\delta)K_{t}+I_{t}\\ K_{t+1}&=(1-\delta)K_{t}+sK_{t}^{1/3}L^{2/3} \end{aligned} \] Dividing both sides by \(L\): \[\frac{K_{t+1}}{L}=(1-\delta)\frac{K_{t}}{L}+s\left(\frac{K_{t}}{L}\right)^{1/3}.\] In steady state, \[\frac{K_{t+1}}{L}=\frac{K_{t}}{L}=\frac{K^{*}}{L},\] This implies: \[\delta\frac{K^{*}}{L}=s\left(\frac{K^{*}}{L}\right)^{1/3} \quad \Rightarrow \quad \boxed{\frac{K^{*}}{L}=\left(\frac{s}{\delta}\right)^{3/2}}\]

  6. Using that: \(Y^{*}=K^{*1/3}L^{2/3}\), we have: \[\frac{Y^{*}}{L}=\left(\frac{K^{*}}{L}\right)^{1/3}\quad\Rightarrow\quad\boxed{\frac{Y^{*}}{L}=\sqrt{\frac{s}{\delta}}}.\]

  7. A straightforward numerical application gives: \[\frac{Y^{*}}{L}=\sqrt{\frac{0.32}{0.08}}=\sqrt{4}=2\]

  8. If the saving rate declines to 16, then: \[\frac{Y^{*}}{L}=\sqrt{\frac{s}{\delta}}=\sqrt{\frac{0.16}{0.08}}=\sqrt{2}\]

K.3 An increase in the depreciation rate in the Solow growth model

  1. The steady-state level of capital per worker is: \[\frac{K^{*}}{L}=\left(\frac{s}{\delta}\right)^{3/2}=\left(\frac{0.10}{0.10}\right)^{3/2}=1.\]

  2. The steady-state level of output per worker is: \[\frac{Y^{*}}{L}=\sqrt{\frac{s}{\delta}}=\sqrt{\frac{0.10}{0.10}}=1.\]

  3. The new steady-state levels of capital per worker and output per worker will be: \[\frac{K^{*}}{L}=\left(\frac{s}{\delta}\right)^{3/2}=\left(\frac{0.10}{0.20}\right)^{3/2} \approx 0.35,\] \[\frac{Y^{*}}{L}=\sqrt{\frac{s}{\delta}}=\sqrt{\frac{0.10}{0.20}}\approx 0.71.\]

  4. We know the evolution of capital per worker is: \[\frac{K_{t+1}}{L}=(1-\delta)\frac{K_{t}}{L}+s\left(\frac{K_{t}}{L}\right)^{1/3}.\] Starting from \(\frac{K_{0}}{L}=1\), with \(s=0.10\), \(\delta=0.20\), we have: \[ \begin{aligned} \frac{K_{1}}{L}&=(1-\delta)\frac{K_{0}}{L}+s\left(\frac{K_{0}}{L}\right)^{1/3}=0.9\\ \frac{K_{2}}{L}&=(1-\delta)\frac{K_{1}}{L}+s\left(\frac{K_{1}}{L}\right)^{1/3} \approx 0.82\\ \frac{K_{3}}{L}&=(1-\delta)\frac{K_{2}}{L}+s\left(\frac{K_{2}}{L}\right)^{1/3}\approx 0.75 \end{aligned} \] For more iterations, you may use Google Sheets: the result is available here. You should see that it indeed converges to the above values. From there, we may calculate the path of output per worker: \[ \begin{aligned} \frac{Y_{1}}{L}&=\left(\frac{K_{1}}{L}\right)^{1/3}\approx 0.97\\ \frac{Y_{2}}{L}&=\left(\frac{K_{2}}{L}\right)^{1/3} \approx 0.93\\ \frac{Y_{3}}{L}&=\left(\frac{K_{3}}{L}\right)^{1/3}\approx 0.91. \end{aligned} \] For more iterations, you may use Google Sheets: the result is available here.

K.4 Deficits and the capital stock

  1. Using the law of motion for the capital stock: \[ \begin{aligned} K_{t+1}&=(1-\delta)K_{t}+I_{t}\\ &=(1-\delta)K_{t}+sY_{t}\\ K_{t+1}&=(1-\delta)K_{t}+s K_{t}^{1/3} L^{2/3}, \end{aligned} \] In the steady-state: \[\delta K^{*} = s (K^{*})^{1/3} L^{2/3}\] This implies that the steady-satte capital stock per worker is such that: \[\frac{K^{*}}{L}=\left(\frac{s}{\delta}\right)^{3/2}\] Therefore, using the intensive form of the production function: \[\frac{Y^{*}}{L}=\left(\frac{K^{*}}{L}\right)^{1/3}=\left(\frac{s}{\delta}\right)^{1/2}\] This implies that the steady-state output per worker is: \[\frac{Y^*}{L} = \sqrt{\frac{s}{\delta}}.\]

  2. The steady-state capital stock per worker is given by: \[\frac{K^{*}}{L}=\left(\frac{s}{\delta}\right)^{3/2}=\left(\frac{15\%}{7.5\%}\right)^{2}=2^{3/2}=2\sqrt{2}\approx 2.828.\] Steady-state output per worker is given by: \[\frac{Y^{*}}{L}=\sqrt{\frac{s}{\delta}}=\sqrt{\frac{15\%}{7.5\%}}=\sqrt{2}\approx 1.414.\]

  3. If the elimination of the government deficit leads to an increase in the saving rate from initially 15% to 20%, then the steady-state capital stock per worker increases since: \[\frac{K^{*}}{L}=\left(\frac{s}{\delta}\right)^{3/2}=\left(\frac{20\%}{7.5\%}\right)^{3/2} \approx 4.35.\] The new steady-state output per worker is: \[\frac{Y^{*}}{L}=\left(\frac{s}{\delta}\right)^{1/2} = \left(\frac{20\%}{7.5\%}\right)^{1/2} \approx 1.63.\] Therefore, both the capital per worker and the output per worker increase.

K.5 U.S. saving and government deficits

  1. According to http://data.worldbank.org/indicator/NY.GDS.TOTL.ZS, the national saving rate was approximately 16.9% in 2016. The steady-state capital stock per worker is given by: \[\frac{K^{*}}{L}=\left(\frac{s}{\delta}\right)^{3/2}=\left(\frac{16.9\%}{7.5\%}\right)^{3/2} \approx 3.38.\]
# [1] 3.382503

The steady-state output per worker is: \[\frac{Y^{*}}{L}=\left(\frac{s}{\delta}\right)^{1/2}=\left(\frac{16.9\%}{7.5\%}\right)^{1/2} \approx 1.5.\]

# [1] 1.501111
  1. For fiscal year 2017, the federal fiscal deficit was 3.5% percent of GDP. Assuming that the federal budget deficit was eliminated and there was no change in private saving, the saving rate would change from 16.9% to 16.9%+3.5%=20.4%. The new steady-state capital stock per worker would be: \[\frac{K^{*}}{L}=\left(\frac{s}{\delta}\right)^{3/2}=\left(\frac{20.4\%}{7.5\%}\right)^{3/2} \approx 4.49\]
# [1] 4.485939

The new steady-state output per worker would be: \[\frac{Y}{L}=\left(\frac{s}{\delta}\right)^{1/2}=\left(\frac{20.4\%}{7.5\%}\right)^{1/2}\approx 1.65\]

# [1] 1.649242
  1. The saving rate in China was 46.54% in year 2016, which is much higher than the saving rate in the United States. This is perhaps not that surprising according to the Solow model, as China is still in the process of catching up. At the same time, it is not clear that what China was lacking before was capital, rather than market-oriented economic reforms.