$\textbf{Full disclaimer:}$ I am studying for my candidacy exams and this question is from one of the exams last year.

$$max\;\sum_{t=0}^\infty \beta^t \frac{c_t^{1-\gamma}}{1-\gamma}$$Subject to the following constraints:

$$c_t+i_t=y_t$$ $$y_t=A_tk_t^\alpha$$ $$k_{t+1}=(1-\delta)k_t+i_t$$ $$ln(A_{t+1})=ln(A_t)+\mu$$ $$c_t,k_{t+1}\geq 0$$ (b) Compute a balanced growth path in which capital, consumption and output grow at constant rates. On the balanced growth path

(i) What is the growth rate of capital?

(ii) What is the growth rate of consumption?

(iii) What is the growth rate of output?

(iv) What is the ratio of capital to output?

(v) What is the ratio of consumption to output?

$\textbf{My work:}$

Combining the constraints we get: $$c_t=A_tk_t^\alpha+(1-\delta)k_t-k_{t+1}$$

Taking two terms from the objective function we get: $$...+\beta^t\bigg(\frac{\big(A_tk_t^\alpha+(1-\delta k_t-k_{t+1}\big)^{1-\gamma}}{1-\gamma}\bigg)+\beta^{t+1}\bigg(\frac{\big(A_{t+1}k_{t+1}^\alpha+(1-\delta k_{t+1}-k_{t+2}\big)^{1-\gamma}}{1-\gamma}\bigg)+...$$

Differentiating w.r.t $k_{t+1}$ and substituting consumption back in we get: $$c_t^{-\gamma}=\beta\bigg[\alpha \bigg(\frac{A_{t+1}k_{t+1}^\alpha}{k_{t+1}}\bigg)+(1-\delta)\bigg]c_{t+1}^{-\gamma}$$ $$\implies \frac{c_{t+1}}{c_t}=\bigg[\beta\bigg(\alpha\bigg(\frac{y_{t+1}}{k_{t+1}}\bigg)+(1-\delta)\bigg)\bigg]^{\frac{1}{\gamma}}\qquad(1)$$ I also derived: $$\frac{k_{t+1}}{k_t}=(1-\delta)+\frac{c_t}{k_t}-\frac{y_t}{k_t}\qquad(2)$$ $$\frac{y_{t+1}}{y_t}=e^\mu\bigg(\frac{k_{t+1}}{k_t}\bigg)^\alpha\qquad(3)$$ My train of thought is the following:

$\frac{k_{t+1}}{k_t}=\frac{c_{t+1}}{c_t}$ because if $\frac{k_{t+1}}{k_t}>\frac{c_{t+1}}{c_t}$ we would not satisfy the transversality condition because we would have capital in the "last" period which means we cannot be maximizing consumption. If $\frac{k_{t+1}}{k_t}<\frac{c_{t+1}}{c_t}$, we would eventually not be able to maintain constant consumption. Using that and the fact that the ratio of output to capital will be constant over time, equating equations (1) and (2) yields the following messy results: $$\bigg(\frac{y_t}{k_t}\bigg)^\gamma+\alpha\beta\bigg(\frac{y_t}{k_t}\bigg)=\bigg((1-\delta)+\frac{c_t}{k_t}\bigg)^\gamma+\beta(1-\delta)$$ As far as I can tell, this equation isn't solvable. Does this mean I cannot say that the growth rate of consumption is equal to the growth rate of capital or have I made a mistake somewhere along the way? Also, how would one solve for the growth rates if not this way??

  • $\begingroup$ You seem to ignore the "hint" provided by the $\ln (A_{t+1})$ equation, (don't you think you should consider why it is given in logarithms?) , which is the critical ingredient that delivers constant growth (otherwise the diminishing marginal product of capital would lead to zero growth). Remember also that $\ln x_{t+1} - \ln x_{t}$ is an acceptable approximation to the growth rate of a variable. Use these two to linearize things (and when you do, you may consider posting an answer to your own question, which is totally acceptable practice). $\endgroup$ Jun 7, 2016 at 0:03
  • $\begingroup$ @AlecosPapadopoulos thanks I'll take a look! $\endgroup$
    – DornerA
    Jun 7, 2016 at 11:27
  • $\begingroup$ @AlecosPapadopoulos can I still assert the growth rate for capital being equal to the growth rate for consumption? Or is this not necessary? $\endgroup$
    – DornerA
    Jun 7, 2016 at 15:29
  • $\begingroup$ It is, and for the reasons you have already explained in your question. $\endgroup$ Jun 7, 2016 at 16:46
  • $\begingroup$ @AlecosPapadopoulos So I get what you were telling me that those two things lead to $ln(y_{t+1})-ln(y_t)=\mu+\alpha(ln(k_{t+1})-ln(k_t))$ and allows output to stay above zero, but I still get stuck there. I can't figure out a way to get an equation that solves for one of my variables like $\frac{y_t}{k_t}$. I don't know if it's too much to ask, but could you give me an idea of any equation manipulations that I may need to do? I tried to look to see what it would look like if I log-linearized all of the constraints, but that didn't seem to help because I cannot split up any of the added terms. $\endgroup$
    – DornerA
    Jun 7, 2016 at 19:50

1 Answer 1


(That one can use the log-difference approximation for the growth rates, can be glimpsed by the fact that while the model apparently is set in discrete time, the log-evolution of technology is expressed using the exponential, which is how we express a constant growth rate in continuous time).

From $$\frac{y_{t+1}}{y_t}=e^\mu\bigg(\frac{k_{t+1}}{k_t}\bigg)^\alpha\qquad(3)$$

Taking logs you get

$$g_y = \mu + \alpha g_k$$

Where the "$g$"s are growth rates. One the balanced growth path, they have to be equal to some constant, say $g^*$, so you have

$$g^* = \mu + \alpha g^* \implies g^* = \frac {\mu}{1-\alpha}$$

which is the growth rate of the economy on the balanced growth path.

You should also be able (by simmply manipulating the equation with which you start showing your work), to arrive easily at

$$\left (\frac {c}{y}\right)^* = 1- (\delta +g^*)\left (\frac {k}{y}\right)^*$$ So you need only to determine one of the two to obtain the other.

Realize how you can write the left-hand side of $(1)$ while on the balanced growth path, then turn the equation around and solve for $k/y$. Don't expect to get something necessarily "simple-looking".


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