Balanced Growth Path (Qualifier Question)

Question

$\textbf{Full disclaimer:}$ 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??

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".