This question is an extension of Endogenous Growth: Balanced Growth Path with CRRA Utility however, this question asks about a specific concept used in that question, and I think it would be helpful to have a question dedicated to this concept.

We will use the model in this question:

$\textbf{Model:}$ $$K_t=\frac{1}{n}\sum_{t=1}^nk_t$$ In this model, $k_t$ is chosen by agents, and $K_t=\bar{k}_t$ (the average of all $k_t$).

Now, agents want to dynamically maximize utility (under certain constraints) and they have CRRA (constant relative risk aversion) utility, so the maximization looks like: $$\sum_{t=0}^\infty\beta^t\bigg(\frac{c_t^{1-\gamma}}{1-\gamma}\bigg)$$ $$s.t.\;Y_t=k_t^\alpha(E_tL)^{1-\alpha}$$ $$c_t+i_t=Y_t$$ $$k_{t+1}=(1-\delta)k_t+i_t$$ $$c_t,i_t\geq0$$

$E_tL$ is effective labor and the rest of the variables are typical (I can give their definitions if requested).

One last addition to the model is that there are two equilibrium constraints: $$E_t=\frac{K_t}{L}$$ $$k_t=K_t$$

In the answers section, I was instructed to use transversality conditions and inada conditions in order to show that a balanced growth path is optimal. Below is my derivation of the transversality condition:

First, we must derive the finite horizon transversality condition. We do this by solving: $$\underset{\{k_t\}_{t=0}^{T+1}}{max}\;\sum_{t=0}^{T}\beta^tU(\frac{}{})$$ $$s.t.\;k_{T+1}\geq0$$ With our model that means: $$\underset{\{k_t\}_{t=0}^{T+1}}{max}\;\sum_{t=0}^{T}\beta^t\bigg(\frac{((1+1-\delta)k_t-k_{t+1})^{1-\gamma}}{1-\gamma}\bigg)$$ $$s.t.\;k_{T+1}\geq0$$

Our Lagrangian is: $$\ell(\frac{}{})=\sum_{t=0}^{T}\beta^t\bigg(\frac{((1+1-\delta)k_t-k_{t+1})^{1-\gamma}}{1-\gamma}\bigg)+\lambda k_{T+1}$$

In order to get the transversality condition, we differentiate with respect to $k_{T+1}$ and use the constraint.

FOC: $$\frac{\beta^T}{((1+(1-\delta))k_T-k_{T+1})^\gamma}=\lambda \qquad (1)$$ $$\lambda k_{T+1}=0 \qquad (2)$$ $$\lambda , k_{T+1}\geq 0 \qquad (3)$$ Solving this system, we get $\textbf{the finite horizon transversality condition is:}$ $$\bigg(\frac{\beta^T}{((1+(1-\delta))k_T-k_{T+1})^\gamma}\bigg)k_{T+1}=0$$ Now, in order to get the infinite horizon transversality condition, we maximize the discounted utility function again, but we set the limit of finite horizon transversality condition as $T\rightarrow \infty$ equal to 0 as the constraint.

This means that we have to solve: $$\underset{\{k_t\}_{t=0}^\infty}{max}\;\sum_{t=0}^\infty\beta^t\bigg(\frac{((1+(1-\delta))k_t-k_{t+1})^{1-\gamma}}{1-\gamma}\bigg)$$ $$s.t.\; \underset{T\rightarrow\infty}{lim}\;\bigg[\bigg(\frac{\beta^T}{((1+(1-\delta))k_T-k_{T+1})^\gamma}\bigg)k_{T+1}\bigg]=0$$

This is where my question comes. How do we solve this? Can we use a lagrangian? If so, how do we take first order conditions in the presence of the limit? Any help would be greatly appreciated!

  • $\begingroup$ I'm a bit confused - that condition you've given will hold for all values of k. It isn't really binding in any way, is it? $\endgroup$
    – 123
    May 6, 2016 at 0:04

1 Answer 1


Your confusion comes from the fact that you treat the Transversality condition as a constraint, while it is a condition for optimality. So the formulation at the end of your question is wrong. What you do is you solve your model as usual, and then check if the solution (here the balanced growth path) satisfies the Transversality condition.

The Transversality condition (TVC) is

$$\lim_{t \rightarrow \infty} \beta^t u'(c_t)k_{t+1} =0$$

In your model, $u'(c_t) = c_t^{-\gamma}$ and $c_t = (2-\delta)k_t - k_{t+1}$

So the TVC becomes indeed

$$\lim_{t \rightarrow \infty} \frac {\beta^tk_{t+1}}{[(2-\delta)k_t - k_{t+1}]^{\gamma}} =0$$

What we want is to check whether the (unique) balanced growth path satisfies the Transversality condition. On the balanced growth path we have $k_{t+1} = (1+g)k_t$ where $g$ is a constant determined by the exogenous parameters of the model and the optimizing conditions. So when on the balanced growth path, the TVC becomes

$$\lim_{t \rightarrow \infty} \frac {\beta^t(1+g)k_t}{[(2-\delta)k_t - (1+g)k_t]^{\gamma}} = \lim_{t \rightarrow \infty}\frac {(1+g)}{(1-\delta-g)^{\gamma}}\beta^t k_t^{1-\gamma}=0$$

The constants are unimportant, so we want that

$$\lim_{t \rightarrow \infty}\beta^t k_t^{1-\gamma}=0$$

holds. Note that if $\gamma \geq 1$ (which is the consensus in the literature) the condition does hold, since capital goes in the denominator, and we're done.

For the case $\gamma <1$, we want the expression to hold at the limit only, i.e. eventually . "Eventually" implies that, if $k^*_s$ is the level of capital at time $s<t$ when the economy arrives at the balanced growth path, we will have $k_t = k_s^*\cdot (1+g)^{t-s}$. Then the TVC becomes

$$\lim_{t \rightarrow \infty}\beta^t \big[k_s^*\cdot (1+g)^{t-s}\big]^{1-\gamma}=0$$

Again, shed things that do not depend on $t$, to get

$$\lim_{t \rightarrow \infty}\beta^t (1+g)^{(1-\gamma)t}=0 \implies \lim_{t \rightarrow \infty}\big[\beta (1+g)^{(1-\gamma)}\big]^t=0$$

So the TVC will hold if and only if

$$\beta (1+g)^{(1-\gamma)} <1$$

In the other question the OP has obtained

$$1+g = [\beta(\alpha+1-\delta)]^{\frac{1}{\gamma}}$$

Substituting, we want

$$\beta \Big([\beta(\alpha+1-\delta)]^{1/\gamma}\Big)^{(1-\gamma)} <1$$

This can be simplified a bit, and it is a restriction on the parameters that must hold in order for the balanced growth path to satisfy the Transversality condition (when $\gamma <1$). For baseline values of the other parameters one can see that the condition is not likely to be satisfied.

  • 1
    $\begingroup$ Ohh I see where my confusion was. I was looking at some lecture notes online and I thought that when they said the solution solved the last equation I have above, they were speaking about the solution of the infinite horizon transversality condition. Thank you so much! This is exactly why I put this question up here because I was mightily confused and (clearly) had no idea what I was doing! $\endgroup$
    – DornerA
    May 6, 2016 at 12:51

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