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

<p>The Transversality condition (TVC) is</p>

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

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

<p>So the TVC becomes indeed</p>

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

<p>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</p>

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

<p>The constants are unimportant, so we want that</p>

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

<p>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. </p>

<p>For the case $\gamma &lt;1$, we want the expression to hold at the limit only, i.e. <em>eventually</em> . "Eventually" implies that, if $k^*_s$ is the level of capital at time $s&lt;t$ 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</p>

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

<p>Again, shed things that do not depend on $t$, to get</p>

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

<p>So the TVC will hold if and only if</p>

<p>$$\beta (1+g)^{(1-\gamma)} &lt;1$$</p>

<p>In the other question the OP has obtained</p>

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

<p>Substituting, we want</p>

<p>$$\beta \Big([\beta(\alpha+1-\delta)]^{1/\gamma}\Big)^{(1-\gamma)} &lt;1$$</p>

<p>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 $\gamma &lt;1$). For baseline values of the other parameters one can see that the condition is not likely to be satisfied.</p>

CC BY-SA 3.0
2016-05-06T12:51:51.073Z
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!
DornerA
CC BY-SA 3.0
2016-05-06T02:52:29.057Z
2016-05-06T02:52:29.057Z
Alecos Papadopoulos
<p>This question is an extension of <a href="https://economics.stackexchange.com/questions/11531/endogenous-growth-balanced-growth-path-with-crra-utility">Endogenous Growth: Balanced Growth Path with CRRA Utility</a>

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. </p>

<p>We will use the model in this question:</p>

<blockquote>
  <p>$\textbf{Model:}$
  $$K_t=\frac{1}{n}\sum_{t=1}^nk_t$$
  In this model, $k_t$ is chosen by agents, and $K_t=\bar{k}_t$ (the average of all $k_t$).</p>
  
  <p>Now, agents want to dynamically maximize utility (under certain constraints) and they have CRRA (constant relative risk aversion) utility, so the maximization looks like:
  $$\sum_{t=0}^\infty\beta^t\bigg(\frac{c_t^{1-\gamma}}{1-\gamma}\bigg)$$
  $$s.t.\;Y_t=k_t^\alpha(E_tL)^{1-\alpha}$$
  $$c_t+i_t=Y_t$$
  $$k_{t+1}=(1-\delta)k_t+i_t$$
  $$c_t,i_t\geq0$$</p>
  
  <p>$E_tL$ is effective labor and the rest of the variables are typical  (I can give their definitions if requested).</p>
  
  <p>One last addition to the model is that there are two equilibrium constraints:
  $$E_t=\frac{K_t}{L}$$
  $$k_t=K_t$$ </p>
</blockquote>

<p>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:</p>

<p>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:
$$\underset{\{k_t\}_{t=0}^{T+1}}{max}\;\sum_{t=0}^{T}\beta^t\bigg(\frac{((1+1-\delta)k_t-k_{t+1})^{1-\gamma}}{1-\gamma}\bigg)$$
$$s.t.\;k_{T+1}\geq0$$</p>

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

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

<p>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.</p>

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

<p>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!</p>

CC BY-SA 3.0
2016-05-06T00:04:51.560Z
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?
123
<macroeconomics><economic-growth><endogenous-growth>
Infinite Horizon Transversality Condition
DornerA
# Infinite Horizon Transversality Condition

(Asked by: username_0 on 2016-05-05)

This question is an extension of [Endogenous Growth: Balanced Growth Path with CRRA Utility](https://economics.stackexchange.com/questions/11531/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!

**Tags:** macroeconomics, economic-growth, endogenous-growth
> **Comments:**
> **username_1**: 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?  

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## 1 answers

### Answer by username_2 on 2016-05-06. Score: 5 | Accepted answer

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

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

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.

> **Comments:**
> **username_0**: 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!  

---

