In the neo-classical growth model there is the following transversality condition:

$$\lim_{t\rightarrow\infty}\beta^{t}u'(c_{t})k_{t+1}= 0,$$ where $k_{t+1}$ is the capital at period $t$.

My questions are:

  1. How we derive this condition?

  2. Why do we require this, if we want to rule out paths with no debt accumulation?

  3. Why are the Lagrange multipliers $\beta^{t}u'(c_{t}) = \beta^{t}\lambda_{t}$ the present discounted value of the capital?

  • $\begingroup$ Check out these answers for the distinction between the transversality optimality condition and the solvency exogenous constraint, economics.stackexchange.com/a/13681/61 and economics.stackexchange.com/a/11866/61 $\endgroup$ Commented Feb 9, 2017 at 2:47
  • $\begingroup$ I tried to give a non-mathematical, plain-language description of the intuition behind the transversality condition in this post: medium.com/@alexanderdouglas/… I'm not a macroeconomist, however, so I might well have got it wrong. If so, I hope some replies will appear soon. $\endgroup$ Commented Apr 24, 2017 at 13:35
  • $\begingroup$ This should be a comment, since you only provide a link to external content. Also, the transverslaity condition does not depend on any assumption about expectations formation, since it is a condition imposed even in deterministic models where uncertainty is absent. And it is not specifically related to government debt, but to any assets in general. The basic point is the following: assuming no bequest motive (we don't care about our offspring or society), it is suboptimal to "leave behind" unconsumed wealth. That's all there is to it. $\endgroup$ Commented Apr 24, 2017 at 14:21
  • $\begingroup$ CONTD It is fairly straightforward with a finite horizon, and, as is usual the case, when the horizon becomes "inifinite" it becomes a bit less straightforward and self-evident. $\endgroup$ Commented Apr 24, 2017 at 14:23

2 Answers 2


The transversality condition may be more easily understood if we start from a problem with finite horizon.

In the standard version, our objective is to $$ \max_{\{c_t,k_{t+1}\}_{t=0}^T} \sum_{t=0}^T\beta^t u(c_t) $$ subject to $$ \begin{aligned} f(k_t)-c_t-k_{t+1}&\ge0,\quad t=0,\dots,T &&\text{(resource/budget constraint)}\\ c_t,k_{t+1}&\ge0,\quad t=0,\dots,T &&\text{(non-negativity constraint)} \end{aligned} $$ with $k_0$ given. The associated Lagrangian (with multipliers $\lambda_t$, $\mu_t$, and $\omega_t$) is $$ \max_{\{c_t,k_{t+1},\lambda_t,\mu_t,\omega_t\}_{t=0}^T} \sum_{t=0}^T \beta^tu(c_t)+\lambda_t(f(k_t)-c_t-k_{t+1})+\mu_tc_t+\omega_tk_{t+1} $$ The FOCs are $$ \begin{align} c_t:&& \beta^tu'(c_t)-\lambda_t+\mu_t&=0,\quad t=0,\dots,T \\ k_{t+1}:&& -\lambda_t+\lambda_{t+1}f'(k_{t+1})+\omega_t&=0,\quad t=0,\dots,T-1 \\ k_{T+1}:&& -\lambda_T+\omega_T&=0,\quad T+1 \tag{1} \end{align} $$ with the Kuhn-Tucker complementary slackness conditions: for $t=0,\dots,T$, $$ \begin{align} \lambda_t(f(k_t)-c_t-k_{t+1})&=0 & \lambda_t&\ge0 \\ \mu_tc_t&=0 & \mu_t&\ge0\\ \omega_tk_{t+1}&=0&\omega_t&\ge0\tag{2} \end{align} $$ Since resource constraint must be binding in all periods, i.e. $\lambda_t>0$ for all $t$, it follows that at the last period $T$, $\omega_T=\lambda_T>0$, which in turn implies $k_{T+1}=0$.

Usually we assume $c_t>0$ for all $t$ (the Inada condition), and this implies $\mu_t=0$ for all $t$. So the consumption FOC becomes $$ \beta^tu'(c_t)=\lambda_t \tag{3} $$

Looking at conditions $(1)$ $(2)$ and $(3)$ in the last period $T$, we get $$\beta^Tu'(c_T)k_{T+1}=0$$ Extending this to the infinite horizon, we get the transversality condition $$\lim_{T\to\infty}\beta^Tu'(c_T)k_{T+1}=0$$

The intuition of the transversality condition is partly that "there is no savings in the last period". But as there is no "last period" in an infinite horizon environment, we take the limit as time goes to infinity.

  • $\begingroup$ Just because the resource constraint is binding, why does it necessarily mean that lambda(t)>0? I get the intuition, but it should be greater than equal to 0 (in other words, the constraint exactly hits the optimum). $\endgroup$ Commented Aug 29, 2022 at 18:52
  • $\begingroup$ @KwameBrown: $\lambda>0$ is what a constraint being binding means. If $\lambda=0$, then the constraint is redundant regardless of whether or not it happens to hold with equality at the optimum. This is because $\lambda=0$ implies that relaxing the constraint won't change the optimal solution, which cannot happen when the constraint is binding. $\endgroup$
    – Herr K.
    Commented Aug 30, 2022 at 1:51

In my opinion, the best derivation is by logic. Think about it this way: If the only thing we are telling the household is to maximize its utility, optimal behaviour would then be just making infinite debt and consume infinitely. This is no sensible solution. We therefore need another optimality condition. This should answer question 2.

In a finite horizon setting, feasibility would be achieved by debt having to be repayed by the last period. This is not possible in a infinite horizon setting. However, "ruling out debt accumulation", as you suggest, is too strict a condition (The transversality condition allows for debt!).

To answer question 3, let us look at the term $\beta^t \lambda_t k_{t+1}$. It stands for the (marginal) utility gain (in present-value utils) of shifting $k_{t+1}$ units of capital to period t and consuming them. If this utility gain were positive at infinity, we could increase overall utility by consuming more at "period infinity", hence our capital path would not be optimal.

To question 1: To derive this condition, you can either make the logical argument I just made, showing that without the transversality condition holding, the capital path is not optimal, or, for a mathematical proof, you can check out, for example, Per Krusell's Notes (although it is rather hard to grasp)

  • $\begingroup$ the link is broken. $\endgroup$
    – garej
    Commented Dec 7, 2021 at 7:19

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