In http://www.uni-hamburg.de/fachbereiche-einrichtungen/fb03/iwwt/makro/slides2.pdf page 8, lagrangian is written as follow: $$L = E_0 \sum_{t=0}^{\infty}\beta^t\{U(C_t,N_t) + \lambda_t(P_tC_t + Q_tB_t - B_{t-1}-W_tN_t+T_t)+\psi_t(\lim_{T \to \infty} B_T)\}$$ where bond $B_t$ has solvency condition $\lim_{T \to \infty} B_T \geq 0$.

In page 9 then all first-order conditions are derived, but I don't see anything related to $\psi_t$ and solvency condition. Why can the first-order condition relating to $\psi_t$ be dropped?


1 Answer 1


The condition is mostly referred to as the No-Ponzi (-scheme) [NP] condition. It is one additional constraint, that prevents Ponzi-schemes: Paying debt with new higher debt, ad infinitum.

By the way: The NP condition is one condition, hence the associated multiplier should be $\psi$ instead of $\psi_t$. While certainly nothing is lost repeating the same condition over and over again (for any $t$), we don't need it more than once, and it is being imprecise.

Think about optimization for finite $T$ periods. Then, you have the condition that $B_T \geq 0$. The Lagrangian optimization gives you the local optimization between $0, 1, 2$... There are many solutions that are locally optimal, but you will only allow solutions that in the end lead to $B_T > 0$.

A simple example

Your example is much too messy to think about these core issues. Look instead at the problem

$$ \max_{\{c_t, a_{t+1}\}_t} \sum_t \beta^t U(c_t) + \lambda_t (a_{t+1} + c_t - Ra_t)$$

That is, a household that choses assets $a$ and consumption $c$ to maximize his utility. You can summarize the FOC as

$$ \beta^t U'(c_t) = \lambda_t \\ \lambda_t = R\lambda_{t+1}\\ \Leftrightarrow U'(c_t) = \beta R U'(c_{t+1}) $$

Look for a moment at the special case where $\beta R = 1$ (what does that imply?). With most preferences, this necessarily leads to $c_t = c_{t+1}$. This is the local optimization that I was referring to, which is what the Lagrangian gives you. There are, however, infinitely many solutions that satisfy $c_t = c_{t+1}$. Next, we try to use the budget constraint:

$$ a_{t+1} + c_t = R a_t\\ \Leftrightarrow R a_0 = \lim_{T\to\infty}\sum_{t=0}^T \frac{c_t}{R^t} + \frac{a_{T+1}}{R^T}$$

This is as far we get using the (infinite) set of local budget constraints, where I have used forward iteration (hopefully correctly), assuming any start date $t=0$.

Now, if the household also has to satisfy the NP condition, this boils down to

$$R a_0 = \lim_{T\to\infty}\sum_{t=0}^T \frac{c_t}{R^t}$$

which, as we showed $c_t$ to be constant, we can solve easily and receive a single budget constraint. The unique solution to the problem that satisfies the NP condition is the solution where $c_t$ is a constant and this last equation holds.

  • $\begingroup$ What happens if there are 2 assets with different returns? Will there be two no ponzi conditions? Would be glad to get some reference on this. $\endgroup$
    – EndLoop
    Feb 6, 2021 at 4:14
  • $\begingroup$ @EndLoop intuitively, you can't go roll forward debt until the end of time. The way you would write it as $\psi(B_T^1 + B_T^2)$, where $1$ and $2$ index the two assets. You could have two conditions separately $\psi^1 B_T^1 + \psi^2 B_t^2$, which would lead to the same result, but it would not be "on the letter" what the NP condition is about. $\endgroup$
    – FooBar
    Feb 10, 2021 at 10:10
  • $\begingroup$ Perhaps I should put a question up for this to continue this discussion. $p_{t}c_{t} + p_{t} b_{t+1} + b_{gt+1} = y_{t} + b_{gt}(1+i_{t}) + p_{t} b_{t} $ where $b_{t}$ denotes an asset which is traded in two countries and $b_{gt}$ is an asset which is traded in only one country. Now the No-ponzi condition just prevents me from borrowing more than what I could pay in one lifetime. So as you said it is going to be a combination of these assets. So the initial value for these assets for the problem would also be a combination of these two assets? $\endgroup$
    – EndLoop
    Feb 12, 2021 at 1:41
  • $\begingroup$ I have posted my question here - economics.stackexchange.com/questions/42524/… $\endgroup$
    – EndLoop
    Feb 12, 2021 at 3:32

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