Abstract
In this paper, we use the recursive saddle point method developed by Marcet and Marimon (1999, 2011) to a simple concave dynamic optimization problem. While the recursive saddle point problem is well defined and delivers the correct value of our optimization problem, it does not generate only optimal policies. Indeed some of the solutions that it produces are either suboptimal or do not even satisfy feasibility. We identify the reasons underlying this failure and discuss its implications for some existing applications.
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Notes
One of the key advantages of this method with respect to standard dynamic programming is that the state space is not itself endogenous but is given exogenously. This can simplify the numerical analysis substantially as costly preliminary computations which the Bellman approach might imply can be avoided (see, for example, [1] and [5]).
Ramsey taxation problems are known to be in general non-concave since [13] (see page 62). In the case of default with capital accumulation Cooley et al. [7] themselves (in footnote 10) point out that the problem is non-concave since the endogeneity of default value creates nonconvexities in the incentive feasibility set.
Notice that throughout we stick to the notation of MM in order to facilitate the comparison.
In the terminology of optimal contracting, one can interpret our problem also as one where a planner maximizes his discounted returns subject to the constraint that the agent should have no incentive to ‘default’, i.e, (1c).
Notice, that, consistently with the results of MM, the value function \(W^*\) is both continuous and homogeneous of degree one (see Proposition 4 of [15]).
The slopes in \(\gamma ^1=\mu ^0-\mu ^1\) given in the following expression should be interpreted as the right-hand and left-hand derivatives, respectively.
Technical feasibility is obviously satisfied. Moreover, setting \(t=0\) in (2c) gives \(a_0\le b\) as required in (6b). From the same condition follows that one must have \(a_t\ge \max \{0, (b-\beta \bar{a})/(1-\beta ) \}\) for all \(t\ge 0\): For any \(a\) smaller than \((b-\beta \bar{a})/(1-\beta )\), not even continuing with the maximal transfers \(\bar{a}\) in all future periods would allow to satisfy the no-default condition (2c).
Only in the initial period, the set of admissible choices is different from \([\max \{0,(b-\beta \bar{a})/(1-\beta ),\bar{a}\}]\).
For a more detailed discussion of what follows see [22].
This seems to be the case of the ‘unrestricted’ steady state in the imperfect enforceability model of [14].
Again, the slopes at the point \(\gamma ^1=\mu ^0-\mu ^1\) are to be interpreted as left-hand and right-hand derivatives respectively.
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For helpful comments and discussions we would like to thank Orazio Attanasio, Alberto Bisin, Antonio Cabrales, Costas Meghir and Mattias Polborn.
Appendix
Appendix
In this appendix, we show that the problem which arises in our linear example might be present also in the non-concave case. In particular, we will consider again problem (1a)–(1c) and assume that \(u(a)=a^2,\, \bar{a}=1\) and \(b=\beta ^n\), where \(n\) is some natural number.
Before we proceed to solve this problem with the RS method, we would like to point out, that its unique ‘true’ solution is given by a stream of transfers which are equal to zero for the first \(2n\) periods and equal to one afterward. It is also important to notice that this optimal stream together with the sequence of Lagrange multipliers \((1,0,0,...)\) constitutes a saddle point of the problem’s associated Lagrangean (hence the failure of the RS method does not derive from nonexistence of a sequential saddle point).
The recursive saddle point functional equation corresponding to the modified problem is given by
We will show in the following that the solution to this problem is given by
Furthermore, we will prove that the corresponding policy correspondences are
Before we continue with the proof, notice that when starting with the initial values \(\mu ^0=1\) and \(\mu ^1=0\), the set of solutions generated with the above policy correspondences is simply the set of all sequences composed of zeros and ones which start with 0. In particular, this set contains also the zero sequence which of course fails to satisfy feasibility for any \(1>b>0\). Hence, just as in the linear case, the RS method allows for non-feasible transfer streams as solutions.
The following proof essentially follows the same steps as the proof for the linear case contained in the text. That is, we will simply show that the above policy correspondences solve the saddle point problem given \(W^*\). Showing that \(W^*\) indeed then solves the functional equation is a simple algebraic exercise and is therefore omitted. We start by substituting the constraints in our saddle point problem which yields
One can immediately see that the above expression is strictly increasing in \(\gamma _0\), independently of \(a\) and the welfare weights \(\mu ^0\) and \(\mu ^1\). Hence, \(\gamma _0\) must be equal to zero.
Next consider the choice of \(a\). The only terms of (12) which depend on \(a\) are
This expression is convex in \(a\). Hence, the only two candidates for a maximizer are the extreme points of the set of technically feasible transfers (i.e. \(a=0\) and \(a=1\)). Which of the two candidates is optimal depends on whether \(\mu ^0\) is larger or smaller than \(\mu ^1+\gamma ^1\). While in the former case we get \(a=0\) the latter case implies \(a=1\) and of course both candidates are optimal if \(\mu ^0=\mu ^1+\gamma ^1\).
Suppose now that \(\mu ^0< \mu ^1\). As \(\gamma ^1\) can never be negative, we know that in this case we must always have \(a=1\) (since the condition \(\mu ^0< \mu ^1 +\gamma _1\) is always satisfied). Therefore the slope of the objective in \(\gamma ^1\) is given by
which implies that the unique minimizing value for \(\gamma ^1\) is zero.
Next, consider the case \(\mu ^0>\mu ^1\). We have to show that there is only one saddle point, namely \(a=0\) and \(\gamma ^1=\mu ^0-\mu ^1\). In order to do so, we will first prove that \(\mu ^0-\mu ^1\) is the unique minimizer of the objective in \(\gamma ^1\) for \(a=0\). Since we have already seen that \(a=0\) maximiezes (12) for \(\gamma ^1=\mu ^0-\mu ^1\) it then only remains to show that there is no saddle point with \(a=1\).
Notice first that the slope of (12) in \(\gamma ^1\) is given byFootnote 14
Plugging in \(a=0\) yields
for the range \([0,\mu ^0-\mu ^1]\) and
if \(\gamma ^1\ge \mu ^0-\mu ^1\). Hence, if \(a=0\) (12) is indeed decreasing in \(\gamma ^1\) up to \(\mu ^0-\mu ^1\) and increasing afterward (remember that \(b=\beta ^n\)).
If instead \(a=1\) the slopes over the two ranges are
(over \([0,\mu ^0-\mu ^1]\)) and
(for \(\gamma ^1\ge \mu ^0-\mu ^1\)). Both expressions are positive (since \(b<1\)) and hence \(\gamma ^1\) would have to be chosen equal to zero. We have already seen though that \(\gamma ^1<\mu ^0-\mu ^1\) is only compatible with \(a=0\).
Finally, consider the case \(\mu ^0=\mu ^1\). It is easily verified that \(\mu ^0-\mu ^1=0\) is still the unique minimizer of (12) if \(a=0\) (and so \(a=0,\, \gamma ^1=\mu ^0-\mu ^1=0\) is a saddle ). But since in this situation \(\gamma ^1\) can never be chosen smaller then \(\mu ^0-\mu ^1\) (14) implies that also \(a=1\) and \(\gamma ^1=0\) is a saddle point, which concludes the prove.
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Messner, M., Pavoni, N. On the Recursive Saddle Point Method. Dyn Games Appl 6, 161–173 (2016). https://doi.org/10.1007/s13235-015-0137-3
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DOI: https://doi.org/10.1007/s13235-015-0137-3