On the Recursive Saddle Point Method

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.

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

$$\begin{aligned} W(\mu ^{0},\mu ^{1})&= \sup _{a\in \left[ 0,\bar{a}\right] }\inf _{\mathop {{\scriptscriptstyle \mu ^{0'},\mu ^{1'} \ge 0}}\limits ^{{\scriptscriptstyle \gamma ^0,\gamma ^1\ge 0,}} } \mu ^{0}\left( y-a\right) +\gamma ^{0}\left( y-a-R\right) +\mu ^{1}a^2+\gamma ^{1}\left( a^2-\frac{b^2}{1-\beta }\right) \nonumber \\&\quad +\beta W(\mu ^{0'},\mu ^{1'} ) \nonumber \\ \hbox {s.t.} \quad \mu ^{0'}&= \mu ^{0}+\gamma ^{0} \nonumber \\ \mu ^{1'}&= \mu ^{1}+\gamma ^{1} \end{aligned}$$

(11)

We will show in the following that the solution to this problem is given by

$$\begin{aligned} W^*(\mu ^0,\mu ^1)=\left\{ \begin{array}{cl} \mu ^0\frac{\displaystyle {y-b^2}}{\displaystyle {1-\beta }}+\mu ^1 \frac{\displaystyle {b}^2}{\displaystyle {1-\beta }}&{}\text{ if } \, \mu ^0 \ge \mu ^1 \\ \mu ^0\frac{\displaystyle {y-1}}{\displaystyle {1-\beta }}+\mu ^1\frac{\displaystyle {1}}{\displaystyle {1-\beta }}&{}\text{ if } \, \mu ^0 < \mu ^1 \end{array} \right. . \end{aligned}$$

Furthermore, we will prove that the corresponding policy correspondences are

$$\begin{aligned} \phi _a(\mu ^0,\mu ^1)&= \left\{ \begin{array}{cl} 1 &{} \text{ if } \,\mu ^0 <\mu ^1 \\ \{0,1\} &{} \text{ if } \,\mu ^0 =\mu ^1 \\ 0 &{} \text{ if } \,\mu ^0 >\mu ^1 \end{array} \right. \\&\text{ and } \\ \phi _\gamma (\mu ^0,\mu ^1)&= \left\{ \begin{array}{cl} (0,0) &{} \text{ if } \,\mu ^0 < \mu ^1 \\ (0,\mu ^0-\mu ^1) &{} \text{ if } \,\mu ^0 \ge \mu ^1. \end{array} \right. \end{aligned}$$

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

$$\begin{aligned}&\sup _{a\in \left[ 0,1\right] }\inf _{\gamma ^0,\gamma ^1\ge 0}\;\mu ^{0}\left( y-a\right) +\gamma ^{0}\left( y-a\right) +\mu ^{1}a^2+\gamma ^{1}\left( a^2-\frac{b^2}{1-\beta }\right) \nonumber \\&\quad +\beta W^*(\mu ^0+\gamma ^0,\mu ^1+\gamma ^1). \end{aligned}$$

(12)

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

$$\begin{aligned} -\mu ^0a +(\mu ^1 +\gamma ^1)a^2. \end{aligned}$$

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

$$\begin{aligned} 1-\frac{b^2}{1-\beta }+\frac{\beta }{1-\beta }=\frac{1-b^2}{1-\beta }>0, \end{aligned}$$

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 by¹⁴

$$\begin{aligned} a^2-\frac{b^2}{1-\beta }+ \beta \left\{ \begin{array}{cc} \displaystyle {\frac{b^2}{1-\beta }}&{} \text{ if } \, \gamma ^1 \le \mu ^0-{\mu ^1} \\ \displaystyle {\frac{1}{1-\beta }}&{} \text{ if } \, \gamma ^1 \ge \mu ^0-{\mu ^1}. \end{array} \right. \end{aligned}$$

(13)

Plugging in $a=0$ yields

$$\begin{aligned} -\frac{b^2}{1-\beta } + \beta \frac{b^2}{1-\beta }= -b^2<0 \end{aligned}$$

for the range $[0,\mu ^0-\mu ^1]$ and

$$\begin{aligned} -\frac{b^2}{1-\beta } + \beta \frac{1}{1-\beta }= \frac{\beta -b^2}{1-\beta }>0 \end{aligned}$$

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

$$\begin{aligned} 1-\frac{b^2}{1-\beta } + \beta \frac{b^2}{1-\beta }= 1-b^2 \end{aligned}$$

(over $[0,\mu ^0-\mu ^1]$) and

$$\begin{aligned} 1-\frac{b^2}{1-\beta } + \beta \frac{1}{1-\beta }= \frac{1-b^2}{1-\beta } \end{aligned}$$

(14)

(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.