Topkis's theorem

Topkis's Theorem is a result that is useful for establishing comparative statics. The theorem allows researchers to understand how the optimal value for a choice variable changes when a feature of the environment changes. The result states that if f is supermodular in (x,θ), and D is a lattice, then $x^{*}(\theta )=\arg \max _{x\in D}f(x,\theta )$ is nondecreasing in θ. The result is especially helpful for establishing comparative static results when the objective function is not differentiable.

An Example

This example will show how using Topkis's Theorem gives the same result as using more standard tools. The advantage of using Topkis's Theorem is that it can be applied to a wider class of problems than can be studied with standard economics tools.

An agent is driving down a highway and must choose her speed, s. Abstract away from the probability of accidents and other motorists. Going faster is desirable, but she is more likely to be arrested if you speed. There is some amount of police presence, p. The presence of police makes you more likely to be arrested. Note that s is a choice variable and p is a parameter of the environment that is fixed from the perspective of the agent. The agent seeks to $max_{s}U(s,p)$.

We would like to understand how the agent's speed (a choice variable) changes with the amount of police: ${\frac {\partial s^{\ast }(p)}{\partial p}}$.

If one wanted to solve the problem with standard tools such as the implicit function theorem, one would have to assume that the problem is well behaved: U(.) is twice continuously differentiable, concave in s, that the domain over which s is defined is convex, and that it there is a unique maximizer $s^{\ast }(p)$ for every value of p and that $s^{\ast }(p)$ is in the interior of the set over which s is defined. Note that the optimal speed is a function of the amount of policing. Taking the first order condition, we know that at the optimum, $U_{s}(s^{\ast }(p),p)=0$. Differentiating the first order condition, with respect to p and using the implicit function theorem, we find that $U_{ss}(s^{\ast }(p),p)(\partial s^{\ast }(p)/(\partial p))+U_{sp}(s^{\ast }(p),p)=0$ or that ${\frac {\partial s^{\ast }(p)}{\partial p}}={\frac {-U_{sp}(s^{\ast }(p),p)}{\underset {{\text{negative since we assumed }}U(.){\text{ was concave in }}s}{\underbrace {U_{ss}(s^{\ast }(p),p)} }}}$. So, ${\frac {\partial s^{\ast }(p)}{\partial p}}{\overset {\text{sign}}{=}}U_{sp}(s^{\ast }(p),p)$. If s and p are substitutes, $U_{sp}(s^{\ast }(p),p)<0$ and hence ${\frac {\partial s^{\ast }(p)}{\partial p}}<0$ and more police causes less speeding. Clearly it is more reasonable to assume that they are substitutes.

The problem with the above approach is that it relies on the differentiability of the object function and on concavity. We could get at the same answer using Topkis's Theorem in the following way. We want to show that $U(s,p)$ is submodular (the opposite of supermodular) in $\left(s,p\right)$. Note that the choice set is clearly a lattice. The cross partial of U being negative, ${\frac {\partial U}{\partial s\partial p}}<0$, is a sufficient condition. Hence if ${\frac {\partial U}{\partial s\partial p}}<0,$ we know that ${\frac {\partial s^{\ast }(p)}{\partial p}}$.

Hence using the implicit function theorem and Topkis's Theorem gives the same result.