Macroeconomics/Math Review

Introduction

We have a Bellman equation and first we want to know if there exists a value function that satisfies the equation and second we want to know the properties of such a solution. In order to answer the question we will define a mapping which maps a function to another function, and a fixed point of the mapping is to be a solution. The mapping we discussed is a mapping on the set of functions, which is a bit abstract. So today we will look at the math review.

So first we consider a set, $S\subset {ℝ}^l$, For us what it will be relevant to describe a sort of distance between any two points in a set. We will use the concept of a metric.

Metric

A metric is a function $\rho :S\times S\to ℝ$ with the properties that it is non-negative, $\rho (x,y)\ge 0$, symmetric, $\rho (x,y)=\rho (y,x)$, and satisfies the triangle inequality,$\rho (x,z)\le \rho (x,y)+\rho (y,z)$,

A common metric is euclidean distance, ${\rho }_E(x,y)=\sqrt{\sum _{i=1}^l(x_i-y_i{)}^2}$, Another is ${\rho }_{max}(x,y)=\underset{1\le i\le l}{\max }{x}_i-y_i$,

Space

A space, is a set of objects equipped with some general properties and structure

We may be interested in a metric space, a space with a metric such as, $(S,\rho )$ where $S$ is the set of all bounded rational functions, and $\rho$ is some distance function. Once we have a metric space we can discuss convergence and continuity.

convergence

A sequence, $\{x_i\}\subset S$, converges to $x$, $x_i\to x$, if $\mathrm{\forall }ϵ>0,\mathrm{\exists }{N}_{ϵ}$ s.t. $\rho (x_n,y_n)<ϵ$ for $n>N_{ϵ}$,

Cauchy sequence

A sequence , $\{x_i\}\subset S$, is called a Cauchy sequence if $\mathrm{\forall }ϵ>0,\mathrm{\exists }{N}_{ϵ}s.t.\rho (x_n,y_m)<ϵ$ for $n,m>N_{ϵ}$,

Question: does every Cauchy sequence converge?

Completeness

The metric space, $(S,\rho )$ is complete if every Cauchy sequence converges.

examples of completeness

• $(ℝ,{\rho }_E)$ is complete.
• $((0,1),{\rho }_E)$ is not complete. Proof: let $x_n=\frac{1}{n+2}$, So $\{x_n\}$ os Cauchy, but does not converge to a point in our set $(0,1)$,
• $([0,1],{\rho }_E)$ is complete. Are all closed sets complete? A closed subspace of a complete space is complete.
• $(\{0,1,2\},{\rho }_E)$ is complete.

Contraction Mapping

A mappting $T:S\to S$ is a contraction mapping on a metric space, $(S,\rho )$, if $\mathrm{\exists }o\ge \beta <1$ such that $\rho (tx,Ty)\le \beta \rho (x,y)\mathrm{\forall }x,y\in S$, Sometimes we write $T(x)$ instead of $TX$,

This means that any two points in our set, $S$, are mapped such that after the mapping the distance between the points shrinks.

examples of contraction mapping

• $Tx=.9x$ is a contraction mapping on $[(0,1],{\rho }_E)$,

Now we state the contraction mapping theorem.

Contraction mapping theorem

If $(S,\rho )$ is complete and $T:s\to S$ is a contraction mapping, then $\mathrm{\exists }!x^{*}$ with $T{x}^{*}=x^{*}$,

We will prove this theorem for a general metric space later on. However, we must remember that it is necessary for this proof that the space be complete.

Let us now look at a criteria to verify that a mapping is a contraction mapping.

Contraction Mapping criteria

For $S\subset {ℝ}^l$ and $\rho ={\rho }_E$, Let $T:S\to S$ satisfy the following two conditions:

• (M, monotonic condition)$\mathrm{\forall }x=(x_1,x_2,\dots ,x_l)\in S$ and $y=(y_1,y_2,\dots ,y_l)\in S$, and $T=(T_1,T_2,\dots ,T_l$, if $x_i\ge y_i\iff x\ge y$ then $T_{i}x\ge T_{i}y\iff Tx\ge Ty$,
• (D, discout condition) $\mathrm{\forall }x=(x_1,x_2,\dots ,x_l)\in S$, for $\underset{\_}{\overrightarrow{a}}=(a,a,\dots ,a)$, $T_i(x_1+a,x_2+a,\dots ,x_l+a)\le T_i(x_1,x_2,\dots ,x_l)+\beta a\mathrm{\forall }i\iff T(x+\underset{\_}{a})\le TX+\beta \underset{\_}{a}$.

Then $T$ is a contraction mapping.