Macroeconomics/Math Review

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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 \mathbb{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.


A metric is a function \rho :S \times S \to \mathbb{R} with the properties that it is non-negative,  \rho( x, y) \geq 0, symmetric, \rho(x,y) = \rho(y,x), and satisfies the triangle inequality,\rho(x,z) \leq \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) = \max_{1\leq i\leq l} x_i-y_i ,


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.


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

Cauchy sequence[edit]

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

Question: does every Cauchy sequence converge?


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

examples of completeness[edit]

  • (\mathbb{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[edit]

A mappting T:S \rightarrow S is a contraction mapping on a metric space, (S, \rho), if \exists o \geq \beta < 1 such that \rho(tx, Ty) \leq \beta \rho(x,y) \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[edit]

  • Tx=.9x is a contraction mapping on [(0,1],\rho_E),

Now we state the contraction mapping theorem.

Contraction mapping theorem[edit]

If (S, \rho) is complete and T:s \rightarrow S is a contraction mapping, then \exists !x^* with Tx^*=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[edit]

For S \subset \mathbb{R}^l and \rho=\rho_E, Let T:S\rightarrow S satisfy the following two conditions:

  • (M, monotonic condition)\forall x=(x_1, x_2, \ldots, x_l)\in S and y=(y_1, y_2, \ldots, y_l)\in S, and T=(T_1, T_2, \ldots, T_l, if x_i \geq y_i\Leftrightarrow x\geq y then T_i x \geq T_i y \Leftrightarrow Tx \geq Ty,
  • (D, discout condition) \forall x=(x_1, x_2, \ldots, x_l)\in S, for \underline{\vec{a}}=(a, a, \ldots, a), T_i(x_1+a, x_2+a, \ldots, x_l + a) \leq T_i(x_1,x_2,\ldots,x_l)+\beta a \forall i \Leftrightarrow T(x+\underline{a}) \leq TX + \beta \underline{a}.

Then T is a contraction mapping.