Possible worlds/beliefs/Probability Matrix/Example 2

Question

I have posted a similar question elsewhere, but this one is different. Take the definition and example as below as copied from the Handbook of Game theory with Economic applications.

I want to design a $3\times3$ matrix $A$ which satisfies the condition $(2.1)$ below, but enjoyes a particlar conditions of asymetry, in particular that it is asymetric, i.e. $A^T\neq A$ and is composed from nonzero elements with small fractions,i.e. with fractions $\frac{a}{b}$ of probabilities with small , say less than $12$, quantities $a+b$, and enjoyes even stronger intuitive assymetry conditions, such as having different columns, and with each row and column summing up to $1$. Is it possible at all to design such a matirx $A$?

Answer 1

There is no such asymmetric matrix.

Let $M = (m_{ij}) \in M_n$ be a matrix with the following properties:

• $[P_1]$: $m_{ij} \neq 0 \Rightarrow m_{zi} = m_{zj} \ \forall z = 1, \ldots,n$
• $[P_2]$: $\sum_{i=1}^n m_{ij} = 1 \ \forall j = 1, \ldots,n$
• $[P_3]$: $\sum_{j=1}^n m_{ij} = 1 \ \forall i = 1, \ldots,n$

(Food for thought: does $[P_3]$ follow from the book definition?)

$[P_1]$ means: if $m_{ij} \neq 0$ then the $i$-th row and the $j$-th row are equal.

For $n=2$ there are no such asymmetric matrices: $M = \begin{pmatrix} a & 1-a \\ 1-a & a \end{pmatrix}$ by $[P_2]$ and $[P_3]$.

Then by induction: $M = \begin{pmatrix} a & u \\ v & M_1 \end{pmatrix}$, where $a$ is a number, $u$ a row-vector, v a column-vector and $M_1$ is a square matrix.

If $a = 0$, then there is an element $m_{i1} \neq 0$ by $[P_2]$, then $0 = a = m_{11} = m_{i1} \neq 0$ by $[P_3]$: contradiction.

If $a=1$, then $\begin{pmatrix} 1 & 0 \\ 0 & M_1 \end{pmatrix}$, so $M_1$ respects the properties $[P_1]$, $[P_2]$, $[P_3]$ and then by induction: $M_1$ is symmetric, so $M$ is symmetric.

Then, if $a \in (0,1)$: if $m_{i1} \neq 0$ then $m_{i1} = a$; if $m_{i1} = 0$ then $m_{1i} = 0$. Call $k$ the number of numbers of the first column different from $0$. Then $a = 1/k$ and the corresponding rows are equal.

Let $m_{1i} \neq 0$, then $1 = \sum_{i=1}^n m_{ij} = \sum_{i \text{ s.t. } m_{ij} \neq 0} m_{ij} = \sum_{i \text{ s.t. } m_{ij} \neq 0} m_{1j} \Rightarrow m_{1j} = a = m_{j1} \Rightarrow u^T = v$.

To verify that $M_1 = M_1^T$: some rows (and the corresponding columns) of $M_1$ sums to $1$, the others to $1-\frac{1}{k}$: call $Q = M_1 + \frac{1}{k(n-1)} N$, where $N_{ij} = 1$ if both the $i$-row and the $j$-column of $M_1$ sums to $1-\frac{1}{k}$ and $N_{ij} = 0$ otherwise. Then $Q$ verify the inductive hypothesis, so

$$Q^T = Q \Rightarrow M_1+N = M_1^T + N^T = M_1^T + N$$

This proof is not pretty, but you can explicitly try (and fail) to build a 3x3 matrix with the aid of $[P_1]$.