Higher order beliefs and coherency in game theory

Question

I am reading about the higher order beliefs. Before getting into the formal definitions, I will define some common terminology which I will need for the formal definitions.

If $X$ and $Y$ are two spaces, denote the set of probabilities over $X$ as $\Delta(X)$, and for $\delta\in \Delta(X\times Y)$ define the marginal probability measure of $\delta$ on $X$ $$ marg(\delta;X)(E)=\delta(E\times Y) $$ for every measurable subset of $E$ of $X$.

The formal definition of higher order order is as follows:

Let $N=\{1,2,\dots, n\}$ be the set of players. For each $i\in N$, $A_i\neq\emptyset$ is the finite set of actions available to player $i$. Denote the set of mixed strategies for the player $i$ as $\Sigma_i=\Delta(A_i)$. As usual define $\Sigma_{-i}=\times_{j\neq i} \Sigma_i$ and $\Sigma=\times_i \Sigma_i$. Define the se set of first order beliefs $B_i^1=\Delta(\Sigma_{-i})$ and $B_{-i}^1=\times_{j\neq i} B_j^1$, and $B^1=\times_i B_i^1$. Define inductively, for each $k\geq 1$ $$ B_i^{k+1}=\Delta(\Sigma_{-i}\times B_{-i}^1\times \dots B_{i}^k)\\ B_{-i}^{k+1}=\times_{j\neq i} B_j^{k+1}\quad B^{k+1}=\times_i B_i^k $$ Finally, $B_i=\times_{k=1}^\infty B_i^k $

The coherent belief is described in the following way:

$b_i=(b_i^1,b_i^2,\dots)\in\times_i B_i^k$ is coherent if for each $k\geq 1$ $marg(b_i^{k+1},\Sigma_{-i}\times B_{-i}^1\times\dots\times B_{-i}^k)=b_i^k$ where $marg$ is the marginal probability measure.

Now according to this definition for $E\subset \Sigma_{-i}$ we have $marg(b_i^2;\Sigma_{-i})(E)=b_i^1(E)$.

I try to understand this definition. So I tried to consider a game in which there are two players $i$ and $j$ and two actions for each player. So

$$ \Sigma_i=\{(p,1-p):p\in[0,1]\}\quad \Sigma_j=\{(q,1-q):q\in[0,1]\} $$ and $b_i^1\in\Delta(\Sigma_j)$ and $b_i^2\in\Delta(\Sigma_j\times B_j^1)$. So $b_i^1$ is a probability measure over $q$, and $b_i^2$ is a joint probability measure over $q$ and the first order beliefs of $j$. Suppose $E$ is the collection of $(q,1-q)$ such that $q\leq 0.5$, which is a subset of $\Sigma_j$. I could not convince myself why $marg(b_i^2;\Sigma_j)=b_i^1(E)$. Apologies for the length of the question and any help is greatly appreciated.

Answer 1

The way hierarchies of beliefs are specified, beliefs on the same events are encoded in different places.

The basic idea is actually quite simple. You have two players, Ann and Bob, say. Ann's first order beliefs specify how likely she thinks each strategy choice of Bob. Ann's second order beliefs specify how likely she thinks each combination of a strategy choice of Bob and a belief of Bob on Ann's strategy choices. Second order beliefs are joint probabilities.

Now suppose Ann's first order beliefs specify that she believes that Bob plays the strategies C and D with probability $1/2$ each.

Suppose also that Ann believes with probability $1/4$ that "Bob plays C and believes with probability 1 that Ann plays D for sure" and believes with probability $3/4$ that "Bob plays D and believes Ann plays D for sure".

If you ask someone who knows only Ann's first order beliefs how likely Ann thinks it is that Bob will play C, that person will answer with $1/2$. If you ask someone who knows only Ann's second order beliefs how likely Ann thinks it is that Bob will play C, that person will answer with $1/4$. Indeed, that is what the marginal of her second order belief on the space of strategies says.

This is absurd and makes no sense. Ann's beliefs on how likely it is that Bob plays C should be the same according to her first order beliefs as to her second order beliefs. Now coherency is exactly the condition that guarantees that. If some belief on an event E is specified in more than one level of the hierarchy of beliefs, each hierarchy should specify that the same belief on the event E.

In terms of the formalism: The marginal of a probability measure is again a probability measure, so $marg(b_i^2;\Sigma_j)=b_i^1(E)$ is comparing apples and oranges; on the left is a probability measure, on the right a number. What you want is $marg(b_i^2;\Sigma_j)(E)=b_i^1(E)$, as prescribed by coherency- and indeed by common sense.