Implications of the concepts of combination/expansion of an information structure I would like your help to understand the concept of combination/expansion of an information structure in the incomplete information game at p.6-9 this paper.



Let me summarise the game as described in the paper. 

There are $N\in \mathbb{N}$ players, with $i$ denoting a generic player.

There is a finite set of states $\Theta$, with $\theta$ denoting a generic state. 

A basic game $G$ consists of


for each player $i$, a finite set of actions $A_i$, where we write $A\equiv A_1\times A_2\times ... \times A_N$, and a utility function $u_i: A\times \Theta \rightarrow \mathbb{R}$.
a full support prior $\psi\in \Delta(\Theta)$.


An information structure $S$ consists of


for each player $i$, a finite set of signals $T_i$, where we write $T\equiv T_1\times T_2\times ... \times T_N$.
a signal distribution $\pi: \Theta \rightarrow \Delta(T)$.


A decision rule of the incomplete information game $(G,S)$ is a mapping 
$$
\sigma: T\times \Theta\rightarrow \Delta(A)
$$



Combination/Expansion:

Consider two information structures, $S^1\equiv (T^1, \pi^1)$ and $S^2\equiv (T^2, \pi^2)$. We say that $S^*\equiv (T^*, \pi^*)$ is a combination of $S^1$ and $S^2$ if 


$T_i^*=T_i^1\times T_i^2$ $\forall i$.
$\pi^*:\Theta \rightarrow \Delta(T^1\times T^2)$ has $\pi^1$ and $\pi^2$ as marginals. 


An information structure $S^*$ is an expansion of an information structure $S^1$ if there exists an information structure $S^2$ such that $S^*$ is a combination of $S^1$ and $S^2$. 



Question 1:  Is it correct to say that $S^*$ is at least as informative as $S^1$?



Question 2: 

The game described above assumes that, before privately receiving the signal, each player $i$ knows nothing about what will be the realisation of the state. I call this "baseline level of information". 

Let $\underline{S}$ denote an information structure that is totally uninformative, i.e., it does not add anything to the baseline level of information assumed (also called DEGENERATE at p.26 of the linked paper). In other words, $\underline{S}$ consists of 

(a) for each player $i$, a finite set of signals $\underline{T}_i$, where we write 
  $\underline{T}\equiv \underline{T}_1\times \underline{T}_2\times ... \times \underline{T}_N$.

(b) a signal distribution $\underline{\pi}: \Theta \rightarrow \Delta(\underline{T})$ such that $\underline{\pi}(\cdot|\theta)=\tilde{\pi}$ $\forall \theta \in \Theta$ for some $\tilde{\pi}\in \Delta(\underline{T})$. In other words, the conditional probability is equal to the unconditional one and players' belief about the probability distribution of the state is not updated. 

Let $\mathcal{S}$ denote the collection of all possible information structures. More precisely,

$$
\mathcal{S}\equiv \{S| T \text{ is a separable metric space}, \text{ $\pi:\Theta \rightarrow \Delta(T)$ is a probability measure on $(T,\mathcal{B}(T))$}\}
$$
where $\mathcal{B}(\cdot)$ denotes Borel sigma algebra.

Note that $\underline{S}\in \mathcal{S}$. 

Is it correct to say that: $\forall S \in \mathcal{S}$ where $S\equiv ((T_i)_{i=1}^N, \pi)$, there exists a combination $S^*\in \mathcal{S}$ of $\underline{S}$ and $S$ such that $S^*$ and $S$ are informationally equivalent. In particular, one can set $S^*\equiv ((\underline{T}_i\times T_i)_{i=1}^N, \underline{\pi}\times \pi)$. Therefore, every $S\in \mathcal{S}$ is at least as informative as $\underline{S}$.
Answer to Question 1: Yes. To see this, note that for every combination of elements of $T^1\times T^2$ in the support of $\pi^*$ you get a, possibly different, posterior. To show $\pi^*$ is at least as informative as $\pi^1$ you need these posteriors to be a mean-preserving spread the posteriors induced by $\pi^1$ (there are actually several characterizations of Blackwell's partial order, I'm thinking there is a more elegant proof, but this is the one I can think of right now). You can easily show this is the case if you average up the posteriors induced by $S^*$, over the realizations of $T^2$.

Answer to Question 2: Yes. For any signal $S$, you can find a "trivial" expansion by combining it with $\underline \pi$, so the resulting signal $S^*$ (which is informationally equivalent to $S$) is at least as informative than $\underline S$.

In that sense, Bayesian decision makers can never have less information than their prior.