In Kamenica and Gentzkow (2017) as well as in Bergemann and Morris (2016) the notion of Blackwell comparioson of experiments is used to compare different information structures. I am trying to find the equivalence between Blackwells order and stochastic dominance order. Let me give a bit notation of the Kamenica and Gentzkow environment.

In Kamenica and Gentzkow environment, there are $n-$ senders indexed by $i$, there exist a finite state space $\Omega$ and $\omega$ is the typical element of the state space. All senders share a commom prior $\mu_0$ about the state of the world and each of them chooses a singan $\pi_i\in \Pi_i$ where $\Pi=\times_{i}\Pi$ is the information environment, then if $\pi\in\Pi$ is a profile of strategies (signals that are chosen by the senders) let $<\pi>$ denote the distribution of beliefs of a Bayesian with prior $\mu_0$ who observes the realization of all signals in $\pi$. So, if $\pi$ is a Nash equilibrium, then $<\pi>$ denotes the equilibrium outcome and we define the distribution of posteriors as $\tau = <\pi>$. In other words, $\tau$ is a feasible outcome if there exists $\pi\in\Pi$ such that $\tau = <\pi>$.

Kamenica and Gentzkow quote that a distribution of posterior beliefs $\tau \succeq \tau^{'}$ if $\tau$ is a mean preserving spread of $\tau^{'}$ in which case this means that $\tau$ is more informatiove than $\tau^{'}$.

From my understanding the relation of Blackwell order with the Stochastic dominance order should be something like the following

$$\tau \succeq \tau^{'} \Leftrightarrow \text{$\tau \leq_{SOSD} \tau^{'}$ ? or $\tau \geq_{SOSD} \tau^{'} $}? \tag{1}$$

because $\tau$ is a mean preserving spread of $\tau^{'}$, and this means that the latter must be less informative with respect to the former which part of $(1)$ is true?

  • $\begingroup$ Usually, second-order stochastic dominance is defined for one-dimensional distributions. But $\tau\succeq\tau'$ is equivalent to $\int f~\mathrm d\tau\leq \int f~\mathrm d\tau'$ for every concave function. $\endgroup$ Oct 9, 2023 at 14:36
  • $\begingroup$ Well, the distrubutions that are defined in Kamenica and Gentzkow are n-dimensional distributions, with $n\geq 2$ so the Blackwell order is a better way of representation than the Stochastic dominance order and that is why they do not choose to use latter ordering, right? $\endgroup$ Oct 9, 2023 at 15:14
  • 2
    $\begingroup$ Depending how you define it, there are versions of the order known as convex order, dilation order, Choquet order... There is no hope for a single all-encompassing name and, thanks to the equivalence theorems, no need. $\endgroup$ Oct 9, 2023 at 16:57


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.