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In game theory, in the attempt to define sequential equilibrium, for every tuple $b$ of behavioral strategies for each player, the corresponding belief system induced by $b$ is defined. A tuple $(b,μ)$ is called an assessment if $μ$ is the belief system induced by $b$.

Then, the criterion of an assessment being sequentially rational is defined.

Finally, sequential equilibrium is defined as an assessment $(b,μ)$ such that:

  1. $(b,μ)$ is sequentially rational.
  2. There exists a sequence of assessments $(b_i,μ_i)$ such that each $b_i$ assigns nonzero probabilities to all decisions and the sequence converges to $(b,μ)$.

My question is: can there be a sequence of assessments $(b_i,μ_i)$ satisfying condition number 2 and converging to a sequential equilibrium but with each element being not sequentially rational (but nevertheless converges to an assessment that is sequentially rational)?

EDIT:

Just in case the answer is YES, what if it is given that: each sequence of assessments satisfying condition number 2 -- and converging to an assessment $(b,μ)$ -- has $(b_i,μ_i)$ being not sequentially rational for all large enough $i$? Can then it be concluded that $(b,μ)$ is not sequentially rational?

I have a strong intuition that the answer must be YES, because if it isn't, then the definition of sequential equilibrium seems pointless.

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The answer is yes. Note that your condition 2 does not require that elements of the sequence be sequentially rational.

Consider the following game:

enter image description here

It is clear that the strategy profile $(b_1,b_2)=(A,L)$ and the belief system \begin{equation} \mu_1(\varnothing)=1 \qquad\qquad\qquad \mu_2(A)=1-\mu_2(B)=1 \end{equation} form a sequential equilibrium.

For any sequence of completely mixed strategies $(\sigma_1^n,\sigma_2^n)\to(b_1,b_2)$ with \begin{equation} \sigma_1^n(A)=p_n \quad\text{and}\quad \sigma_2^n(L)=q_n,\qquad p_n,q_n\in(0,1)\;\forall n, \end{equation} Player 2's consistent beliefs are $\mu_2^n(A)=p_n$. However, given $p_n\in(0,1)$, it is not sequentially rational for Player 2 to play $\sigma_2^n(L)=q_n\in(0,1)$; Player 2 should best respond with the pure strategy $L$, which is not feasible unless in the limit.

Thus, the above example shows that, by forcing players to use proper mixed strategies, the sequence of assessments $(b_i,\mu_i)$ may prevent a player from playing the unique pure strategy best response to the other player's mixed strategy. As a result of such prevention, sequential rationality cannot be satisfied.

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  • $\begingroup$ I'm sorry... I should've said "just in case the answer is YES"... (If the answer to the second question is NO, then, as a direct corollary, the answer to the first question is YES.) So let's move on to the second question. I need this property (if the answer is YES) to prove that every sequential equilibrium is a subgame-perfect equilibrium, and also a result on a generalized version of backward induction. $\endgroup$ – Mauri Ericson Sombowadile May 29 '18 at 6:38
  • $\begingroup$ @MauriEricsonSombowadile: I think my example answers your second question as well. $\endgroup$ – Herr K. May 29 '18 at 15:41
  • $\begingroup$ Oh, I see. Your example works for the second question as well (answering NO). $\endgroup$ – Mauri Ericson Sombowadile May 29 '18 at 16:21

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