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)?


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.


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.

  • $\begingroup$ 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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