用中文解释By contradiction, suppose (s1 ,s2 , ...sn ) is a N.E., but it doesnít survive ISD. Suppose s i is the Örst equilibrium strategy that is deleted in the process of ISD. Suppose deletion is done in the kth round, then s i is striclty dominated given A k
时间: 2024-03-30 12:40:03 浏览: 25
"By contradiction, suppose (s1, s2, ..., sn) is a N.E., but it doesn't survive ISD. Suppose si is the first equilibrium strategy that is deleted in the process of ISD. Suppose deletion is done in the kth round, then si is strictly dominated given Ak." 这句话的意思是:通过反证法假设(s1, s2, ..., sn)是一个纳什均衡,但它不能在独立自主决策过程中生存下来。假设si是在ISD过程中被删除的第一个均衡策略,假设删除发生在第k轮,那么在第k轮中,给定Ak的情况下,si是严格被支配的。
在博弈论中,一个策略被称为严格被支配的,是指在所有其他参与者采取的策略组合下,它对于该参与者来说都不是最优的。因此,如果一个策略被严格支配,那么在理性的前提下,该参与者将不会选择这个策略。
这句话中提到的假设是为了证明纳什均衡的稳定性。如果一个策略被删除了,那么它就不可能是一个稳定的纳什均衡,因为有其他的策略可以替代它。因此,如果假设si是在独立自主决策过程中被删除的第一个均衡策略,那么它就是被其他策略所严格支配的,这就说明了这个纳什均衡是不稳定的。
相关问题
Prove the following by deductive system: ⊨ φ ∨ ¬φ (i.e., ∅ ⊨ φ ∨ ¬φ)
Proof:
1. Assume ¬(φ ∨ ¬φ) for the sake of contradiction.
2. By De Morgan's Law, ¬(φ ∨ ¬φ) is equivalent to ¬φ ∧ ¬¬φ.
3. By Double Negation, ¬¬φ is equivalent to φ.
4. Therefore, ¬φ ∧ φ.
5. By the principle of Non-Contradiction, a statement and its negation cannot both be true.
6. Therefore, ¬(φ ∨ ¬φ) leads to a contradiction.
7. By the principle of Explosion, a contradiction implies anything.
8. Therefore, φ ∨ ¬φ must be true, as assuming its negation leads to a contradiction.
9. Therefore, ⊨ φ ∨ ¬φ, as it is true under all interpretations.
Note: This proof uses the principle of Non-Contradiction and the principle of Explosion, which are two fundamental principles of classical logic.
Show that the columns of X are linearly independent if the constraint α1 = 0 is imposed.
To show that the columns of X are linearly independent if the constraint α1 = 0 is imposed, we need to show that the only solution to the equation Xα = 0 (where α is a vector of coefficients) is the trivial solution α = 0.
Let X be an n x p matrix with columns x1, x2, ..., xp. If we impose the constraint α1 = 0, then the equation Xα = 0 becomes:
0*x1 + α2*x2 + α3*x3 + ... + αp*xp = 0
This can be rewritten as:
α2*x2 + α3*x3 + ... + αp*xp = 0
Assume that the columns of X are linearly dependent, so there exist coefficients α2, α3, ..., αp, not all zero, such that α2*x2 + α3*x3 + ... + αp*xp = 0. Without loss of generality, we can assume that α2 ≠ 0.
Then we can solve for x2 in terms of x3, ..., xp as:
x2 = (-α3/α2)*x3 - ... - (αp/α2)*xp
Substituting this expression into the equation α2*x2 + α3*x3 + ... + αp*xp = 0, we get:
α2*(-α3/α2)*x3 - α2*(α4/α2)*x4 - ... - α2*(αp/α2)*xp + α3*x3 + ... + αp*xp = 0
Simplifying this expression, we get:
-α3*x3 - α4*x4 - ... - αp*xp + α3*x3 + ... + αp*xp = 0
This is a contradiction, since we assumed that not all of the coefficients α2, α3, ..., αp are zero, but we ended up with the equation 0 = 0. Therefore, our assumption that the columns of X are linearly dependent must be false, and we can conclude that the columns of X are linearly independent if the constraint α1 = 0 is imposed.