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This problem is of Fernando Vega Redondo(Economics and the theory of games)

Exercise 2.1 Let G be a game in strategic form. Prove that, for every player $i\in N$, every mixted strategy $\sigma_{i}\in \Sigma_{i}$ that assigns positive weight to a pure strategy $s_{i}\in S_{i}$ that is dominated can be itself always be improved by another strategy $\sigma_{i}'$.

This is if $s_{i}\in S_{i}$ is strongly dominated for some $\sigma_{i}$$\in \Sigma$ with $\sigma_{i}(s_{i})>0$, $\exists$ $\sigma_{i}'\in \Sigma_{i}$ such that $\forall$$s_{-i}\in S_{-i}$$: \pi_{i}(\sigma_{i}',s_{-i})>\pi_{i}(\sigma_{i},s_{-i})$.

Q1: why they said that affirmation is obvious?

*I would like to know if anyone could tell me how to build that mixed strategy that dominates that strategy that assigns positive probability, since it is not obvious to me.

Thanks!

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The short answer is that a mixed strategy $\sigma_i$ that uses with positive probability a dominated pure strategy $s_i$ can always be improved by excluding $s_i$ from its mixing support and redistributing its "probability weight" to one of its dominating strategies.

More formally, suppose that strategy $s_i'$ dominates $s_i$ (you can think of $s_i'$ as a pure or mixed strategy itself, this does not make any difference for the argument; for simplicity I suppose $s_i'$ is a pure strategy), that is $\forall s_{-i}\in S_{-i}: \pi(s_i',s_{-i})>\pi(s_i,s_{-i})$.

Also, suppose that strategy $\sigma_i$ plays strategy $s_i$ with positive probability $p_{s_i}\in(0,1)$. Build strategy $\sigma_i'$ as follows: identical to $\sigma_i$ but replace $s_i$ with $s_i'$ , that is play $s_i$ with probability 0 and play $s_i'$ with (additional) probability $p_{s_i}$.

By linearity of the payoff function you have, $\forall s_{-i}\in S_{-i}$: $$ \pi(\sigma_i',s_{-i})=p_{s_i}\pi(s_i',s_{-i}) + (1-p_{s_i})\pi(\Lambda_i,s_{-i}) > p_{s_i}\pi(s_i,s_{-i}) + (1-p_{s_i})\pi(\Lambda_i,s_{-i}) = \pi(\sigma_i,s_{-i}) $$ where $\Lambda_i$, loosely speaking, denotes the residual combination of strategies played in $\sigma_i$, apart from $s_i$, inherited by $\sigma_i'$ too.

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