strictly dominated strategy

Question

PROBLEM

Proof that if a pure strategy is strictly dominated, then any mixed strategy that assigns positive probability will also be strictly dominated by some other mixed strategy.

i.e

If $s_{i}\in S_{i}$ is strictly dominated $\forall \widetilde{\sigma_{i}}\in \sum_{i}$ , $\widetilde{\sigma_{i}}(s_{i})>0$,  $\exists{\sigma_{i}}\in \sum_{i}$ s.t

$\forall{s_{-i}}\in S_{-i}$:  $\prod_{i}(\sigma_{i},s_{-i})>\prod_{i}(\widetilde{\sigma_{i}},s_{-i})$

Solution

Suppose...

$\forall{\sigma_{i}}\in \sum_{i} \ , \ \exists {s_{-i}}\in S_{-i}$:  $\prod_{i}(\sigma_{i},s_{-i})\leq \prod_{i}(\widetilde{\sigma_{i}},s_{-i})$

in particular we choose to be strictly dominated by $\widetilde{\sigma_{i}}$

$\prod_{i}(\sigma_{i},s_{-i})\leq \prod_{i}(\widetilde{\sigma_{i}},s_{-i})$

if $A=\prod_{i}(\sigma_{i},s_{-i})=\sum_{s_{i}\in S_{i}}\sigma_{i}\prod_{i}(s_{i},s_{-i})$...................$(\alpha)$

But $s_{i}\in S_{i}$ is strictly dominated it is : $\exists \sigma_{i}\in \sum_{i} , \forall s_{-i}\in S_{-i}$ s.t

$\prod_{i}(s_{i},s_{-i})<\prod_{i}(\sigma_{i},s_{-i})$ using $(\alpha)$

$\prod_{i}(s_{i},s_{-i})<\sum_{s_{i}\in S_{i}}\sigma_{i}\prod_{i}(s_{i},s_{-i})$ multiplication $\widetilde{\sigma_{i}}$

$\widetilde{\sigma_{i}}\prod_{i}(s_{i},s_{-i})<\widetilde{\sigma_{i}}A$

$\sum_{s_{i}\in S_{i}}\widetilde{\sigma_{i}}\prod_{i}(s_{i},s_{-i})<\sum_{s_{i}\in S_{i}}\widetilde{\sigma_{i}}A$

But $\sum_{s_{i}\in S_{i}}\widetilde{\sigma_{i}}\prod_{i}(s_{i},s_{-i})=\prod_{i}(\widetilde{\sigma_{i}},s_{-i})$

Finally $A\leq \prod_{i}(\widetilde{\sigma_{i}},s_{-i})<\sum_{s_{i}\in S_{i}}\widetilde{\sigma_{i}}A$

$A<A\sum_{s_{i}\in S_{i}}\widetilde{\sigma_{i}}$

$A<A$ (Absurd)

This is correct?

The teacher indicated that it is a direct consequence of the definition. But I do not see it as direct, in fact it is proposed as an exercise. I would like to know if someone has experience with this topic, so that you can help me.

Thanks!

Answer 1

I'm not familiar with the notation you're using, but the basic idea is pretty simple. Suppose there is some strategy that is dominated. Call it strategy1. Since it's dominated, there is some strategy that dominates it. Call it strategy2. Suppose there is some mixed strategy ms1 that assigns strategy1 non-zero probability p, and probability 1-p to some other mix of strategies. Then define ms2 as gibing probability 1-p to that same mix of other strategies, and assigning probability p to strategy2 instead of strategy1. Then ms2 dominates ms1.