<p>No. Trembling hand perfection would be an additional consideration.</p>

<p>Consider this case:</p>

<p>Two agents engage in a two stage game. In stage one, player one selects Rock, Paper or Scissors. In stage two, player two selects Rock, Paper or Scissors. They receive payouts depending on whether or not they match. They both have identical payout matrices.</p>

<p>Payoffs (Match, Unmatched)</p>

<ul>
<li>Rock (1,0)</li>
<li>Paper (1,0) </li>
<li>Scissors (1,-1)</li>
</ul>

<p>In a mixed strategy nash equillibrium consider player one. He can choose any mixed strategy such that</p>

<p>p(Rock)+p(Paper)+p(Scissors)=1</p>

<p>and they would all be equally rational. Since even if he chooses scissors, a rational player two would match him. However, taking into account trembling hand perfection, the subset of strategies where</p>

<p>p(Scissors)=0</p>

<p>dominate the previous strategy set. This is the case because we are concerned that player two has a chance of making a mistake if we choose scissors and not matching us. Therefore, mixed strategy and trembling hand mixed strategy give us different sets of strategies which fulfill our Nash equilibrium criteria.</p>

CC BY-SA 3.0
2016-11-07T21:16:51.820Z
It may be worth noting that Nash equilibria with *completely* mixed strategies are always trembling hand perfect.
Herr K.
CC BY-SA 3.0
2016-11-08T16:46:40.907Z
@HerrK I'm pretty certain this is not the case. If you look at the example I listed and give each of Player 1's choices an equal probability (fulfilling the completely mixed condition) it fails to be trembling hand perfect.
Lee Sin
CC BY-SA 3.0
2016-11-08T19:10:04.873Z
A NE $\sigma$ is THP if and only if there exists a sequence of completely mixed strategies $\sigma^k\to\sigma$ and that $\sigma_i$ is a best response to $\sigma_{-i}^k$ for every $k$. If $\sigma$ is a profile of completely mixed strategies, then we can let $\sigma^k=\sigma$ for each $k$. Then, $\sigma^k\to\sigma$ trivially and $\sigma_i$ best responds to each $\sigma_{-i}^k(=\sigma_{-i})$ since $(\sigma_i,\sigma_{-i})$ is a NE. This shows that a NE with completely mixed strategies is necessarily THP.
Herr K.
CC BY-SA 3.0
2016-11-07T14:48:53.427Z
2016-11-07T14:56:10.300Z
2016-11-07T14:56:10.300Z
Lee Sin
<p>Moreover, does it fulfill other criteria automatically, like Myerson's "Proper equilibrium" refinement?</p>

CC BY-SA 4.0
2018-06-19T23:48:55.673Z
I'd never heard of a Nash equilibria described like this. Did Reinhard Selten come up with the name himself? Seems a bit odd choice for a mathematical economist.
Mozibur Ullah
<game-theory>
Do all mixed strategy Nash Equilibria satisfy Trembling Hand perfection?
Matt R
# Do all mixed strategy Nash Equilibria satisfy Trembling Hand perfection?

(Asked by: username_0 on 2016-11-07)

Moreover, does it fulfill other criteria automatically, like Myerson's "Proper equilibrium" refinement?

**Tags:** game-theory
> **Comments:**
> **username_1**: I'd never heard of a Nash equilibria described like this. Did Reinhard Selten come up with the name himself? Seems a bit odd choice for a mathematical economist.  

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## 1 answers

### Answer by username_2 on 2016-11-07. Score: 4 | Accepted answer

No. Trembling hand perfection would be an additional consideration.

Consider this case:

Two agents engage in a two stage game. In stage one, player one selects Rock, Paper or Scissors. In stage two, player two selects Rock, Paper or Scissors. They receive payouts depending on whether or not they match. They both have identical payout matrices.

Payoffs (Match, Unmatched)

* Rock (1,0)
* Paper (1,0)
* Scissors (1,-1)

In a mixed strategy nash equillibrium consider player one. He can choose any mixed strategy such that

p(Rock)+p(Paper)+p(Scissors)=1

and they would all be equally rational. Since even if he chooses scissors, a rational player two would match him. However, taking into account trembling hand perfection, the subset of strategies where

p(Scissors)=0

dominate the previous strategy set. This is the case because we are concerned that player two has a chance of making a mistake if we choose scissors and not matching us. Therefore, mixed strategy and trembling hand mixed strategy give us different sets of strategies which fulfill our Nash equilibrium criteria.

> **Comments:**
> **username_3**: It may be worth noting that Nash equilibria with *completely* mixed strategies are always trembling hand perfect.  
> **username_2**: @HerrK I'm pretty certain this is not the case. If you look at the example I listed and give each of Player 1's choices an equal probability (fulfilling the completely mixed condition) it fails to be trembling hand perfect.  
> **username_3**: A NE $\sigma$ is THP if and only if there exists a sequence of completely mixed strategies $\sigma^k\to\sigma$ and that $\sigma_i$ is a best response to $\sigma_{-i}^k$ for every $k$. If $\sigma$ is a profile of completely mixed strategies, then we can let $\sigma^k=\sigma$ for each $k$. Then, $\sigma^k\to\sigma$ trivially and $\sigma_i$ best responds to each $\sigma_{-i}^k(=\sigma_{-i})$ since $(\sigma_i,\sigma_{-i})$ is a NE. This shows that a NE with completely mixed strategies is necessarily THP.  

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