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Suppose that we have a two players game, where $(S^i)_{i=1}^2$ denotes the set of pure strategies for each one. The set of mixed strategies of player $i$ is denoted by $\Sigma^i=\Delta(S^i)$ while $\Sigma=\Sigma_1\times\Sigma_2=\Delta(S_1)\times \Delta(S_2)$. My questions are the following:

  1. Suppose that $S_1=\{x,y\}$ and $S_2=\{u,v\}$, then the set $\Sigma^1=\Delta(S^1)=\{\{x\},\{y\},(p(x),(1-p)(y))\}$ and $\Sigma^2=\Delta(S^2)=\{\{u\},\{v\},(q(u),(1-q)(v))\}$, where $p$ and $q$ denote the probability distributions upon the set of pure strategies. If I am not mistaken until here, then what is the set $\Sigma$ look like?
  2. With the notation of $\Delta$ in front of a set we denote the simplex of the set? If not, how you can find it and how is it related to the pure and mixed strategies sets?
  3. Does the following hold $\Delta(\underbrace{S_1\times S_2}_{\Delta(\Pi_{i=1}^2 S^i)})\underbrace{=}_{=}\underbrace{\Delta(S_1)\times\Delta(S_2)}_{\Pi_{i=1}^2 \Delta(S^i)}$ an can this result generalized for $k(<+\infty)$ player games?

Thank you in advance! Though the calculations or my answers may look simple, I am a bit confused with all these notation.

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    $\begingroup$ @Giskard you are right I am mistaken... let me fix this...in fact $\Sigma=\Sigma_1\times\Sigma_2=\Delta(S_1)\times\Delta(S_2)$ $\endgroup$ Sep 12 at 11:27
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  1. then what is the set $\Sigma$ look like?

$\Sigma^1$ is the set of all probability distributions over $\{x,y\}$, as such it is given by the one dimensional unit simplex: $$ \Sigma^1 = \{(p_1, p_2) \in \mathbb{R}^2_+| p_1 + p_2 = 1\}. $$ Here $p_1$ is the probability of playing $x$ and $p_2$ the probability of playing $y$. If you look at $\Sigma^1$ as a subset of $\mathbb{R}^2$ it is the set of all vectors in the positive orthonant whose components sum to 1 as given in the picture below: 1 dimensional simplex

It is called the one dimensional simplex as the line is of dimension 1.

If there would be three strategies then we would get the two dimensional unit simplex: $$ \Sigma^1 = \{(p_1, p_2, p_3) \in \mathbb{R}^3_+|p_1 + p_2 + p_3 = 1\}. $$ In $\mathbb{R}^3$ this would look like the surface of a triangle spanned by the vectors $(1,0,0), (0,1,0), (0,0,1)$. We get a 2 dimensional simplex as the surface is a 2 dimensional object.

  1. With the notation of $\Delta$ in front of a set we denote the simplex of the set? If not, how you can find it and how is it related to the pure and mixed strategies sets?

Yes $\Delta(S)$ is usually used to denote the set of all probability distributions on the set $S$. If $S$ is finite, then $\Delta S$ is simply a unit simplex of dimension 1 lower than the number of elements in $S$. In particular if $|S| = n$ then $$ \Delta(S) = \left\{(p_1, \ldots, p_n) \in \mathbb{R}^n_+\middle| \sum_{i = 1}^n p_i = 1\right\}. $$ If $S$ is the set of pure strategies of a player then $\Delta(S)$ is the set of mixed strategies of the same player.

  1. Does the following hold $\Delta(S1 \times S2) = \Delta(S_1)\times \Delta(S_1)$ and can this result generalized for $k(< +\infty)$ player games?

No, the two sets are not the same. For example, let $S_1 = \{x,y\}$ and $S_2 = \{u,v\}$ Then the elements of $S_1 \times S_2$ are: $$ S_1 \times S_2 = \{(x,u), (x, v), (y,u), (y,v)\}. $$ As such, $\Delta(S_1 \times S_2)$ is the 3 dimensional unit simplex with elements $(p_{(x,u)}, p_{(x,v)}, p_{(y,u)}, p_{(y,v)})$

On the other hand, $\Delta(S_1) \times \Delta(S_2)$ consists of all combinations $((p_x, p_y), (p_u, p_v))$ where $(p_x, p_y) \in \Delta(S_1)$ and $(p_u, p_v) \in \delta(S_2)$. Although you could also see these as 4 dimensional vectors they are not in the 3 dimensional unit simplex as $p_x + p_y = 1$ and $p_v + p_w = 1$, so their sum equals 2 (and not 1).

In some sense the set $\Delta(S_1 \times S_2)$ contains more 'distributions' than the set $\Delta(S_1) \times \Delta(S_2)$. By this, I mean that there are vectors $(p_{(x,u)}, p_{(x,v)}, p_{(y,u)}, p_{(y,v)}) \in \Delta(S_1 \times S_2)$ for which it is impossible to find $(p_x, p_y) \in \Delta(S_1)$ and $(p_u, p_v) \in \Delta(S_2)$ such that: $$ p_x \cdot p_u = p_{(x,u)},\\ p_x \cdot p_v = p_{(x,v)},\\ p_y \cdot p_u = p_{(y,u)},\\ p_y \cdot p_v = p_{(y,v)}. $$

For example, $p_{(x,u)} = 0.5, p_{(x,v)} = 0, p_{(y,u)} = 0, p_{(y,v)} = 0.5$ is impossible to decompose in this way. In other way the set of all distributions that can be obtained by a profile of mixed strategies is a strict subset of the set of all probability distributions over the pure strategy profiles.

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