Given the definition of the mixed extension of a finite game as in the link below (only first 7 lines): How to find perfect equilibria in a finite game?
We define the expected utility function in the mixed extension of a finite game (with the same notation in the link) as
$\forall \sigma=(\sigma_1,\cdots,\sigma_n)\in\Delta S,~u_i'(\sigma)=\Sigma_{s\in S}[\Pi_{i\in N}\sigma_i(s_i)]\cdot u_i(s)$, where $s=(s_1,\cdots,s_n)\in S=S_1\times\cdots\times S_n$ and $\sigma_i(s_i)$ is the chance using $s_i$ in $\sigma_i$.
I wonder if such a expected utility function is linear, though others say so. I literally tried to prove it:
Let $\sigma=(\sigma_1,\cdots,\sigma_n)$, $\gamma=(\gamma_1,\cdots,\gamma_n)$, and $\alpha\in\mathbb{R}$, $u_i'(\alpha\sigma+\gamma)=\Sigma_{s\in S}[\Pi_{i\in N}(\alpha\sigma+\gamma)_i(s_i)]\cdot u_i(s)=\Sigma_{s\in S}[\Pi_{i\in N}(\alpha\sigma(s_i)+\gamma(s_i))]\cdot u_i(s)=\Sigma_{s\in S}[\Pi_{i\in N}\alpha\sigma_i(s_i)]\cdot u_i(s)+\Sigma_{s\in S}[\Pi_{i\in N}\gamma_i(s_i)]\cdot u_i(s_i)$.
The problem is in the first part of the last equality $\Sigma_{s\in S}[\Pi_{i\in N}\alpha\sigma_i(s_i)]\cdot u_i(s)$. We can't simply take $\alpha$ out of this expression. So it can't be linear.
Is there any problem above?
Let me give an example here. Suppose we have 2 players having strategy sets $S_1=\{a_1,a_2\}$ and $S_2=\{b_1,b_2\}$. We then have $\Delta S_1=\{(x,1-x)\in\mathbb{R}^2|0\leq x\leq1\}$, where $x$ is the chance that player 1 uses $a_1$ and $\Delta S_2=\{(y,1-y)\in\mathbb{R}^2|0\leq y\leq1\}$, where $y$ is the chance that player 2 uses $b_1$. For each player $i$, given a strategy profile $\sigma=(x,1-x,y,1-y)\in\Delta S_1\times\Delta S_2$, $u_i'(\sigma)=[x\cdot y\cdot u_i(a_1,b_1)]+[x\cdot (1-y)\cdot u_i(a_1,b_2)]+[(1-x)\cdot y\cdot u_i(a_2,b_1)]+[(1-x)\cdot(1-y)\cdot u_i(a_2,b_2)]$.
Is this illustration of the expected utility function clear to you?
And no, the expressions I mentioned at the beginning does not have any miscalculation like $(x+y)^2=x^2+y^2$. :)
A different idea comes to me:
Does this linearity of $u_i'(\sigma_1,\cdots,\sigma_n$ mean the linearity only in $\sigma_i$?
That is, for all $i$, if $\sigma_i,\sigma_i'\in \Delta S_i$ and for all $\alpha\in\mathbb{R}$, $u_i'(\sigma_1,\cdot,\alpha\sigma_i+\sigma_i',\cdots,\sigma_n)=\alpha u_i'(\sigma_1,\cdots,\sigma_i,\cdots,\sigma_n)+u_i'(\sigma_1,\cdots,\sigma_i',\cdots,\sigma_n)$, which is proved to be right.
The expected utility is multilinear in the sense that: for every $\sigma \in \Delta(S)$, every $ \alpha_{i}$ and $\beta_{i}$ in $\Delta(S_{i})$ and every $\lambda \in [0,1]$ you have $U_{i}(\sigma_{-i},\lambda \alpha_{i} + (1-\lambda ) \beta_{i}) = \lambda U_{i}(\sigma_{-i},\alpha_{i}) + (1-\lambda )U_{i}(\sigma_{-i},\beta_{i})$.