I'd like to check algebraically a result from Osborne and Rubinstein's A Course in Game Theory. Here's it:
Let $G:=\langle N,(A_i),(u_i)\rangle$ be a finite strategic game. Then, $$t:=(t_1,...,t_n)\in \Delta(A_1)\times...\times \Delta (A_n)$$ is a Nash equilibrium of $G$ if and only if for every player $i\in N$, every pure strategy in the support of $t_i$ is a best response to $t_{-i}$.
The proof presented in the book is the following:
First suppose that there is an action $a_i$ in the support of $t_i$ that is not a best response to $t_{-i}$ . Then by linearity of $U_i$ in $t_i$ player $i$ can increase his payoff by transferring probability from $a_i$ to an action that is a best response; hence $t_i$ is not a best response to $t_{-i}$.
I know $U_i:\Delta(A_1)\times...\times \Delta (A_n) \rightarrow \mathbb R$ given by $$U_i(s):=\sum_{a\in A}u_i(a)\prod_{j=1}^ns_j(a_j)$$ is linear but I didn't understand that transfer of probability. So I could not formalize the demonstration. If someone could help me, I'd be grateful. Thanks in advance!
An example might help. Consider the following simple game, where only the payoffs of player 1 are shown $$\begin{array}{ccc} & L & R \\ T & 2 & 0 \\ B & 1 & 1 \\ \end{array}$$ Suppose player 2 is playing $L$, while Player 1 is playing $T$ with probability $p<1$ and $B$ with probability $1-p >0$. (Note that $B$ is not a best response for Player 1.)
Using this strategy, his expected playoff is $$2p + 1(1-p) = 1 + p$$ which is of course increasing in $p$. So Player 1 can increase his payoff by transferring probability from $B$ to $T$ by increasing $p$.
If you want to be more formal, interpret a "transfer" as replacing $p$ with $p^\prime = p + \varepsilon$, where $0< \varepsilon \le 1-p$. Then compare Player 1's payoff using $p$ versus using $p^\prime$.