I was reading the solution of a $2\times 2$ games with mixed strategies. My book gives answer like this:
The pay-off matrix is \begin{matrix}
& &C_1 &C_2\\
&R_1 &a_{11} &a_{12}\\
&R_2 &a_{21} &a_{22}
\end{matrix}
with the probability vector $\bar p=(p_1,p_2), \bar q=(q_1,q_2)$ for $R$ and $C$ respt.
When $C$ chooses his move $C_1$, the expected gain of $R$ is $E_1(\bar p)=a_{11}p_1+a_{21}p_2=a_{11}p_1+a_{21}(1-p_1)$
and when $C$ chooses his move $C_2$, the expected gain of $R$ is $E_2(\bar p)=a_{12}p_1+a_{22}p_2=a_{11}p_1+a_{21}(1-p_1)$
Similarly $E_1(\bar q)=a_{11}q_1+a_{12}q_2=a_{11}q_1+a_{12}(1-q_1)$ and $E_2(\bar q)=a_{21}q_1+a_{22}q_2=a_{21}q_1+a_{22}(1-q_1)$
If $v$ be the values of the game, then clearly
$E_1(\bar p)\ge v, E_2(\bar p)\ge v$ and $E_1(\bar q)\le v, E_2(\bar q)\le v$.
This is not clear to me. How do these inequalities hold for $\bar p$ and $\bar q$?
Game has value, if lower value equals upper value. Recall that lower value is $$ \underline v = \max_{\bar p} \min_{\bar q}r(\bar p, \bar q) $$ where $r$ stands for revenue function (a convex combination of $a_{ij}$ with coefficient obtained from $\bar p$ and $\bar q$) - your expected income.
Fix strategy for first player $\bar p^*$. We have $$ \underline v \geq \min_{\bar q}r(\bar p^*, \bar q) $$ directly from definition. Two of those strategies are deterministic ones: $\bar p^* = (1, 0)$ which correspond to $E_1(\bar q)\leq v$ and $\bar p^* = (0, 1)$ corresponding to $E_2(\bar q) \leq v$.
Similar reasoning can be done with upper value to obtain remaining inequalities.