Q: Finding ALL pareto optimal profiles in a bimatrix game I have a classic Prisoner Dilemma game with the following matrix:
$$
\begin{pmatrix}
(4,4) & (1,10) \\
(10,1) & (2,2)
\end{pmatrix}
$$
pure profiles $(1,10)$ and $(10,1)$ are p.o. because it is the best payoff for second and first player, respectively. Now, $(4,4)$ is not pareto optimal because mixed profile $((1/2,1/2),(1/2,1/2))$ pareto dominates it. It gives both players expected payoff of $4.25$.
It is clear that if given profile is best for some player, then every randomization will yield at most the payoff he is given in his best pure strategy.
Many internet sources on game theory say that some (pure) profiles are pareto optimal even if they do not give the best payoff to some player. If we tweaked the numbers in the game a bit, the left up corner would be pareto optimal
(as claimed for example here on page 5). But I don't see why it is not dominated by some mixed profile.
So, how can I prove that some pure strategy profile is pareto optimal?
How do I find all other (mixed) pareto optimal profiles and show that I indeed have all of them ?

A: The general idea is that a Pareto Optimal strategy profile is such in which an improvement to the payoff of one player cannot be done without decreasing the payoff of the other. This is true in the $(10,1)$ case: you can't improve the $1$ without damaging the $10$.
In this case, you can draw the 4 payoffs and their convex hull. The Pareto Optimal payoffs are all those that are "on the right and upper line". In this case, it would be all the payoffs of the form $\alpha (10,1)+ (1-\alpha) (1,10)$ for $\alpha\in [0,1]$.
The intuition is clear: any interior point is not Pareto (you can add $\epsilon$ to all coordinates and get a better payoff for all) while this line is decreasing (so one's gain is another's loss) and optimal (no possible movement in the $(1,1)$ direction, not even infinitesimal).
