Remember that player y is lying: the blue segment lies below y=0.95, but only for the fake values that y is claiming. In actual fact, that blue line is always above 0.95 (you can see this on the first diagram).
Possibly my confusion lies in the way values are being re-normalized after player y lies.
In diagram 2, consider the outcome (.5,.6). Even if we re-normalize that outcome by multiplying by the sum of y’s real utilities and dividing by the sum of y’s fake utilities, .6 * (3.1 / 2.55) =~ .73, well below the default outcome of .95. Am I doing that wrong?
There’s no need to renormalise: any outcome on the blue line is a probabilistic mixture between the (0,1) and (0.95,0.95) choices (to use the genuine utilities of these outcomes). This is better for y than the pure (0.95,0.95) option.
Remember that player y is lying: the blue segment lies below y=0.95, but only for the fake values that y is claiming. In actual fact, that blue line is always above 0.95 (you can see this on the first diagram).
Possibly my confusion lies in the way values are being re-normalized after player y lies.
In diagram 2, consider the outcome (.5,.6). Even if we re-normalize that outcome by multiplying by the sum of y’s real utilities and dividing by the sum of y’s fake utilities, .6 * (3.1 / 2.55) =~ .73, well below the default outcome of .95. Am I doing that wrong?
There’s no need to renormalise: any outcome on the blue line is a probabilistic mixture between the (0,1) and (0.95,0.95) choices (to use the genuine utilities of these outcomes). This is better for y than the pure (0.95,0.95) option.
Oh, I see. That’s why the straight lines are significant: they show that no mixture involving the (.6,.6) point is optimal. Thanks for explaining.