It doesn't have anything to do with displacement, it pressure and force we're talking about.
An example...
Assume we have a master cylinder that has a piston area of 1 square inch, we have a pedal/fulcrum arrangement that has a mechanical advantage of 5:1 and we apply 100lbs of force to the pedal, the fluid (which is incompressible) "sees" a line pressure of 500lb per square inch.
This pressure (which is equal in all parts of the system) is transferred to our caliper piston, which for this exercise we will say has an area of 2 square inches, it's "seeing" a line pressure behind it of 500lbs PER SQUARE INCH, it has an area of 2 square inches, so it's applying 1000lbs of force to the rear of the brake pad... it hydraulic multiplication.
So if we up the caliper piston size to say 3 square inches, and all else stays the same, the fluid pressure stays the same (with the same MC size and pedal ratio and force applied) at 500 lbs per square inch but the force applied to the back of the pad increases to 1500lbs of force.
Increasing the master cylinder size and displacing more fluid per inch of stroke doesn't equate to more hydraulic force, it actually lessens it....
Another example, using the same pedal ratio of 5:1 and pedal force of 100lbs as the first, if we double the master cylinder size to 2 square inches, the force applied to the master cylinder piston is the same at 500lbs, but the line pressure will halved to 250 lbs per square inch because the piston area has doubled, even though for the same stroke we will shift more fliud.
SteveC