Plot a function under constraint

Question

Let's say I have a function $F(q,p)$ and I want to plot the function in 3D view for $p$ and $q$ where $p+q=1$ As you probably notice that $p$ and $q$ are the probability of occurring/not occurring for an event.

I tried the following:

 f = (p^2 (0.25 - 0.375 q) - 0.5 (1. - 1. q)^2 q + 
 p (-1. + 0.875 q + 0.125 q^2))/(p (-1. + q) (1.5 + 1. q))
 Plot3D[f, {p, 0, 1}, {q, 0, 1},RegionFunction -> Function[{p, q}, p + q = 1]]

Which gave me the plot for all $p,q \in [0,1]$ and this is not the thing I wanted. I tried to set the constraint to zero, where no values satisfy this except for zeros. But I also gave me the same thing.

I also tried

Plot3D[ConditionalExpression[f,  p + q = 1], {p, 0, 1}, {q, 0, 1}, BoundaryStyle -> Directive[Thick, Red], Mesh -> None]

But it gave me an empty graph.

Answer 1

I've found that the simplest way is to use the RegionFunction command. For example, here is a simple plot of the Gini Impurity function:

Plot3D[ z, {x, 0, 1}, {y, 0, 1},
 RegionFunction -> Function[{x, y}, 0.99 < x + y < 1.01 ],
 BaseStyle -> {FontSize -> 16, FontFamily -> "Latin Modern Roman"},
 AxesStyle -> Black,
 PlotRange -> {0, .6},
 AxesLabel -> Automatic
 ]

which produces the following:

Note: if the constraint was simply x+y == 1, then it appears as if you get an empty graph. That is because that line is infinitesimally thin. Its best if you give the plot some "width" by putting in an inequality as a constraint.