Suppose a cake is divided between two players using the Nash bargaining solution. Also, assume that both players have positive and continuous utility functions, and each one of them has utility $1$ for the whole cake and utility $0$ for receiving nothing. Also, assume that the set of all possible utility pairs is closed, bounded, and convex. Show that the final utility for each player is greater than or equal to $\frac{1}{2}$.
I got this problem from my game theory TA. It doesn't even seem intuitive, and I don't know whether I should use the nash bargain solution using a purely analytic approach and find the max of $z=xy$ over the defined set or use the four axioms.
Thanks for this question! Somehow, despite being quite interested in game theory, I hadn’t come across the Nash bargaining solution before. So there may well be a more elegant solution to this, but here’s how I’d prove it.
Draw a line $L$ from the solution $(x,y)$ to $(1,0)$. Since both $(x,y)$ and $(1,0)$ are feasible pairs and the set of feasible pairs is convex, all of $L$ is feasible. Now consider the isolines of $z=xy$. For $(x,y)$ to be optimal, the slope of the isoline through $(x,y)$ must be greater (less negative) than the slope of $L$ (since otherwise points on $L$ near $(x,y)$ would have greater $z$ than $(x,y)$).
So the set of possible solutions is bounded by the locus of points at which the tangent through the isoline goes through $(1,0)$. If you work it out, that turns out to be the line $x=\frac12$, so the solution must have $x\ge\frac12$. By symmetry, analogously $y\ge\frac12$.