User:IssaRice/Moral public goods example

working out the general optimal tax for the example given in https://www.greaterwrong.com/posts/pqKwra9rRYYMvySHc/moral-public-goods

each noble has utility function $u(x_{n},x_{p})=\log x_{n}+\log x_{p}$, where $x_{n}$ is the noble's own wealth and $x_{p}$ is the average wealth of a peasant (since the donation/tax is distributed equally among peasants, this is the same as saying the wealth of a single peasant).

a tax of amount $a$ shifts the current utility amount $u(x_{n},x_{p})$ to ${\textstyle u(x_{n}-a,x_{p}+{\frac {N_{n}}{N_{p}}}a)}$, where $N_{n}$ is the number of nobles and $N_{p}$ is the number of peasants. This is because each noble loses $a$, so there is $N_{n}a$ to distribute, and this is divided by the number of peasants.

If each noble starts out with $\alpha$ times as much wealth as the average peasant, we have $x_{n}=\alpha x_{p}$.

So to solve for the highest tax that the nobles are willing to pay, we want to find $a>0$ such that ${\textstyle u(x_{n},x_{n}/\alpha )=u(x_{n}-a,x_{n}/\alpha +{\frac {N_{n}}{N_{p}}}a)}$.

This is ${\textstyle \log x_{n}+\log(x_{n}/\alpha )=\log(x_{n}-a)+\log(x_{n}/\alpha +{\frac {N_{n}}{N_{p}}}a)}$ which reduces to ${\textstyle x_{n}(x_{n}/\alpha )=(x_{n}-a)(x_{n}/\alpha +{\frac {N_{n}}{N_{p}}}a)}$, which is quadratic in $a$. Using the quadratic formula and simplifying, the variable $x_{n}$ actually drops out, and we get ${\textstyle a=1-{\frac {1}{\alpha }}\cdot {\frac {N_{p}}{N_{n}}}}$ as the non-zero solution (the other solution is always zero).

Now put $A=N_{n}\alpha$ (the total wealth initially owned collectively by the nobles, normalized to one peasant owning 1) and $B=N_{p}$ (the total wealth initially owned collectively by the peasants). Then the tax is ${\textstyle \%t=1-{\frac {B}{A}}}$. We also know that the fraction of wealth the nobles initially control is ${\textstyle \%n={\frac {N_{n}x_{n}}{N_{n}x_{n}+N_{p}x_{p}}}={\frac {A}{A+B}}}$. Can we find the optimal tax in terms of $\%n$? Yes. ${\textstyle 1-\%t={\frac {B}{A}}={\frac {1}{\%n}}-1}$, so ${\textstyle \%t=2-{\frac {1}{\%n}}}$.