# A New Tax Policy

Consider a simple tax policy where you plug your income into a function, and it spits your tax rate out.  The idea is to do away with the discrete tax bracket system and instead have a continuous system.  Consider the generic function

$t_0(x)=1-e^{-\alpha x^2}.$

We have $t(0)=0$ and $t(\infty)=1$.  This satisfies the boundary conditions we want on a continuous progressive tax policy where $t(x)$ represents the tax rate ($1$ being $100$%) if your income is $x$.  To gain some control over this generic function, we will wish to fix a point on it (i.e. assign a value for $\alpha$).  Suppose the average income is $\sigma$, and that we want $t(\sigma)=.25$.  Then we have that $\alpha=\sigma^{-2}\log\frac{4}{3}$.  Substituting this gives

$\displaystyle t(x)=1-e^{\log(\frac{3}{4})\sigma^{-2}x^2}=1-\left(\frac{3}{4}\right)^{\sigma^{-2}x^2}.$

Differentiating once shows the slope starts and “ends” at $0$.  Differentiating again shows that the maximal rate of tax rate increase is at $x=\sigma/\sqrt{\log(\frac{16}{9})}$ (that is, tax rates increase the fastest around this income).

We may now attempt to construct another function $\mu$ where $\mu(x)$ is the number of people in a system who have income $x$.  We can generalize to define $\mu(A)$ for some $A\subseteq [0,\infty)$, where it will denote the number of people whose income falls in $A$$\mu([0,\infty))=P$ where $P$ is a constant called the population.  We could attempt to start with a generic function

$\mu_0(x)=P_0e^{-\alpha x^2}$

where $P_0$ is called the unemployment number ($\mu_0(0)=P_0$).  In particular we could attempt to determine $\alpha$ by requiring

$\displaystyle\sigma=\frac{1}{P}\int_0^\infty\mu_0(x)\,dx.$

But evaluating the integral poses a problem; and hence so does expressing $\alpha$ in terms of the mean (or even median).  Thus it is difficult to start with the mean or median.  Let $m$ denote the mode income and $P_m$ be the number of people with income $m$.  Consider the function

$\displaystyle\mu(x)=P_m\left(\frac{P_m}{P_0}\right)^{-m^{-2}(x-m)^2}.$

Here we have $\mu(m)=P_m$, $\mu(0)=P_0$, and $\lim_{x\to\infty}\mu(x)=0$ since $P_m\geq P_0.$  With tax and distribution functions $t$ and $\mu$ we can define the revenue function as the Riemann-Stieltjes integral

$\displaystyle R(t,\mu)=\int_0^\infty xt(x)\,d\mu(x).$

Similarly we have the gross product

$\displaystyle\Pi=\int_0^\infty\mu(x)\,dx=P\sigma.$