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.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: