Editing Executive governance (section)

=== Infinite lifetime REP ===
Whenever a validation pool is opened with a fee of [[Validation Pool#Vote|<math>f</math>]] in fungible currency units, then [[Validation Pool#Vote|<math>mf</math>]] reputation tokens are minted, and [[Validation Pool#Vote|<math>c_3mf</math>]] tokens are staked in the poster’s name and [[Validation Pool#Vote|<math>(1-c_3)mf</math>]] tokens are given to the bench for policing. Here [[Validation Pool#Vote|<math>0\leq c_3\leq 1</math>]] gives the '''work-to-policing reward ratio''' of [[Validation Pool#Vote|<math>c_3/(1-c_3)</math>]]. Default is [[Validation Pool#Vote|<math>c_3=1/2</math>]] when work and policing are equal. When [[Validation Pool#Vote|<math>c_3<1/2</math>]] policing is augmented. When [[Validation Pool#Vote|<math>c_3>1/2</math>]] work is encouraged. The term “policing” may give the wrong impression, since automated policing may have a greater effect on how much entrenched power is protected, with [[Validation Pool#Vote|<math>c_3\approx 1</math>]] encouraging new members to join and earn REP through work, and [[Validation Pool#Vote|<math>c_3\approx 0</math>]] favoring older members who run the automated policing algorithms to protect their power. These claims will become more specific and obvious from the mathematical analysis that follows. In the end, these calculations will dictate how to set the policing reward parameter to motivate optimal policing and also whether to encourage new members to participate or to protect established members’ power.

As in the previous calculations, the rate of fees [[Validation Pool#Vote|<math>f'</math>]] determines the present value of a policing token. The total number of tokens in a DAO is again given by [[Validation Pool#Vote|<math display="block">R(t)=\int_{-\infty}^{t} m*f' (s)ds=R_0+\int_{0}^t m*f' (s)ds</math>]]<nowiki/>where [[Validation Pool#Vote|''<math>R_0=R(0)</math>'']]. Assuming a single token was minted at time [[Validation Pool#Vote|<math>t=0</math>]] the fees it earns is again given by[[Validation Pool#Vote|<math display="block">f_0^1 (t)=\int_0^t\frac{f'(s)}{R(s)}  ds.</math>]]The [[present value]] at time [[Validation Pool#Vote|<math>t_0=0</math>]] when 1 token was minted is[[Validation Pool#Vote|<math display="block">PVf_0^1=\int_0^\infty e^{-rt} \frac{d}{dt} f_0^1 (t)dt=\int_0^\infty e^{-rt}\frac{f'(t)}{R(t)}dt.</math>]]

But the new REP tokens the single token earns by policing is governed by the parameter [[Validation Pool#Vote|<math>c_3</math>]]. For each fee that enters the DAO, the first REP token earns the fraction [[Validation Pool#Vote|<math>p_31/R^p(t)</math>]] of the [[Validation Pool#Vote|<math>mf'</math>]] newly minted REP tokens, where the policing parameter is simply [[Validation Pool#Vote|<math>p_3:=1-c_3</math>]] and [[Validation Pool#Vote|<math>R^p</math>]] is the number of tokens that participate in the policing actions. So [[Validation Pool#Vote|<math>R^p\leq R</math>]]. For simplicity, we will conservatively assume [[Validation Pool#Vote|<math>R^p=R</math>]] which is reasonable if [[Validation Pool#Vote|<math>p_3</math>]] is large enough to justify participation. However, this means our estimates will undervalue reliably active tokens, since all DAO members’ continual participation is not likely.

Now each token that the first token earns by policing will also earn further policing tokens at the same rate, ''ad infinitum''. The tokens that result from this policing process from a single initial REP token is denoted by [[Validation Pool#Vote|<math display="inline">P_0^1(t)</math>]] which satisfies the initial condition [[Validation Pool#Vote|<math display="inline">P_0^1(0)=1</math>]]. So [[Validation Pool#Vote|<math display="inline">P_0^1(t)</math>]] grows according to the ODE[[Validation Pool#Vote|<math display="block">\frac{dP_0^1(t)}{dt}  =p_3mf'(t)\frac{P_0^1(t)}{R(t)}  .</math>]]''<small>Equation 10</small>''

which is separable and can be explicitly solved as[[Validation Pool#Vote|<math display="block">\int \frac{dP}{P}  =p_3m\int \frac{f'(t)}{R(t)}  .</math>]]<nowiki/>assuming [[Validation Pool#Vote|<math>p_3</math>]] and [[Validation Pool#Vote|<math>m</math>]] are constant. Then[[Validation Pool#Vote|<math display="block">P_0^1(t)  =exp\Biggl( p_3m\int _0^t \frac{f'(t)}{R(t)}\Biggr) .</math>]]''<small>Equation 11</small>''

1 token which is actively policing then produces [[Validation Pool#Vote|<math display="inline">P_0^1</math>]] coins which together have the reputational salary [[Validation Pool#Vote|<math display="inline">f_0^{1,P}</math>]] given by the formula[[Validation Pool#Vote|<math display="block">f_0^{1,P}(t)=\int_0^t f'(s)\frac{P_0^1(s)}{R(s)}ds.</math>]]1 token which is actively policing then has present value[[Validation Pool#Vote|<math display="block">PVf_0^{1,P}=\int_0^\infty e^{-rt} \frac{d}{dt} f_0^{1,P} (t)dt</math>]]<nowiki/>similar to the present value of a passive token.

To improve our intuition for these formulas and gain some mathematical perspective about how important policing is, next we make simplifying assumptions, such as constant or exponential fee rates [[Validation Pool#Vote|<math>f'</math>]].

==== Constant fees ====
For constant fees [[Validation Pool#Vote|<math>f'(t):=f_0'</math>]] we get[[Validation Pool#Vote|<math display="block">R(t)=R_0+\int_{0}^t m*f' (s)ds=R_0+mf_0't </math>]]<nowiki/>and so Equation 11 gives[[Validation Pool#Vote|<math display="block">P_0^1(t)  =exp\Biggl( p_3m\int _0^t \frac{f_0'}{R_0+mf_0's}\Biggr)ds=exp\Biggl( p_3 \biggl(ln(R_0+mf_0's)-ln(R_0)\biggr)\Biggr) =\biggl(1+\frac{mf_0'}{R_0}t\biggr)^{p_3}.</math>]]If (as we are assuming) all members police all the time, then[[Validation Pool#Vote|<math display="block">P_0^{R_0}(t)=R_0 \biggl(1+\frac{mf_0'}{R_0}t\biggr)^{p_3}.</math>]]<nowiki/>gives the formula for the collection of all policing tokens [[Validation Pool#Vote|<math display="inline">P_0^{R_0}</math>]] generated from the original tokens [[Validation Pool#Vote|<math display="inline">R_0</math>]] from time [[Validation Pool#Vote|<math display="inline">t=0</math>]]. The policing tokens from a single token  grow without bound, however the relative power that all original tokens [[Validation Pool#Vote|<math display="inline">R_0</math>]] maintain under automated policing is[[Validation Pool#Vote|<math display="block">\frac{P_0^{R_0}(t)}{R(t)}=\frac{R_0 \biggl(1+\frac{mf_0'}{R_0}t\biggr)^{p_3}}{R_0+mf_0t}=\biggl(1+\frac{mf_0'}{R_0}t\biggr)^{p_3-1}\rightarrow 0</math>]]<nowiki/>as <math>t\rightarrow \infty</math> anytime [[Validation Pool#Vote|<math>p_3<1</math>]]. This proves policing alone cannot maintain your relative power in the DAO, nor the power of the original cohort . The inflationary minting mechanism will dilute your power if you don’t continue to contribute to the DAO with further work. So the entire starting DAO members will eventually be usurped if they don’t add more original contributions beyond automated policing.

We also can calculate the income stream [[Validation Pool#Vote|<math display="inline">f_0^{1,P}(t)</math>]] of 1 token with policing [[Validation Pool#Vote|<math display="inline">P_0^1(t)</math>]] and its present value under the assumption of constant fees:[[Validation Pool#Vote|<math display="block">f_0^{1,P}(t)=\int_0^t f'(s)\frac{P_0^1(s)}{R(s)}ds=\int_0^t f_0'\frac{\biggl(1+\frac{mf_0'}{R_0}s\biggr)^{p_3}}{R_0+mf_0's}ds=\frac{f_0'}{R_0}\int_0^t \biggl(1+\frac{mf_0'}{R_0}s\biggr)^{p_3-1}ds=\frac{1}{mp_3}\Biggl(\biggl(1+\frac{mf_0'}{R_0}t\biggr)^{p_3}-1\Biggr).</math>]]<nowiki/>and<math display="block">PVf_0^{1,P}=\int_0^\infty e^{-rt} \frac{d}{dt} f_0^{1,P} (t)dt=\frac{f_0'}{R_0}\int_0^\infty e^{-rt}\biggl(1+\frac{mf_0'}{R_0}t\biggr)^{p_3-1}dt</math>This last expression is finite, but not expressible using elementary functions. This is common for any expression that derives from a complicated continuous process—most elementary functions lack elementary antiderivatives. Technically it is an incomplete gamma function. When necessary, we can make tables of values for any relevant parameters we happen to be using in order to gain intuition for how it behaves.

==== Exponential fees ====
Now we assume, as above, the fees follow an exponential rate [[Validation Pool#Vote|<math display="inline">f'(t):=f_0'e^{ct}</math>]]

This gives[[Validation Pool#Vote|<math display="block">R(t)=R_0+\int_{0}^t m*f' (s)ds=R_0+mf_0'\frac{e^{ct}-1}{c} </math>]]Using Equation 11 we get[[Validation Pool#Vote|<math display="block">P_0^1(t)  =exp\Biggl( p_3m\int _0^t \frac{f'(t)}{R(t)}\Biggr)=\Biggl(1+\frac{mf_0'}{R_0}\frac{e^{ct}-1}{c} \Biggr)^{p_3}</math>]]The single policing token gives the reputational salary[[Validation Pool#Vote|<math display="block">f_0^{1,P}(t)=\int_0^t f'(s)\frac{P_0^1(s)}{R(s)}ds=\frac{1}{mp_3}\Biggl(\biggl(1+\frac{mf_0'}{R_0}\frac{e^{ct}-1}{c}\biggr)^{p_3}-1\Biggr).</math>]]

Notice policing makes a qualitative difference under exponential fees, since a passive token had salary [[Validation Pool#Vote|<math display="inline">f_0^1(t) \sim \frac {1}{m}ct  </math>]] asymptotically linear, whereas policing gives [[Validation Pool#Vote|<math display="inline">f_0^{1,P}(t) \sim e^{cp_3t}</math>]] asymptotically exponential salary.

This salary’s present value is<math display="block">PVf_0^{1,P}=\int_0^\infty e^{-rt} \frac{d}{dt} f_0^{1,P} (t)dt</math>which is not expressible using elementary functions, but is given by the hypergeometric function <math>2F_1</math>. Notice <math display="inline">PVf_0^{1,P}</math> is finite when <math>cp_3<r</math> and infinite otherwise. Therefore, we see that when the policing ratio <math>p_3</math> and the exponential growth rate <math>c</math> overcome the interest rate <math>r</math> we can expect explosive returns, which is a threat to the stability of the DAO since it is likely a hype cycle would form, generating unreasonable expectations of future earnings. 

Even though exponentially growing fees gives an exponentially growing salary, notice that the <math>R_0</math> original tokens, even if they all participate reliably in policing will be diluted in relative power according to <math display="block">\frac{R_0P_0^1(t)}{R(t)}=\frac{R_0\Biggl(1+\frac{mf_0'}{R_0}\frac{e^{ct}-1}{c} \Biggr)^{p_3}}{R_0+mf_0'\frac{e^{ct}-1}{c}}=\Biggl(1+\frac{mf_0'}{R_0}\frac{e^{ct}-1}{c} \Biggr)^{p_3-1}\rightarrow0</math>

as <math>t\rightarrow \infty</math> so the DAO will predictably decentralize, even if the founders continually police the newcomers. The only way founders can maintain majority power is by bring the majority of new fees to the DAO by continually performing the majority of productive work.