Editing Executive governance (section)

== Policing valuation ==
In this section we use [[Reputation tokenomics|tokenomics]] formulas to find the [[present value]] of automated policing, based on the work-to-policing ratio set in the validation pool. This will help guide DAO governance decisions for how to decide this parameter’s value to motivate more or less policing.

=== 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.

=== Finite lifetime automated policing ===
For the case of REP tokens with finite lifetime, the ODE of Equation 7 becomes a delay differential equation (DDE)[[Validation Pool#Vote|<math display="block">\frac{dP_0^1}{dt}(t)  =p_3mf'(t)\frac{P_0^1(t)}{R(t)} - \frac{dP_0^1}{dt}(t-L) .</math>]]''<small>Equation 12</small>''

with initial condition [[Validation Pool#Vote|<math display="block">\frac{dP_0^1}{dt}(t)  = \begin{cases} 0 & \text{if} \quad -L\leq t<0 \\ 1 & \text{if } \quad t=0.  \end{cases}</math>]]As before, <math>R</math>  represents the total number of tokens in a DAO with finite lifetime <math>L</math> and is given by [[Validation Pool#Vote|<math display="block">R(t)=\int_{t-L}^t m*f' (s)ds </math>]]

Again, [[Validation Pool#Vote|<math>p_3:=1-c_3</math>]] is the policing parameter, and the tokens that result from this policing process from one REP token is denoted [[Validation Pool#Vote|<math display="inline">P_0^1(t)</math>]] with [[Validation Pool#Vote|<math display="inline">P_0^1(0)=1</math>]] which 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>]]<nowiki/>which 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>]] 

To explain why the policing token holdings [[Validation Pool#Vote|<math display="inline">P_0^1</math>]] evolve according to the DDE of Equation 12, we focus on the two terms on the right-hand side. The first term [[Validation Pool#Vote|<math display="inline">p_3mf'(t)\frac{P_0^1(t)}{R(t)}</math>]] is the same as Equation 10, coming from the fact that the rate [[Validation Pool#Vote|<math display="inline">dP_0^1/dt</math>]] of newly generated policing tokens is proportional to the newly minted tokens as a fraction of the total active policing tokens [[Validation Pool#Vote|<math display="inline">R(t)</math>]]. These new tokens are minted in proportion to fees [[Validation Pool#Vote|<math display="inline">f'</math>]] and the minting ratio [[Validation Pool#Vote|<math display="inline">m</math>]] and the amount policing is rewarded [[Validation Pool#Vote|<math display="inline">p_3</math>]]. The second term [[Validation Pool#Vote|<math display="inline"> - \frac{dP_0^1}{dt}(t-L)</math>]] comes from the fact that tokens [[Validation Pool#Vote|<math display="inline">L</math>]] units of time old are expiring at the same rate they were added [[Validation Pool#Vote|<math display="inline">L</math>]] units in the past. 

The term [[Validation Pool#Vote|<math display="inline"> - \frac{dP_0^1}{dt}(t-L)</math>]] makes Equation 12 a delay differential equation<ref>Technically this is a '''neutral delay differential equation''' (NDDE), since the delay comes from the function’s past derivative, not directly from the function itself.</ref>, because the way [[Validation Pool#Vote|<math display="inline">P_0^1</math>]] changes at time [[Validation Pool#Vote|<math display="inline">t</math>]] depends on the moment  [[Validation Pool#Vote|<math display="inline">L</math>]] units before the present. Such DDEs can be solved numerically for any practical choice of parameters [[Validation Pool#Vote|<math display="inline">p_3</math>]], [[Validation Pool#Vote|<math display="inline">m</math>]], and [[Validation Pool#Vote|<math display="inline">f'</math>]].

=== Attenuation function & policing ===
In this section we detail the formulas for a more complex tool that DAOs have at their disposal. The tokens can be programmed to diminish in value in much more complicated ways than merely programming an expiration date. We can instead design the DAO so that the tokens’ power shrinks (or grows) in value as time goes on. To do this we specify an attenuation function <math>L:\Bbb R \rightarrow \Bbb R</math> which means that a token minted at time <math>0</math> will have potency <math>L(t)</math> at time <math>t</math>. So <math>L(t)</math> is a multiplier for the value of any token minted at time <math>0</math>. This means that one token minted at time <math>0</math> pays out the fraction <math>L(t)/R(t)</math> of the fees the DAO earns at time <math>t</math> instead of the usual fraction <math>1/R(t)</math>. Typically <math>L</math> will naturally be constrained <math>0 \leq L(t) \leq 1</math>, but the formulas are valid for more general functions. In particular, these formulas can be used by a DAO’s governing body to account for the reputation tokenomics consequences from diminishing or enhancing older power according to any chosen design, which amounts to choosing a time formula for <math>L</math>.

Here as before, <math>R(t)</math> represents the total amount of token power that exists in the DAO at time <math>t</math>. This is different from <math>R^T(t)</math> which is defined to be the total number of tokens ever minted (ignoring attenuation). As before <math>R^T(t)</math> satisfies[[Validation Pool#Vote|<math display="block">R^T(t)=R^T_0+\int_{0}^t m*f' (s)ds </math>]]<nowiki/>where [[Validation Pool#Vote|<math display="inline">R^T(0)=:R^T_0 </math>]] which gives[[Validation Pool#Vote|<math display="block">\frac{dR^T}{dt}(t)= m*f' (t). </math>]]This time, however, we have [[Validation Pool#Vote|<math display="block">R(t)=\int_{-\infty}^t L(t-s)\frac{dR^T}{dt}(s)ds. </math>]]''<small>Equation 13</small>''

To justify the form of <math>R(t)</math> in Equation 13, consider how at time <math>t</math> the term <math>L(t-s)</math> attenuates the tokens that have aged <math>t-s</math> units of time. The integral in Equation 13 sums all the tokens ever minted, with the term <math>dR^T/dt</math> over all times <math>s</math> before time <math>t</math> modified by how much they have attenuated in their current age <math>L(t-s)</math>.

Assuming a single token was minted at time <math>t=0</math> the fees it earns is given by [[Validation Pool#Vote|<math display="block">f_0^{1,L}(t)=\int_0^t f'(s)\frac{L(s)}{R(s)}ds</math>]]Then as before, the present value of a single token with attenuation (but without policing) has present value [[Validation Pool#Vote|<math display="block">PVf_0^{1,L}=\int_0^\infty e^{-rt} \frac{d}{dt} f_0^{1,L} (t)dt</math>]]


Finally, we add automated policing with parameter [[Validation Pool#Vote|<math display="inline">p_3</math>]] and the analogous quantity of active tokens under attenuation <math>L</math> is now denoted by [[Validation Pool#Vote|<math display="inline">P_0^{1,L}</math>]]. It is easier to keep track of all policing tokens earned from one token [[Validation Pool#Vote|<math display="inline">P_0^{1,U}=P_0^1</math>]] unadjusted by <math>L</math> which evolve according to the DDE[[Validation Pool#Vote|<math display="block">\frac{dP_0^{1,U}(t)}{dt}  =p_3mf'(t)\frac{P_0^{1,L}(t)}{R(t)}  .</math><math display="block">P_0^{1,L}(t)  =L(t)P_0^1(0)+\int_{0}^{t}L(t-s)\frac{dP_0^{1,U}}{ds}(s)ds</math>]][[Validation Pool#Vote|<math display="block">P_0^1(0)=1</math>]]This is a DDE since <math>\frac{dP_0^{1,U}}{dt}(t)</math> depends on the past of <math>P_0^{1,U}(t)</math> for <math>s<t</math>.


Using an attenuation function with non-zero derivative will improve the non-fungibility of the REP tokens. Technically two tokens minted at the exact same moment will be fungible if you ignore the possibility of slashing from review. However, similar to financial derivatives, any token minted at a different time is non-fungible with any token minted at any other time. This makes the market for REP tokens much shallower, which would result in REP tokens being more closely tied to its owner, decreasing the need for KYC protocols.

=== Policing conclusion ===
The policing parameter determines the reward for policing, which motivates members to participate in the automated function or not. If it is set high enough, then we can expect everyone in the DAO to participate, because it will be profitable. If it is on the edge of profitability, some will drop off, which will give greater reward to those who remain.

Secondly, the higher the policing parameter is set, the more expensive it is to pay for a 51% attack arbitrage, which diminishes the arbitrage possibilities.

Thirdly, the higher the policing parameter, the more old REP remains in the DAO, even with REP with bounded lifetime, because policing creates new REP from old REP. Even though policing REP diminishes in time, some subsequently generated policing REP will persist for all time, even though individual REP tokens have a finite lifetime.

This leads to the problems identified with infinite life-time tokens, albeit in a diminished form. Older members have some advantage over younger members. However, in this respect it is more justified. Older members only have the minor advantage as long as they continue policing. If they ever turn off their machines, then all their REP eventually will be extinguished.