Venice 1866, citizens of San Marco going to vote <cit.>

Dealing with quadratic payments, marginal probability is usually considered ideally constant, maybe for the sake of initial simplicity. Considering the voting scenario depicted in <cit.>, firstly its math foundations are made explicit. Developing a simple referendum model, more realistic outcome probability and marginal probability qualitative shapes are introduced. Enforcing seemingly reasonable assumptions, quadratic payments are then generalized to take into account these new functions shapes, and the way they are still quadratic is discussed. Closing remarks underline the emerging of trade-off constraints not existing in ideal case.

Missing 'biblatex' package
The bibliography requires the 'biblatex' package.

family=Andrea Barontini,

titleQuadratic Influence misc stuff
family=Andrea Barontini,

titleQuadratic payments with constrained probabilities
family=Raffaele Pontremoli,

family=Le Monde Illustré n.499,

titleLes habitants du quartier Saint-Marc, $4^{re}$ section, se rendent à l'Athénée pour voter le plébiscite.
family=Vitalik Buterin,

titleQuadratic Payments: A Primer
family=Wikipedia contributors,

titleBig O notation
This article is the rewrite -to brush up my  and to use a typesetting environment more suitable for this kind of content- of the one I published on Medium on June 10th, 2020 <cit.>.

Given some new typographic peculiarities, the contents are substantially unchanged apart from a couple of minor additions and some typos corrections:

* subtitle and tl;dr substituted by Abstract;

* sentences citing external URLs slightly modified to use bibliographic engine;

* added this Version history section;

* in A generic marginal probability & generalized quadratic payments section, inversion notation for $f(i)$ and $c(i)$ is now coherent with the rest of the article;

* in equation (7), added  previously missing $_{max}$ subscript;

* in last equation before And the name? section, generic $\Delta p(i)$ is now never used in favor of explicit special case $\Delta p$;

* grey typos corrected with gray;

* in the second last equation of And the name? section, changed system members order and restored the correct strict inequality

§ AN INSPIRING ARTICLE

Some months ago my geek-attention has been caught by a Vitalik Buterin’s article about quadratic payments and how they could be applied to some everyday choices, e.g. voting <cit.>. In many moments of my life I have wondered how much effective democracy and universal suffrage are for ballots on topics whose evaluation could be influenced by specific knowledge and awareness (for example in Italy in 1987 we had a referendum about the use of nuclear energy for civil purposes). So, starting from the article, I have felt the need to elaborate more by myself, to better understand and -why not- try to go deeper.

Before continuing here, I strongly suggest you to take a look at Vitalik’s words if you haven’t done yet, I think they are really inspiring <cit.>.

I just quickly recap the concepts about vote pricing, to have a common ground from which going on and to introduce a slightly different notation:

* $i$ is the whole number representing the votes you express, i.e. the votes you have bought;

* $c(i)$ is the price, the cost of the $i$-th vote;

* $p(i)$ is the probability the referendum will result in your desired outcome when you express i votes (so $p(0)$ is the probability when you don’t take part);

* $\Delta p(i)$ is the marginal probability: $\Delta p(i)=p(i)-p(i-1)$, the gain in probability you get when you buy one more vote, the $i$-th, after the already bought $i-1$ votes;

* $V$ is how much you value your desired outcome (e.g. if you prefer the yes winning in a referendum, it represents how much important that result is).

A reasonable condition is that:

\begin{equation}\label{eq:1}
\Delta p(i) \cdot V \geq c(i)
\end{equation}

stating that the potential value gain you get for $i$-th vote has to be higher or equal to the price of the $i$-th vote: it models the buying threshold of a rational player. Putting it in layman terms:

* $V$ is the value you assign to your desired outcome,

* so if the probability of your preferred final result is $1$ then the whole
referendum values exactly $V$ as well;

* if instead the probability of your preferred final result is $0$ then the referendum has no value for you (you could argue that referendum has negative value, but it can be absorbed with an higher $V$ for the case resulting in your desired outcome);

* each time you buy a new vote you increase the probability of getting the preferred final result (which, remember, corresponds to the whole referendum valuing $V$) by a factor $\Delta p(i)$, so you don’t want to pay that increase more than the fraction of $V$ (given by $\Delta p(i)V$) that you have potentially gained.

Of course the correspondence between actual (not the perceived) value and probability lies in frequency interpretation of the latter and return of investment cannot be guaranteed: a referendum happens only once, you really don’t have many occurrences of it determining success cases over total cases, and outcome is always binary: yes wins or no wins, no fuzziness there! That’s why I have used the potential and potentially words earlier. Nevertheless the goal here is to build incentives to make influence of each stakeholder proportional to their perceived value V, and previous mathematical stuff works for that.

I also want to underline a couple of Vitalik’s assumptions, I guess for the sake of simplicity given the introductory nature of his article:

* $\Delta p(i)$ is always considered constant: $\Delta p(i) \equiv \Delta p$

You can suspect that things get more complicated outside of ideal case when he writes: [...] though eventually the gains will decrease as the probability approaches 100% [...] , but it seems to be the only reference about it;

* influence isn’t actually defined to keep the scope of the reasoning wide, however we can observe that with constant $\Delta p$ we are in a special case where the influence is simply the number of votes $i$ that you have bought.

That said, the constraint <ref> becomes:

\begin{equation}\label{eq:2}
\Delta p \cdot V \geq c(i)
\end{equation}

The goal is to enforce <ref> to act as incentive to get an influence proportional to $V$, so in our special case we’ll want the maximum number of bought votes $i_{max}$ to be proportional to $V$.

Different types of cost function $c(i)$ are considered; First case:

\begin{equation*}
c(i) \equiv c \quad \Longrightarrow \quad \Delta p \cdot V \geq c \quad \Longrightarrow \quad V \geq \frac{c}{\Delta p}
\end{equation*}

so when $V$ reaches $c/\Delta p$ every vote we want to buy is worth its price: this means $i_{max}$ is unlimited if you can afford $c$ price for each of them: everyone who has a lot of funds to spend has a lot of influence, regardless of how much the desired outcome is valued ($V$).

Second case:

\begin{equation*}
\begin{cases}
c & i = 1\\
\infty & i > 1
\end{cases}
\end{equation*}

After the first vote, all the others have infinite cost, meaning you can buy just one, again regardless of $V$.

Plutocratic & Democratic Cost Functions

We could say first case is too plutocratic, the second one too democratic!

That’s where quadratic payments magic comes on stage! If we use a cost
function linear in $i$ (Vitalik uses a smart heuristic reasoning to derive it):

\begin{equation}\label{eq:3}
c(i) = c \cdot i
\end{equation}

from the constraint <ref> we get:

\begin{equation*}
\Delta p \cdot V \geq c \cdot i \quad \Longrightarrow \quad i \leq \Delta p \frac{V}{c} \quad \Longrightarrow \quad i_{max} = \left \lfloor \Delta p \frac{V}{c}\right \rfloor
\end{equation*}

Linear Cost Function

the stairs-like profile derives from $i_{max}$ needing to be a whole number, but linearity in whole $\Delta p \cdot V/c$ is plain! And if we calculate how much we spend to buy $i_{max}$ votes:

\begin{equation*}
c(1)+c(2)+\dots+c(i_{max})=c(1+2+\dots+i_{max})=c \frac{i_{max}(i_{max}+1)}{2} \in \mathcal{O}(i_{max}^2)
\end{equation*}

which is why we call them quadratic payments.

More or less, very summarized and with a little of more math here and there, this is the core of what you can find about voting in Vitalik’s article... now let’s try to make a few steps ahead.

§ LET’S PLAY WITH MATH!

Let’s try to derive in a formal way the linear cost function <ref>. From <ref> we know:

\begin{equation*}
c(i) \leq \Delta p \cdot V
\end{equation*}

let’s make two assumptions (that of course will have to be verified and will anyway limit the generality of the result):

* $c(i)$ is an invertible function

* it's inverse $c^{-1}(i)$ is monotonically increasing (to avoid problems with inequality sign)

which, by the way, can be condensed requiring $c(i)$ to be monotonically strictly increasing; so we can write:

\begin{equation*}
c^{-1}(c(i)) \leq c^{-1}(\Delta p\cdot V) \quad \Longrightarrow \quad i \leq c^{-1}(\Delta p\cdot V)
\end{equation*}

we have a condition limiting $i$, what we wanted. Then we require the $i$-upper bound to be proportional to $V$ (remember that we are seeking a $c(i)$ which makes $i_{max} \propto V$):

\begin{equation*}
i \leq c^{-1}(\Delta p\cdot V) = K \cdot V \quad \textrm{,} \quad K>0
\end{equation*}

from simple algebra (multiplying and dividing by $\Delta p$) it follows that:

\begin{equation*}
c^{-1}(\Delta p\cdot V) = \left(\frac{K}{\Delta p}\right) (\Delta p\cdot V)
\end{equation*}

so we can define:

\begin{equation*}
c^{-1}(\bullet) \triangleq \frac{K}{\Delta p}\bullet
\end{equation*}

we invert and get:

\begin{equation}\label{eq:4}
c(i)=\frac{\Delta p}{K}i
\end{equation}

Last step, we note <ref> is monotonically strictly increasing, so it’s an acceptable result because it respects the previous assumptions. And, as expected, it confirms <ref> setting $K= \Delta p/c$. So we have:

\begin{equation}\label{eq:5}
\begin{cases}
i \leq K \cdot V \quad \Longrightarrow \quad i_{max} = \lfloor KV \rfloor\\
c(i) = \displaystyle\frac{\Delta p}{K}i
\end{cases}
\end{equation}

It’s also useful to underline a few technicalities of the formal derivation:

* $K$, by which we have imposed the proportionality between influence and perceived value, also acts as a result’s degree of freedom defining both the maximum number of bought votes and the cost of the first one (the cheaper)... and it seems to be an unbounded degree of freedom (apart from, of course, being positive).

* The derivation cannot say anything about not monotonically increasing cost functions because only if it’s monotonically increasing we can, inverting, transform condition on cost $c(i)$ into condition on vote index $i$; however…
* …it’s not a huge limit by itself because we have an ab-initio more serious lack of generality given by $\Delta p(i) \triangleq \Delta p$ (if not, we should invert ${f(i) \triangleq c(i) / \Delta p(i)}$ and we couldn’t obtain a strictly defining condition for $c(i)$)

By now, it seems obvious we have to dig into the shape of $\Delta p(i)$ in real life to make any educated guess on how to proceed.

§ A SIMPLE REFERENDUM MODEL

So I brushed up my old interest in voting effectiveness and I tried to come up with a simple ballot model.

Let’s imagine that we are close to a referendum (a ballot with only two possible outcomes: yes or no): supporting the yes, we have done a statistical research on voters and we have got a voting prediction for the average voter. What I mean is that, instead of dealing with many different voters, each one with a different probability to vote yes, in our calculations we will use the average voter and the voting prediction associated to him (ok, it’s a rough model, but we have to start from somewhere). We have:

* $n$: number of voters

* $y$: probability the average voter will vote yes (our voting prediction)

* $p(y,n)$: probability the referendum outcome will be yes

resulting in:

\begin{equation*}
p(y,n)=\sum_{d=\left \lfloor n/2 \right \rfloor +1}^{n} \binom{n}{d} y^{d} (1-y)^{n-d}
\end{equation*}

Let me try to convince you this is a reasonable model.

Each addend of the summation takes into account cases in which yes supporters are more than no supporters, starting from minimal difference (1 or 2 votes, depending if $n$ is odd or even) and ending with all voters choosing the yes.

In each iteration $d$ is the number of yes voters, the remaining $n-d$ the number of no voters, and the binomial coefficient returns the number of different order combinations of yes and no votes (e.g. $yynyn \dots$, $yyynn\dots$, $nynyy\dots$, $\dots$).

Plotting this function gives some by-itself interesting insights into the model:

Simplest referendum model

We can see that when the number of voters $n$ grows, the curve tends to a step function polarizing the outcome (yes or no) on the sides of a discontinuity of $p(y,n)$ at $y=0.5$: when there are a lot of voters, also a very small bias in vote preference will cause the referendum result to be yes or no for certain, depending on the direction of the bias. Luckily, it’s what we expect from a referendum: even if there’s a lot of uncertainty, we want one party to win even if by few votes.

So let’s look what happens for exactly $y=0.5$ (and a couple of other values) when $n$ grows:

Growing number of voters

As we have seen earlier, for $y \neq 0.5$, $p(y,n)$ quickly tends to 1 or 0; for $y = 0.5$ it seems converging to 0.5 but with an at-first-unexpected sawtooth-like profile (visible in the leftmost parts of the other two curves as well): what is that? Just a tip: think to roulette-gambling$\dots$

Ok, time’s up: it depends on $n$ being alternatively odd and even: when even, the referendum final outcome could be yes, no... but a break-even could also be possible, and of course votes combinations leading to it will lower the yes probability, for every $y$ (like zero in roulette makes betting on a color a less than 50% affair).

Good, our model seems reasonable: so, remembering our purpose of exploring a more real (or at least a less unreal) $\Delta p(i)$, let’s take a step further introducing vote buying. Our function becomes:

\begin{equation*}
\begin{cases}
\displaystyle
\sum_{d=\left \lfloor n/2 \right \rfloor +1-i}^{n-i} \binom{n-i}{d} y^{d} (1-y)^{n-i-d} & \quad i \leq \left \lfloor n/2 \right \rfloor +1\\
\qquad \qquad \qquad \qquad 1 & \quad i > \left \lfloor n/2 \right \rfloor +1
\end{cases}
\end{equation*}

where, as earlier, $i$ is the number of votes bought for yes. The new expression reflects that:

* the total number of votes whose probabilities need to be taken into account is not $n$ anymore, but $n-i$ in fact (each bought vote has probability 1 so no need to appear in calculus);

* if yes outcome has an $i$ votes initial treasure, we also need to sum the probabilities of cases which lack -compared to previous expression- up to $i$ yes votes (that’s why the change in summation lower bound);

* the inequality condition on $i$ is the mathematical way to guarantee all quantities involving it are positive or zero, but also the formalization of the obvious fact that it’s enough to buy half of the votes plus one to be sure of yes winning.

Fixing $n=100$ and plotting between $0 \leq i \leq 60$:

Referendum model with votes-buying

Of course higher $y$ (probability the average voter will choose yes), sooner $p(y,100,i)$ reaches 1 while $i$ grows; for $y > 0.6$ buying votes is almost useless (remember discontinuity around $y = 0.5$ for high number of voters), and by $i = 51$ ($= n/2 +1$) all curves have reached highest probability.

So if a very uncertain referendum involves 100 voters and our statistical research can only establish that their penchant for yes falls between 40% and 50%, then graph above tells us we have to buy at least 9 votes (and no more than 23 needed) to have an higher than 80% probability of yes winning (just check for which i the curves for $y=0.4$ and $y=0.5$ reach the 0.8 height).

By the way, it’s obvious that $\Delta p(i) = p(i)-p(i-1)$ cannot be constant, but let’s plot it to see its shape:

Real marginal probability

Definitely not constant!

We could play a lot this way: if you want on my GitHub there’s a repository <cit.> where I have uploaded some unleashed Octave code I have used to generate the above graphs and a quite big archive (almost 1GB overall) of precomputed values of $p(y,n,i)$ for the $101 \times 1000 \times 502$ lattice domain defined by:

\begin{equation*}
\begin{cases}
y \in(0, 0.01, 0.02, \dots, 1)\\
n \in(1, 2, 3, \dots, 1000)\\
i \in(0, 1, 2, \dots, 501)
\end{cases}
\end{equation*}

Feel free to play with it, if you want. But here it’s time to extrapolate a few general properties of $p(i)$ and $\Delta p(i)$ (inspired by, but independent from, our referendum model).

§ A GENERIC MARGINAL PROBABILITY & GENERALIZED QUADRATIC PAYMENTS

Let’s recap what we have discovered about our probabilities:

* $p(i)$ is monotonically strictly increasing between 0 and 1;

* $\Delta p(i)$ is consequently positive but in general neither constant nor monotonically increasing (that’s all we know about our generic marginal probability).

It seems now we haven’t a lot to try a formal derivation as we did earlier when marginal probability $\Delta p(i) \equiv \Delta p$, because (check technicalities after <ref>):

* $\Delta p(i)$ isn’t constant in $i$ so we should invert $f(i) \triangleq c(i)/\Delta p(i)$ …
* ...but, wanting to deal with a general case, then we haven’t enough constraints on $\Delta p(i)$ to be able to derive (please note that here $[\quad]^{-1}$ is the inversion, not a power):

\begin{equation*}
f^{-1}(i) = \left[ \frac{c(i)}{\Delta p(i)}\right]^{-1} \longrightarrow \quad c^{-1}(i) \quad \longrightarrow \quad c(i)
\end{equation*}

We have to try to proceed in a wily way. Let’s begin trying to find an expression for $i_{max}$.

When marginal probability was constant, degree of influence was simply given by the number of bought votes $i$, so making influence proportional to perceived value $V$ led to:

\begin{equation*}
i \leq KV \quad \Longrightarrow \quad i_{max} = \lfloor KV \rfloor
\end{equation*}

Now each bought vote brings an influence increase $\Delta p(i)$, so to impose influence-value proportionality we should write:

\begin{equation}\label{eq:6}
\sum_{i=1}^{i_{max}} \Delta p(i) = K_{2}V \quad \textrm{,} \quad K_{2}>0
\end{equation}

The above expression isn’t rigorous as-is, other constraints will apply apart from $K_{2}$ positiveness, we will deal with them in a few lines.

By the way, now we use $K_{2}$ because when earlier we introduced $K$ we omitted $\Delta p$ from influence evaluation; however adding just one more line to that formal derivation we could have written instead:

\begin{equation}\label{eq:7}
\Delta p \cdot i \leq K_{2}V \quad \Longrightarrow \quad i_{max} = \left\lfloor \frac{K_{2}}{\Delta p} V \right\rfloor
\end{equation}

From which it follows (it will be useful later) that $K = K_{2} / \Delta p$.

Returning to our summation, to expand it we note that:

\begin{equation*}
\sum_{i=1}^{m} \Delta p(i) = \big(p(1)-p(0)\big) + \big(p(2)-p(1)\big) + \dots + \big(p(m)-p(m-1)\big) = p(m)-p(0)
\end{equation*}

\begin{equation*}
\sum_{i=1}^{i_{max}} \Delta p(i) = K_{2}V \quad \Longrightarrow \quad p(i_{max})-p(0) =  K_{2}V \quad \Longrightarrow \quad p(i_{max}) = K_{2}V + p(0)
\end{equation*}

Having a probability on the left side of the equation limits the permitted values of right side: here it is one more constraint on $K_{2}$ (differently from $K$ which was unbounded instead):

\begin{equation*}
K_{2}V+p(0)<1 \quad \Longrightarrow \quad K_{2} < \frac{1-p(0)}{V}
\end{equation*}

it’s a strict inequality (equality not allowed!) because we will want to invert $p(i)$ and, as we have discovered with our referendum model, probability function saturates when it reaches 1 (remember, to be sure of yes winning buying $n/2 + 1$ votes was enough): so invertibility of $p(i)$ is possible only in $[0,1[$.

So, recapping, we got:

\begin{equation*}
p(i_{max}) = K_{2}V + p(0) \quad \textrm{,} \quad 0 < K_{2} < \frac{1-p(0)}{V}
\end{equation*}

However we need another constraint because $i$ and $i_{max}$ are whole numbers and:

* $p^{-1}(p(i_{max})) \equiv i_{max}$ for sure …
* …but $p^{-1}(K_{2}V+p(0))$ isn’t necessarily a whole number

Unluckily we cannot introduce a further explicit constraint on $K_{2}$ because we don’t actually know $p^{-1}(\quad)$, so we cannot calculate which values for $K_{2}$ would make the right side of the equality a whole number. However, instead of searching which values of $K_{2}$ should be discarded, we can choose to accept all of them and to lead the outcomes of this permissiveness back to the allowed values. How? With the floor function:

\begin{equation*}
i_{max} = p^{-1}(p(i_{max})) = \left\lfloor p^{-1}(K_{2}V+p(0)) \right\rfloor \quad \textrm{,} \quad 0 < K_{2} < \frac{1-p(0)}{V}
\end{equation*}

Needing to insert (in an order-preserving way) each real number $p^{-1}(K_{2}V + p(0))$ into one of a set of equivalence classes labeled by whole numbers, each real value has two main whole numbers it can be mapped to: the greatest less than itself (given by the floor function) and the least greater than itself (given by the ceiling function); why have we chosen the first one? Because we are seeking the maximum whole $i$, so the real number we get is a sort of upper-bound, a value which cannot be exceeded by the whole number we need.

If not yet convinced, let’s check if this $i_{max}$ expression falls back to <ref> when ${\Delta p(i) \equiv \Delta p}$:

\begin{equation*}
p(i): \rho = \Delta p \cdot i + P \quad \Longrightarrow \quad p^{-1}(\rho): i = \frac{\rho - P}{\Delta p}
\end{equation*}

\begin{equation*}
i_{max}=\left\lfloor p^{-1}(K_{2}V+p(0)) \right\rfloor = \left\lfloor \frac{(K_{2}V+P)-P}{\Delta p} \right\rfloor = \left\lfloor \frac{K_{2}V}{\Delta p} \right\rfloor = \left\lfloor KV \right\rfloor
\end{equation*}

which confirms that floor function is the right choice, allowing particular $i_{max}$ expression to be derived from the general one.

By the way, you have perhaps noted that the way we have inverted $p(i)$ implies both $p^{-1}(i)$ and $p(i)$ being $\mathbb{R} \rightarrow \mathbb{R}$ functions; no problem even if $p(i)$ was initially defined as $\mathbb{N} \rightarrow \mathbb{R}$ function, because we can always extend it to a $\mathbb{R} \rightarrow \mathbb{R}$ one (with a polyline if nothing better is suitable, we don’t need to calculate derivatives).

Now that we know how to calculate $i_{max}$ for a generic marginal probability, it’s time to focus on cost function $c(i)$.

Because of fairness considerations it would be sound for $c(i)$, once we consider a specific value $i$, to be proportional to $\Delta p(i)$: a vote price should be linear in probability gain it allows, given all other conditions. This means the structure of cost function should be: $c(i)= \Delta p(i) g(i)$.

Cost function enforces $i_{max}$ limit: it must be unfavorable to buy more than $i_{max}$ votes, so we want $g(i)$ to keep $c(i)$ increasing with $i$, until we get:

\begin{equation*}
c(i)=\Delta p(i) \cdot g(i) > \Delta p(i) \cdot V \quad \textrm{,} \quad i > i_{max}
\end{equation*}

Note we are just saying that we want to satisfy -even if in its dual reformulation- condition <ref>. So:

\begin{equation*}
c(i)=\Delta p(i) \cdot g(i) \leq \Delta p(i) \cdot V \quad \Longrightarrow \quad g(i) \leq V \quad \textrm{,} \quad i \leq i_{max}
\end{equation*}

Applying $g^{-1}(\quad)$ to both sides we obtain:

\begin{equation*}
i \leq g^{-1}(V) \quad \textrm{,} \quad i \leq i_{max}
\end{equation*}

(inversion and inequality sign are ok because sought $g(i)$ will be an increasing function with positive domain and codomain). We now have two inequalities with $i$ on the left side, comparing them:

\begin{equation*}
\left\lfloor g^{-1}(V) \right\rfloor = i_{max}
\end{equation*}

with floor function appearing, as usual, to handle $ i_{max}$ whole-ness.
Remembering  $i_{max}$ formula, we get an explicit expression for $g^{-1}(V)$:

\begin{equation*}
g^{-1}(V): i = p^{-1}(K_{2}V+p(0))
\end{equation*}

Inverting it:

\begin{equation*}
\end{equation*}

strictly increasing, with positive domain and codomain. Great!

We have discovered that:

\begin{equation*}
c(i)=\Delta p(i) \frac{p(i)-p(0)}{K_{2}}
\end{equation*}

Let’s double-check that <ref> is satisfied (double because it should, having started the derivation imposing it):

\begin{equation*}
c(i)=\Delta p(i) \frac{p(i)-p(0)}{K_{2}} \leq \Delta p(i) \cdot V \quad \Longrightarrow \quad \frac{p(i)-p(0)}{K_{2}} \leq V
\end{equation*}

So, as expected:

\begin{equation*}
p(i) \leq K_{2}V+p(0) \quad \Longrightarrow \quad i \leq \left\lfloor p^{-1}(K_{2}V+p(0)) \right\rfloor = i_{max}
\end{equation*}

Good, and now last step, evaluating $c(i)$ when $\Delta p(i) \equiv \Delta p$:

\begin{align*}
p(i) &= \Delta p \cdot i + P\\
c(i) &= \Delta p \frac{p(i)-p(0)}{K_{2}}=\Delta p \frac{(\Delta p \cdot i + P)-P}{K_{2}} = \frac{\Delta p^{2}}{K_{2}}i = \frac{\Delta p}{K}i
\end{align*}

§ AND THE NAME?

In the beginning, calculating how much we spend to buy $i_{max}$ votes when
$\Delta p(i) \equiv \Delta p$, we got:

\begin{equation*}
\sum_{i=1}^{i_{max}} c(i) = c \frac{i_{max}(i_{max}+1)}{2} \in \mathcal{O}(i_{max}^2)
\end{equation*}

Which, considering <ref>, for the sake of thoroughness becomes:

\begin{equation}\label{eq:8}
\sum_{i=1}^{i_{max}} c(i) = \frac{\Delta p}{K} \cdot \frac{i_{max}(i_{max}+1)}{2} = \frac{\Delta p^{2}}{K_{2}} \cdot \frac{i_{max}^{2}+i_{max}}{2}\in \mathcal{O}(i_{max}^2)
\end{equation}

Now it seems interesting to check if the quadratic nature of the payment is preserved by our brand new $c(i)$:

\begin{equation*}
\sum_{i=1}^{i_{max}} c(i) = \sum_{i=1}^{i_{max}} \Delta p(i) \frac{p(i)-p(0)}{K_{2}} = \frac{1}{K_{2}} \sum_{i=1}^{i_{max}} \Delta p(i) \cdot \big[ p(i)-p(0) \big]
\end{equation*}

as previously seen, the difference inside square brackets can be expressed as summation of marginal probabilities $\Delta p(i)$, so:

\begin{equation}\label{eq:9}
\sum_{i=1}^{i_{max}} c(i) =  \frac{1}{K_{2}} \sum_{i=1}^{i_{max}} \Delta p(i) \sum_{j=1}^{i} \Delta p(j) =  \frac{1}{K_{2}} \sum_{i=1}^{i_{max}} \sum_{j=1}^{i} \Delta p(i) \Delta p(j)
\end{equation}

We need to untie that double summation. Let’s study $i-j$ plane to understand what we are adding up:

$i-j$ plane

The double summation regards the gray cells; more, values of cells are symmetric with respect to the diagonal, so when we deal with summation of values:

\begin{equation}\label{eq:10}
\textrm{gray cells} = (\textrm{whole plane}+\textrm{diagonal}) / 2
\end{equation}

Summation of diagonal is given by: $\sum \Delta p(i)^{2}$

To calculate the sum of whole plane we note that, for example, for the
lowest row we have:

\begin{equation*}
\Delta p(1) \cdot \big[ \Delta p(1)+ \Delta p(2) + \dots + \Delta p(i_{max}) \big] = \Delta p(1) \sum_{i=1}^{i_{max}}p(i)
\end{equation*}

and likewise for all other rows as well. So when we add all rows to get the whole plane:

\begin{equation*}
\big[ \Delta p(1)+\Delta p(2)+\dots+\Delta p(i_{max}) \big] \sum_{i=1}^{i_{max}}p(i) = \sum_{i=1}^{i_{max}}p(i) \sum_{i=1}^{i_{max}}p(i) = \left[ \sum_{i=1}^{i_{max}}p(i) \right]^{2}
\end{equation*}

Putting together <ref> and <ref> we get:

\begin{equation}\label{eq:11}
\sum_{i=1}^{i_{max}} c(i) = \frac{\left[ \displaystyle\sum_{i=1}^{i_{max}}p(i) \right]^{2} +  \displaystyle\sum_{i=1}^{i_{max}}p(i)^{2}}{2 K_{2}}
\end{equation}

\begin{equation}\label{eq:12}
\sum_{i=1}^{i_{max}} c(i) = \frac{A(i_{max})^{2} \cdot i_{max}^{2} + B(i_{max}) \cdot i_{max}}{2 K_{2}}
\end{equation}

where we have defined the average values in $[1,i_{max}]$:

\begin{equation}\label{eq:13}
\begin{aligned}
A(i_{max}) &\triangleq \left< \Delta p(i) \right> \quad \leq 1\\
B(i_{max}) &\triangleq \left< \Delta p(i)^{2} \right> \quad \leq 1
\end{aligned}
\end{equation}

Not surprisingly, when $\Delta p(i) \equiv \Delta p$ we obtain $A \equiv \Delta p$ , $B \equiv \Delta p^{2}$ and <ref> falls back exactly to <ref>.

In general case, using <ref> inequalities in <ref>, we obtain:

\begin{equation*}
\sum_{i=1}^{i_{max}} c(i) \leq \frac{i_{max}^{2} + i_{max}}{2 K_{2}}
\end{equation*}

Applying Wikipedia formal definition of Big-$\mathcal{O}$ notation <cit.>:

barobackcolorcolback=block-gray, boxrule=0pt,boxsep=0pt,breakable

That is, $f(x)=\mathcal{O}\big(g(x)\bigr)$ if there exists a positive real number $M$ and a real number $x_{0}$ such that

$\quad \quad \left| f(x) \right| \leq Mg(x) \quad \textrm{for all } x \geq x_{0}$

we get:

\begin{equation*}
\sum_{i=1}^{i_{max}} c(i) \leq \frac{i_{max}^{2} + i_{max}}{2 K_{2}} \leq M \cdot i_{max}^{2} \quad \Longrightarrow \quad i_{max} \geq \frac{1}{2K_{2}M-1}
\end{equation*}

So we could still talk of quadratic payments in the sense that:

\begin{equation*}
\sum_{i=1}^{i_{max}} c(i) \in \mathcal{O}(i_{max}^2)
\end{equation*}

But remembering that 1 is just an upper-bound (a very high upper-bound probably) for both $A$ and $B$, above result is perhaps too coarse: let’s check with some graphs.
These are four different probability functions $p(i)$ and their cost functions $c(i)$ for $1 \leq i \leq 1000$:

Miscellaneous probability functions & related cost ones

Their total cost functions for a given $i_{max}$ is:

Miscellaneous total cost functions

The dashed line is a reference $M \cdot i_{max}^{2}$ : as we have been suspecting, it’s definitely not a good estimator of total cost $\sum c(i)$ behavior for a generic probability $p(i)$.
However it jumps to the eye that total cost functions aren’t too different from their own probability functions. This suggests that perhaps Big-$\mathcal{O}$ equivalence classes should be driven by $p(i)$ and not by $i$: let’s take <ref> and rewrite it as:

\begin{equation*}
\sum_{i=1}^{i_{max}} c(i) = \frac{\left[ \displaystyle\sum_{i=1}^{i_{max}}p(i) \right]^{2} +  \displaystyle\sum_{i=1}^{i_{max}}p(i)^{2}}{2 K_{2}} = \frac{\big[ p(i_{max})-p(0)\big]^{2} +  \displaystyle\sum_{i=1}^{i_{max}}p(i)^{2}}{2 K_{2}}
\end{equation*}

Remembering that $\Delta p(i) \leq 1 \Longrightarrow \Delta p(i)^{2} \leq \Delta p(i)$ :

\begin{equation*}
\sum_{i=1}^{i_{max}} c(i) \leq \frac{\big[ p(i_{max})-p(0)\big]^{2} +  \displaystyle\sum_{i=1}^{i_{max}}p(i)}{2 K_{2}} = \frac{\big[ p(i_{max})-p(0)\big]^{2} + p(i_{max})-p(0)}{2 K_{2}}
\end{equation*}

From algebra:

\begin{equation*}
\sum_{i=1}^{i_{max}} c(i) \leq \frac{p(i_{max})^{2} + \big[1-2p(0)\big] p(i_{max}) +p(0)^{2} - p(0)}{2 K_{2}}
\end{equation*}

But, again, $p(0)^{2} \leq p(0)$ , so:

\begin{equation*}
\sum_{i=1}^{i_{max}} c(i) \leq \frac{p(i_{max})^{2} + \big[1-2p(0)\big] p(i_{max})}{2 K_{2}} \leq M \cdot p(i_{max})^{2}
\end{equation*}

If $[1-2p(0)] \leq 0$ the above inequality will be easily satisfied, for example, by $M=1/2K_{2} \quad \forall \ i_{max}$

When instead [1-2p(0)] > 0 we will have:

\begin{equation*}
p(i_{max}) \geq \frac{1-2p(0)}{2K_{2}M-1} \quad \Longrightarrow \quad i_{max} \geq p^{-1}\left( \frac{1-2p(0)}{2K_{2}M-1} \right)
\end{equation*}

with, for the inverted probability argument, the constraint:

\begin{align*}
0 \leq \frac{1-2p(0)}{2K_{2}M-1} \leq 1 \quad &\Longrightarrow \quad
\begin{cases}
2K_{2}M-1 > 0 &\Longrightarrow M > 1/2K_{2}\\
1-2p(0) \leq 2K_{2}M-1 &\Longrightarrow M \geq \displaystyle\frac{1-p(0)}{K_{2}}
\end{cases}\\
&\Longrightarrow \quad M \geq \displaystyle\frac{1-p(0)}{K_{2}}
\end{align*}

So we can say:

\begin{equation*}
\sum_{i=1}^{i_{max}} c(i) \in \mathcal{O}\big(p(i_{max})^2\big)
\end{equation*}

which is a far better estimator than earlier one. So legacy quadratic payments (QP) actually seem a particular case of more general family which could perhaps be named quadratic success payments (QSP) - because, remember, $p(i)$ is the probability of desired outcome given $i$ involvement (e.g. bought votes).

§ RECAPPING

Our long journey can be summarized by following table:

generic $\Delta p(i)$     $\Delta p(i) \equiv \Delta p$

$i_{max}=\left\lfloor p^{-1}(K_{2}V+p(0)) \right\rfloor$             $i_{max}=\left\lfloor KV \right\rfloor$

with $K=K_{2}/\Delta p$

$c(i)=\Delta p(i) \displaystyle\frac{p(i)-p(0)}{K_{2}}$
$c(i) = \displaystyle\frac{\Delta p}{K} i$

$\displaystyle\sum_{i=1}^{i_{max}} c(i) \in \mathcal{O}\big(p(i_{max})^2\big)$             $\displaystyle\sum_{i=1}^{i_{max}} c(i) \in \mathcal{O}\big(i_{max}^2\big)$

Recapping QSP & QP

These constraints apply:

* $i$ and $i_{max}$	are whole numbers

* $p(i)$ is monotonically strictly increasing

* $p(i):\mathbb{R} \rightarrow \mathbb{R}$ (but also $p(i):\mathbb{N} \rightarrow \mathbb{R}$ extended by a polyline is ok)

* $K_{2}>0$

* $K_{2}<\big(1-p(0)\big)/V$

Constraint a) is a way of saying that involvement of stakeholders has its own intrinsic granularity: e.g. you can buy 1 vote, 2 votes, $n$ votes but you cannot buy 1.5 votes. Note that it doesn’t rule out decimals altogether: there could be contexts in which involvement envisages fractions of unit, but those cases can fall back to whole numbers via rescaling. However continuity of real numbers isn’t contemplated;

constraints b) and c) formalize obvious characteristics of a probability function linking desired outcome to involvement, and guarantee its invertibility;

constraint d) is what makes influence proportional to perceived value $V$ (and being both positives, $K_{2}$ also has to);

constraint e) tells us that exercisable infuence is a finite asset: from <ref> and <ref> we know that influence is strictly linked to product $K_{2}V$, so having the upper-bound of $K_{2}$ inversely proportional to $V$ means maximum influence is limited as well (to 1-p(0) not surprisingly, given our definition of influence related to marginal probability).

And there’s also a consequence on market definition because it’s a two-way constraint between $K_{2}$ and $V$:

* $V$ is a per-stakeholder parameter, meaning that each player will have a different perception of desired outcome value;

* given $p(i)$, $K_{2}$ has an important role in defining cost function $c(i)$, which hasn’t to be per-stakeholder-defined: otherwise incentives mechanics would not succeed in inducing fair ratios between players’ influences, given their perception of value (if cost function changes, two stakeholders with the same V could have different $i_{max}$ and, consequently, different acquired influences). So we need a single $K_{2}$ for the entire market;

all of that seems to mean we should choose $K_{2}$ as small as we can to raise the bar of allowed $V$, however:

* a hugely motivated player we haven’t foreseen (aka one with a very high perceived value $V$, higher than our predictions) will always be able to fall out of our game rules;

* prices would increase ($K_{2}$ is at denominator of $c(i)$), meaning that stakeholders with small $V$s would be compressed towards small $i_{max}$ values: being whole numbers, the risk of too much compression is to flatten players with different perceived values to the same $i_{max}$.

Elaborating a bit more: when you choose $K_{2}$ , e) gives you an upper-bound for $V$, meaning that a stakeholder with that maximum $V$ will be incentivized to acquire all available influence ($1-p(0)$): so we should choose a $K_{2}$ corresponding to the sum of all $V$s of all players (if predictable) plus some guard value to take into account unexpected stakeholders appearance. How small $K_{2}$ (or big the guard value) will be is a trade-off needing to balance:

* how much protection is needed against a player with a very big $V$ coming earlier than others and hoarding influence;

* the need to keep prices low enough to have a good spread of $i_{max}$ values for actual players.

So $K_{2}$ also affects which difference between two values of $V$ is big enough to guarantee two different values of $i_{max}$, i.e. it chooses the granularity of $V$: because no rational player will pay more than the minimum needed for a certain level of influence.

big $K_{2}$     small $K_{2}$

More spread of $i_{max}$/influence (finer-grained $V$)     Bigger $V$ allowed

$K_{2}$ vs $V$ trade-off

In optimal case it will be:

\begin{equation*}
\min_{m \neq n}\Big| \left\lfloor p^{-1}(K_{2}V_{m}+p(0)) \right\rfloor - \left\lfloor p^{-1}(K_{2}V_{n}+p(0)) \right\rfloor \Big| = 1
\end{equation*}

Of course we have always been assuming that $p(i)$ saturates to 1 for an $i$ big enough to have space permitting fair involvement of all interested players: it doesn’t seem a limiting condition because if there isn’t enough space to act then there isn’t a market either.

§ NOT ONLY REFERENDUMS

During last thoughts about constraints I have mainly used general terms: e.g. market and stakeholders instead of ballot and voters: that’s because formulas scope seems wider than the initial referendum context.

That’s why -for example- previous words about maximum allowed sum of $V$s seem to suggest stakeholders could be in mutual competition buying as much influence as possible, so to be worried about voters with high $V$s: it’s a general possibility emerging from the framework, but it isn’t always that case, and probably it isn’t when we are modeling voters hoping for the same event (yes winning): granularity of $V$ seems much more important to exploit each voter commitment fairly.

So, reconsidering what we are dealing with from a more general point of view, we have:

* a stochastic process $S$ whose execution could lead or not to $D$ outcome;

* an $i$ metric of involvement in favor of $D$;

* a generic $p(i)$ probability of $D$ outcome;

* a $V$ value assigned to $D$ outcome by a stakeholder;

* a pricing of $i$ with a cost function $c(i)$ incentivizing the stakeholder to influence the outcome in favor of $D$ (by means of $i$ buying) in a proportional-to-$V$ way: greater involvement would be anti-economic.

The whole picture seems to suggest many applications... but that’s enough for now! Thanks for reading!


Proceedings of the Fifth Annual LHCP CMS CR-2017/217

ELECTROWEAK PRECISION MEASUREMENTS IN CMS

MILOS DORDEVIC

On behalf of the CMS Experiment,
Vinca Institute of Nuclear Sciences
University of Belgrade, Serbia

> ABSTRACT
>
>
>
>
> An overview of recent results on electroweak precision measurements from the
> CMS Collaboration is presented. Studies of the weak boson differential
> transverse momentum spectra, Z boson angular coefficients, forward-backward
> asymmetry of Drell-Yan lepton pairs and charge asymmetry of W boson
> production are made in comparison to the state-of-the-art Monte Carlo
> generators and theoretical predictions. The results show a good agreement
> with the Standard Model. As a proof of principle for future W mass
> measurements, a W-like analysis of the Z boson mass is performed.

> PRESENTED AT
>
>
>
>
> The Fifth Annual Conference
>
> on Large Hadron Collider Physics
>
> Shanghai Jiao Tong University, Shanghai, China
>
> May 15-20, 2017

##  1 Introduction

Precision measurements of the electroweak observables of the Standard Model
(SM) provide one of the most stringent test of its consistency. The data
recorded at the highest available energies by the experiments such as ATLAS
[1] and CMS [2] allows to tune the Monte Carlo (MC) generators and provide
state-of-the-art theory calculations. Any deviation from the SM prediction
could be a sign of new physics. Differential measurement of the weak boson
transverse momentum spectrum, angular coefficients of the Z boson, forward-
backward asymmetry of Drell-Yan lepton pairs, W boson charge asymmetry and
W-like analysis of Z boson mass are presented in this overview.

##  2 Weak vector boson transverse momentum spectra

The W and Z boson differential cross section was measured [3] using a low
pile-up run with an average of 4 interactions per bunch crossing. The results
were compared to the prediction of RESBOS (P and CP versions) [4, 5, 6],
POWHEG [7] and FEWZ [8, 9, 10] generators. The ratio of the predicted to the
measured differential cross section for the W boson is presented in Fig. 2. In
the case of RESBOS-P the agreement is good up to the W boson transverse
momentum of 110 GeV, while POWHEG shown an increase of 12$\%$ at 25 GeV. The
NNLO FEWZ generator reported a difference of 10$\%$ at the transverse momentum
of 60 GeV. The corresponding results for the Z boson shows a good agreement
with the RESBOS-CP, the 30$\%$ lower estimate of data in the 0 to 2.5 GeV
range and 18$\%$ excess between 7.5 and 10 GeV. In the low transverse momentum
region below 20 GeV, the FEWZ generator has a divergent behaviour, as shown in
Fig. 2. The $W^{+}/W^{-}$ and $Z/W$ ratios are measured as well and all three
generators listed above describe data well within the experimental
uncertainties. The ratio of differential Z boson cross section was measured at
7 and 8 TeV in the muon channel and theory prediction provides a good
description of data, within the uncertainties.

Figure 1: Normalized differential cross section for charge independent W boson
production as a function of $p_{T}^{W}$ for muon decay [3].

Figure 2: Comparison of the normalized dimuon differential transverse momentum
distribution with different theoretical predictions [3].

##  3 Angular coefficients of Z boson

The first CMS measurement [11] of the angular coefficients of Z boson decaying
to muon pair is presented. These coefficients govern the decay of Z boson and
thus also the kinematics of the lepton. The values of the coefficients follow
from the vector and axial vector structure of boson-fermion couplings. In
proton-proton collisions the dominant contribution is the quark annihilation
process, while the contribution of the qg Compton process is larger than in
the proton-antiproton collisions. The coefficients were measured as a function
of the transverse momentum of the two lepton system ($q_{T}$) and in two
rapidity bins. The data was compared to the prediction of MadGraph (LO) [12],
POWHEG (NLO) and FEWZ (NNLO) generators. The MadGraph generator predicts
higher $A_{4}$ coefficient, as shown in Fig. 4 as it calculates the
$\Theta_{W}$ without considering the radiative corrections. A violation of the
Lam-Tung relation is observed in the $A_{0}-A_{2}$ distributions, as shown
Fig. 4, in particular for high $q_{T}$ due to missing higher order
calculations. The measurement of these coefficients is important for the high-
precision measurements of the W mass and electroweak mixing angle.

Figure 3: The $A_{4}$ coefficient measured in bins of $q_{t}$ and rapidity up
to 1 [11].

Figure 4: The $A_{0}-A_{2}$ measured in bins of $q_{t}$ and rapidity up to 1
[11].

##  4 Forward-backward asymmetry of Drell-Yan lepton pairs

The forward-backward asymmetry $A_{FB}$ arises from the vector and axial-
vector couplings of the electroweak bosons to fermions and it is measured [13]
for dilepton masses between 40 GeV and 2 TeV and dilepton rapidity up to 5.
The invariant mass distribution for the dimuon events in shown in Fig. 6. The
events are assigned either to ”forward” or ”backward” region based on the
positive or negative value of $cos\Theta^{*}$, where $\Theta^{*}$ represents
emission angle of the negatively charged lepton relative to the quark momentum
in the rest frame of the dilepton system. Corrections were applied for
detector resolution, acceptance and final state radiation. The effective weak
mixing angle of $sin^{2}\Theta_{lept}^{eff}=0.2312$ is used for comparison
with the prediction from POWHEG. For all rapidity regions the measured
$A_{FB}$ values are in a good agreement with this prediction. The unfolded
$A_{FB}$ distribution in the forward rapidity region is shown in Fig. 6.

Figure 5: Invariant mass distribution for dimuon events, rapidity up to 2.4
[13].

Figure 6: The unfolded $A_{FB}$ distribution for the forward rapidity region
[13].

##  5 W boson differential cross section and charge asymmetry

The differential cross section is measured [14] as a function of the muon
pseudorapidity and found to be well described by the prediction from FEWZ
generator and several parton density functions (PDF). No electroweak
corrections are included for this measurement. Measured charge asymmetry also
agrees well with the tested PDF sets, as shown in Fig. 8 and Fig. 8. This
measurement is able to provide constrains on the proton PDFs. The results are
incorporated into QCD NNLO analysis and combined with the deep inelastic
scattering data recorded by the HERA experiment [15]. A significant
improvement in the accuracy of the valence quark distributions in the low
range of Bjoerken variable, $0.001<x<0.1$, has been observed.

Figure 7: Comparison of asymmetries to NNLO predictions calculated using
FEWZ3.1 generator interface with different PDF sets [14].

Figure 8: Differences between the theoretical predictions and the measured
asymmetries [14].

##  6 Z boson inclusive and differential cross section

The inclusive and differential Z boson cross section produced via Drell-Yan
process is measured [16] as a function of kinematic variables. The transverse
momentum distribution of the Z boson is a probe of the strong nuclear
interaction. The total inclusive cross section times branching fraction is
measured for the di-muon mass in the range of 60 to 120 GeV and found to have
a value of $1870\pm 2(stat)\pm 35(syst)\pm 51(lumi)pb$. This result is in a
good agreement with the NNLO QCD and NLO electroweak calculations from FEWZ
generator and various sets of PDFs used. The comparison of the measured
differential cross section to MadGraph5aMC@NLO, POWHEG and FEWZ generators is
shown in Fig. 10 and Fig. 10. A good agreement with FEWZ is found, but a
deviation at the low transverse momentum is observed due to absence of
resummation. The MadGraph5 and POWHEG slightly over predict the distribution,
within the uncertainties.

Figure 9: Transverse momentum distribution of the Z boson decaying to dimuons
[16].

Figure 10: The MC prediction to data ratio for the dimuon transverse momentum
[16].

##  7 W-like measurement of Z boson mass

Precise measurement of the W, top and Higgs mass represent a crucial test of
the SM. In this measurement [17] the dimuon events from the Z boson decay were
selected and one of the two muons removed to mimic the W boson decay topology.
This is proof of principle and a test of the analysis methods for the future W
mass measurement. The muon momentum was calibrated using the $J/\Psi$ and
$Y(1s)$ dimuon events, as shown in Fig. 12. To reach an accurate control over
the missing transverse energy, the hadronic recoil calibration is performed
using dimuon events from Z boson decay. The results of the fits to data are
shown in Fig. 12 for the lepton transverse momentum, transverse mass and
missing transverse energy, in the positive and negative ”W-like” mass cases,
where the experimental uncertainties on the recoil and the lepton are quoted
separately from the others. The observable expected to provide the most
precise measurement of the W mass, the transverse mass, provides a fitted mass
that differs from the PDG value by 18 MeV or -20 MeV when using positive or
negative samples, respectively. The future W mass measurement requires these
calibration methods to be refined and the presented analysis methods to be
extrapolated to the W mass case.

Figure 11: Closure of the calibration of the relative scale (data with respect
to MC) for $J/\Psi$,$\Upsilon(1S)$ and Z dimuons, as a function of transverse
momentum of the positive muon [17].

Figure 12: Difference between the fitted mass and $M_{Z}^{PDG}$ obtained with
each of the three observables [17].

##  8 Conclusions

In the past years a lot of work has been done to understand the SM at the
highest possible precision. The measurements clearly profited from the
advances in theory calculations and MC generators. In this overview the
differential cross sections and their ratios were compared to the NNLO
predictions. The angular Z boson coefficients were measured and forward-
backward asymmetry studied in the extended range of rapidity up to 5. The
accuracy of valence quark distributions was improved with the measurement of
the charge asymmetry of the W boson production. The proof of principle and
validation of analysis methods for W mass measurement was established.

ACKNOWLEDGEMENTS

I thank the CMS Collaboration for the work presented here and for the support
in preparing this talk, the CERN accelerator division for the excellent
operation of the LHC and all funding agencies that made this experiment
possible.

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