Deriving the Geometric Utilitarian Weights

post by StrivingForLegibility · 2024-08-07T01:39:58.686Z · LW · GW · 0 comments

Contents

  Overview
  Parameterizing the Harsanyi Hyperplane
  How Does G Change as We Change These Parameters?
    How does our point on H change as we change these parameters?
    How Does G Change as We Move Around on H?
    Putting These Terms Together
  Solving for the Geometric Weights
  How to Calculate Weights for p
  Checking Our Solution
  P Is the Unique Optimum of G When Weights Are Positive
  Interactive Implementations
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This is a supplemental post to Geometric Utilitarianism (And Why It Matters), in which I show how I derived the weights $\psi$ which make any Pareto optimal point $p$ optimal according to the geometric weighted average. This is a subproblem of the proof laid out in the first post [LW · GW] of this sequence, and the main post [LW · GW] describes why that problem is interesting.

Overview

So how are we going to calculate weights $\psi$ which make $p$ optimal among $F$?

The idea here is to identify the Harsanyi hyperplane $H$, which contains all of the joint utilities $u\in R^n$ which satisfy $H(u,\varphi )=H(p,\varphi )$. Where $\varphi$ are the weights which make our chosen point $p\in R^n$ optimal with respect to $H(\_,\varphi )$. And we're going to calculate new weights $\psi$ which make $p$ optimal with respect to $G(\_,\psi )$. It turns out it's sufficient to make $p$ optimal among $H$, and $p$ will also be optimal across our entire feasible set $F$.

In terms of calculus, we're going to be constructing a function $G\circ H$, which tells us about how moving around on $H$ changes $G$. And we're going to choose weights $\psi$ which make the gradient ${\nabla }_v(G\circ H)$ equal 0 at $p$. This makes it a local optimum, and it will turn out to be a global maximum across $H$, which in turn will make it a global maximum across $F$.

Geometrically, we can think of that as the surface gradient of $G$ across $H$. And so in terms of the overall gradient ${\nabla }_{u}G$, we're designing $\psi$ so that ${\nabla }_{u}G$ is perpendicular to $H$ at $p$.

Parameterizing the Harsanyi Hyperplane

When thinking about moving around on the Harsanyi hyperplane $H$, we have a linear constraint that says no matter which $u\in H$ we pick, we know that $u\cdot \varphi =p\cdot \varphi =H(p,\varphi )$. If we know $u$ lies on $H$, we can calculate the the $n$-th agent's utility $u_n$ from the first $n-1$ utilities. We'll be referring to these first n-1 utilities a lot, so let's call them $v\in R^{n-1}$. So $v_i=u_i$ for all $i<n$.

$H$ and $G$ are both symmetrical with respect to shuffling the indices of agents around, so without loss of generality we'll assume that the n-th agent is one we're assigning positive Harsanyi weight to: ${\varphi }_n$ > 0. This is necessary for the reconstruction to work for all $\varphi$.

So we can think of $H$ as a function $H:R^{n-1}\to R^n$, where the $j$-th output is $H_j\left(v\right)=u_j$ for $j<n$. We can use the $n$-th output to reconstruct $u_n$ given $v$ like this:  $H_n\left(v\right)=\frac{H(p,\varphi )-{\sum }_{i=1}^{n-1}{v}_i{\varphi }_i}{{\varphi }_n}$. This lets us move around $R^{n-1}$ to pick $v$ however we want, and the function $H$ will map that to its image, helpfully also called $H$!

Alright, now we have $H:R^{n-1}\to R^n$ and we also have $G:R^n\times [0,1{]}^n\to R$, which is the geometric weighted average whose gradient we're trying to design through our choice of $\psi$. So let's compose them together to form $G\circ H:R^{n-1}\times [0,1{]}^n\to R$. And since we want $p$ to be an optimum of $G$ across the hyperplane $H$, we can set the gradient $\nabla (G\circ H)(q,\psi )=0$, where $q\in R^{n-1}$ are the first $n-1$ utilities of our target joint utility $p$. Solving this equation for $\psi$ will give us the weights we need! 

This looks like to solving a family of $n-1$ equations$\frac{\partial (G\circ H)}{\partial v_i}(q,\psi )=0$. Where we're holding the weights constant for the purposes of differentiation, but we'll be solving for the weights that make the derivative 0 at $p$.

How Does G Change as We Change These Parameters?

 Ok, we've built up a few layers of abstraction, so let's start unpacking. By the chain rule, and using the notation that $H_j$ is the $j$-th element of the output of $H$:

$\frac{\partial (G\circ H)}{\partial v_i}(v,\psi )={\sum }_{j=1}^n\frac{\partial G}{\partial H_j}\left(H\right(v),\psi )\frac{\partial H_j}{\partial v_i}\left(v\right)$

How does our point on H change as we change these parameters?

Let's start by computing $\frac{\partial H_j}{\partial v_i}\left(v\right)$. 

For the first n-1 terms this is simple, because $H_j$ simply returns $v_j$. So $\frac{\partial H_j}{\partial v_i}$ is 1 when $i=j$, and 0 otherwise, which we can represent using the Kronecker delta ${\delta }_i^j$. And $\frac{\partial H_n}{\partial v_i}=-\frac{{\varphi }_i}{{\varphi }_n}$.

Geometrically, this is telling us about the slope of $H$. Note that:

• $\frac{\partial H_n}{\partial v_i}$ is constant and doesn't depend on our choice of $v_i$
• $\frac{\partial H_n}{\partial v_i}\le 0$ (We can never increase agent $n$'s utility by increasing another agent's utility. This is always true at the Pareto frontier.)

How Does G Change as We Move Around on H?

We can start solving for $\frac{\partial G}{\partial H_j}$ by substituting in the definition of G:

$\frac{\partial G}{\partial H_j}=\frac{\partial }{\partial H_j}\left({\prod }_{k=1}^n{H}_k^{{\psi }_k}\right)$. 

From here we can apply the n-factor product rule:

 $\frac{\partial G}{\partial H_j}=\left({\prod }_{k=1}^n{H}_k^{{\psi }_k}\right)\left({\sum }_{k=1}^n\frac{\frac{\partial }{\partial H_j}\left(H_k^{{\psi }_k}\right)}{H_k^{{\psi }_k}}\right)$.

Thankfully, $\frac{\partial }{\partial H_j}\left(H_k^{{\psi }_k}\right)=0$ whenever $k\ne j$, leaving just $\frac{\partial }{\partial H_j}\left(H_j^{{\psi }_j}\right)={\psi }_j{H}_j^{{\psi }_j-1}$. We can also notice ${\prod }_{k=1}^n{H}_k^{{\psi }_k}=G$, leaving us with the much nicer $\frac{\partial G}{\partial H_j}=G\frac{{\psi }_j{H}_j^{{\psi }_j-1}}{H_j^{{\psi }_j}}=G\frac{{\psi }_j}{H_j}=\frac{{\psi }_j}{H_j}G$.

It will be important later that this partial derivative is undefined when $H_j=0$, aka wherever any agent is receiving their least feasible utility.

Writing function arguments explicitly:

$\frac{\partial G}{\partial H_j}\left(H\right(v),\psi )=\frac{{\psi }_j}{H_j\left(v\right)}G\left(H\right(v),\psi )$

Putting These Terms Together

Let's start putting these together. We can start by breaking apart the two cases of $\frac{\partial H_j}{\partial v_i}$, like this:

 $\frac{\partial (G\circ H)}{\partial v_i}\left(H\right(v),\psi )={\sum }_{j=1}^n\frac{\partial G}{\partial H_j}(u,\psi )\frac{\partial H_j}{\partial u_i}\left(v\right)$

 $\frac{\partial (G\circ H)}{\partial v_i}\left(H\right(v),\psi )=\left[{\sum }_{j=1}^{n-1}\frac{\partial G}{\partial H_j}\right(H\left(v\right),\psi \left)\frac{\partial H_j}{\partial v_i}\right(v\left)\right]+\frac{\partial G}{\partial H_n}\left(H\right(v),\psi )\frac{\partial H_n}{\partial v_i}\left(v\right)$

$\frac{\partial (G\circ H)}{\partial v_i}(v,\psi )=\left[{\sum }_{j=1}^{n-1}\frac{\partial G}{\partial H_j}\right(H\left(v\right),\psi \left){\delta }_i^j\right]-\frac{\partial G}{\partial H_n}\left(H\right(v),\psi )\frac{{\varphi }_i}{{\varphi }_n}$

Here's one reason why it's useful to know about the Kronecker delta ${\delta }_i^j$: it filters out all but the $i$-th element of a sum: ${\sum }_j{a}_j{\delta }_i^j=a_i$. When you're working in Einstein notation (which is great by the way), you just write it as $a_j{\delta }_i^j=a_i$ and you can think of the $j$'s as "cancelling".

That leaves us with:

$\frac{\partial (G\circ H)}{\partial v_i}(v,\psi )=\frac{\partial G}{\partial H_i}\left(H\right(v),\psi )-\frac{\partial G}{\partial H_n}\left(H\right(v),\psi )\frac{{\varphi }_i}{{\varphi }_n}$

And we know $\frac{\partial G}{\partial H_i}\left(H\right(v),\psi )$, so let's plug that in:

$\frac{\partial (G\circ H)}{\partial v_i}(v,\psi )=\frac{{\psi }_i}{H_i\left(v\right)}G\left(H\right(v),\psi )-\frac{{\psi }_n}{H_n\left(v\right)}G\left(H\right(v),\psi )\frac{{\varphi }_i}{{\varphi }_n}$

$\frac{\partial (G\circ H)}{\partial v_i}(v,\psi )=(\frac{{\psi }_i}{H_i\left(v\right)}-\frac{{\psi }_n{\varphi }_i}{H_n\left(v\right){\varphi }_n})G\left(H\right(v),\psi )$

And that is the family of $n-1$ equations that we want to all be 0 when $v=q$. (This causes $H\left(v\right)=u=p$.) We'll call this gradient ${\nabla }_v(G\circ H)$ to remind ourselves that this is the gradient of $(G\circ H)(v,\psi )$ where we're holding the weights $\psi$ constant.

Solving for the Geometric Weights

Ok, now we can set $v=q$, $\frac{\partial (G\circ H)}{\partial v_i}=0$ and solve for ${\psi }_i$, for $i<n$:

$(\frac{{\psi }_i}{H_i\left(q\right)}-\frac{{\psi }_n{\varphi }_i}{H_n\left(q\right){\varphi }_n})G\left(H\right(q),\psi )=0$

$\frac{{\psi }_i}{H_i\left(q\right)}G\left(H\right(q),\psi )-\frac{{\psi }_n{\varphi }_i}{H_n\left(q\right){\varphi }_n}G\left(H\right(q),\psi )=0$

$\frac{{\psi }_i}{H_i\left(q\right)}G\left(H\right(q),\psi )=\frac{{\psi }_n{\varphi }_i}{H_n\left(q\right){\varphi }_n}G\left(H\right(q),\psi )$

$\frac{{\psi }_i}{p_i}G(p,\psi )=\frac{{\psi }_n{\varphi }_i}{p_n{\varphi }_n}G(p,\psi )$

$\frac{{\psi }_i}{p_i}=\frac{{\psi }_n{\varphi }_i}{p_n{\varphi }_n}$

${\psi }_i={\psi }_n\frac{p_i{\varphi }_i}{p_n{\varphi }_n}$

This is still a system of linear equations we need to solve, since each ${\psi }_i$ for $i<n$ depends on ${\psi }_n$, which in turn satisfies ${\psi }_n=1-{\sum }_{i=1}^{n-1}{\psi }_i$. So let's solve it for ${\psi }_n$!

${\psi }_n=1-{\sum }_{i=1}^{n-1}{\psi }_i$

${\psi }_n=1-{\sum }_{i=1}^{n-1}{\psi }_n\frac{p_i{\varphi }_i}{p_n{\varphi }_n}$

${\psi }_n=1-{\psi }_n{\sum }_{i=1}^{n-1}\frac{p_i{\varphi }_i}{p_n{\varphi }_n}$

${\psi }_n+{\psi }_n{\sum }_{i=1}^{n-1}\frac{p_i{\varphi }_i}{p_n{\varphi }_n}=1$

${\psi }_n(1+{\sum }_{i=1}^{n-1}\frac{p_i{\varphi }_i}{p_n{\varphi }_n})=1$

${\psi }_n=\frac{1}{1+{\sum }_{i=1}^{n-1}\frac{p_i{\varphi }_i}{p_n{\varphi }_n}}$

Remembering that $H(p,\varphi )={\sum }_{i=1}^n{p}_i{\varphi }_i={\sum }_{i=1}^{n-1}{p}_i{\varphi }_i+p_n{\varphi }_n$, we can notice that:

$\frac{H(p,\varphi )}{p_n{\varphi }_n}=\frac{{\sum }_{i=1}^{n-1}{p}_i{\varphi }_i+p_n{\varphi }_n}{p_n{\varphi }_n}$

$\frac{H(p,\varphi )}{p_n{\varphi }_n}={\sum }_{i=1}^{n-1}\frac{p_i{\varphi }_i}{p_n{\varphi }_n}+1$

This lets us simplify ${\psi }_n$ down to

${\psi }_n=\frac{1}{\left(\frac{H(p,\varphi )}{p_n{\varphi }_n}\right)}$

${\psi }_n=\frac{p_n{\varphi }_n}{H(p,\varphi )}$

${\psi }_n={\varphi }_n\frac{p_n}{H(p,\varphi )}$

And now we can plug that back into the formula for all the other ${\psi }_i$!

${\psi }_i={\varphi }_n\frac{p_n}{H(p,\varphi )}\frac{p_i{\varphi }_i}{p_n{\varphi }_n}$

${\psi }_i=\frac{p_i{\varphi }_i}{H(p,\varphi )}$

${\psi }_i={\varphi }_i\frac{p_i}{H(p,\varphi )}$

Well isn't that convenient! The formula for all ${\psi }_i$ has the same form, and we can think of it like starting with the Harsanyi weights $\varphi$ (which make p optimal according to $H(\_,\varphi )$, along with anything else with the same Harsanyi score $H(p,\varphi )$), and then tweaking them to get $G(\_,\psi )$ to target $p$ in particular.

We can simplify our formula by noting that $H(p,\varphi )=p\cdot \varphi ={\sum }_{j=1}^n{p}_j{\varphi }_j$

${\psi }_i=\frac{p_i{\varphi }_i}{p\cdot \varphi }$

To make the formula a little prettier, and to get some extra geometric insight, we can introduce the element-wise product $\odot$, where $(p\odot \varphi {)}_i=p_i{\varphi }_i$. 

$\psi =\frac{p\odot \varphi }{p\cdot \varphi }$

Here's a good opportunity to make sure our weights $\psi$ sum up to 1:

${\sum }_{i=1}^n{\psi }_i={\sum }_{i=1}^n\frac{p_i{\varphi }_i}{p\cdot \varphi }$

${\sum }_{i=1}^n{\psi }_i=\frac{1}{p\cdot \varphi }{\sum }_{i=1}^n{p}_i{\varphi }_i$

${\sum }_{i=1}^n{\psi }_i=\frac{1}{p\cdot \varphi }(p\cdot \varphi )$

${\sum }_{i=1}^n{\psi }_i=1$

Great! $p\cdot \varphi$ is acting like a normalization term, and we can think of $p\odot \varphi$ as telling us which direction $\psi$ points in. This vector of weights is then scaled to land on the hypersurface of weights that sum to 1, known as the standard simplex ${\Delta }^n$, which we'll discuss more later [LW · GW].

We can also think of $\varphi$ as a function $\varphi :R^n\times C^n\to R^n$ denoted as $\varphi (p,F)$ which returns the Harsanyi weights $\varphi$ for $p$ in the context of a compact, convex subset $F\subset R^n$. This is it, so let's make a new heading to find it later!

How to Calculate Weights for p

We now have a formula for $\psi :R^n\times [0,1{]}^n\to R^n$, which we can write as

 $\psi (p,\varphi (p,F\left)\right)=\frac{p\odot \varphi (p,F)}{p\cdot \varphi (p,F)}$

Or we can suppress function arguments and simply write

$\psi =\frac{p\odot \varphi }{p\cdot \varphi }$

Where $p\odot \varphi \in R^n$ is the element-wise product of $p$ and $\varphi$: $(p\odot \varphi {)}_i=p_i{\varphi }_i$ and $p\cdot \varphi \in R^n$ is the dot product $p\cdot \varphi ={\sum }_{j=1}^n{p}_j{\varphi }_j$

For a single component ${\psi }_i$, we have

${\psi }_i=\frac{p_i{\varphi }_i}{p\cdot \varphi }$

Note that $\psi$ isn't defined when $p\cdot \varphi =0$.

Is this a problem? Not really! $p\cdot \varphi =H(p,\varphi )$, the Harsanyi aggregate utility of p when $\varphi$ has been chosen to make p optimal under $H(\_,\varphi )$. When this is 0, it means the individual utilities must all be 0 and the entire feasible set $F$ must be a single point at the origin. When that happens, any weights will make $p$ optimal according to $G(\_,\psi )$ or $H(\_,\varphi )$. Feel free to use any convention that works for your application, if we're in a context where $\varphi \left(\overrightarrow{0}\right)$ is defined we can inherit

$\psi (\overrightarrow{0},\varphi (\overrightarrow{0}\left)\right)=\varphi \left(\overrightarrow{0}\right)$

If $F$ is shrinking towards becoming a single point, we can use $\psi \left(\overrightarrow{0}\right)=\underset{p\to \overrightarrow{0}}{lim}\psi \left(p\right)$.

Checking Our Solution

Assuming we calculated ${\nabla }_v(G\circ H)(q,\psi )$ correctly, we can verify that these weights lead to ${\nabla }_v(G\circ H)(q,\psi )=0$. This requires $\frac{\partial (G\circ H)}{\partial v_i}(q,\psi )=0$ for the first $n-1$ utilities, so let's check that:

$\frac{\partial (G\circ H)}{\partial v_i}(q,\psi )=0$

$(\frac{{\psi }_i}{H_i\left(q\right)}-\frac{{\psi }_n{\varphi }_i}{H_n\left(q\right){\varphi }_n})G\left(H\right(q),\psi )=0$

$(\frac{{\psi }_i}{p_i}-\frac{{\psi }_n{\varphi }_i}{p_n{\varphi }_n})G(p,\psi )=0$

$\frac{{\psi }_i}{p_i}G(p,\psi )=\frac{{\psi }_n{\varphi }_i}{p_n{\varphi }_n}G(p,\psi )$

$\frac{1}{p_i}\frac{p_i{\varphi }_i}{p\cdot \varphi }G(p,\psi )=\frac{{\varphi }_i}{p_n{\varphi }_n}\frac{p_n{\varphi }_n}{p\cdot \varphi }G(p,\psi )$

$\frac{{\varphi }_i}{p\cdot \varphi }G(p,\psi )=\frac{{\varphi }_i{\varphi }_n}{{\varphi }_{n}p\cdot \varphi }G(p,\psi )$

$\frac{{\varphi }_i}{p\cdot \varphi }G(p,\psi )=\frac{{\varphi }_i}{p\cdot \varphi }G(p,\psi )$

Success! $p$ is an optimum of $G$ among $H$. But is it unique?

P Is the Unique Optimum of G When Weights Are Positive

Let's see how $\psi$ and $p$ influenced the outcome here, and keep track of the critical points which can make ${\nabla }_v(G\circ H{)}_i=0$ or undefined. These are the only points which can be extrema, and for each we need to check if it is a minimum or maximum among $H$. ($G$ doesn't have any saddle points, and $H$ doesn't have any boundaries of its own to worry about. Where $H$ meets the boundary of $G$'s domain, the $u_i=0$ axes, $G=0$ there.)

For example, whenever any individual utility $u_i=0$, $\frac{\partial G}{\partial H_i}$ is undefined, which causes ${\nabla }_v(G\circ H{)}_i$ to be undefined. But note that these will be minimal points of $G$, unless ${\psi }_i=0$. To find maximal points $u\in H$ of $G$ across $H$ we need

$\frac{{\psi }_i}{u_i}-\frac{{\psi }_n{\varphi }_i}{u_n{\varphi }_n}=0$

$\frac{{\psi }_i{u}_n{\varphi }_n}{u_i{u}_n{\varphi }_n}-\frac{u_i{\psi }_n{\varphi }_i}{u_i{u}_n{\varphi }_n}=0$

$\frac{{\psi }_i{u}_n{\varphi }_n-u_i{\psi }_n{\varphi }_i}{u_i{u}_n{\varphi }_n}=0$

If $u_i$ or $u_n$ are 0, then ${\nabla }_v(G\circ H{)}_i$ is undefined, and we'll check later if these can still be optimal. We assumed that the index $n$ refers to an agent with ${\varphi }_n>0$, in order to prevent that exact case from breaking our entire solution.

${\psi }_i{u}_n{\varphi }_n-u_i{\psi }_n{\varphi }_i=0$

$u_n{\varphi }_n{\psi }_i-u_i{\varphi }_i{\psi }_n=0$

If we were handed $\psi$ from some external source, we could solve this equation to see which $u\in H$ happened to be optimal. But we designed $\psi$, so let's see what we caused to be optimal.

$u_n{\varphi }_n\frac{p_i{\varphi }_i}{p\cdot \varphi }-u_i{\varphi }_i\frac{p_n{\varphi }_n}{p\cdot \varphi }=0$

$\frac{u_n{\varphi }_n{p}_i{\varphi }_i-u_i{\varphi }_i{p}_n{\varphi }_n}{p\cdot \varphi }=0$

If $p\cdot \varphi =0$ then ${\nabla }_v(G\circ H{)}_i$ is undefined. This only happens when $F$ is a single point, in which case $p$ is indeed the unique optimum of $G$.

$u_n{\varphi }_n{p}_i{\varphi }_i-u_i{\varphi }_i{p}_n{\varphi }_n=0$

$u_i{p}_n{\varphi }_i{\varphi }_n=u_n{p}_i{\varphi }_i{\varphi }_n$

Here we're going to be careful about which weights can be 0. We'll again use the fact that ${\varphi }_n>0$ to safely divide it from both sides.

$u_i{p}_n{\varphi }_i=u_n{p}_i{\varphi }_i$

Here again we can see that $u=p$ solves this family of n-1 equations. And this is very exciting because this is our first maximum of $G$! Are there any other solutions?

Each of these equations is satisfied when one of the following is true:

${\varphi }_i=0$

$u_i{p}_n=u_n{p}_i$

In other words, assigning an agent 0 Harsanyi weight ${\varphi }_i$ (and thus geometric weight ${\psi }_i$) can allow $G(\_,\psi )$ to have multiple optima among $H$, which can give it multiple optima among $F$.

What about when all geometric weights $\psi$ are positive? Are there any other solutions to that second family of n-1 equations?

Having all positive geometric weights $\psi$ implies having all positive Harsanyi weights $\varphi$, and all positive individual utilities $p$. It also implies that any optimum of $G(\_,\psi )$ will have all positive individual utilities $u$. This lets us freely divide by any of these terms, without needing to worry that we might be dividing by 0.

$\frac{u_i}{u_n}=\frac{p_i}{p_n}$

Since $u_n$ and $p_n$ are both positive, we can think of $u_n$ as a scaled version of $p_n$.

$u_n=\lambda p_n$

How does this scalar influence the other terms in these equations?

$\frac{u_i}{\lambda p_n}=\frac{p_i}{p_n}$

$u_i=\lambda p_i$

$u=\lambda p$

This forms a line from the origin to $p$, which only intersects $H$ at $p$. (Since scaling $p$ up or down changes $H(p,\varphi )$.) So when all geometric weights $\psi$ are positive, $p$ is the unique optimum of $G(\_,\psi )$ among $H$!

When ${\varphi }_i=0$, ${\psi }_i$ is also 0, so $u_i$ doesn't affect $G(\_,\psi )$. We can start with $p$, and then freely vary the utilities of any agent with 0 weight and remain optimal.

Interactive Implementations

We can also check our entire calculation, including those pages of calculus, by actually implementing our solution and seeing if it works! A graphing calculator is sufficient to check this in 2 and 3 dimensions. We can show all the points which satisfy $G(s,\psi )=G(p,\psi )$ and they should trace out the contours of $G$, showing all the joint utilities which have the same $G$ score as $p$.

In 2 dimension, the graph looks like this:

The Harsanyi hyperplane is a line, and the contour curves are skewed hyperbolas. 

As expected, taking $p$ out of the positive quadrant violates our assumption that utilities are non-negative, leading to invalid settings for $\psi$. Similarly, if $H$ has a positive slope, this violates our assumption that $p$ is on the Pareto frontier. (A positive slope implies that we can make both players better off simultaneously. If we calculate $\varphi$ anyway, a positive slope implies that ${\varphi }_i$ is negative for the agent on the x axis.) This allows $H$ to pass up through the hyperbola at another point other than $p$, but this never happens when ${\varphi }_i\ge 0$.

With 3 agents, the graph looks like this: 

In 3 dimensions, the Harsanyi hyperplane is a plane, and the contour surfaces are skewed hyperboloids.

We can move $p$ around on the hyperplane, and this changes $\psi$, which changes where the contour touches $H$. We can see that $p$ always lies at the intersection of this contour curve and $H$, and this is a visual proof that $p$ maximizes $G(\_,\psi )$ among $H$. And when $H$ corresponds to all agents having positive Harsanyi weight $\varphi$, this intersection only happens at $p$!

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