Wolfgang Schwarz

Harsanyi's trick

Harsanyi (1955) famously showed that a few seemingly harmless assumptions, when combined, entail the utilitarian doctrine that the goodness of a state of the world is the sum of the state's goodness for each individual. In other words, moral value is additive across people.

Recently, I've argued that value is additive on the grounds that its components are "separable", in the sense that if two states s and s' differ only with respect to some components, then the betterness ranking of s and s' does not depend on the respects in which s and s' agree. Debreu (1960) showed that, under some modest further assumptions, separability entails additive representability. I've never had a close look at Debreu's theorem, since the result isn't surprising.

Harsanyi's result, by contrast, is very surprising. It's a trick.

To understand how the trick works, let's apply it to a different example.

Imagine we would like to go on holiday. A number of holiday packages are on offer. We are going to assign a numerical value U(X) to each package X. The higher U(X), the better X.

We will make a few assumptions.

First, the only aspects of a holiday package that we care about are remoteness (how far we will travel) and duration (how long we'll be at the destination). We can therefore represent every package X by a pair of numbers (r,d) in which the first number represents remoteness and the second duration. U is a function of these pairs.

Let's also assume, plausibly enough, that there is an upper limit to remoteness and duration. Perhaps the furthest distance is 10,000 kilometres and the longest duration is 1 month. This isn't technically needed for the result, so I won't list it as an assumption, but it simplifies the proof below.

With a maximum for remoteness and duration, we can scale the r and d numbers so that their maximum is 1. Thus (1,1) represents a 1-month holiday that is 10,000 km away; (0,1) represents a 1-month holiday at home; (1,0) represents a trip where we travel 10,000 km to our destination and then immediately return back home. (0,0) represents no holiday at all.

My second assumption is that we like the (1,0) package and the (0,1) package equally well, and prefer both to the (0,0) package:

\[\begin{align*} (0)\quad & U(0,0) < U(1,0) = U(0,1). \end{align*} \]

We can implement this by scaling the value function U so that

\[\begin{align*} (1)\quad& U(0,0) = 0.\\ (2)\quad& U(1,0) = U(0,1) = 1. \end{align*} \]

Assuming that U is only determinate up to positive linear transformation, (1) and (2) don't go beyond assumption (0).

Now I have to tell you an oddity about our travel agency. Besides ordinary holiday packages, our agency offers lottery packages. For any two ordinary holiday packages A and B, and any number x between 0 and 1, they offer a lottery package that would send us on holiday A with chance x and on B with chance 1-x.

My third assumption is that the U function can be extended to lottery packages. Specifically, I assume that the U-value of a lottery package is the expectation of the value of its component packages. That is, if L is a lottery package that sends us to (r,d) with probability x and to (r',d') with probability 1-x, then

\[\begin{align*} (3)\quad& U(L) = x \cdot U(r,d) + (1-x) \cdot U(r',d'). \end{align*} \]

Note that the assumption isn't that we actually evaluate lottery packages in this manner. The assumption is only that the function U that represents our preferences over pure packages can be extended to lottery packages in accordance with (3). As such, the assumption looks almost empty.

My fourth and final assumption is that we can extend our remoteness and duration measures to lottery packages. Specifically, we say that the remoteness of a lottery between two trips is the average of the remoteness of the two trips, weighted by their chances. Likewise for duration.

Like the third assumption, this doesn't look like an assumption at all. It looks like a terminological stipulation that helps us to assign remoteness and duration values to every package, including lottery packages.

To sum up, the only substantive assumptions we seem to have made are (i) that we only care about remoteness and duration, and (ii) that we rank (0,1) and (1,0) equally, and prefer both to (0,0).

And now – boom – Harsanyi's theorem tells us that the value U(X) of any package X must equal the sum of X's remoteness and duration:

\[\begin{align*} (H)\quad& U(r,d) = r + d. \end{align*} \]

How did this happen? One might have thought that our assumptions are compatible with lots of other ways in which the value of a package might be determined. Indeed, one might have thought that U could be any function of r and d at all, as long as it respects (1) and (2).

For example, I never said that we generally prefer longer holidays to shorter holidays. I never said that we generally like holidays that take us further away. Nor did I say that remoteness and duration are separable. For all I've said, there could be all sorts of interaction effects: perhaps longer holidays are preferred more strongly if they are further away, so that U(r,d) = (r + d) / (r-d)²? (Note that this is compatible with (1) and (2).)

Even if we assume separability, where did the linear dependence on r and d come from? (H) doesn't merely say that U(r,d) is the sum of some function of remoteness and duration. No, we're adding up r and d themselves. Why couldn't the value of a package increase with, say, the log of remoteness and the square of duration, so that U(r,d) = log₂(1+r) + d²? (Again, this is compatible with (1) and (2).)

In fact, I haven't even said how remoteness is measured, except that the maximum is 1. It could be measured in kilometres (divided by the maximum distance). Or it could be measured by the log of kilometres, so that an added kilometre contributes less to remoteness if a trip has already taken us 5000 km away than if a trip is only 1 km away. Why not? But U can't be a (non-trivial) linear function of both of these! If (H) is correct, then our ranking of packages depends on an arbitrary convention for how we measure remoteness. This makes no sense.

I hope you agree that we have been tricked. But how?

Fortunately, the proof of Harsanyi's theorem is easy, at least for our current application. Let's work through the proof of (H).

As you might have guessed, lottery packages play a key role. Let L be a lottery package that leads to (r,d) with probability x and to (r',d') with probability 1-x. The crucial point is that there are two ways of expressing L's U-value. The first uses (3):

\[\begin{align*} (3)\quad& U(L) = x \cdot U(r,d) + (1-x) \cdot U(r',d'). \end{align*} \]

This says that the value of L is the weighted average of the value of its components, (r,d) and (r',d').

The second way to express U(L) exploits the fact that we can directly assign a remoteness and duration to L. The "remoteness" of L is xr + (1-x)r'. The "duration" of L is xd + (1-x)d'. Since the value of any package (including a lottery package) is a function of its remoteness and duration, we have

\[\begin{align*} (4)\quad& U(L) = U(xr + (1-x)r', xd + (1-x)d'). \end{align*} \]

Combining equations (3) and (4), we get

\[\begin{align*} (5)\quad& U(xr + (1-x)r', xd + (1-x)d') = x \cdot U(r,d) + (1-x) \cdot U(r',d'). \end{align*} \]

This equation will be doing all the work.

To begin, consider a case where r' and d' are both 0. We then have

\[\begin{align*} (6)\quad U(xr, xd) &= x \cdot U(r,d) + (1-x) \cdot U(0,0) &\text{ [by (5)]}\\ &= x \cdot U(r,d) &\text{ [by (1)]} \end{align*} \]

This allows us to compute U(r,0), for any r between 0 and 1:

\[\begin{align*} (7)\quad U(r,0) &= U(r\cdot1, r\cdot0) & \\ &= r \cdot U(1,0) & \text{ [by (6), swapping x and r]}\\ &= r & \text{ [by (2)]} \end{align*} \]

Similarly, we get

\[\begin{align*} (8)\quad& U(0,d) = d. \end{align*} \]

Now instantiate (5) with r' = d = 0 and x = 1/2. We get

\[\begin{align*} (9)\quad U(r/2 + d'/2) &= U(r,0)/2 + U(0,d')/2 & \\ &= r/2 + d'/2 & \text{ [by (7) and (8)].} \end{align*} \]

By (6), U(r/2 + d'/2) = U(r,d')/2. So

\[\begin{align*} (10)\quad& U(r,d')/2 = r/2 + d'/2. \end{align*} \]

Relabelling d' as d and multiplying both sides by 2 yields (H).

(This proof is inspired by the proof in Resnik (1987), which however doesn't go via (5).)

You can see how (5) is doing all the work. Informally, (5) says that the average value of two packages equals the value of the average between the two packages.

This rules out a non-linear dependence of value on remoteness and duration. Suppose, for example, that remoteness has "declining marginal value". The expected value of a fair lottery between (0,d) and (1,d) is then less than the value of (.5,d). But (5) says that the expected value of the lottery equals the value of the average of the packages, which is (.5,d). So remoteness can't have declining marginal value. For similar reasons, (5) rules out any interaction effects between remoteness and duration.

How did we get to (5)? Let me repeat the assumptions from which this follows:

1. The value U(X) of a package X is a function of X's remoteness and duration.

2. U can be extended to lottery packages by the expectation rule.

3. The remoteness and duration of a lottery can be defined as the average of the relevant aspects of the component holidays.

In isolation, all of these look harmless. The last two in particular look like mere conventions. The trick is that the last assumption twists the meaning of the first, in a way that is incompatible with the second unless (5) holds. In the presence of 3, assumption 1 implies that the value of a lottery between two packages equals the value of the average between the two packages. This is a very demanding assumption. If we add 2, it is clear that we get (5).

It's a great trick. But it's a trick nonetheless. Debreu's Theorem establishes a plausible route to value additivity. Harsanyi's theorem does not.

Many utilitarians disagree. John Broome, for example, has argued that Harsanyi's theorem shows something deep and important about moral value. (See, for example, Broome (1987), Broome (1991), and Broome (2015).)

Let's briefly look at Broome's application of Harsanyi's theorem – which is close to Harsanyi's own application.

We're now interested in the moral value, or "goodness", U(X) of a possible state of the world X. We assume that any such state can be evaluated with respect to how good it is for each individual. Assume, for concreteness, that there are just two individuals, R and D.

We assume that the value of a state is determined by how good it is for the individuals. By itself, this isn't a strong assumption, since we haven't said much about how "good for R" and "good for D" are interpreted. We could stipulate that worlds where animals suffer, or where R's rights are violated, are bad for both R and D. (Here Broome departs from Harsanyi.)

Next, we assume that the (0,1) state and the (1,0) state are equally good, and better than the (0,0) state. The first of these assumptions reflects an "anonymity" condition, suggesting that R and D contribute equally to overall goodness. The second is an instance of a weak Pareto condition, saying that if a state is better for some people and not worse for anybody, then it is better overall.

Next, we assume that U can be extended to lotteries, and that the extension evaluates lotteries by the expectation rule.

Finally, we assume that the "good for R" and "good for D" measures can also be extended to lotteries by the expectation rule.

And then, boom, it follows that the goodness of any state X equals the sum of X's goodness for R and D. We get (H). Utilitarianism!

All the assumptions look harmless and plausible. Can't we all agree on them? If so, we're all implicitly committed to utilitarianism.

As before, the trick lies in how the first, third, and fourth assumptions interact. Together, they imply that the average goodness of two states (r,d) and (r',d') equals the goodness of the average state (xr+(1-x)r', xd+(1-x)d'). This rules out many plausible ways in which overall goodness might depend on individual goodness.

It's a great trick, because we can't blame any of the assumptions on its own. "Which of the assumptions do you reject?", Broome might ask (well, and does ask). Whichever we pick, he can say that the assumption looks eminently plausible.