# 6 Transitivity

A relation \(R\subseteq X\times X\) is **transitive** when for all \(x, y, z\in X\), if \(x\mathrel{R} y\) and \(y\mathrel{R} z\), then \(x\mathrel{R}z\).

An key assumption in many rational choice models is that the decision maker’s preference on a set \(X\) is transitive. That is, the assumption that a decision maker’s preferences are transitive means that:

- the decision maker’s strict preference relation \(P\) is transitive,
- the decision maker’s indifference relation \(I\) is transitive, and

- the decision maker’s non-comparability relation \(N\) is transitive.

## 6.1 Transitivity of Indifference and Non-Comparability

There are reasons to reject the assumption that the decision maker’s indifference relation and non-comparability relation are transitive.

**Example 6.1 (Transitivity of Indifference) **Suppose that you are indifferent between a curry with \(x\) amount of cayenne pepper, and a curry with \(x\) plus one particle of cayenne pepper for any amount \(x\). That is, if \(x\) is the amount of cayenne pepper in the curry, then we have that \(x\mathrel{I} (x+1)\), where \(x\) is the amount of cayenne pepper added to the curry and \(x+1\) represents adding \(x\) plus 1 additional particle of cayenne pepper to the curry. In particular, you have the following preferences: \[0\mathrel{I}1 \qquad\mbox{and}\qquad 1\mathrel{I}2\] Then, assuming that \(I\) is transitive, we reason as follows:

- \(0\mathrel{I}1\) and \(1\mathrel{I}2\).
- Assuming \(I\) is transitive implies that \(0\mathrel{I}2\).
- Given item 2 and \(2\mathrel{I}3\), assuming that \(I\) is transitive implies that \(0\mathrel{I}3\)
- Given item 3 and \(3\mathrel{I}4\), assuming that \(I\) is transitive implies \(0\mathrel{I}4\)
- And so on…

This implies that for any number \(x\) of particles of cayenne peper, you have the preference \(0\mathrel{I}x\). But you are not indifferent between a curry with no cayenne pepper and a curry with 1 pound of cayenne pepper in it!

**Example 6.2 (Transitivity of Non-Comparability) **Suppose that you cannot compare having a job as a professor with having a job as a programmer. Furthermore, you cannot compare having a job as a programmer with having a job as a professor with an extra $1,000. More formally, let \(p\) mean that you have a job as a professor, \(c\) mean you have a job as a programmer, and \(\langle p, \$1000\rangle\) mean you have a job as a professor plus you have an extra \(\$1,000\). Then, you have the following preferences: \[p\mathrel{N}c\qquad\mbox{and}\qquad c\mathrel{N} \langle p, \$1000\rangle.\] Assuming that \(N\) is transitive implies that \(p\mathrel{N}\langle p, \$1000\rangle\). However, you do strictly prefer having a job as a professor with an extra $1,000 to having a job as a professor!

Setting aside the issues raised in Example 6.1 and Example 6.2, we assume the following:

- Transitivity of Indifference and Non-Comparability
- Suppose that \(I\subseteq X\times X\) represents a decision maker’s indifference relation and that \(N\subseteq X\times X\) represents a decision maker’s non-comparability relation. We assume that \(I\) and \(N\) are both transitive.

- For all \(x, y, z\in X\), if \(x\mathrel{I} y\) and \(y\mathrel{I} z\), then \(x\mathrel{I} z\).
- For all \(x, y, z\in X\), if \(x\mathrel{N} y\) and \(y\mathrel{N} z\), then \(x\mathrel{N} z\).

## 6.2 Transitivity of Strict Preferences

While there are some experiments that raise doubts about whether transitivity is a good description of people’s strict preferences^{1}, it is common to assume that a decision maker’s strict preference is transitive.

There are two ways that a decision maker’s strict preference \(P\) on \(X\) may fail transitivity:

- The decision maker lacks a strict preference: There are \(x, y, z \in X\) such that \(x \mathrel{P} y\) and \(y \mathrel{P} z\), but \(x \mathrel{N} z\) (i.e., \(x\) and \(z\) are incomparable).
- There is a cycle in the decision maker’s preferences: There are \(x, y, z \in X\) such that \(x \mathrel{P} y\), \(y \mathrel{P} z\), and \(z \mathrel{P} x\).

To justify the assumption that a strict preference relation is transitive, we need to argue that there is something irrational about both of the above situations. In the next section, we explain how to rule out cycles in a decision maker’s strict preferences. The first situation is ruled out with an additional assumption about a decision maker’s preferences (see Chapter 7).

### 6.2.1 Ruling out Cycles

A **cycle** (of length 3) in a relation \(P\subseteq X\times X\) is a sequence \((x, y, z)\) where \(x P y\), \(y P z\) and \(z P x\) (recall Definition 2.3). There are two main arguments that rule out preferences with cycles.

**Argument 1**\(\quad\) We cannot make sense of a decision maker with a strict preference that has a cycle. This argument is nicely expressed with the following quote from Donald Davidson:

I do not think we can clearly say what should convince us that a [person] at a given time (without change of mind) preferred \(a\) to \(b\), \(b\) to \(c\) and \(c\) to \(a\). The reason for our difficulty is that we cannot make good sense of an attribution of preference except against a background of coherent attitudes…My point is that if we are intelligibly to attribute attitudes and beliefs, or usefully to describe motions as behaviour, then we are committed to finding, in the pattern of behaviour, belief, and desire, a large degree of rationality and consistency. (Davidson 2001, 237)

**Argument 2: The Money-Pump Argument**\(\quad\) The Money-Pump Argument is a thought experiment demonstrating that a decision maker with a cycle in her strict preferences can end up paying an indefinite amount of money without gaining anything new. For an item \(x\in X\), we write \(\langle x, \$u\rangle\) to mean that the decision maker has \(x\) and \(\$u\) and write \(\langle x, -\$u\rangle\) to mean that the decision maker has \(x\) and pays \(\$u\). There are three key assumptions about a decision maker’s strict preference \(P\) and the decision maker’s opinion about money:

- If \(x \mathrel{P} y\), then the decision maker will always take \(x\) when \(y\) is the only alternative.
- If \(x \mathrel{P} y\), then there is some \(v > 0\) such that for all \(u\), \(\langle x, -\$u\rangle \mathrel{P} y\) if and only if \(0\le u\le v\).

- The items and money are
*separable*and the decision maker prefers more money to less: For all \(x\in X\) and \(w,z\), we have that \(\langle x,\$w\rangle \mathrel{P} \langle x,\$z\rangle\) if and only if \(w>z\); and, for all \(x, y \in X\) and \(w\), if \(x \mathrel{P} y\), then \(\langle x,\$w\rangle \mathrel{P} \langle y,\$w\rangle\).

Suppose that Ann has a cycle in her strict preferences over the set \(\{r, w, b\}\): \(r P w\), \(w P b\), and \(b P r\). Furthermore, in line with assumption 2, assume that Ann is willing to pay \(\$1\) to swap \(w\) for \(r\), \(\$1\) to swap \(b\) for \(w\), and \(\$1\) to swap \(r\) for \(b\). That is, Ann has the following strict preferences: \[\langle r, -\$1\rangle \mathrel{P} w \qquad \langle w,-\$1\rangle \mathrel{P} b \qquad \langle b, -\$1\rangle \mathrel{P} r.\]

Suppose that Ann currently has item \(b\). Given assumptions 1-3, we argue as follows:

- Since \(\langle b,-\$1\rangle\mathrel{P} r\), by assumption 1, she will accept an offer to trade \(r\) for \(b\) plus she must pay \(\$1\). After the trade, she has \(b\) and has paid \(\$1\).

- Now, suppose she is offered a chance to trade \(w\) for \(b\) plus she must pay \(\$1\). Since \(\langle w - \$1\rangle \mathrel{P} b\), by assumption 1, she will accept that offer. So, she now has \(w\) and has paid \(\$2\).
- Suppose she is offered a chance to trade \(w\) for \(r\) plus she must pay \(\$1\). Since \(\langle r, -\$1\rangle \mathrel{P} w\), by assumption 1, she will accept that offer. Now she has \(r\) and has paid \(\$3\).

But she started with \(r\) and paying \(\$0\) and ended up with \(r\) and paying \(\$3\)! By assumption 3, this is a strictly worse situation for Ann: \(\langle r, \$0\rangle \mathrel{P}\langle r, -\$3\rangle\). But it does not end here, Ann will continue to accept the offers resulting in her paying an indefinite amount of money. Ann can avoid such a money-pump argument by ensuring that there are no cycles in her strict preferences.

## 6.3 Exercises

Suppose that \(X=\{a,b,c,d\}\). Which of the following relations are transitive? If the relation is not transitive, explain why.

- \(R=\{(a,b)\}\)
- \(R=\{(a,b), (c,b), (b,a)\}\)
- \(R=\{(a,b), (b, c), (a,c)\}\)
- \(R=\{(a,b), (b,a), (a,a), (b,b), (a,c)\}\)
- \(R=\{(a,b), (b,a), (a,a), (b,b), (a,c), (b,c)\}\)
- \(R=\{(a,b), (b,c), (a,c), (c,d)\}\)
- \(R=\{(a,b), (b,c)\}\)
- \(R=\{(a, b), (b, a), (a, a)\}\)

True or False: The Money-Pump argument shows that a rational decision maker’s strict preferences must be transitive.

Suppose that \(X=\{a,b,c,d\}\). Which of the following relations are transitive? If the relation is not transitive, explain why.

\(R=\{(a,b)\}\)

This relation is transitive.

\(R=\{(a,b), (c,b), (b,a)\}\)

This relation is not transitive since \((a,b)\in R\), \((b,a)\in R\) but \((a,a)\notin R\).

\(R=\{(a,b), (b, c), (a,c)\}\)

This relation is transitive.

\(R=\{(a,b), (b,a), (a,a), (b,b), (a,c)\}\)

This relation is not transitive since \((b,a)\in R\), \((a,c)\in R\) but \((b, c)\notin R\).

\(R=\{(a,b), (b,a), (a,a), (b,b), (a,c), (b,c)\}\). This relation is transitive.

\(R=\{(a,b), (b,c), (a,c), (c,d)\}\)

This relation is not transitive since \((a,c)\in R\), \((c,d)\in R\) but \((a,d)\not\in R\).

\(R=\{(a,b), (b,c)\}\)

This relation is not transitive since \((a,b)\in R\), \((b,c)\in R\) but \((a,c)\not\in R\)

\(R=\{(a, b), (b, a), (a, a)\}\)

This relation is not transitive since \((b,a)\in R\), \((a,b)\in R\) but \((b,b)\not\in R\)

True or False: The Money-Pump argument shows that a rational decision maker’s strict preferences must be transitive.

This is false. The Money-Pump argument shows that a rational decision maker’s strict preferences cannot contain a cycle. We need an additional assumption to rule out situation in which a decision maker strictly prefers \(x\) to \(y\) and \(y\) to \(z\), but cannot compare \(x\) and \(z\).

See A. Tversky’s classic paper Intransitivity of Preferences and M. Regenwetter, J. Dana, and C. P. Davis-Stober, Transitivity of Preferences for a critique of these experiments.↩︎