4  Transitivity

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

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

  1. the decision maker’s strict preference relation \(P\) is transitive,
  2. the decision maker’s indifference relation \(I\) is transitive, and
  3. the decision maker’s non-comparability relation \(N\) is transitive.

4.1 Transitivity of Indifference and Non-Comparability

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

Example 4.1 (Transitivity of Indifference) Suppose that you are indifferent between a curry with \(x\) amount of cayenne pepper and a curry with \(x\) plus one additional particle of cayenne pepper, for any amount \(x\). This means that if \(x\) is the amount of cayenne pepper in the curry, then \(x \mathrel{I} (x + 1)\), where \(x\) represents the amount of cayenne pepper already in the curry, and \(x + 1\) represents adding one more particle of cayenne pepper.

Given this, your preferences would look like this:

  • \(x \mathrel{I} (x + 1)\)
  • \((x + 1) \mathrel{I} (x + 2)\)
  • \((x + 2) \mathrel{I} (x + 3)\)
    \(\vdots\)

According to transitivity, if you are indifferent between \(x\) and \(x + 1\), and between \(x + 1\) and \(x + 2\), then you should also be indifferent between \(x\) and \(x + 2\). Extending this reasoning, you would be indifferent between \(x\) and any larger amount of cayenne pepper, even when the difference becomes significant. However, this outcome seems counterintuitive, as adding a large number of particles will eventually change the taste of the curry. This example illustrates why the assumption that indifference is transitive may not always hold.

Example 4.2 (Transitivity of Non-Comparability) Suppose you are unable to compare having a job as a professor with having a job as a programmer. Additionally, you cannot compare having a job as a programmer with having a job as a professor plus an extra $1,000.

More formally, let \(p\) represent having a job as a professor, \(c\) represent having a job as a programmer, and \(\langle p, \$1000\rangle\) represent having a job as a professor with an additional \(\$1,000\). Given these conditions, your preferences are:

\[p\mathrel{N}c\qquad\mbox{and}\qquad c\mathrel{N} \langle p, \$1000\rangle.\] Assuming that the non-comparability relation \(N\) is transitive would imply that \(p \mathrel{N} \langle p, \$1000\rangle\). However, this leads to a contradiction because you do have a strict preference for having a job as a professor with an extra \(\$1,000\) over simply having a job as a professor.

Setting aside the issues raised in Example 4.1 and Example 4.2, 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 \(x\mathrel{N} z\).

4.2 Transitivity of Strict Preferences

While some experiments raise doubts about whether transitivity accurately describes people’s strict preferences1, it is still common to assume that a decision maker’s strict preference is transitive.

There are two ways in which a decision maker’s strict preference \(P\) on \(X\) may fail to be transitive:

  1. Lack of 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).
  2. Preference Cycles: 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 must argue that both of the above situations are irrational. In the next section, we will explain how to rule out cycles in a decision maker’s strict preferences. The first situation can be ruled out with an additional assumption about the decision maker’s preferences (see Chapter 5).

4.2.1 Ruling out Cycles

A cycle (of length 3) in a relation \(P \subseteq X \times X\) is a sequence \((x, y, z)\) where \(x \mathrel{P} y\), \(y \mathrel{P} z\), and \(z \mathrel{P} x\) (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 having a strict preference that involves a cycle. This argument is well-articulated by Donald Davidson in the following quote:

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 that demonstrates how a decision maker with a cycle in their 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 we write \(\langle x, -\$u\rangle\) to mean that the decision maker has \(x\) and pays \(\$u\). The argument relies on three key assumptions about a decision maker’s strict preference \(P\) and their attitude toward money:

  1. Strict Preference Guides Choice: If \(x \mathrel{P} y\), then the decision maker will always take \(x\) when \(y\) is the only alternative.
  2. Willingness to Pay: 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 \(0\le u\le v\).
  3. Separation of Items and Money: 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 \(\langle x,\$w\rangle \mathrel{P} \langle y,\$w\rangle\).

Now, suppose Ann has a cycle in her strict preferences over the set \({r, w, b}\): \(r \mathrel{P} w\), \(w \mathrel{P} b\), and \(b \mathrel{P} r\). Furthermore, in line with Assumption 2, Ann is willing to pay $1 to swap \(w\) for \(r\), $1 to swap \(b\) for \(w\), and $1 to swap \(r\) for \(b\). This means 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 Ann currently has item \(b\). Given Assumptions 1-3, we can argue as follows:

  1. Since \(\langle b, -\$1\rangle \mathrel{P} r\), by Assumption 1, Ann will accept an offer to trade \(r\) for \(b\) plus pay $1. After the trade, she has \(r\) and has paid $1.
  2. Now, suppose she is offered the chance to trade \(w\) for \(r\) plus pay $1. Since \(\langle w, -\$1\rangle \mathrel{P} r\), by Assumption 1, she will accept the offer. She now has \(w\) and has paid \(\$2\).
  3. Suppose she is offered the chance to trade \(b\) for \(w\) plus pay \(\$1\). Since \(\langle b, -\$1\rangle \mathrel{P} w\), by Assumption 1, she will accept the offer. Now she has \(b\) and has paid $3.

But Ann started with \(b\) and \(\$0\), and she ended up with \(b\) and has paid $3! According to Assumption 3, this is a strictly worse situation for Ann: \(\langle b, \$0\rangle \mathrel{P} \langle b, -\$3\rangle\). But the problem does not end here—Ann will continue to accept these offers, resulting in her paying an indefinite amount of money.

Ann can avoid such a money-pump scenario by ensuring that there are no cycles in her strict preferences.

4.3 Exercises

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

    1. \(R=\{(a,b)\}\)
    2. \(R=\{(a,b), (c,b), (b,a)\}\)
    3. \(R=\{(a,b), (b, c), (a,c)\}\)
    4. \(R=\{(a,b), (b,a), (a,a), (b,b), (a,c)\}\)
    5. \(R=\{(a,b), (b,a), (a,a), (b,b), (a,c), (b,c)\}\)
    6. \(R=\{(a,b), (b,c), (a,c), (c,d)\}\)
    7. \(R=\{(a,b), (b,c)\}\)
    8. \(R=\{(a, b), (b, a), (a, a)\}\)

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

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

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

      This relation is transitive.

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

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

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

      This relation is transitive.

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

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

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

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

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

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

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

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

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

  2. 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 \(z\).

  1. See A. Tversky’s classic paper (Tversky 1969) and Regenwetter, Dana, and Davis-Stober (2011) for a critique of these experiments.↩︎