Exercises · 2. Preference and Choice

Easy

1Easy

A consumer reports $(5, 1) \succ (3, 3)$ and $(3, 3) \succ (2, 4)$ .

If the preference relation $\succsim$ is complete and transitive, what can be said about $(5, 1)$ versus $(2, 4)$ ?

x \succ y \iff x \succsim y \text{ and not } y \succsim x

Preference Relations and Axioms

Show solution

Strict preference $\succ$ is transitive whenever $\succsim$ is complete and transitive. From $(5,1) \succ (3,3)$ and $(3,3) \succ (2,4)$ we get $(5, 1) \succ (2, 4)$ .
Proof sketch of transitivity: let $x \succsim y$ , $\lnot(y \succsim x)$ , $y \succsim z$ , $\lnot(z \succsim y)$ .
Transitivity gives $x \succsim z$ .
If $z \succsim x$ held, then $y \succsim z \succsim x$ would give $y \succsim x$ , a contradiction.
Hence $\lnot(z \succsim x)$ , i.e. $x \succ z$ .

Medium

2Medium

Consider a perfect-substitutes consumer with $u(x_1, x_2) = x_1 + x_2$ .

Find the marginal rate of substitution $MRS = \frac{\partial u / \partial x_1}{\partial u / \partial x_2}$ and write the equation of the indifference curve through $(6, 6)$ .

Is $B = (9, 4)$ preferred to $A = (6, 6)$ ?

u(x_1, x_2) = x_1 + x_2

Preference Relations and Axioms

Perfect substitutes,

A = (6, 6)

,

B = (9, 4)

— straight-line indifference curves of slope

-1

.

Show solution

$\partial u / \partial x_1 = 1$ and $\partial u / \partial x_2 = 1$ , so $MRS = 1/1 = 1$ everywhere — giving up one unit of good 1 is exactly compensated by one unit of good 2.
Utility at $(6, 6)$ is $12$ , so the indifference curve is $x_1 + x_2 = 12$ , i.e. $x_2 = 12 - x_1$ . The whole indifference map is the family of parallel lines with slope $-1$ .
$u(B) = 9 + 4 = 13 > 12 = u(A)$ , so $B \succ A$ . In the figure below, check that the dashed line through $B$ lies outside the solid line through $A$ .

3Medium

The consumer chose $x = (20, 40)$ at $(p, w) = ((1, 1), 60)$ and $y = (25, 25)$ at $(p', w') = ((3, 1), 100)$ .

Compute the revealed-preference relations and decide whether WARP is violated.

p \cdot x = 60, \quad p' \cdot y = 100

Choice Rules and WARP

Show solution

Budget check: $p \cdot x = 20 + 40 = 60 \le 60$ and $p' \cdot y = 75 + 25 = 100 \le 100$ — both bundles are affordable.
In situation 1, $y$ costs $p \cdot y = 25 + 25 = 50 \le 60 = p \cdot x$ . Since $y$ was affordable, $x \succsim^* y$ .
In situation 2, $x$ costs $p' \cdot x = 60 + 40 = 100 \le 100 = p' \cdot y$ .
Since $x$ was affordable, $y \succsim^* x$ .
Revealed preference holds both ways between two distinct bundles, so WARP is violated.
$x \succsim^* y \text{ and } y \succsim^* x,\ x \ne y \quad\Rightarrow\quad \text{WARP violated}$

Hard

4Hard

Pairwise choices over the three bundles $A = (4, 4)$ , $B = (5, 2)$ , $C = (2, 5)$ came out as follows (each pair has a unique winner): $A$ beats $B$ , $B$ beats $C$ , and $C$ beats $A$ .

Prove that no complete and transitive preference relation $\succsim$ rationalizes all three choices.

A \text{ chosen over } B, \quad B \text{ chosen over } C, \quad C \text{ chosen over } A

Show solution

Suppose $\succsim$ rationalizes the three choices.
That $A$ alone was chosen from $\{A, B\}$ means $A \succsim B$ and $\lnot(B \succsim A)$ , i.e. $A \succ B$ .
Likewise $B \succ C$ and $C \succ A$ .
Since $\succsim$ is complete and transitive, so is $\succ$ (the proof in exercise 1). From $A \succ B$ and $B \succ C$ we get $A \succ C$ .
But $C \succ A$ means $C \succsim A$ and $\lnot(A \succsim C)$ , while $A \succ C$ requires $A \succsim C$ — a contradiction.
Hence no such $\succsim$ exists.
This is what breaks when transitivity fails: cycling choices are explained by no rational preference, and in revealed-preference terms they are a WARP violation $A \succsim^* B \succsim^* C \succsim^* A$ .