Question

2. An agent's preference over a set of alternatives X can be represented in the following manner: there is a function V:X → R such that the agent strictly prefers r to r' (denoted by 2 x') if V(x) > V:') + 1 and is indifferent between x and Z denoted by I ') if V(x) - V(') <1. (i) Show that if > and > ", then I r". (ii) Provide a counterexample to demonstrate that the following statement is false: If >x' and ' ", then I r". (iii) Modify the conclusion of the statement in (ii) in a way that makes it true.

Answer 1

Set of Alternatives function & VX R X . X > X XNX i if V(x) > VIX') +1. (u) von 1 10) It x>x' implies Vix) > VIX!) + L If x' > X" implicy VIX'> V(x") + L This implies, VIX) 2vQ) +1 > VIX!) > V(X") + 1. = V(x) > V(X") + L Hence, x> X (i) Counterenample: Suppose X X is VIX) VEX')+L true which _ implies - & X NX" is true which implies VIX") – V(x) | - V(x) = V(x) + K V(XD > VX let v(x) = V(X")-1 . Putting in 0 we get VIX) > G (X)-1) til

Vix> V (X"). hel OSV (X) – V(X) =) Hence the statement is tabe ini By changing XNX" the conclusion statement from X X " will become to true.
The statement 1 is true for all the values of x. Statement 2 is false for which counterexample is given in the solution. Also, changing conclusion from strictly increasing to indifferent makes the statement 2 true.