Proving Expected Utility Theorem

Question

I am struggling to understand the proof of the second step in the Expected Utility Theorem, particularly the part that deals with preferences over weighted sums of lotteries. The statement I am trying to prove is:

For the best possible lottery $ L_h $ and the worst possible lottery $ L_l $, the compound lottery $ bL_h + (1-b)L_l $ is preferred over $ aL_h + (1-a)L_l $ if and only if $ b > a $.

I've attempted to prove it by contradiction but have encountered difficulties in rigorously showing how having 'more' of $ L_h $ in one lottery over another implies it should be preferred. I of course understand the intuition, but I am struggling with the formal proof using the independence axiom.

Thank you in advance for your help

Answer 1

We have that $L_h \succ L_l$.

Then as $L_h \succ L_l$ we have by IIA that for $b \in ]0,1[$ $$ b L_h + (1-b) L_l \succ b L_l + (1-b) L_l = L_l. $$ Next as $b L_h + (1-b) L_l \succeq L_l$ we have again by IIA that for all all $q \in ]0,1[$ $$ q(b L_h + (1-b) L_l) + (1-q) (b L_h + (1-b) L_l) \succ q L_l + (1-q)(b L_h + (1-b) L_l)$$ Simplifying gives: $$ b L_h + (1-b) L_l \succ (1-q) b L_h + [(1-b) + bq] L_l $$ Now let $q = \frac{b- a}{b} \in ]0,1[$ (because $b > a$). This gives: $$ b L_h + (1-b) L_l \succ a L_h + (1-a) L_l. $$