<p>We have that <span class="math-container">$L_h \succ L_l$</span>.</p>
<p>Then as <span class="math-container">$L_h \succ L_l$</span> we have by IIA that for <span class="math-container">$b \in ]0,1[$</span>
<span class="math-container">$$
b L_h + (1-b) L_l \succ b L_l + (1-b) L_l = L_l.
$$</span>
Next as <span class="math-container">$b L_h + (1-b) L_l \succeq L_l$</span> we have again by IIA that for all all <span class="math-container">$q \in ]0,1[$</span>
<span class="math-container">$$
q(b L_h + (1-b) L_l) + (1-q) (b L_h + (1-b) L_l) \succ q L_l + (1-q)(b L_h + (1-b) L_l)$$</span>
Simplifying gives:
<span class="math-container">$$
b L_h + (1-b) L_l \succ (1-q) b L_h + [(1-b) + bq] L_l
$$</span>
Now let <span class="math-container">$q = \frac{b- a}{b} \in ]0,1[$</span> (because <span class="math-container">$b &gt; a$</span>). This gives:
<span class="math-container">$$
b L_h + (1-b) L_l \succ a L_h + (1-a) L_l.
$$</span></p>

CC BY-SA 4.0
2023-10-06T05:24:36.723Z
2023-10-06T05:24:36.723Z
tdm
<p>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:</p>
<p>For the best possible lottery <span class="math-container">$ L_h $</span> and the worst possible lottery <span class="math-container">$ L_l $</span>, the compound lottery <span class="math-container">$ bL_h + (1-b)L_l $</span> is preferred over <span class="math-container">$ aL_h + (1-a)L_l $</span> if and only if <span class="math-container">$ b &gt; a $</span>.</p>
<p>I've attempted to prove it by contradiction but have encountered difficulties in rigorously showing how having 'more' of <span class="math-container">$ L_h $</span> 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.</p>
<p>Thank you in advance for your help</p>

<microeconomics><utility><proof>
Proving Expected Utility Theorem
Lorena_dok
# Proving Expected Utility Theorem

(Asked by: username_0 on 2023-10-05)

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

**Tags:** microeconomics, utility, proof


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## 1 answers

### Answer by username_1 on 2023-10-06. Score: 1 | Accepted answer

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:
$$
b L\_h + (1-b) L\_l \succ a L\_h + (1-a) L\_l.
$$


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