# Lotteries and expected utility

Suppose we have the following four lotteries:

$L_{1}=[(1,\$1)]L_{2}=[(0.01,\$0),(0.89,\$1),(0.1,\$5)]$

$L_{3}=[(0.9,\$0),(0.1,\$5)]$

$L_{4}=[(0.89,\$0),(0.11,\$1)]$

If our agent says that he prefers $L_{1}$ to $L_{2}$ and prefers $L_{3}$ to $L_{4}$, then he is not following the expected utility assumption of independence (also known as substitutability).

Anyone know how to explain this?

• Your question is unclear. Your phrasing makes it sound like you're asking for an explanation for why someone would have these preferences, rather than why these preferences would violate the assumption of independence. Commented Apr 16, 2018 at 22:05

Your example is the classic Allais paradox.

I think the best way to see how the preference pattern $L_1\succ L_2$ and $L_3\succ L_4$ violates independence is to visualize it geometrically. Consider the following figure:

Observe that \begin{align} L_2&=L_1+(0.01,-0.11,0.1)\\ L_3&=L_4+(0.01,-0.11,0.1). \end{align} Therefore, the indifference curves passing through $L_1$ and $L_4$ should be parallel. Since $L_1\succ L_2$, the latter lottery should lie below the IC through $L_1$. By implication of the parallelism of the ICs, $L_3$ should also lie below the IC through $L_4$. But we're given that $L_3\succ L_4$. This means that $L_4$ lies below the IC through $L_3$ (the blue dashed line). Since parallelism of ICs is a necessary condition of independence, the agent's preference is inconsistent with independence, because it can only be justified by a set of non-parallel ICs.

Alternatively, you can think of the inconsistency between the agent's preference and expected utility theory in the following way. Suppose expected utility theory holds. Then $L_1\succ L_2$ is equivalent to the inequality $$u(\1)>0.01u(\0)+0.89u(\1)+0.1u(\5).$$ Adding $0.89u(\$0)-0.89u(\$1)$ on both sides, we get $$0.89u(\0)+0.11u(\1)>0.9u(\0)+0.1u(\5).$$ But this second inequality is simply saying $L_4\succ L_3$, which is contradictory to the agent's preference.

I've been working on it and I believe this is the way to approach it:

Write the lotteries as follows:

$L_{1}=[(0.89,\$1),(0.11,\$1)]$

$L_{2}=[(0.89,\$1),(0.11,L')]$where we compound$L'$inside$L_{2}$with$L'=[(\frac{1}{11},\$0),(\frac{10}{11},\$5)]L_{3}=[(0.89,\$0),(0.11,L')]$ using the same $L'$

$L_{4}=[(0.89,\$0),(0.11,\$1)]$

We have $L_{1}$ is preferred to $L_{2}$.

So, by the assumption of independence or substitutability, changing the lottery value from $(0.89,\$1)$to$(0.89,\$0)$ with all else remaining constant, should not change our preferences. So, if the assumption is followed, we should have $L_{4}$ is preferred to $L_{3}$.

However, $L_{3}$ is preferred to $L_{4}$, so it is not the case.

What do you think of this approach? Is it valid?

• Check out my answer. Despite the downvotes I'm not sure whats wrong with its validity Commented Apr 15, 2018 at 17:25
• @Sadem, you are correct. See my answer below in which I formalize your arguments fully.
– user11305
Commented Apr 16, 2018 at 0:54

This is just @Sadem's argument formalized. Note that we do not need to use any utility function representations, we can use just the preference over lotteries in conjunction with the axiom of independence.

The axiom of independence, formally stated is: $L'' \succeq L'$ $\Leftrightarrow$ $$\alpha L'' + (1-\alpha) L \succeq \alpha L' + (1-\alpha) L$$ $\forall L, L', L'' \in \mathcal{L}$ and $\forall \alpha \in (0,1)$.

Define $L'' := L_{1}$, $L' := \big((0,1/11), (5,10/11)\big)$, $L := L_{1}$, and $\alpha = .11$. Thus, the first statement implies $$\alpha L'' + (1-\alpha) L \succ \alpha L' + (1-\alpha) L$$

Now, define $L''' := (0,1)$. Hence, the second statement implies $$\alpha L'' + (1-\alpha) L''' \prec \alpha L' + (1-\alpha) L'''$$

Thus the independence axiom is violated.

• Shouldn't $\alpha=0.11$ instead of $0.89$? Otherwise $\alpha L'+(1-\alpha)L\ne L_2$. Commented Apr 16, 2018 at 4:15
• @HerrK. You are correct. Fixed!
– user11305
Commented Apr 16, 2018 at 4:40

The expected utility of $L_1$ and $L_2$ are:

$E(L_1)=\sum_{i=1}^{n=1}{p_iu_i}=1*1= 1$ E(L_2)=\sum_{i=1}^{n=3}p_iu_i=0.01*0+0.89*1+0.1*5= 1.39

Therefore if the agent says that he prefers $L_1$ to $L_2$ then he is not following the expected utility assumption of independence as $E(L_2) > E(L_1)$.

Similarly with $L_3$ and $L_4$

$E(L_3)=\sum_{i=1}^{n=2}p_iu_i=0.05$ E(L_4)=\sum_{i=1}^{n=2}p_iu_i= 0.11

Because $E(L_4)>E(L_3)$, if the agent says he prefers $L_3$ to $L_4$ then he is also not following the expected utility assumption on independence.

Hope this is helpful!

• We don't know that there's a utility function representation.
– user11305
Commented Apr 15, 2018 at 17:07
• @WaitakereCity what do you mean? i don't know why this is wrong Commented Apr 15, 2018 at 17:18
• I think you are confusing expected utility and expected value. Even though the expected value of a lottery is bigger than another, that does not mean that its expected utility must also be bigger. Also, it has nothing to do with the independence assumption. Commented Apr 15, 2018 at 17:26
• @Sadem huh? The expected value of the utility is the expected utility! Commented Apr 15, 2018 at 18:29
• @FreakconFrank, the expected value of the lottery is the summation of probabilities times the values. The expected utility of the lottery is the summation of probabilities times the expected utility of the values. Take for example the lottery $[(0.5,0),(0.5,100)]$ of expected value $50$ and the lottery $[(1,36)]$ of expected value $36$, although the first one has higher expected value then the second one, if you take the utility function $u(x)=\sqrt{x}$, then the agent will prefer the second lottery over the first. Commented Apr 15, 2018 at 18:39