# Proof that $U(\sum_{n=1}^{N}{p_nL_n})=\sum_{n}^{N}{p_nU(L_n)}$

I understand the expected value of a lottery is $$\sum_n^N{p_nL_n}$$ where there are $$N$$ possible outcomes, each with a probability $$p_n$$ with $$n=1,...,N$$ and $$\sum_{n}p_n=1$$ (that's rather trivial I believe).

Now we say $$U(\sum_{n=1}^{N}{p_nL_n})=\sum_{n}^{N}{p_nU(L_n)}$$ or in other words $$U(L)=u_1p_1+...+u_np_N$$ where $$U:\mathbb{L} \mapsto \mathbb{R}$$ with $$L=(p_1,...,p_n) \in \mathbb{L}$$.

Why is that? Is that just the "definition" of expected utility?

Why is, for a two-outcomes-case,

$$U(pL_1+(1-p)L_2)=pU(L_1)+(1-p)U(L_2)$$?

Wouldn't this imply that the utility of the expected value equals the expected utility since the expression $$pL_1+(1-p)L_2$$ is the expected value? This should not be true though (except if U(x)=x) but it's not stated anywhere this is assumed.

I saw this in Varian, Microeconomic Analysis, chapter 11.3 "uniqueness of the expected utility function":

Shouldn't this be $$v(u(px+(1-p)y)) = ...$$ on the first line?

What's wrong here? I'm confused.

$$p\circ x\oplus(1-p)\circ y$$ is a lottery that gives you the prize $$x$$ with probability $$p$$ and the price $$y$$ with probability $$(1-p)$$. Unless $$x,y$$ can be identified with numbers, such as amounts of money, it makes no sense to take the expectation of this lottery. What is the expected value of a lottery that gives you a cow with a probability of $$0.5$$ and a sheep with a probability of $$0.5$$?
If $$U$$ gives your the expected utility of each lottery, then you have $$U(p_1\circ L_1\oplus p_2 L_2\oplus\cdots\oplus p_n\circ L_n)=p_1U(L_1)+p_2 U(L_2)\cdots+p_n U(L_n).$$ On the right, you are actually taking an expectation, on the left you apply $$U$$ to a compound lottery. Sometimes people write down the compound lottery on the left as if it were a weighted sum and hope that the context makes the difference clear, but these are really different things that get related by expected utility theory.
• Oh I see. But what if $x,y$ are numbers, such as amounts of money? Then wouldn't this equal the expected value? Jun 8, 2021 at 22:00
• A lottery that gives you $0$ dollars with a probability of $0.5$ and $1000$ dollars with a probability of $0.5$ is still different than a lottery that gives you $500$ dollars with probability one. Jun 8, 2021 at 22:10
• Oh lord. Yeah that is horribly confusing tbh. One should write $EU(L)$ or use a capital U for expected utility and lowercase u for the actual utility function or similar. So I understood that the utility of the second lottery (your latter example) should be higher than the utility of the first lottery. This would be the case for f.ex. CobbDouglas preferences where $u(x)=lnx$ except if $u(x)=x$ (perfect substitutes). Thanks (you're great, the example you gave in your comment was especially helpful and clear)! Jun 8, 2021 at 23:02
• @j3141592653589793238, almost everybody gets confused by this in the beginning. It might be better to consult MCWG on expected utility theory. They distinguish between the Bernoulli-utility function $u$ (defined on prizes) and the vNM-utility function $U$ defined on lotteries over prizes, where $U=Eu$. That helps. Jun 10, 2021 at 19:33