I'm stuck with the following question:

Let's say that C1, C2 and C3 represent the certainty equivalents and (x,p,y) the prospects.

C1 ~ (x, p, 0) C2 ~ (x, p, C1) C3 ~ (C1, p, 0)

What is C3 such that no random variables are left?

I suppose that it can be solves usiong the certainty equivalent method but im stuck. When normalizing u(x)=1, I find p^2 for u(c3) but it still doesn't give any information about c3 itself.

  • $\begingroup$ Hi! What exactly are you asking here? Are you expected to express $C_3$ without the utility function and preference relation, or did @tdm answer your question? $\endgroup$
    – Giskard
    Dec 30, 2023 at 10:29
  • $\begingroup$ I'm expected to express C3 without the utility function. $\endgroup$
    – EcoSTUD233
    Jan 2 at 12:15
  • $\begingroup$ If you can use the preference relation, tdm's answer works. If you cannot use any such think, you can easily come up with a numerical example where $C_3$ will depend on the utility function, i.e. it will have different values if $u(c) = \sqrt{c}$ and if $u(c) = c^2$. $\endgroup$
    – Giskard
    Jan 2 at 12:37

1 Answer 1


If you receive $C3$, you get lottery $C_1$ with probability $p$ and $0$ with probability $(1-p)$.

$C_1$ is the lottery that gives $x$ with probability $p$ and $0$ with probability $(1-p)$.

So in total, you receive $x$ with probability $p^2$ and zero with probability $(1-p) + p(1-p) = 1 - p^2$.

As such, $C_3 \sim (x, p^2, 0)$.


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