<p>If you receive <span class="math-container">$C3$</span>, you get lottery <span class="math-container">$C_1$</span> with probability <span class="math-container">$p$</span> and <span class="math-container">$0$</span> with probability <span class="math-container">$(1-p)$</span>.</p>
<p><span class="math-container">$C_1$</span> is the lottery that gives <span class="math-container">$x$</span> with probability <span class="math-container">$p$</span> and <span class="math-container">$0$</span> with probability <span class="math-container">$(1-p)$</span>.</p>
<p>So in total, you receive <span class="math-container">$x$</span> with probability <span class="math-container">$p^2$</span> and zero with probability <span class="math-container">$(1-p) + p(1-p) = 1 - p^2$</span>.</p>
<p>As such, <span class="math-container">$C_3 \sim (x, p^2, 0)$</span>.</p>

CC BY-SA 4.0
2023-12-30T08:29:10.557Z
2023-12-30T08:29:10.557Z
tdm
<p>I'm stuck with the following question:</p>
<p>Let's say that C1, C2 and C3 represent the certainty equivalents and (x,p,y) the prospects.</p>
<p>C1 ~ (x, p, 0)
C2 ~ (x, p, C1)
C3 ~ (C1, p, 0)</p>
<p>What is C3 such that no random variables are left?</p>
<p>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.</p>

CC BY-SA 4.0
2023-12-30T10:29:33.780Z
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](https://economics.stackexchange.com/a/57289/1601) your question?
Giskard
CC BY-SA 4.0
2024-01-02T12:15:09.923Z
I'm expected to express C3 without the utility function.
EcoSTUD233
CC BY-SA 4.0
2024-01-02T12:37:12.017Z
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 $u(c) = c^2$.
Giskard
<decision-theory><expected-utility><uncertainty>
Decision theory: elicitation method
EcoSTUD233
# Decision theory: elicitation method

(Asked by: username_0 on 2023-12-28)

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.

**Tags:** decision-theory, expected-utility, uncertainty
> **Comments:**
> **username_1**: 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](https://economics.stackexchange.com/a/57289/1601) your question?  
> **username_0**: I'm expected to express C3 without the utility function.  
> **username_1**: 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 $u(c) = c^2$.  

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## 1 answers

### Answer by username_2 on 2023-12-30. Score: 3

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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