<h1>Hint</h1>
<p>Let <span class="math-container">$\{1,2,\dots,N\}$</span> be the set of outcomes. Without loss of generality, let <span class="math-container">$1\succsim 2\succsim\cdots\succsim N$</span>. Let <span class="math-container">$\mathbf e_i$</span> denote the degenerate lottery that assigns probability <span class="math-container">$1$</span> on outcome <span class="math-container">$i$</span> and <span class="math-container">$0$</span> on the rest. Note that any lottery <span class="math-container">$L=(p_1,\dots,p_N)$</span> can be written as a compound lottery <span class="math-container">$p_1\mathbf e_1+\cdots+p_N\mathbf e_N$</span>.</p>
<p>Since lottery <span class="math-container">$\mathbf e_i$</span> amounts to getting outcome <span class="math-container">$i$</span> for sure, we must have
<span class="math-container">\begin{equation}
\mathbf e_1\succsim \mathbf e_2\succsim \cdots \succsim \mathbf e_N.\tag{1}
\end{equation}</span>
Successively applying the independence axiom to <span class="math-container">$(1)$</span>, you'll obtain <span class="math-container">$\mathbf e_1$</span> as (one of) the best lottery and <span class="math-container">$\mathbf e_N$</span> as (one of) the worst.</p>

CC BY-SA 4.0
2018-12-02T02:30:03.817Z
2018-12-02T02:30:03.817Z
2020-06-17T09:40:54.643Z
Herr K.
<p>How can <strong>'the best and worst lotteries exist when the set of outcome is finite and the rational preference relation satisfies independence axiom'</strong> be proven?</p>

CC BY-SA 4.0
2018-12-01T22:32:44.127Z
And have you considered accepting the answer to your [previous question](https://economics.stackexchange.com/questions/20898/axiomatic-bargaining-kalai-smorodinsky-and-nash-solutions-coincide)?
Giskard
CC BY-SA 4.0
2018-12-02T02:56:33.193Z
@denesp: Since this is not an elementary question, it is conceivable that the OP genuinely does not even know how to start approaching it. In light of this consideration, I've provided a hint, though not a full answer, on how to start thinking about the question.
Herr K.
CC BY-SA 4.0
2018-12-02T07:02:14.593Z
@HerrK. I would not characterize your answer as "not a full answer", but perhaps I am wrong. Anyway, you have every right to do as you see fit.
Giskard
CC BY-SA 4.0
2019-09-04T11:45:08.083Z
How to do this using induction? I have read in my textbook that gives a hint about using induction on the size of support of p.
Shauryagoel
<expected-utility><uncertainty>
Existence of 'best' and 'worst' lottery
John Stern
# Existence of 'best' and 'worst' lottery

(Asked by: username_0 on 2018-12-01)

How can **'the best and worst lotteries exist when the set of outcome is finite and the rational preference relation satisfies independence axiom'** be proven?

**Tags:** expected-utility, uncertainty
> **Comments:**
> **username_1**: And have you considered accepting the answer to your [previous question](https://economics.stackexchange.com/questions/20898/axiomatic-bargaining-kalai-smorodinsky-and-nash-solutions-coincide)?  
> **username_2**: @denesp: Since this is not an elementary question, it is conceivable that the OP genuinely does not even know how to start approaching it. In light of this consideration, I've provided a hint, though not a full answer, on how to start thinking about the question.  
> **username_1**: @HerrK. I would not characterize your answer as "not a full answer", but perhaps I am wrong. Anyway, you have every right to do as you see fit.  
> **username_3**: How to do this using induction? I have read in my textbook that gives a hint about using induction on the size of support of p.  

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## 1 answers

### Answer by username_2 on 2018-12-02. Score: 3 | Accepted answer

# Hint

Let $\{1,2,\dots,N\}$ be the set of outcomes. Without loss of generality, let $1\succsim 2\succsim\cdots\succsim N$. Let $\mathbf e\_i$ denote the degenerate lottery that assigns probability $1$ on outcome $i$ and $0$ on the rest. Note that any lottery $L=(p\_1,\dots,p\_N)$ can be written as a compound lottery $p\_1\mathbf e\_1+\cdots+p\_N\mathbf e\_N$.

Since lottery $\mathbf e\_i$ amounts to getting outcome $i$ for sure, we must have
\begin{equation}
\mathbf e\_1\succsim \mathbf e\_2\succsim \cdots \succsim \mathbf e\_N.\tag{1}
\end{equation}
Successively applying the independence axiom to $(1)$, you'll obtain $\mathbf e\_1$ as (one of) the best lottery and $\mathbf e\_N$ as (one of) the worst.


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