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?
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$\begingroup$ And have you considered accepting the answer to your previous question? $\endgroup$– GiskardCommented Dec 1, 2018 at 22:32
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$\begingroup$ @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. $\endgroup$– Herr K.Commented Dec 2, 2018 at 2:56
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$\begingroup$ @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. $\endgroup$– GiskardCommented Dec 2, 2018 at 7:02
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$\begingroup$ 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. $\endgroup$– ShauryagoelCommented Sep 4, 2019 at 11:45
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1 Answer
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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.
