# Existence of 'best' and 'worst' lottery

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?

• And have you considered accepting the answer to your previous question? – Giskard Dec 1 '18 at 22:32
• @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. Dec 2 '18 at 2:56
• @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 Dec 2 '18 at 7:02
• 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. – sudddddd Sep 4 at 11:45

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.