# Proof that independence implies monotonicity in Osborne and Rubinstein

I'm struggling to understand a proof in Osborne and Rubinstein's Models in Microeconomic Theory (p. 35). The relevant lemma is

Let $$Z$$ be a set of prizes. Assume that $$\succeq$$ is a preference relation over $$L(Z)$$ and satisfies the independence property. Let $$a$$ and $$b$$ be two prizes with $$[a] \succ [b]$$, and let $$\alpha$$ and $$\beta$$ be two probabilities. Then $$\alpha > \beta \iff \alpha \cdot a \oplus (1 - \alpha) \cdot b \succ \beta \cdot a \oplus (1 - \beta) \cdot b.$$

The proof they provide is the following:

Let $$p_{\alpha} = \alpha \cdot a \oplus (1 - \alpha) \cdot b$$. Because $$\succeq$$ satisfies the independence property, $$p_{\alpha} \succ \alpha \cdot b \oplus (1 - \alpha) \cdot b = [b]$$. Using again the independence property, we have $$p_{\alpha} = (\beta/\alpha) \cdot p_{\alpha} \oplus (1 - \beta/\alpha) \cdot p_{\alpha} \succ (\beta/\alpha) \cdot p_{\alpha} \oplus (1 - \beta/\alpha) \cdot b = \beta \cdot a \oplus (1 - \beta) \cdot b.$$ $$\square$$

I don't understand how this proves both directions of the if-and-only-if statement in the lemma. Doesn't it prove only one direction?

You're right. The proof is provided for the proposition $$\alpha > \beta \implies \alpha \cdot a \ \oplus (1-\alpha) \cdot b \succ \beta \cdot a \ \oplus (1-\beta) \cdot b$$

But note that proving the converse

$$\alpha \cdot a \ \oplus (1-\alpha) \cdot b \succ \beta \cdot a \ \oplus (1-\beta) \cdot b\implies \alpha >\beta$$

is same as proving its contrapositive

$$\beta \geq \alpha \implies \beta \cdot a \ \oplus (1-\beta) \cdot b \succsim \alpha \cdot a \ \oplus (1-\alpha) \cdot b$$

and here $$\beta =\alpha$$ case is trivial, and the proof for the case when $$\beta > \alpha$$ holds is exactly the same as for $$\alpha > \beta$$ (provided in the question). We just need to reverse the roles of $$\alpha$$ and $$\beta$$.

• Oh, I see. Many thanks! Commented Jun 22 at 23:33