# Question About Stochastic Choice - MWG Exercise 1.D.5

I am studying microeconomic theory using MWG. I got stuck on Exercise 1.D.5, specifically part (c), but I would also like to have my part (a) and (b) checked by someone. Here is the exercise and my attempt so far:

## Problem

Let $$X = \{x,y,z\}$$ and $$\mathscr{B} = \{ \{x,y\},\{y,z\},\{z,x\} \}$$. Suppose that choice is now stochastic in the sense that, for every $$B \in \mathscr{B}$$, $$C(B)$$ is a frequency distribution over alternatives in $$B$$. For example, if $$B = \{x,y\}$$, we write $$C(B) = (C_x(B), C_y(B))$$, where $$C_x(B)$$ and $$C_y(B)$$ are nonnegative numbers with $$C_x(B) + C_y(B) = 1$$. We say that the stochastic choice function $$C(\cdot)$$ can be rationalized by preferences if we can find a probability distribution $$Pr$$ over the six possible (strict) preference relations on $$X$$ such that for every $$B \in \mathscr{B}$$, $$C(B)$$ is precisely the frequency of choices induced by $$Pr$$. For example, if $$B = \{x,y\}$$, then $$C_x(B) = Pr(\{\succ:x \succ y\})$$.

(a) Show that the stochastic choice chunction $$C(\{x,y\}) = C(\{y,z\}) = C(\{z,x\}) = (\frac{1}{2},\frac{1}{2})$$ can be rationalized by preferences.

(b) Show that the stochastic choice chunction $$C(\{x,y\}) = C(\{y,z\}) = C(\{z,x\}) = (\frac{1}{4},\frac{3}{4})$$ is not rationalizable by preferences.

(c) Determine the $$0 < \alpha < 1$$ at which $$C(\{x,y\}) = C(\{y,z\}) = C(\{z,x\}) = (\alpha,1-\alpha)$$ switches from rationalizable to nonrationalizable.

## My Attempt

Define the probability distribution on the six mutually exclusive events as \begin{align*} p_1 & = Pr(x \succ y \succ z) \\ p_2 & = Pr(y \succ z \succ x) \\ p_3 & = Pr(z \succ x \succ y) \\ p_4 & = Pr(x \succ z \succ y) \\ p_5 & = Pr(y \succ x \succ z) \\ 1-p_1-p_2-p_3-p_4-p_5 & = Pr(z \succ y \succ x). \end{align*}

(a) Let $$p_i = \frac{1}{6}$$ for all $$i = 1,\dots,5$$ and we are done.

(b) Assume to the contrary that $$C(\{x,y\}) = C(\{y,z\}) = C(\{z,x\}) = (\frac{1}{4},\frac{3}{4})$$ is rationalizable. Then \begin{align*} p_1 + p_4 + p_3 & = \frac{1}{4}\tag1 \\ p_1 + p_2 + p_5 & = \frac{1}{4}\tag2 \\ p_2 + p_3 + (1-p_1-p_2-p_3-p_4-p_5) & = 1 - p_1 - p_4 - p_5 = \frac{1}{4}\tag3. \end{align*} Then, $$(1) + (3)$$ yields \begin{align*} p_5 = \frac{1}{2}+p_3\tag4, \end{align*} and $$(2) + (3)$$ yields \begin{align*} p_4 = \frac{1}{2}+p_2\tag5. \end{align*} Plug $$(4)$$ and $$(5)$$ into $$(3)$$ yields \begin{align*} p_1 = -p_2-p_3-\frac{1}{4} < 0 \end{align*} which is impossible. Thus, it is not rationalizable.

(c) Suppose $$C(\{x,y\}) = C(\{y,z\}) = C(\{z,x\}) = (\alpha,1-\alpha)$$ is rationalizable. Then, similarly as in part (b), we have \begin{align*} p_1 + p_4 + p_3 & = \alpha \\ p_1 + p_2 + p_5 & = \alpha \\ 1 - p_1 - p_4 - p_5 & = \alpha. \end{align*}

But I got stuck here and don't know how to proceed next.

## My Question

• You have a collection of equalities and inequalities (as for every $p_i$ you need to add the condition that $0 \le p_i \le 1$. These can be solved (painstakingly) using quantifier elimination. Eliminate one by one all variables $p_1, \ldots, p_5$ by combining lower and upperbounds on these variables.