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:


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

Could someone please help me complete the argument for (c)? Thanks a lot in advance!

(I can understand the answer to this question from the online pdf solution manual, but I feel their approach is not "natural" to me and would like to try it in a different way.)

  • $\begingroup$ 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. $\endgroup$
    – tdm
    Nov 2 at 19:03


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