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1. "Liking more of everything"

Given $$X = \mathbb{R}^L$$, for every two bundles $$x, y \in X$$, $$x \succeq y$$ if and only if $$x_k \ge y_k \forall k$$.

a) Preferences are complete if for every two bundles $$x, y \in X$$, either $$x \succeq y$$ or $$x \preceq y$$ or $$x \sim y$$. Consider two bundles $$x,y$$, with $$x_i &gt; y_i$$ and $$x_j &lt; y_j$$. Then $$x \npreceq y$$ and $$x \nsucceq y$$, so they preferences are not complete.

b) Preferences are transitive if for any bundles $$x,y,z \in X$$, if $$x \succeq y$$ and $$y \succeq z$$, then $$x \succeq z$$.

Assume that $$x \succeq y$$ and $$y \succeq z$$. Then by the preference relation above, $$x_k \ge y_k \forall k$$ and $$y_k \ge z_k \forall k$$. Substituting, we can see that $$x_k \ge y_k \ge z_k \Rightarrow x_k \ge z_k \forall k$$, hence $$x \succeq z$$.

c) Preferences are strongly monotonic if $$x_k \ge y_k \forall k$$ and $$x \neq y$$ implies $$x \succ y$$. But by the above preference, $$x_k \ge y_k \forall k$$ implies only that $$x \succeq y$$, so the preferences are not strongly monotonic.

d) Preferences are strictly convex if for any three bundles $$x, y, z \in X$$, with $$x \preceq z$$, $$y \preceq z$$, and $$x \neq z$$, $$ax + (1-a)y \prec z \forall a \in (0,1)$$.

Since $$x \preceq z$$, $$y \preceq z$$, we know that $$x_k \ge z_k$$ and $$y_k \ge z_k \forall k$$. In addition, $$x \neq z \Rightarrow x_i &gt; z_i$$ for some $$i$$. Also, since $$a \in (0,1)$$, $$ax_k + (1-a)y_k \ge z_k \forall k$$ and $$ax_i + (1-a)y_i &gt; z_i$$ for at least one $$i$$. This implies that $$ax + (1-a)y \ge z$$, and $$ax + (1-a)y \neq z$$, so $$ax + (1-a)y \succeq z but the reverse is not true. Therefore, $$ax + (1-a)y \succ z$$.

2. Checking properties of a preference relation

Consider a consumer with $$x = (x_1, x_2) \succeq y = (y_1, y_2)$$ if and only if $$\max{x_1, x_2} \ge \min{y_1, y_2}$$.

a) The consumer weakly prefers $$x$$ to $$y$$ if the maximum component of $$x$$ is at least as valuable as the minimum component of $$y$$. That is, if a single component of $$x$$ is larger than the smallest component of $$y$$, the consumer prefers $$x$$ to $$y$$, or is indifferent between $$x$$ and $$y$$.

b) The preferences have the following properties:

• Completeness: The preferences are complete, since $$x \succeq y \Rightarrow \max(x) \ge \min(y)$$, and $$x \preceq y \Rightarrow \max(x) \le \min(y)$$, so any two bundles $$x,y \in X$$ are comparable.
• Transitive: Preferences are not transitive, as an example will show. Let $$x = (1,5)$$, $$y = (2,8)$$, and $$z = (6,7)$$. Then $$x \succeq y$$ since $$\max{x_i} \ge \min{y_i}$$ and $$y \succeq z$$ since $$\max{y_i} \ge \min{z_i}$$, but $$\max{x_i} \le \min{z_i}$$, so $$x \preceq z$$.
• Monotone: By the definition of monotone, $$x_k \ge y_k \forall k \Rightarrow x \succeq y$$ and $$x_k &gt; y_k \forall k \Rightarrow x \succ y$$. Assume $$x_k \ge y_k \forall k$$. Then $$\max(x) \ge \min(y)$$, so $$x \succeq y$$. Now assume $$x_k &gt; y_k \forall k$$. Then $$\max(x) &gt; \min(y)$$, but $$\max(y)$$ can be greater or equal to $$min(x)$$, so the second requirement of monotinicity does not hold. Hence the preferences are not monotonic. An example is illustrative. Let $$x = (2,10)$$ and $$y = (1,9)$$. Then $$x_k &gt; y_k \forall k$$ and $$max(x) &gt; min(y)$$ so $$x \succeq y$$. But $$max(y) &gt; min(x)$$, so $$y \succeq x$$. Since $$x \succeq y$$ and $$y \succeq x$$, it does not hold that $$x_k &gt; y_k \Rightarrow x \sim y$$, so the preferences are not monotonic. XXX * Convex: Let $$x \succeq z$$ and $$y \succeq z$$, with $$x \neq z$$ and $$y \neq z$$. If the preferences are convex, then $$\alpha x + (1 - \alpha)y \succeq z \Rightarrow \max(\alpha x + (1-\alpha)y) \ge \min(z)$$. Since We can rewrite this as $$\max(\alpha x + (1-\alpha)y) \ge \min(z) Then $$\max(\alpha x + (1-\alpha) y) \ge \min(z)$$. $$. Since $$x \succeq z \Rightarrow \max(x) \ge \min(z)$$ and $$y \succeq z \Rightarrow \max(y) \ge \min(z)$$, and with $$\alpha \in (0,1)$$, $$\alpha
• Locally Non-Satiated: The preference relation is not LNS. A preference relation is LNS if for any scalar $$\epsilon &gt; 0$$ you can define a new bundle $$y$ where $$||x - y|| &lt; \epsilon$$, such that $$y \succ x$$. In this case, $$y \succ x \Rightarrow \max(y) \ge \min(x)$$ and $$\max(x) &lt; \min(y)$$. Choose $$y = x + \frac{\epsilon}{2}$$ for some $$\epsilon$$. Then

$$\begin{align} ||x - y|| &= \sqrt{\sum_i(x_i - y_i)^2}\ &= \sqrt{\sum_i\left(x_i - (x_i + \frac{\epsilon}{2})\right)^2}\ &= \sqrt{\sum_i\left(\frac{\epsilon}{2}\right)^2}\ &= \frac{\epsilon}{\sqrt{2}} \end{align}$$,

where $$\frac{\epsilon}{\sqrt{2}} &lt; \epsilon$$. But choose $$\epsilon = \frac{\max(x) - \min(x)}{2}$$. But choose $$\epsilon = \frac{\max(x) - \min(x)}{2}$$. Then $$y \succ x$$ implies

$$\begin{align} \max(x) &< \min(y)\\ \max(x) &< \min(x + \frac{\epsilon}{2})\\ \max(x) &< \min(x) + \frac{\epsilon}{2}\\ \max(x) &< \min(x) + \frac{max(x) - min(x)}{4}\\ \max(x) &< \min(x) \end{align}$$

which is false, so the relation is not LNS.

3. Strictly Convex Preferences

Consider preferences defined on $$X \in \mathbb{R}^2_{+}$$.

XXX a) Preferences are convex if the convex combination of bundles is preferred over a single bundle alone, or given $$x = (x_1,x_2)$$ and $$y = (y_1,y_2)$$, $$x \succeq y \Rightarrow ax + (1-a)y \succeq y$$. Solving, we get

$$\begin{align} \frac{ax}{y} + 1 - a &\ge 1 \\ \frac{ax}{y} &\ge a \\ \frac{x}{y} &\ge 1 \\ x &\ge y \end{align} $$

Since $$x \ge y$$, the preferences are convex.

b) Given $$U(x) = x_1 x_2$$,

4. Quasi-Linear Preferences

XXX

5. Rubinstein PS 2

3). Given $$\succeq$$ be continuous preferences on $$X \subseteq \mathbb{R}^n$$ that contains the interval connecting points $$x$$ and $$z$$. Show that if $$y \in X$$ and $$x \succeq y \succeq z$$, then there exists a point $$m$$ on that interval such that $$y \sim m$$.

Following the Bolzano-Weierstrass theorem, let $$I \subseteq X$$ be the interval connecting $$x$$ and $$z$$, and $$y \in X$$ a third point with $$x \succeq y \succeq z$$. Inductively, define two sequences of points in $$I$$, $${x_t}$$ and $${z_t}$$. Construct $$x_t$$ and $$z_t$$ in the following way. Define $$x_0 = x$$ and $$z_0 = z$$. Since $$x \succeq z$$ and $$x_t \in I$$, $$x \succeq x_t \forall t$$. Similarly, $$z \preceq z_t \forall t$$. Now choose a point $$m$$ such that $$x_t \succeq m \succeq z_t$$. Then since $$x \succeq y \succeq z$$, either $$m \succeq y$$ or $$y \succeq m$$ or both. In the first case, define $$x_{t+1} = m$$ and $$z_{t+1} = z_t$$. In the second, define $$z_{t+1} = m$$ and $$x_{t+1} = x_t$$. The sequences $${x_t}$$ and $${z_t}$$ are conversing, and they converge to $$y$$ because $$\lim_{t \to \infty} |x_t - z_t| = 0$$. Since the preferences are continuous, at that point we have $$m \succeq y$$ and $$y \succeq m$$, so $$m \sim y$$.

XXX 4). Consider $$(\succeq^n){n=1,2,...}$$, defined on $$\mathbb{R}^2{+}$$, where $$\succeq^n$$ is represented by $$u_n(x_1, x_2) = x^n_1 + x^n_2$$. Say that the sequence $$\succeq^n$$ converges to preferences

5). 6). Given $$X \in \mathbb{R}^2_{+}$$, assume that $$\succeq$$ satisfies additivity, strong monotonicity, and continuity. Define $$x = (x_1, x_2)$$ and $$y = (y_1, y_2)$$, where $$x,y \in X$$.

a) Given $$\succeq$$ is linear, that is, $$u(x) = \alpha x_1 + \beta x_2, \alpha, \beta &gt; 0$$, show that it has the following properties:

• Additivity: Let $$u(x) \succeq u(y)$$. Then $$\alpha x_1 + \beta x_2 \ge