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4

For completeness, let me illustrate this in the continuous time framework. The Solow equation, in the simplest of cases, is $\dot{k} = s f(k) - \delta k = \phi(k)$ Then we have $\frac{\partial \phi}{\partial k} = s f'(k) - \delta = \frac{sf'(k)k - \delta k }{k}$. In steady state (i.e., $\dot{k} = \phi(k^{\ast}) = 0$), we have $\delta k = s f(k)$, hence ...


4

Let $x(w, q)$ denote the solution to the cost minimization problem : \begin{eqnarray*} \min_{x} & \ w\cdot x \\ \text{s.t.} & \ \ f(x) \geq q \end{eqnarray*} where $f$ is the production function. Since $x(w, q)$ minimizes cost at $(w, q)$, following holds for all $w$ and for all $q$ : \begin{eqnarray*} w\cdot x(w, q) \leq w\cdot x(w', q) \ \ \ \...


3

@HerrK. got it right in his comment (he should have deleted the somewhat confusing "yes" from the beginning and then posted it as an answer) It is possible that no pairwise improvements are possible but general Pareto-improvements are still possible. A simple counterexample for three actors and three goods is as follows. Let the utility functions be the same ...


3

In order to do that, you need to define $u(·)$ as a utility function on "sure things" rather than on lotteries. In your example, you need to think in terms of the set of possible prizes to the lotteries. Say the set of possible prizes is given by $R$ and asume that is finite. For any $r\in R$, define $w_r$ as a lottery that pays $r$ in every state of nature. ...


3

I don't think it is true in a standard pure exchange economy the question is referring to. Consider the following counterexample: Suppose $I = \{1,2\}$ and $u_1(x_1, m_1) = \sqrt{x_1} + m_1$ and $u_2(x_2, m_2) = \sqrt{x_2} + m_2$. and let the set of feasible allocations be $\{((x_1, m_1), (x_2, m_2))\in\mathbb{R}^2_+\times\mathbb{R}^2_+: x_1+x_2 = 2, ...


3

For stability, we want $$\frac{\partial k_{t+1}}{\partial k_t}\Big|_{\bar k} <1 \implies \frac{(1-\delta) + \sigma A_0 f'(\bar k)}{1+n} <1$$ $$ \implies f'(\bar k) < \frac {\delta+n}{\sigma A_0 } = \frac {f(\bar k)}{\bar k}$$ So we need the marginal product of capital to be smaller than the average product at the steady state. Equivalently,...


2

Consider any $x_2'$ and $x_2''$ in $\mathbb{R}_+$. Without loss of generality, let $x_2'' > x_2'$. We can choose $x_1'=f(x_2'')-f(x_2') > 0$ so that $U(0,x_2'')=U(x_1',x_2')$. Let $\lambda(x_1',x_2')+(1-\lambda)(0,x_2'')$ be a convex combination of $(x_1',x_2')$, and $(0,x_2'')$. Since $\succsim$ is strictly convex and $U(0,x_2'')=U(x_1',x_2')$, ...


2

We want to show that for $\succcurlyeq$ on $X$, Def 1 $\iff$ Def 2 $\boxed \Longrightarrow$ Assume that $\succcurlyeq$ is continuous by Def 1. Let us say $x \succ y$. Denote our open-balls as $B(x, r)$, an open ball around $x$ of radius $r$. Suppose $\forall n, \ \exists \ x^n \in B(x, \frac{1}{n}), \ y^n \in B(y, \frac{1}{n})$ such that $y^n \succcurlyeq ...


2

Edit: Edge cases suck; see comments. See also MWG Chapter 10 section C, D. Suppose $(\vec x^*, \vec m^*)$ solves $$\max \sum^I_{i=1} m_i + \phi_i(x_i)$$ but is not Pareto optimal. $$\begin{align} \implies \exists \ (x_i', m_i') \quad \text{s.t.} \quad & u_i(x_i', m_i') \geq u_i(x_i^*, m_i^*) \quad \forall \ i = 1,\cdots,I \\ & u_i(x_i', m_i') >...


2

While it is true that a function has the expected utility form if and only if it is linear (in probabilities), it is not the case that any linear function can represent a preference that satisfies the vNM axioms. The expected utility theorem simply says that when a preference satisfies the vNM axioms, there exists a linear utility function that represents it....


2

It would suffice to show that $U$ is linear. But is $U$ necessarily linear if it satisfies the vNM axioms? Hint: No.


1

An example with two agents and two goods: let $$ U_1(x) = 0, \hskip 20pt U_2(x) = x_1+x_2, \hskip 20pt w = (1,1). $$ In this case allocating all the goods, so (1,1) to the first consumer solves the above problem. Even though any other feasible allocation fulfills the conditions, none of them gives a higher utility to the first consumer. Yet this allocation ...


1

You're asked to prove that $u(x)\ge u(y)\;\Leftrightarrow\;x\succsim y$ for any $x,y\in X$, where $u(x)=|\{z\in X:z\prec x\}|$, i.e. the utility of $x$ is measured by the number of other alternatives that rank strictly below it. Since $X$ is finite, let's suppose without loss of generality that $X=\{1,2,\dots,N\}$ where $N$ is some finite number. I'll ...


1

You have $$ \begin{align} u(\lambda a + (1 - \lambda) b, \lambda b + (1 - \lambda)a) &= \sqrt{\lambda a + (1 - \lambda)b} + \sqrt{\lambda b + (1 - \lambda)a} \\ &\geq \lambda\sqrt{a} + (1 - \lambda)\sqrt{b} + \lambda\sqrt{b} + (1 - \lambda)\sqrt{a} \\ &= \sqrt{a} + \sqrt{b} \\ &= u(a, b) \end{align}$$ The inequality in the second line ...


1

I believe you are referring to the following result: Any PE allocation maximizes $\sum_{i=1}^{I}\phi_{i}(x_{i})$, but it is hard to know precisely since you are not specific about feasibility. Let me be more specific. For each $i\in\{1,\ldots,I\}$, $(x_{i},m_{i})\in\mathbb{R}_{+}\times\mathbb{R}$. An allocation is $a=(x_{i},m_{i})_{i=1}^{I}$. The set of ...


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