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7

No function that is homogeneous of degree one, is at the same time strictly concave in its arguments. If the function is differentiable (or non-differentiable at a finite number of points), then the Hessian of a linear homogeneous function is singular. So if you want to end up with a unit cost function that is strictly concave, you have to drop at the same ...


5

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) \ \ \ \...


5

As Bertrand pointed out, strict-concavity will necessarily fail along any rays through the origin. But one can have strict concavity for normalized price systems. So let $f:\mathbb{R}^n_+\to\mathbb{R}_+$ be a production function. We let $\Delta^n_{++}$ be the set of all points in $\mathbb{R}^n$ with all coordinates being strictly positive and summing to one. ...


4

You have to show that something does not hold universally true. To show this, you just have to show that there is at least one exception- a counterexample. For this counterexample, you can make any assumption that does not contradict the assumptions of your problem.


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

From FOC, we know that: \begin{align} \nabla_x\pi(\mathbf{x},\mathbf{w})=p\nabla f(\mathbf{x})-\mathbf{w}=\mathbf{0} \tag{1} \end{align} This will be true at equilibrium, i.e. for any given $\mathbf{w}$, the input vector $\mathbf{x}$ will adjust so that the above holds. Now consider $d\pi(\mathbf{x},\mathbf{w})/d w_i$ (and using $(1)$): \begin{align} \frac{d\...


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

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') >...


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,...


3

Assume that, towards a contradiction, that both $\succeq$ and $\succeq^\ast$ rationalise the choice function and that they are different. The fact that $\succeq$ and $\succeq^\ast$ are different means that there should exist options $x, y$ such that $\succeq$ and $\succeq^\ast$ disagree on the preference over $x$ and $y$. So, for example, $x \succeq y$ and $...


2

@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 ...


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

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.


2

(Without using differentiation) When $w \leq w'$ it follows that $pf(x) − w · x \geq pf(x) − w' · x$ and so $\pi(p,w) \geq \pi(p,w')$. EDIT 1. The last inequality (first left as an exercise) can be justified as follow: $w \leq w'$ implies that $$pf(x) − w · x \geq pf(x) − w' · x$$ for any $x \geq 0$ and admissible. The inequality is in particular true for $x=...


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

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 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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