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9 votes
Accepted

Cost Minimization and Karush-Kuhn-Tucker

The term $\lambda_2(x_1-1)$ in your Lagrangian is incorrect; it treats the second constraint as an equality rather than an inequality. To allow for the constraint being an inequality you can include a ...
Adam Bailey's user avatar
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6 votes

Cost Minimization and Karush-Kuhn-Tucker

Here is the cost minimisation problem that we need to solve: \begin{eqnarray*} \min_{x_1,x_2} & w_1x_1+w_2x_2 \\ \text{s.t. } & \sqrt{x_1x_2}=\overline{y} \\ \text{and } & x_1\geq 1, x_2\...
Amit's user avatar
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5 votes

Cost Minimization and Karush-Kuhn-Tucker

Let's setup the optimization problem first, $$min_{\{x_1,x_2\}} \omega_1x_1+\omega_2x_2 $$ $$ s.t \hspace{5 mm} (\bar{y}=x_1^{\frac{1}{2}}x_2^{\frac{1}{2}}) \wedge(x_1 \ge 1)\wedge (x_2 \ge 0)$$ ...
SGP's user avatar
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3 votes
Accepted

What is the formula for price

A price is just a monetary transfer from buyers to sellers. To find the price in any given setting, you need much much more structure. What is the market structure you're working with? What are the ...
matthewoulton's user avatar
3 votes
Accepted

Question About Proving $\alpha(\cdot)$ is Continuous in the Proof of Proposition 3.C.1 from MWG

The sequence you proposed $0,1,0,2,0,3,\ldots$ i.e. $x_n=\begin{cases}0 &\text{if $n$ is odd} \\ \frac{n}{2} & \text{if $n$ is even} \end{cases}$ is not a bounded sequence and is therefore, ...
Amit's user avatar
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2 votes
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Proof: Let $\epsilon>0$ and $x'\in\mathbb{R}^L_+$ be such that $\|x'-x\|\geq\epsilon$. Then $\alpha(x')$ belongs to some $[\alpha_0,\alpha_1]$

We don't need $X$ (domain of the utility) to be equal to the set of its limit points to use the sequential characterisation of continuity. In other words, the sequential characterisation of continuity ...
Amit's user avatar
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2 votes
Accepted

Question About the Step Proving $\alpha(x)$ Represents Preferences in the Proof of Proposition 3.C.1 from MWG

Def 3.B.2: The preference relation $\succsim$ on $X$ is monotone if $x\in X$ and $y\gg x$ implies $y\succ x$. [1] If $\alpha(x)\geq \alpha(y)$, then there are two possibilities: (i) $\alpha(x)>\...
Amit's user avatar
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2 votes
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Market signaling with a national exam

I think your reasoning is correct. There is no need to have an ''expected pay'' as in a separating equilibrium, everyone ''knows'' that good types will take the test and bad types will not. The only ...
tdm's user avatar
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2 votes
Accepted

Using lagrange on a quasi-concave utility function

As you can see in this post, that there are also "corner" solutions to this problem under some conditions. These are solutions where $x_1=0$ or $x_2=0$. For this reason, you may use Kuhn-...
Amit's user avatar
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2 votes
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Solving utility maximization, and finding demand function

Check that the following utility function also represents the same preference as the one in the question: $u(x_1,x_2)=(x_1+1)(2x_2+1)$ Solving $\max_{(x_1,x_2)\in\mathbb{R}^2_+} (x_1+1)(2x_2+1)$ ...
Amit's user avatar
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1 vote

When is argmax increasing in some multiplier of the objective function?

First, we do need additional conditions on $f$ beyond the ones you supplied in the question. Consider $c(x)=\frac{1}{2}x^2$ and $f(x)=\tanh x$, both restricted to the domain $\mathbb{R}_+$. These ...
Joseph Basford's user avatar
1 vote

Proving quasi-concavity for a utility function

Consider the upper level set \begin{eqnarray*}S_a&=&\left\{(x_1,x_2)\in\mathbb{R}^2_+|2x_1x_2+x_1+2x_2\geq a\right\}\\ &=& \left\{(x_1,x_2)\in\mathbb{R}^2_+|2(x_1+1)\left(x_2+\frac{1}{...
Amit's user avatar
  • 8,811
1 vote

Revealed preferences - Commuter preferences

It depends on the extent to which your satisfaction data take into account: Heterogeneity of preferences Price The intuition behind (strong) revealed preference is that you can tell a consumer ...
matthewoulton's user avatar
1 vote

Intuition of sign used for Lagrange multiplier and corresponding constraint function in constrained optimization

Let $\mathbf x \in \mathbb{R^n} \quad \mathbf x= (x_1 , x_2 , \cdots , x_n ) $ $A$, $B$ $m$x$n$ matrix, $\mathbf b \in \mathbb{R^m}$, $\mathbf c \in \mathbb{R^m}$. Let indicate with $< . , . >$ ...
marco tognoli's user avatar
1 vote
Accepted

Intuition of sign used for Lagrange multiplier and corresponding constraint function in constrained optimization

The Lagrange multiplier provides the "shadow price" of the constraint. In the case where the Lagrangian takes the form: $$ L = f(x,y) - \lambda(g(x,y) - c), $$ then $\lambda$ measures the ...
tdm's user avatar
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