# Tag Info

22

The proofs I will present are based on techniques relevant to the fact that the CES production function has the form of a generalized weighted mean. This was used in the original paper where the CES function was introduced, Arrow, K. J., Chenery, H. B., Minhas, B. S., & Solow, R. M. (1961). Capital-labor substitution and economic efficiency. The Review ...

11

The regular method of obtaining Cobb-Douglas and Leotief is L'Hôpital's rule. Another methods should be used too. Setting $\gamma=1$ will be return $Q=[a K^{-\rho} +(1-a) L^{-\rho} ]^{-\frac{1}{\rho}}$ and $$Q^{-\rho}=[a K^{-\rho} +(1-a) L^{-\rho} ]$$ By The total derivative via differentials we will have $$-\rho Q^{-\rho-1}dQ=- a\rho K^{-\rho-1}dK -(... 6 We know that if u represents \succeq on X, then for any strictly increasing function f: \mathbb{R} \rightarrow \mathbb{R}, then v(x) = f(u(x)) represents \succeq on X (X in this case is \mathbb{R^n}) Consider v(x, \rho) = \ln(u(x, \rho)) - \frac{\ln\left[\sum^n_{i=1}\alpha_i \right]}{\rho}, which is strictly increasing.$$v(x, \rho) = \...

5

This is not a weird case, but a Leontief production function which is not homogeneous of degree one, but homogeneous of degree $b$. You can see this if you use the connection between a C.E.S. production function and the Leontief one. Consider $$Q_b=[a K^{-\rho} +(1-a) L^{-\rho} ]^{-\frac{b}{\rho}},\;\; b>0$$ $$\Rightarrow Q_b = \frac 1{[a (1/K^{\rho}) ... 5 You are missing a \min operator just before the bracket. The utility maximization problem is as follows,$$\max \ \min [\alpha x_1, ..., \gamma x_3] \\ \ \ \text{such that} \ \ \lambda_1x_1 + ... + \lambda_3x_3 = M$$Consider the case of two goods with utility u given by u(x) = \min[\alpha x_1, \beta x_2]. At the optimum, what do you know about the ... 4 Since you are interested in labour, let's assume for simplicity that the stock of capital is fixed at \bar{K}. Then, the optimal choice of capital and labour is given by:$$\frac{L^*}{a}=\frac{\bar{K}}{b}$$Therefore, optimal labour is:$$L^* = \frac{a}{b}\bar{K}$$The marginal product of labour depends on how actual labour relates to optimal labour: ... 4 We are given the production function Q = \left(\min(K, L)\right)^\beta. Cost minimization problem of the producer is defined as finding the labor-capital combination that minimizes the cost of producing at least y units of output given that the price of labor is w, and price of capital is r. \begin{eqnarray*} \min_{L, K} && wL + rK \\ \text{... 4 Hint: Second Welfare Theorem: Suppose that for every individual i = 1,...,N the utility function U^i(c_i) is continuous, locally non-satiated, and quasi-concave. If \left\{c_1^P,...,c_N^P\right\} represents a Pareto optimal allocation in which c_i^P >> 0 for every i = 1,...,N, then there exists a set of prices, p^E \geq 0 and p^E \neq 0... 4 These are standard mathematical results for generalized means. For example,for the \rho \rightarrow 0 result, write (setting without loss of generality \sum_{i=1}^na_i =1),$$U = \left[\sum^n_{i=1} \alpha_i x^\rho_i \right]^\frac{1}{\rho} = \exp\left\{\frac 1\rho\ln \left(\sum^n_{i=1} \alpha_i x^\rho_i\right)\right\}$$Apply L'Hopital's rule on$$\...

3

You can re write $u(x, y)$ as \begin{eqnarray*} u(x, y) = \begin{cases} ax + y & \text{ if } x > y\\ ay + x & \text{ if } x \leq y\end{cases} \end{eqnarray*} When one plot the indifference curves, this is how they will look: Here $\mu_1 > \mu_2 > \mu_3$.

3

The bottom right origin is actually not in the Pareto set. At that point, $(x_A,y_A)=(8,0)$, so $U_A(x_A,y_A)=0$. Similarly, $(x_B,y_B)=(0,4)$, so $U_B(x_B,y_B)=0$. As an example, $B$ could give one unit of $y$ to $A$ and thereby raise $A$'s utility to 8 without hurting $B$'s own utility (which would remain zero). As a matter of fact, any allocation ...

3

No problem. $$Q =\left(\min\{K, L\} \right)^b$$ It just means that first you compare $K$ and $L$ and your quantity $Q$ will be equal the lower one to the power $b$. Example: $b=2$, $K=3$ and $L=7 \implies Q = 3^2 = 9$.

2

One needs to go case-by-case and arrive at a utility function with branches. To get you started, if $x_1 < x_2/2 \implies \min(2x_1,x_2) = 2x_1$, but then also $\min(x_1,2x_2) = x_1$. Therefore in this case, $u(x_1,x_2) = 2x_1$. etc There are two other intervals to consider as regards the relation between $x_1$ and $x_2$, so in all you will obtain a ...

2

I believe you are correct. The points for which Ya = 0 and $$0\le Xa \le 4$$ will be all Pareto efficient points. Proof: Consider an allocation like (2,0). The indifference curve through this point for Individual A is the positive x-axis. (Look at the graph for the IC of Individual B) Increasing the satisfaction level of A would ...

2

Something you didn't mention is that $x = Ax + y$ means that to produce one unit of $x$, you use $A$ unit of $x$. E.g. you need electricity to produce electricity. Under the condition that $1>|A|\geq0$ $(1-A)^{-1} = \sum_{k=0}^\infty A^k$ Which allows us to write $x = Ly = \left(\sum_{k=0}^\infty A^k \right)y$ Thus, for one unit of $y$, your (...

2

Leontief did not use total capital stock for his calculation of capital/labor requirements in the export/import sector. He measured 'requirements per million dollars of exports and imports replacements', and this for various industries (Leontief, 1953, section III, table 2). So he also did not use total capital stock, he used capital requirements AKA capital ...

2

Your reasoning that $L_1=L_2^{1/3}$ is valid for any $K$. Indeed if this equality does not hold you can lower the cost by reducing the excess input of either skilled or unskilled labor. Thus you can rewrite the cost minimization problem in two dimensions instead of three $$\min_{K,L_1}{200 K + 5 L_1 + 6 L_1^3}$$ subject to \begin{...

2

$p$ represents total production. $Ap$ represents the intermediate goods and services used in production, i.e. intermediate consumption. So $p-Ap=(I-A)p$ represents net production, i.e. output minus input.

1

So, I have not really worked out the maths behind solving this. I prefer to take a shortcut and just compute it numerically. Here is what I have. These are the contour surfaces and the budget constraint. Notice how the solution will be at the "kinked" curve in the middle of each surface. The levels 1 and 2 are only for example. The third level, u = 3....

1

The Lagrangian method wouldn't be of any use, because Leontief function is not differentiable at the point of optimality/kink. However, you can consider the following approach. We know that for $U(x,y)=min\{x,y\}$, optimalilty occurs at the point where $x=y$. Let the budget correspondence be $p_1x+p_2y\leq w$, where $w$ is the income level. Optimal ...

1

What intuition can the mathematical concept of an inverse (function, matrix) have? In a single-input monotonic production function $F(x) = q \implies x= F^{-1}(q)$ the operator $F^{-1}$ is the transformation mechanism that translates output to required input. A "change of units" calculator if you wish. Isn't this what $(I-A)^{-1}$ does in the case of the ...

1

Define $Q=q^b$, where $q=min\{K,L\}$. For $b>0$, $Q$ is a monotonic transformation of $q$. As such, the solution for $q$ is equivalent to the solution for $Q$. Simply solve your problem for $q$, and then rework it in terms of $Q$.

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