# Tag Info

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Does Modern Monetary Theory (MMT) provide a useful insight into how to manage the economy? That depends on your definition of MMT, because it is not generally agreed on what it even is. You will find some arguing it is just a macro/monetary theory (such as the Wikipedia page) but then I seen MMT proponents on this site arguing it is a whole new paradigm ...

7

I don't believe those two terms are used in the same spheres. To me, an economic theorist, signaling plays a role in models with asymmetric information when the informed party moves first and the uninformed player reacts, treating the first action as a signal about the private information. This idea goes back to Spence, and also plays a role in biology with ...

6

First, you need a vector space in order for convex combinations to be well-defined. However, not every metric on a vector space works. Indeed, under the discrete metric, the result will trivially fail unless space consists of the point $0$ alone. What works is a metric on the vector space such that the vector operations of addition and scalar multiplication ...

6

The utility function under consideration is $v(c,q)$ and then $$MRS(c,q) = \frac{\partial v/\partial q}{\partial v/\partial c} = v_2/v_1$$ make the functional denpendency of on $u$ explicit then you have $$\frac{\partial}{\partial u}MRS(c(u),q(u)) = \frac{\partial MRS(c(u),q(u))}{\partial c} \frac{\partial c(u)}{\partial u} + \frac{\partial MRS(c(u),q(u))}{\... 6 In general, it will not represent the same preferences. There seems to be confusion on what "monotonic transformation" means in this context. It does not have much to do with monotonic preferences. We say that the utility function v:X\to\mathbb{R} is a monotonic transformation of the utility function u:X\to\mathbb{R} if there exists a strictly ... 5 If you remember, in a two-dimensional curve, its concavity or convexity (the slope of its slope) is given by the second derivative. For a three-dimensional function, you want to look at the Hessian table (the table of all second derivatives). If the Hessian is negative definite for all values then the function is strictly concave, and if the Hessian is ... 5 To solve$$\max_{x \geq 0} \ (x_1+1)^\alpha(x_2 + 1)^\betas.t. \ \ I \geq p_1x_1 + p_2x_2,$$I would define y_1 = x_1+1 and y_2 = x_2 + 1 to get the problem$$\max_{y \geq 0} \ y_1^\alpha y_2^\betas.t. \ \ \bar I \geq p_1y_1 + p_2y_2,$$where \bar I := I + p_1 + p_2. For \alpha + \beta = 1 the solution is well known to be$$y_1^* = \frac{\...

5

Except for the extreme cases $\alpha=0$ and $\alpha=1$, there is no boundary solution with $x_1=0$. Note that the marginal utility of good 2 is constant, while the limiting marginal utility for $x_1\to 0$ is infinite. So the consumer will always consume a bit of $x_1$.

5

It's a matter of choice how one writes the Lagrangian in the context of Lagrange/KKT. Depending on how it's written, the gradients of the objective and constraint functions are either parallel or anti-parallel at a (suitable) optimum, and the Lagrange multiplier is neither negative or positive. At the end of the day, it is the same (subset of) optima that ...

5

People have different tastes, needs, and budgets. So it doesn't make much sense to speak of the willingness to pay (WTP) of "the consumers". Each consumer has his or her own WTP for, let's say, a cup of coffee. If Alice' WTP is 1€ and the price is 2€ she will not buy. If Bob's WTP is 5€ then he will buy and enjoy the surplus. From Bob's ...

5

The first order condition for individual $a$ when the price of $y$ equals $p_y(1+t)$ is given by: $$\frac{\partial u_a}{\partial y_a} = p_y(1+t) \left(\frac{1}{p_x} \frac{\partial u_a}{\partial x_a} \right)$$ Rewriting the first order condition for the social equilibrium gives: $$\frac{\partial u_a}{\partial y_a} = p_y\left[1 + \left(\frac{p_x}{p_y} \... 5 Confirmation bias While there could be economic reasons for some of these phenomena, I think you are very likely experiencing confirmation bias. You are retelling personal stories that fit the narrative of the proper/standard ones always seem to get sold out first, and the ugly/disgusting ones are always in stock. I have many experiences where I went to ... 5 Yes if you assume that the sub-utility functions are concave. Notice that this is a standard assumption as otherwise, the utility function u = \sum_i f_i is not guaranteed to be concave (nor quasi-concave). Let denote by u_i = \dfrac{\partial u}{\partial x_i} and by u_{i,j} = \dfrac{\partial^2 u}{\partial x_i \partial x_j}. By additivity u_{i,j} = 0 ... 4 In general the demand for a certain good (say from a consumer) can be written as a function of the prices of all available goods and the total amount of money that the consumer has available. Take the setting of two goods, q_1 and q_2 with prices p_1 and p_2 and total income y. Then the demands can be written as:$$ q_1 = d_1(p_1, p_2, y)\\ q_2 = ...

4

Perhaps I misunderstand the question, because it seems trivial coming from such an established researcher. As @HerrK. points out, utility functions that represent intertemporal discounting are generally of the form $$U\left((x_i)_{i=1}^T\right) = u(x_1) + \delta_1 u(x_2) + \delta_2^2 u(x_3) + \dots$$ where $\delta_i$ is the discount factor and $x_i$ is the ...

3

You question can be answered using a revealed preference argument. Let $B = \{q \in \mathbb{R}^n_+| p' q \le m\}$ be some budget set of a consumer (i.e. $B$ gives all possible bundles that the consumer can choose). Let $q^\ast$ be the optimal choice from $B$, i.e. the bundle that optimizes the utility. Then for any other bundle $q \in B$, it must be that $u(... 3 Below is the sketch of the solution. From the Lagrangian $$\max _{\left\{c_{t}, s_{t+1}\right\}} \Pi_{t=0}^{\infty} c_{t}^{\beta^{t}} - \Pi_{t=0}^{\infty}\lambda_t(c_{t}+s_{t+1}-y_{t}-(1+r) s_{t}) \\ \text{s.t.} \ s_0 = 0, c_t > 0$$, you can get the Euler equation $$\frac{\beta_{t+1}c_{t+1}^{\beta_{t+1}-1}}{\beta_{t}c_{t}^{\beta_{t}-1}}= \frac{\beta_{t+... 3 According to advocates of MMT, the primary risk once the economy reaches full employment is inflation, which can be addressed by gathering taxes to reduce the spending capacity of the private sector. This statement is in accord with MMT, and it can be traced back to the concept of Functional Finance. One could do a search for Abba Lerner’s articles on ... 3 A function f:D\rightarrow \mathbb{R} is said to be quasiconcave if the following set is a convex set for every value of a\in\mathbb{R}: P_a = \{x\in D: f(x) \geq a\} To show that f(x,y) =\min(x, 2y) is quasiconcave, we just need to show that P_a = \{(x,y)\in \mathbb{R}^2: \min(x, 2y) \geq a\} is a convex set. For that we consider arbitrary (x', y')... 3 You can easily plot these using Desmos. On the left side, define the utility function with the equation$$ U\left(x,y\right)=\min\left(x,y\right)+\max\left(\frac{x}{2},\frac{y}{2}\right) $$Then ask for the set of points (x,y) which satisfy the equation for a utility level, e.g. 2,$$ 2 = U\left(x,y\right). $$It is important to place the function on the ... 3 Note that \exp(\ln(x)\ln(y))\ne \exp(\ln(x+y)). Instead, $$\exp(\ln(x)\ln(y))=x^{\ln y}=y^{\ln x}.$$ The MRS of the monotonically transformed utility is still \frac{y\ln y}{x\ln x}. 2 The budget set is always defined given a price vector p=(p_i)_{i\leq l} (it seems like l is the number of goods in your problem) and an income w. We usually implicitly assume that prices are strictly positive and the income is finite. Otherwise, as you correctly point out, we end up with infinite consumption being optimal and we don't even have an ... 2 An interesting issue arose while exchanging comments related to @MichaelGreineckeranswer. So let's explore this formally Claim: There is no boundary solution with x_1 = 0. Budget exhaustion holds, so with x_1 = 0 we will have U=x_2 = w/p_2. So in an attempt to disprove the claim, we examine under which conditions the following inequality will hold:$$... 2 Your assumption 3 is compatible with corner solutions of the kind$y_i=0$for some$i, and is not sufficient to avoid corner solutions for some prices small enough. With production functions it is common to assume Inada's conditions to avoid corner solutions. With cost functions, such conditions are quite naturally expressed as $$\lim_{y_i\rightarrow 0} \... 2 To solve for competitive equilibrium, we can first find the demand : Demand for commodity X by A is x_A = \frac{5}{p_x} if p_x < 1, x_A \in [0,5] if p_x = 1, x_A = 0 otherwise. Demand for commodity X by B is x_B = \frac{(30p_x+5)}{2p_x}. Now we can equate demand and supply and solve for p_x. x_A + x_B = 30 yields p_x = \frac{1}{2}. 2 there are 3 goods. If she buys more of x_1, she will buy less of the composite good k. Notice that as x_1 = a-p_1 and x_2 = a - p_2, you have that:$$ \begin{align*} k &= I - p_1 x_1 - p_2 x_2,\\ &= I - p_1 a + (p_1)^2 - p_2 a + (p_2)^2, \end{align*}$So, for example, an increase in$x_1$due to a price decrease of$p_1$will cause a ... 2 For your second example$ax + by = cx + dy$is the line where the kink will be. However, this will not be L shaped (Assuming a,b,c,d > 0) the indifference curves may have an outward or inward kink depending on the values of a,b,c,d values. Try it out in the Desmos environment Giskard linked. 2 Yes, it is. First, what you describe is not as much binary choice, but situation where you have discrete quantities where any quantity higher than 1 does not bring any benefits (a person can have 2 beds just the second bed is useless). You can calculate marginal utility for such case normally how you would do for other goods that come in discrete quantities (... 2 For part (i), in complete rigor, you should also check the determinants of all the leading principal minors of the bordered Hessian and make sure that they have alternating signs. Your final conclusion looks correct though. For part (ii), recall that Walrasian demand is the solution to utility maximization subject to budget constraint. So you should setup a ... 1 First, the fact that$MRS=\frac{1}{4}$does not tell you by itself that the consumer will only buy$x_2$. We need to go back to the 2nd Gossen's law:$\frac{Um{x_1}}{p_1}=\frac{Um{x_2}}{p_2}$(this is from where the$MRS$comes by the way) which is not the case since$\frac{1}{3}<\frac{4}{8}=\frac{1}{2}\$. So you are right the consumer in the first place ...

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