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5

Does quasi-concave utility function imply convex indifference curve? No that is not true. Consider $u(x, y) = -x^2 - y^2$ defined on $\mathbb{R}^2_+$. Since $u$ is concave it is quasiconcave. Observing the graph of the indifference curves, we see that ICs of $u$ are not "convex".


4

To provide some real world(ish) interpretation, you could consider the following: Wallace enjoys eating cheese on its own. He doesn't much care for crackers on their own, but he especially loves eating crackers and cheese together, he makes nice little cracker n cheese sandwiches. In this example, we can think of cheese (x) and crackers (y) as perfect ...


4

I assume you know how does $\min\{x,y\}$ look like? In order to draw utility function of interest, you need to consider cases: $u(x,y)=x+\min\{x,y\}=\begin{cases}2x, \;\; \mathrm{for} \;\; x \leq y \\ x+y, \;\; \mathrm{for} \;\; x > y\end{cases}$ With $x$ on horizontal and $y$ on vertical axis: Not sure about the "usual" perfect complements. It is more ...


4

You seem to be confusing some things. The indifference curve and utility function both have the same equation The utility function has a formula, not an equation. $U(x_1,x_2)$ is a utility function. The points $(x_1,x_2)$ for which $U(x_1,x_2) = 13$ form an indifference curve. (13 was picked as a random number) The graph of a utility function has ...


4

One way to think about this utility function is "I can only use one of these two goods, and I will use the one that I have more of" Some examples: Choosing between two sets of equally-valued but non-matching dishware. Once you have one set, the other one is useless to you. If your budget constraint is such that you can only have one kind of platform of ...


3

Assuming certain regularity conditions, the first order conditions for $$ \max_{x, \lambda} U(x) - \lambda (p \cdot x - m) $$ are \begin{align*} &D_{x}U(x(p, m)) - \lambda p = 0 \\ \text{and} \quad & p \cdot x(p, m) - m = 0. \end{align*} Moreover $x(p, m)$ will be differentiable with respect to $m$ at $(p, m)$, and this fact together with the ...


3

If you have quadratic preferences then your utility function is: $$ U(W) = W - \lambda W^2$$ this implies your expected utility function looks like: $$ E[U(W)] = E[W - \lambda W^2] = E[W] - \lambda E[W^2]$$ $$ = E[W] - \lambda E[W^2 - E[W]^2 + E[W]^2]$$ $$ = E[W] - \lambda E[W^2 - E[W]^2] + \lambda E[E[W]^2]]$$ $$ = \mu_w - \lambda \sigma_w^2 + \lambda ...


3

What you're asking for is equivalent to finding an injective function $f:\mathbb R^n\to\mathbb R$ that is monotone in the sense that if $x$ is coordinate-wise at most as big as $y$, then $f(x)\le f(y)$. As a first step, notice that this is equivalent to finding such a function from $(0,1)^n\to\mathbb (0,1)$. And this is easy by interleaving the digits of the ...


2

It is possible to use a Lagrangian to obtain your Marshallian demands, provided you break each min function up into two different pieces and remain wary of boundary solutions. So for example, if you impose the condition $x_1<\frac{a}{2}, x_2<\frac{b}{2}$ you can simplify your utility function to $x_1 x_2 + x_3$, and then solve for Marshallian demands ...


2

This seems like a homework question, so I'll just give hints. By definition, a quasilinear utility has the form $u(x, y) = x + v(y)$ where $y$ is a vector of all other goods and $v(\cdot)$ is strictly concave. In this case, $x$ is called the numeraire. From utility maximization, what's the first order condition that relates $MU_x$, $MU_y$, $p_x$, and $p_y$?...


2

First, assume risk aversion. By the definition of the certainty equivalent and Jensen's: $$u(CE(u,F))=E(u(x))<u(E(x))$$ Now, from monotonicity: $$CE<E(x)$$ Second, assume $CE<E(x)$. By monotonicity and the definition of $CE$: $$u(E(x))>u(CE)=E(u(x))$$


2

Assumptions $1$ to $3$ are sufficient to obtain a linear representation when $X$ is open and convex. We proceed in two steps. Step $1$: We will repeatedely use the following consequence of continuity and $A1$: If $x \sim x^{\prime}$, then $x \sim x + \lambda (x^{\prime} - x)$ for every $\lambda \in \mathbb{R}$ such that $x + \lambda (x^{\prime} - x) \in X$....


1

Take $(1,1)$ and $(-1,1)$: we have that $U(1,1)=U(-1,1)=1$. However, $U(\frac12(1,1) + \frac12(-1,1)) = U(0,1) - 0 < 1 = \min\{U(1,1),U(-1,1)\}$. Hence the function, at least defined globally over $\mathbb R^2$ is not quasi-concave.


1

Both are goods, so each provides utility if consumed alone. The utility function $$u(x_1, x_2) = \max\{x_1, x_2\}$$ or more generally $$u(x_1, x_2) = \max\{f(x_1), g(x_2)\}$$ for some functions $f,g$, may describe either a) a situation where it is impossible to consume the two goods together. For example, concerts happening at the same time. Note that ...


1

Utility functions and indifference curves are the same object considered from different visual/conceptual angles. Below an illustration on $U(x_1, x_2) = x_1^{0.5} x_2^{0.5}$ (representing Cobb-Douglas preferences), done with excel:


1

In the original problem, for $L=2$, the consumption set was $(-\infty,\infty) \times \mathbb{R}_{+}$. Now, the consumer is restricted to $[0,\infty)\times \mathbb{R}_{+} = \mathbb{R}_{+}^2$. Fix a price $p = (1,p_2)$, and now see what happens to Walrasian demand (i.e. $x(p,w) = (x_1,x_2))$ as we vary $w$. You saw from part (a) that a consumer with these ...


1

It depends what you know about your utility function and the "space" of goods $x_1$ and $x_2$. You ask in your question what the change of optimal choice is with respect to the parameter information. This analysis is called comparative statics, and you can use a number of mathematical tools to achieve this. The most common are The Implicit Function Theorem ...


1

One reason I could think of is regarding convexity of the function (and hence risk aversion). For example, $u(x) = x$ is risk neutral, but $v(x) = \ln(u(x)) = \ln(x)$ is risk averse. See also this.


1

This is an edit By definition,for a convex subset $C\subseteq \mathbb{R}^n$ a convex function $f:C \to \mathbb{R}$ satisfy $$ f((1-t)x+ty)\leq (1-t)f(x)+tf(y)$$ for every $x,y\in C$ and $t \in (0,1)$. We can see if we let $x=(1,3)$ and $y=(3,1)$ and $t=\frac{1}{2}$ that $$ u((1-t)x+ty)=u(2,2)=6 \not \leq 5=\frac{u(1,3)}{2}+\frac{u(3,1)}{2}=(1-t)u(x)+tu(...


1

Consider lotteries over $\{x,y,z\}$. Let $u(x)=0, u(y)=\frac{1}{2}, u(z)=1$. Consider the nonlinear transformation f(t)=t^2. Let $v:=f\circ u$, so $v(x)=0,v(y)=\frac{1}{4}, v(z)=1$. Consider two lotteries, $P=(0,1,0)$ and $Q=(\frac{1}{2},0,\frac{1}{2})$. $$E_P[u]=\frac{1}{2}=E_Q[u]$$ $$E_P[v]=\frac{1}{4}<\frac{1}{2}=E_Q[v]$$ In general, let $\succeq$...


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For case 1, you can argue $z$ will be unique by contradiction: Suppose ad absurdum there is another $z'$ that is feasible (i.e. $p^Tz' =w$) optimal and $z'\neq z$. Then you can consider a convex combination of $z$ and $z'$: $\bar z = \beta z + (1-\beta) z'$, for $\beta\in(0,1)$. Notice that $\bar z$ is still feasible (because it is a combination of two ...


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