5

Given that the utility maximization problem is: \begin{eqnarray*} \max_{\{y_j \geq 0 : 1\leq j\leq J\}} && \sum_{j=1}^{J} u(y_j) \\ \text{s.t.} && \sum_{j=1}^{J} p_jy_j \leq m\end{eqnarray*} where $p_j > 0$ for every $j$, and $m > 0$. If $u$ is convex and strictly increasing, then in optimum consumer will always spend all his money on ...


3

We have two Consumers 1 and 2, and two goods 1 and 2 pure exchange economy. The following utility functions can be used to represent their preferences: $u_1(x_{11}, x_{12}) = x_{11}$ $u_2(x_{21}, x_{22}) = x_{22}$ Equilibrium price vector $(p_1, p_2=1)$ and allocation $((x_{11}, x_{12}), (x_{21}, x_{22}))$ satisfy the following: Optimality Conditions (...


3

Definition 1. Excess supply is supply minus demand. Definition 2. Excess demand is demand minus supply. Example 1. A baker posts a sale price of \$2 per loaf of bread. At this price, he is willing to sell up to 300 loaves of bread (per day), but consumers are willing to buy only 200. We say that quantity supplied and quantity demanded (at the price of \$...


2

In standard proofs of the existence of Walrasian equilibrium, after one applies the appropriate fixed point theorem (typically Kakutani's, but sometimes Brouwer's), one then shows that the given fixed point is, in fact, a Walrasian equilibrium. Depending on the exact construction of the proof, sometimes the fixed point will only give you the equilibrium ...


2

So WARP says that if bundle x' you choose under budget B' is affordable at budget B'' (and you don't choose x' - i.e x'$\not=$x''), then x'' cannot be affordable under budget B'. The picture you showed gives the 2 bundles that the person chooses. Those are given in the picture. As you can see, x' is not affordable at budget B'' and bundle x'' is not ...


2

Since demand equals supply holds for every price $p$, this simply means that every $p$ is an equilibrium price. However, the equilibrium allocation that $p$ supports varies with $p$. To be precise, price $p$ supports the allocation in which 1 consumes $\left(\frac{1+3p}{1+p}, \frac{1+3p}{1+p}\right)$ and 2 consumes $\left(\frac{3+p}{1+p}, \frac{3+p}{1+p}\...


2

An indirect proof. Suppose $$ x(p,w) = w\cdot x(p,1) $$ does not hold. This is equivalent with stating $$ U(x(p,w)) \neq U(w\cdot x(p,1)). $$ (To be precise: $x(p,w)$ and $x(p,1)$ may be set valued. In this case we are talking about two elements at least one of which is not included in both sets.) Case 1. $$ U(x(p,w)) > U(w\cdot x(p,1)) $$ As $U$ is ...


2

I am not sure whether the question asks about the function being convex, or simply the demand correspondence/function being a convex set. The latter is often asked and if that is the case here is an answer. This is a general answer to your question which may be more useful to you and future readers than a specific one. The following theorem holds in ...


2

It is almost true. There are examples of demand that have a negative definite Slutsky matrix but fails the Weak Axiom. However, if we ask that $$v \cdot S(p,w) v <0 $$ whenever $v \not = \alpha p$ for any scalar $\alpha$ (i.e. $S$ is negative definite for all vectors except those proportional to price), then the Weak Axiom holds.


2

Coalition formation is not the same as majority voting. Say, in a single good economy, individual 1 has the entire endowment, so $\mathbf e=(1,0,0,0,0,0)$. Then individual 1 forms a blocking coalition to any allocation $\mathbf x=(x_1,\dots,x_6)$ where $x_1<1$.


1

The second problem is probably for comparison - it is the problem of a benevolent social planner who can freely allocate resources between the two agents.


1

Under what constraints is $u_1+u_2$ maximized? The exercise may be about the First Welfare Theorem: a competitive equilibrium is Pareto efficient. In this case, maximizing the sum may be designed to highlight that the equilibrium you found also maximizes the sum of utilities. In general, the maximum of a sum of functions need not equal the maximum of any ...


1

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 ...


1

for this problem you must conciser two possible branches of the utility function: $$u(\text{x})=x_1+x_3\ \ \text{if} \ \ x_1<x_2$$ $$u(\text{x})=x_2+x_3\ \ \text{if} \ \ x_1>x_2$$ The demands of these then proceed how you would for any case of perfect substitutes. However you must list them for each case. Hope this helps


1

Hint: Imagine that there are two coffee bars, $A$ and $B$. There is only one type of coffee in the world. My preferences are such that I always want 1 unit of sugar with 1 unit of coffee; if I consume units of coffee and sugar in the ratio $1:1$, additional units of only one of the two don't give me extra utility. At coffee bar $A$ they sell coffee and ...


1

Walras' law describes market equilibria conditions, which states roughly speaking that if there exists within an exchange economy market equilibria for $n-1$ good markets, then the last good market $n$ is also in equilibrium. Thus, we have a market system with $n$ equations and $n-1$ independent variables, where the last market is linear dependent from the ...


1

You seem to be looking for the Second Welfare Theorem. The Second Theorem states that out of all possible Pareto optimal outcomes one can achieve any particular one by enacting a lump-sum wealth redistribution and then letting the market take over.


1

Excess demand is the function describing the amount of quantity demanded above quantity supplied at each price level. Mathematically, it is allowed to be negative (so then we have excess quantity supplied actually). Remember that the Walrasian construct imagines an "auction" situation when transactions do not take place as long as the market doesn't clear....


1

If the prices $p_1$ and $p_2$ are positive than as you pointed out the equations \begin{eqnarray*} \frac{30p_1}{p_1+p_2} + \frac{20p_2}{p_1+4p_2} & = & 30 \\ \\ \frac{30p_1}{p_1+p_2} + \frac{4 \cdot 20p_2}{p_1+4p_2} & = & 20. \end{eqnarray*} hold. This is troublesome, because subtracting the first equation from the second yields \...


Only top voted, non community-wiki answers of a minimum length are eligible