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Some confusion and incorrect statements in the answers already given, including the "accepted" answer. (E.g. obvious distinctions between necessity and sufficiency of different conditions for KKT are mixed up.) There are different formulations of KKT theorem. The KKT Theorems one might consider for your example are the following. For notational simplicity,...


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This idea is known as the Fisher separation theorem. Without the investment opportunity to transfer $h$ units of present day value into $w(h)$ units of future value, the perfect credit market gives us the intertemporal budget constraint of $$ c_1 + \frac{c_2}{1+r} = y, $$ which can be represented by a straight line. Without knowledge of the consumer's ...


3

Here is the formulation of the problem : \begin{eqnarray*} \max_{c, h, l} \ & \ln (c - \gamma) + \beta l + \theta h \\ \text{s.t.} & l + h = 1, \\ & c \leq \omega h + \rho, \\ \text{and} & l, h \geq 0, c \geq \gamma \end{eqnarray*} Substituting $l = 1 - h$, we can rewrite the above problem as : \begin{eqnarray*} \max_{c, h} \ & \ln (c ...


3

I assume the notation $\mathbb R^2_+$ refers to $[0,\infty)^2$. Note that the set on which a function is defined need not be the same set on which a function is differentiable. In particular, it's typical that differentiability requirements are imposed on open sets (see e.g. the fundamental theorem of calculus). This is because defining differentiability on ...


2

Objective function of Lagrangian can be set up either with $+\lambda$ or $-\lambda$, depending on how you solve the budget constraint. Actually, for the solution it does not matter if $\lambda$ has negative or positive sign in the equation. You can clearly see it from the formula if you expand the second term: $$ \mathcal{L}(x,y, \lambda) = U(x,y) + \lambda(...


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After substituting for $c_{1t}$ and $c_{2t}$, the problem can be rewritten as : \begin{eqnarray*} \max_{q_t^e} \ & \frac{(1- \pi) (n\omega_t)^{-\gamma} }{-\gamma} \left((1-q_t^e) + z_tq_t^e\right)^{-\gamma} + \frac{\pi (AR\theta\omega_t)^{-\gamma } }{-\gamma z_t^{ -\gamma }} \left((1-q_t^e) + z_tq_t^e\right)^{-\gamma} \\ \text{s.t. } & q_t^e \in [0,...


2

Your value function is as follows: $$ V_t[w] = \max_{c_t \in[0,w]} \left\{u(c_t) + \frac{1}{2}V_{t+1}[\alpha(w_t - c_t)] + \frac{1}{2}V_{t+1}[\beta(w_t-c_t)] \right\} $$ with the terminal condition $$ V_{T}[w_T] = \max_{c_T \in [0,w_T]} u(c_T) $$ So, we can solve this via backward induction. Clearly, at the final period $T$, since $u$ is monotonic, we ...


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The problem is that you are ignoring the division $ \frac {0} {0} $, which is in $ \frac {\partial L} {\partial v} $. Before looking at the solution, and seeing that indeed when $ \theta_ {2} = 0 \ \Rightarrow \ v ^ {*} = 0 $, I want to note that the first constraint $ c_ {1} = w - (1 -v) \theta_ {1} h_ {1} ^ {a1} $ can be more realistic and logical. It is ...


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Take $\Omega_1=\Omega_2=[0,0.5]$. Let $p(x)=x^4$, so that $p''(x)=12x^2>0$ on $(0,0.5)$. Let $q(x)=0.5x$, which is linear in $x$, and $K=0.2>0$. For $\bar\gamma=1$, both objective functions attain their respective maximum at $x=0.5$. As the following figure shows, objective function $(1)$ (blue curve) has a higher maximum than that of objective ...


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Argue that, given your assumptions on the utility function, $x^*$ is the essentially unique (and hence global) maximum. (You need this because there may be local maxima when the assumptions on the utility function is relaxed - this will violate the proposition you're trying to prove). Now simply use the definition of global optima: for any $x\leq x^*$, $u(x)...


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I would leave this as a comment but I cant. You are on the right track. Once you know $V_2(k)$ then you can plug that into to the first hjb and solve. To solve for $V_2$ you need to find the optimal $i$ as a function of $k$. Then plug $i(k)$ into the 2nd HJB. That will give you a second order ode. Solving that will give you $V_2(k)$ and you go to 1.


1

The equation $100\sqrt{x_1} = 8x_1^{3/2} - 5$ is cubic in $\sqrt{x_1}$. Unless you have been asked for an analytic solution (which is possible but complex), the approximate method I suggest is first to ignore the minus $5$, enabling division by $\sqrt{x_1}$ yielding: $$100 = 8x_1$$ The first approximation is then $x_1 = 100/8 = 12.5$. Since $100\sqrt{12.5} \...


1

I assume $x_i$ represents the quantity and belongs to $\mathbb R_{+}$. You can form the constraints as follows: $$ x_i \geq0 \quad\forall i \in [3] \\ \sum_{i=1}^3p_ix_i \leq I $$ You can simplify the objective by noting that for the utility to be the maximum, $x_2 =x_3$. Try to reason why this is true. Hence, the final problem becomes, $$\ \max_{x_1, x_2, ...


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These are many questions. O.k., so let's go step by step: (Q1) What is a mapping actually? A map is just another term for a function. Here, every "law of motion", the actual one (ALM) and the perceived one (PLM), is characterized by its parameters $a$ and $b$. The ALM depends on the PLM, and the function mapping the PLM-parameters to the ALM-parameters is ...


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Right now, the issue is not so much a dollar cost, rather the physical feasibility of either testing, or the availability of N95 masks. 1) To run a testing programme, a jurisdiction needs the testing capacity, as well as the organisational capacity to administer tests. In many jurisdictions, neither capacity exists at the time of writing (mid-April), ...


1

Thunstrom et al. at Wyoming have a working paper on this issue although they look at social distancing rather than masks or testing. The basic reasoning is the same though: look at costs of measures for the economy and benefits in terms of lives saved (using the value of a statistical life) and calculate the net benefits. The working paper is here. From the ...


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Your claim that "MRS can only be calculated on for a point on the indifference curve and not for a point on the budget line" reveals a misunderstanding. Note that every point lies on some indifference curve. So each point on the budget line also lies on some indifference curve. We also know that all indifference curves are convex, since the utility function ...


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The corner solution is not $c=a$ it cannot be because the marginal utility of even a tiny bit of consumption is unbounded there. However, you can have a corner solution where $h=0$. Since the agent has non-labor income $p$, the budget line has a kink. That is, if the agent receives a lot of income even without working, they may choose not to work, and fully ...


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You need to differentiate with respect to Y, as you said! You substitute the phillips curve as you did (because it is a constraint for the central bank's optimization problem), then differentiate w.r.t $\bar{Y}$ (which I assume here stands for gap) and set this to 0 to obtain the optimality condition, which in turn will give you the monetary rule. Recall ...


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Your production function is basically a Cobb-Douglas function of the form $y=A(t_m-a)^\alpha(t_l-b)^\beta$. Therefore, the parameters $\alpha$ and $\beta$ measure the intensity with which inputs are needed for production. The smaller the value of such parameters, the smaller the marginal productivity of the inputs is. In your context, the company is more ...


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To solve this with the exact constraint given in the question, one could set up the following Lagrangian: $\mathcal{L}(c_{1},\lambda(\cdot))=u\left(c_{1}\right)+\int u\left(c_{2}\left(z\right)\right)g(z)\mathrm{d}z-\int\lambda(z)\left\{ f\left(c_{1},z\right)-c_{2}\left(z\right)\right\} \mathrm{d}z,$ where $g(z)$ is the probability density function for the ...


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