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An economics question? It might belong in a different exchange, but i find myself partial to this one... we have the demand function for good x: QxD = 5 - 2Px + .03I - .4P + 2P and we want to find the own price elasticity of demand, ExD, where Px = 2 and QxD = 1. ExD = 2QxD / 2Px = 2 * 2/1 = -4 The"2" in the elasticity in formula (2$Q_x^D$) is really the ...

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Probably because: The utility of the lowest type is being normalized to zero $u(t,\underline v)=0$ ? $G(\overline v |t)=1 \,\forall t$ ? $G(\underline v |t)=0\, \forall t$ ? But it is hard to say because the question did not define what $G$ is (I am assuming is a CDF with support on $[\underline v, \overline v]$).

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$M=M(p_1,p_2,\overline U)$: The minimum income needed to buy the quantities of commodity 1 and commodity 2 that gives utility $\overline U$. $M=p_1 \cdot x_1 + p_1 \cdot x_2 \Rightarrow p_1=\frac{M(p_1,p_2 , \overline U)-p_2 \cdot x_2}{x_1}$ $\frac{\partial p_1}{\partial p_2}=\frac{M'-x_2}{x_1}\Rightarrow p_1'\cdot x_1=M'-x_2$ Thus $p_1'\cdot ... 2 If "breaking even" means you spend as much as you receive, then you need to set those quantities equal. There are (at least) two ways you could measure this: cost per item = price per item total cost = total sales In your problem,$p(x)$is the price per item, but$C(x)$is the total cost, so you're comparing apples to oranges: note that looking ... 2 The point is that the investor does not know the value of$x$when buying the contract, so$x$cannot be part of the decision criterium. If the investor is lucky, the contract will be repaid and the investor will make a profit of$1-P$. If he is unlucky, the homeowner will go bankrupt and the investor will lose$P$. The expected utility of buying the ... 1 The way in which you compute the marginal cost of the next unit is based on the idea of first derivative of the cost function. When you calculate the marginal cost (the cost of an additional unit) you calculate how the cost function varies if you add another unit of input. But the point is that you are calculating the variation of your cost function using ... 1 The result that $$V\big(\hat{\beta} \big) = V\big(\hat{\beta} \mid X \big),\;\;\text{??}$$ (because the variance of$\beta$is zero,$\beta\$ being a vector of constants), would hold only if the regressor matrix was considered deterministic -but in which case, conditioning on a deterministic matrix is essentially meaningless, or at least, useless. The ...

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When a derivative is written, it represents an instantaneous rate of change of some dependent variable with respect to another (independent) variable. In your equation, it is not clear what the variable of differentiation might be; neither is it clear how the other quantities might depend on it. I suppose such a lack of notational precision and ...

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