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The general definition of vectors do not require that a vector has a defined magnitude. It just so happens that in many of the common vector spaces that are useful for physics and sciences that there is a natural choice for a norm, making these spaces normed spaces (which are a special kind of vector space). In general a vector is merely an element of a ...


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Google Books and Scholar attest that this is sometimes called the dimensionless logarithmic derivative of $f$, as it is a dimensionless version of the better-known logarithmic derivative.


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That is not a very general definition of a vector, no. But it is hardly "wrong" for that reason. Direction and magnitude are intimately tied to spatial intuition. The magnitude referred to is usually a "length" measured by a real number, but there are vector spaces where that idea makes no sense. The same goes for "direction" too. You can, at best, only ...


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The profit function, $\pi (p,\mathbf{w})$, is convex in prices $\left ( p,\mathbf{w} \right )$: $$\pi (tp+(1-t)p',t\mathbf{w}+(1-t)\mathbf{w'})\leq t\pi (p,\mathbf{w}) + (1-t)\pi (p',\boldsymbol{w'}),$$ where $p$ is the price of a single output, $\mathbf{w}$ is a vector of price of inputs and $t \in [0,1]$. See Hal Varian, Microeconomic Analysis, 3rd ...


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Whenever you see $$\bar{X}\mid \mathbf{x}$$ or some other form of it, it means to fix the value of $\mathbf{x}$. So, in your case, we seek $$\text{Var}\left[\bar{X}\mid X_1, \cdots, X_n\right]\text{.}$$ But here's the thing. By definition, $$\bar{X} = \dfrac{1}{n}\sum\limits_{i=1}^{n}X_i$$ so thus $\bar{X}$ is constant because $X_1, \cdots, X_n$ are fixed ...


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When you know all your $X_i$ then you know your $\bar{X}$, hence $E(\bar{X}|X_1,X_2,...,X_n)=\bar{X}$ as your sample mean is no longer a random variable but a constant now. Thus, $Var(\bar{X}|X_1,X_2,...,X_n)=E(\bar{X}^2|X_1,...,X_n)-(E(\bar{X}|X_1,...,X_n))^2=\bar{X}^2-\bar{X}^2=0$


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According to the Law of Demand, if the price of a commodity goes up then the quantity demanded will go down.See Law of Demand Definition here. If the two goods are substitutes then an increase in $p_{1}$ will decrease $D_{1}$, so $$\frac{\partial D_{1}}{\partial p_{1}}<0.$$ On the other hand, an increase in $p_{1}$ will increase $D_{2}$, so ...


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What do you mean with "utility is negatively correlated with wealth"? The utility function $$1- 1/w$$ is increasing in w. See http://www.wolframalpha.com/input/?i=Plot[1-1%2Fx%2C{x%2C0%2C100}] However, it is not defined at $w=0$. So you need some initial wealth You can calculate $$\frac{1}{2} u(w+0) + \frac{1}{2} u(w+100) = u (w+x)$$ Then, $x$ is the ...


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Hint: First, start off with the constrained maximization problem and then use the Lagrangian method to get the first-order conditions. Then use the Kuhn-Tucker method (see Mathematics for Economist by Simon and Blume) to solve for your optimal demands. The consumer solves the following maximization problem: \begin{align*} \max_{q_1,q_2} \alpha_1q_1 + ...



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