7

For the 2x2 case being considered, write $$\mathbf{B}=\left[\begin{array}{cc} b_{1,1} & b_{1,2}\\ b_{2,1} & b_{2,2} \end{array}\right].\quad$$ It follows that the element (1,1) in $B^{-1}$ is given by $\frac{b_{2,2}}{b_{1,1}b_{2,2}-b_{1,2}b_{2,1}}$. Notice that $$\frac{\partial q_1(p_1,p_2)}{\partial p_1}=(\frac{\partial p_1(q_1,q_2)}{\partial q_1 }...


5

Let matrix $A = \begin{bmatrix} p_1 & p_2 & \ldots & p_n \end{bmatrix}$. Let $\mathbf{x}^*$ be a fixed solution to $A \mathbf{x} = c$. Then for any vector $\mathbf{u}$ that belongs to the null space of $A$, we have $A \mathbf{u} = 0$ hence $\mathbf{x} = \mathbf{x}^* + \mathbf{u}$ is also a solution (furthermore, all solutions $\mathbf{x}$ can be ...


3

This concerns the partial derivatives of the utility function with respect to goods, and not the partial derivative of the Lagrangian of the maximization problem. So a zero derivative, and moreover at the optimum, would imply a threshold quantity after which utility diminishes. In the real world, we all know that consuming excessively may result in ...


2

I am not sure exactly what you divided by what, but suppose your input-output table looks like this: $$ A = \left[ \begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right] $$ If you proceed to divide the first column by $\sum\limits_i a_{i,1}$ the second by $\sum\limits_i ...


2

Let the $n$ states of the finite-state markov chain be denoted by $\{x_1,...,x_n\}$ and let $\vec e = [e(x_1), ..., e(x_n)]'$. Now, first note that because $X_{t+1} \mid X_t$ is independent of $W_{t+1}$, we can write \begin{align*} \exp(\eta) e(x) &= E[\exp(D'x + x' F W_{t+1})] E[ e(X_{t+1}) \mid X_t = x] \\ &= \exp(D'x + x' F F' x)\, E[ e(X_{t+1}) ...


2

This kind of projections in econometrics are usually employed for partialling out some covariates from a linear regression. Observe that in general $P_X\neq I$. Consider $X=\begin{bmatrix}1&5\\1&0\\1&1\end{bmatrix}$. Then $X'X=\begin{bmatrix}3&6\\6&26\end{bmatrix}$ and $(X'X)^{-1}=\begin{bmatrix}26/42&-6/42\\-6/42&3/42\end{...


2

First, note that the growth rate of $\mu$ is defined as $\dot\mu = \frac{ d\mu }{ \mu }$. Therefore, you will have to take the total differential of the equation; and divide by $\mu$. For the first step, as in your previous question, simply use the rules of total differentials, specifically the product rule that states that for variables $X$ and $Y$, the ...


2

For a variable $X$, let $dX$ denote its total differential. Let $k$ be a constant, and $X$ and $Y$ variables. You'll need the following rules: $$dk = 0$$ (constant rule), $$d(X + Y) = dX + dY$$ (sum rule), $$d(XY) = Y \cdot dX + X \cdot dY$$ (product rule) and $$d\left(\frac{X}{Y}\right) = \frac{ Y \cdot dX - X \cdot dY }{ Y^2 }$$ (quotient rule). ...


1

First of all, I think that 'linear and homogeneous' is a typo of 'linearly homogeneous.' Indeed, it can be shown that if the production function $V$ is linearly homogeneous, Allen Elasticity of Substitution between $K$ and $L$ can be expressed as $$ \sigma = \frac{V_K V_L}{V \cdot V_{KL}} $$ by using the fact that $V_K$ and $V_L$ are homogeneous of degree 0, ...


1

I don't really understand your optimization problem as it stands. Just as a few examples, when you write out $$\min \sum_m \sum_v \left( \frac{\partial \vec f_v (\vec Q_v)}{\partial \vec f_m (\vec Q_m)} - \frac{\partial \vec f_m (\vec Q_m)}{\partial \vec f_m (\vec Q_m)} \right)$$ I have questions about why you are minimizing the difference between optimal ...


1

I got some outside help for the ending of the proof I was attempting. I'll leave this question if by chance someone else finds it useful. So if we want to show $p_1x_1 + \cdots + p_{n−1}x_{n−1} = 0 \implies p_i = 0 \quad \forall i$, then assume without loss of generality that $p_1 \neq 0$. We have $$x_1 = \left(-\frac{p_2}{p_1}\right) x_2 + \cdots + \left(-...


1

All right kids. You gave me the answer when Dismalscience wrote "this is still not a problem as long as the value of the good produced differs from the sum of the value of its inputs" ... actually, one of the branches had a complete set of zeros in the input table exept for one input and a zero in terms of production ... and another one had a complete set ...


1

I didn't divided each column by the sum of the column. This would make no sence since the goal is to produce a technical coefficient matrix that links each input (row) needed by the industry (column) in order to produce their output. I divided each element of the input-output table by the total output of the branch, according to the method presented in ...


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