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You need to start by knowing the following result: If $X_1$ and $X_2$ are orthogonal (i.e., $X_1'X_2 = 0$), then we could get $\beta_1$ by regressing $y$ on $X_1$ alone and $\beta_2$ by regressing $y$ on $X_2$ alone. Please try to figure out why this is so. Now, in the general case, rewrite the regression equation as $$y = M_{X_2}X_1\beta_1 + X_2\gamma + ... 2 The first difference of your covariate will not be exogenous in the first-differenced model if, e.g., the original model has a lagged dependent variable as a regressor. Consider e.g.$$y_{it}=\gamma y_{i,t-1}+x_{it}'\beta+\mu_i+v_{it}, \quad t=1, \ldots, T,$$where v_{it} is independent of \mu_i and any x_{is}. Also assume that v_{it} is ... 2 To show that the relation is complete, let x, y \in X. Because \succeq_B is complete, you know that either x \succeq_B y or y \succeq_B x. If the latter is true, we can swap the labels on x and y. So we can assume without loss of generality that x \succeq_B y. There are still two cases: Either x \succ_B y or x \sim_B y. If the former is ... 2 The answer is basically given in the comments. I write them up just to give the question some closure. It makes only sense to speak about convexity / concavity of a real function f if the domain is an interval. So the question is, is there a system \mathscr I of intervals such that f restricted on each interval in \mathscr I is either convex or ... 1 A clever way in such exercises (optimization with Cobb-Douglas function) is to divide one equation by another. But first you have to put w_1 and w_2 on the RHS.$$p \alpha x_{1}^{\alpha-1}x_{2}^{\beta}=w_{1}\quad (1)p \beta x_{1}^{\alpha}x_{2}^{\beta-1}=w_{2}\quad (2)$$Dividing first equation by the second equation gives ... 1 In general you are right, but your (partial) derivatives seem not right. MRS =\frac{U_F(F,G)}{U_G(F,G)} U_F(F,G)=G^{1/2} U_G(F,G)=\frac12F\cdot G^{-1/2} \frac{U_F(F,G)}{U_G(F,G)}=\frac{G^{1/2}}{\frac12F\cdot G^{-1/2}}=2\frac{G}{F} Is it comprehensible ? If not, feel free to ask. 1 You simply take the minimum of the two values. 1 The case when c_t=0 for all t gives k_t={k_0}^{\alpha^t}. For k_0<1, the sequence is increasing but is bounded by 1. Otherwise the sequence is decreasing and is bounded by k_0. The case when when c_t \geq 0 for all t can run into trouble if k_t ever becomes negative since powers can be complex. Assuming k_t \geq 0, the sequence can be ... 1 There is a typo. It has to be 25,000\cdot 0.\color{red}15. Therefore Si=3750. It follows that A=20,000\cdot 0.29832+3750\approx 9716.4 This is the right result. In general an equation can be set up. The future value of the costs has to be equal to the future value of the savings plus the selling price of the machine. C_0: Present value of the ... 1 Hint: The equation is not the result of implicit differentiation. It is the result of total differentiation:$$dQ(K,L)=\frac{\partial Q}{\partial L}\cdot dL+\frac{\partial Q}{\partial K}\cdot dK=0\frac{\partial Q}{\partial L}\cdot dL=-\frac{\partial Q}{\partial K}\cdot dK Dividing the equation by $\frac{\partial Q}{\partial L}$ and $dK$ gives the ...