TA session 701

• Methods to determine quasi-concavity

1. By defination, all the upper contour sets are convex. That is if $f(x)\ge a,f(y)\ge a,\Rightarrow f(tx+(1-t)y)\ge a$$f(x)\geq a, f(y)\geq a, \Rightarrow f(tx+(1-t)y)\geq a$

2. By defination

   for $\mathrm{\forall }\lambda \in [0,1],f(\lambda x+(1-\lambda )y)\ge min\{f(x),f(y)\}$$\forall \lambda\in \left[0,1\right], f(\lambda x+(1-\lambda)y)\geq \min \{f(x),f(y)\}$,

3. Check borded hessian metrix

   If the largest leading principal minors have the same sign as $(-1{)}^n$$(-1)^n$, and the last $n-m$$n-m$ leading principal minor alternate in sign.

4. Monotonic transformation of a concave function is quasi-concave.

5. For utility function with 2 goods,

   $U(x_1,x_2),U_1,U_2\ge 0$$U(x_1,x_2), U_1,U_2 \geq0$

   Then the absolute of diminishing marginal rate of substitution $\iff$$\Leftrightarrow$ quasi-concave.

   Note that the boardered hessian matrix is

   $$\begin{array}{}\text{(1)}& \begin{array}{c}H=\left[\begin{array}{ccc}0& U_1& U_2\\ U_1& U_{11}& U_{12}\\ U_2& U_{21}& U_{22}\end{array}\right]\end{array}\end{array}$$

   The the determinant of $H$$H$ is then

   $$\begin{array}{}\text{(2)}& det(H)=2U_1{U}_2{U}_{12}-U_2^2{U}_{11}-U_1^2{U}_{22}\end{array}$$

   As long as this determinant is positive, we can claim that the utility function is quasi-concave.

   $$\begin{array}{}\text{(3)}& \frac{{\partial }^2{x}_2}{\partial x_1^2}=\frac{\partial \frac{\partial x_2}{\partial x_1}}{\partial x_1}=\frac{2U_1{U}_2{U}_{12}-U_2^2{U}_{11}-U_1^2{U}_{22}}{U_2^3}>0\end{array}$$

   which indicates that the utility is quasi-concave.

   Since we have

   $$\begin{array}{}\text{(4)}& U(x_1,x_2)=C\end{array}$$

   , which means that

   $$\begin{array}{}\text{(5)}& U_{1}d{x}_1+U_{2}d{x}_2=0\end{array}$$

   Then,

   $$\begin{array}{}\text{(6)}& \frac{d{x}_2}{d{x}_1}=-\frac{U_1}{U_2}\end{array}$$

   Therefore, we only need to check the decreasing MRS to check for 2 commodities-utility function's concavity.

    

   Note that the absolute value of marginal rate of substitution $MR{S}_{12}=-\frac{U_1}{U_2}$$MRS_{12} = - \frac{U_1}{U_2}$, which is to check $\frac{\partial MR{S}_{12}}{\partial x_1}>0$$\frac{\partial MRS_{12}}{\partial x_1}>0$

• Duality between $u(q)$$u(q)$ and $v(p)$$v(p)$ - An example

  Show that Leontief preference, which implies L-shaped indifference curves, in the quantative space are parallel straight indifference curves in the price space.

  Leonitief preference:

  $$\begin{array}{}\text{(7)}& U(x_1,x_2)=min\{a{x}_1,b{x}_2\},a,b>0\end{array}$$

  Indifference curve in price space:

  We have to derive the indirect utility function, and let income level $M$$M$ to be a constant.

  $$\begin{array}{}\text{(8)}& V(p_1,p_2,M)=C\end{array}$$

  By observing from the graph, we are able to see that utility is maximized when

  $$\begin{array}{}\text{(9)}& \begin{array}{rl}{x}_2& =\frac{a}{b}{x}_1\\ p_1& x_1+p_2{x}_2=M\end{array}\end{array}$$

  Therefore, we are able to find the optimal value (marshalian demand for this utility).

  $$\begin{array}{}\text{(10)}& \begin{array}{rl}{x}_1^{\ast }& =\frac{bM}{b{p}_1+a{p}_2}\\ x_2^{\ast }& =\frac{aM}{b{p}_1+a{p}_2}\end{array}\end{array}$$

  Therefore, the indirect utility is then,

  $$\begin{array}{}\text{(11)}& V(p_1,p_2,M)=\frac{abM}{b{p}_1+a{p}_2}\end{array}$$

  this implies that the indifference curve in price space is straight lines and parallel,

  $$\begin{array}{}\text{(12)}& p_1=-\frac{b}{a}{p}_2+\frac{bM}{u^0}\end{array}$$

• JR1.20 & 1.21

  Suppose preferences are represented by the Cobb-Douglas utility function, $u(x_1,x_2)=A{x}_1^{\alpha }{x}_2^{1-\alpha }$$u\left(x_1, x_2\right)=A x_1^\alpha x_2^{1-\alpha}$, $0<\alpha <1$$0<\alpha<1$, and $A>0$$A>0$. Assuming an interior solution, solve for the Marshallian demand functions.

  Set up the lagrange function:

  $$\begin{array}{}\text{(13)}& \mathcal{L}=A{x}_1^{\alpha }{x}_2^{1-\alpha }-\lambda (p_1{x}_1+p_2{x}_2-M)\end{array}$$

  FOCs: (KKT)

  $$\begin{array}{}\text{(14)}& \begin{array}{rl}\frac{\partial \mathcal{L}}{\partial x_1}& =A\alpha x_1^{\alpha -1}{x}_2^{1-\alpha }-p_1\lambda =0\\ \frac{\partial \mathcal{L}}{\partial x_2}& =A(1-\alpha )x_1^{\alpha }{x}_2^{-\alpha }-p_2\lambda =0\\ \lambda & \ge 0,p_1{x}_1+p_2{x}_2\le M\end{array}\end{array}$$

  Then,

  $$\begin{array}{}\text{(15)}& x_1^{\ast }=\frac{\alpha M}{p_1},x_2^{\ast }=\frac{(1-\alpha )M}{p_2}\end{array}$$

  Here, actually we can discuss the products are positive, and if marginal utility of both products are positive, we can claim that utility function must be binding, so that we can get rid of the inequality constraints.

  And if we the utility function is being monotonic transformed, we will still be able to have a same result of marshalian demand.

• A more general case

  $$\begin{array}{}\text{(16)}& maxU(x_1,...,x_n),s.t.\sum _{i=1}^n{p}_i{x}_i=M\end{array}$$

  Set up a lagrange function:

  $$\begin{array}{}\text{(17)}& \mathcal{L}=U+\lambda (M-\sum p_i{x}_i)\end{array}$$

  Then FOCs:

  $$\begin{array}{}\text{(18)}& \begin{array}{rl}{U}_i& =\lambda p_i\\ M& =\sum p_i{x}_i\end{array}\end{array}$$

  $\Rightarrow \frac{U_i}{U_j}=\frac{p_i}{p_j}$$\Rightarrow \frac{U_i}{U_j} = \frac{p_i}{p_j}$ for any $i,j$$i,j$.

  BUt if we have the monotonic transformation $F$$F$, we will have the same $\Rightarrow \frac{U_i}{U_j}=\frac{p_i}{p_j}$$\Rightarrow \frac{U_i}{U_j} = \frac{p_i}{p_j}$. because $F^{\prime }$$F^\prime$ cancels out.

   

• MWG 3.C.6

  Suppose that in a two-commodity world, the consumer's utility function takes the form $u(x)={[{\alpha }_1{x}_1^{\rho }+{\alpha }_2{x}_2^{\prime }]}^{1/\rho }$$u(x)=\left[\alpha_1 x_1^\rho+\alpha_2 x_2^{\prime}\right]^{1 / \rho}$. This utility function is known as the constant elasticity of substitution (or CES) utility function. (a) Show that when $\rho =1$$\rho=1$, indifference curves become linear. (b) Show that as $\rho \to 0$$\rho \rightarrow 0$, this utility function comes to represent the same preferences as the (generalized) Cobb-Douglas utility function $u(x)=x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}$$u(x)=x_1^{\alpha_1} x_2^{\alpha_2}$. (c) Show that as $\rho \to -m$$\rho \rightarrow-m$, indifference curves become "right angles"; that is, this utility function has in the limit the indifference map of the Leontief utility function $u(x_1,\mathrm{x},)=$$u\left(x_1, \mathrm{x},\right)=$ $\mathrm{Min}\{x_1,x_2\}$$\operatorname{Min}\left\{x_1, x_2\right\}$

  CES utility function:

  $$\begin{array}{}\text{(19)}& u(x)={[{\alpha }_1{x}_1^{\rho }+{\alpha }_2{x}_2^{\rho }]}^{1/\rho }\end{array}$$

  when $\rho =1$$\rho = 1$ , indifference curves are linear. it is straight forward.

  $u(x)={\alpha }_1{x}_1+{\alpha }_2{x}_2=u^0$$u(x) = \alpha_1 x_1 + \alpha_2x_2 = u^0$, MRS is $\frac{d{x}_2}{d{x}_1}=-\frac{{\alpha }_1}{{\alpha }_2}$$\frac{dx_2}{dx_1} = -\frac{\alpha_1}{\alpha_2}$, then MRS is the slope of indifference curve.

  when $\rho \to 0$$\rho \rightarrow 0$, we have to show that CES utility function degenerate to a cobb-douglus.

  take log to both sides of CES.

  $$\begin{array}{}\text{(20)}& \log u(x)=\frac{\log ({\alpha }_1{x}_1^{\rho }+{\alpha }_2{x}_2^{\rho })}{\rho }\end{array}$$

  We can apply the 洛必达法则 to derive its limitation

  $$\begin{array}{}\text{(21)}& \underset{\rho \to 0}{lim}\frac{\log ({\alpha }_1{x}_1^{\rho }+{\alpha }_2{x}_2^{\rho })}{\rho }=\underset{\rho \to 0}{lim}\frac{}{{\alpha }_1{x}_1^{\rho }+{\alpha }_2{x}_2^{\rho }}\end{array}$$

  When $\rho \to -inf$$\rho \rightarrow -\inf$ , CES becomes $u(x)=min\{x_1,x_2\}$$u(x) = \min\{x_1,x_2\}$.

  when $x_1>x_2$$x_1>x_2$,

  $$\begin{array}{}\text{(22)}& \begin{array}{rl}& \underset{\rho \to -\mathrm{\infty }}{lim}{({\alpha }_1{x}_1^{\rho }+{\alpha }_2{x}_2^{\rho })}^{1/\rho }\\ =& \underset{\rho \to -\mathrm{\infty }}{lim}{x}_2{({\alpha }_1(\frac{x_1}{x_2}{)}^{\rho }+{\alpha }_2)}^{1/\rho }\\ =& \underset{\rho \to -\mathrm{\infty }}{lim}{x}_2{\alpha }_2^{1/\rho }\\ =& x_2\end{array}\end{array}$$

  For the similar reason, we can show when $x_1<x_2$$x_1<x_2$, $\underset{\rho \to -\mathrm{\infty }}{lim}{({\alpha }_1{x}_1^{\rho }+{\alpha }_2{x}_2^{\rho })}^{1/\rho }=x_1$$\lim_{\rho\rightarrow -\infin} \left(\alpha_1 x_1^\rho+\alpha_2 x_2^{\rho}\right)^{1/\rho} = x_1$.

   

Sep.15

• JR 2.7

  Derive the consumer's inverse demand functions, $p_1(x_1,x_2)$$p_1\left(x_1, x_2\right)$ and $p_2(x_1,x_2)$$p_2\left(x_1, x_2\right)$, when the utility function is of the Cobb-Douglas form, $u(x_1,x_2)=A{x}_1^{\alpha }{x}_2^{1-\alpha }$$u\left(x_1, x_2\right)=A x_1^\alpha x_2^{1-\alpha}$ for $0<\alpha <1$$0<\alpha<1$.

  Two ways solving this question.

  1. Soving UMP

     $$\begin{array}{}\text{(23)}& \begin{array}{c}{x}_1^{\ast }=\frac{\alpha M}{p_1}\\ x_2^{\ast }=\frac{(1-\alpha )M}{p_2}\end{array}\end{array}$$

     And we are able to know the answer to this question.

     Actually from this problem, we know two conditions that helps solve this question:

     $$\begin{array}{}\text{(24)}& \begin{array}{c}{p}_1{x}_1+p_2{x}_2=M\\ \frac{p_1}{p_2}=\frac{x_2}{x_1}\frac{\alpha }{1-\alpha }\end{array}\end{array}$$

     And we can get the solution for $p$$p$.

  2. We can use Hotelling-Wold Lemma

     Let $u(\mathrm{x})$$u(\mathrm{x})$ be the consumer's direct utility function. Then the inverse demand function for good $i$$i$ associated with income $y=1$$y=1$ is given by

     $$\begin{array}{}\text{(25)}& p_i(\mathrm{x})=\frac{\partial u(\mathrm{x})/\partial x_i}{\sum _{j=1}^n{x}_j(\partial u(\mathrm{x})/\partial x_j)}\end{array}$$

• JR 1.25

  A consumer with convex, monotonic preferences consumes non-negative amounts of $x_1$$x_1$ and $x_2$$x_2$.

  • (a) If $u(x_1,x_2)=x_1^{\alpha }{x}_2^{(1/2)-\alpha }$$u\left(x_1, x_2\right)=x_1^\alpha x_2^{(1 / 2)-\alpha}$ represents those preferences, what restrictions must there be on the value of parameter $\alpha$$\alpha$ ? Explain.

  • (b) Given those restrictions, calculate the Marshallian demand functions.

  For (a),

  A preference is monotonic is just equivalent to $U_i()>0$$U_i()>0$.

  So, $U_1=\alpha x_1^{\alpha -1}{x}_2^{1/2-\alpha }>0$$U_1 = \alpha x_1^{\alpha-1}{x_2^{1/2-\alpha}}>0$, $U_2=(1/2-\alpha )x_1^{\alpha }{x}_2^{-1/2-\alpha }>0$$U_2 = (1/2 - \alpha)x_1^\alpha x_2^{-1/2-\alpha}>0$

  Which implies that $0<\alpha <1/2$$0<\alpha<1/2$.

  Preference is convex represents that $U$$U$ is quasi-concave, and MRS is diminishing.$\frac{dMRS}{d{x}_1}<0$$\frac{dMRS}{dx_1}<0$.

• How do we determine a function is homothetic or not?

  1. monotonic transformation of a homogeneous function is nomothetic.

     So, if we can find a homogeneous function $g(x)$$g(x)$ and a monotonic transformation $F$$F$, then $f(x)=F(g(x))$$f(x) = F(g(x))$ is homothetic.

  2. For First-order continuous differentible function $f(x)$$f(x)$, it is homethetic iff MRS is homogeneous of degree 0.

     MRS 这种方法实际上有点像Homothetic function 的定义或者性质

   

• SS Chap 3 Prob 2

  $$\begin{array}{}\text{(26)}& \begin{array}{rl}y& =\log x_1+\log x_2\\ & =\log x_1{x}_2\end{array}\end{array}$$

  Since $g=x_1{x}_2$$g = x_1x_2$ is homogeneous function, and $f=\log (x)$$f = \log(x)$ is monotonic transiformation. Hence this function is homothetic.

  And we can also use the second method to determine.

  $$\begin{array}{}\text{(27)}& MRS=-\frac{f_1}{f_2}=-\frac{x_2}{x_1}\end{array}$$

  , which is H.D.0. $\Rightarrow$$\Rightarrow$ Homothetic.

  Similarly, for this function,

  $$\begin{array}{}\text{(28)}& y=(x_1{x}_2{)}^2-x_1{x}_2\end{array}$$

  MRS: it is also H.D.0.

   

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