701 TA session 2

Midterm Question about revealed preference

Say the original price and consumption bundle is

$$\begin{array}{}\text{(1)}& \begin{array}{c}{p}^0=\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right],x=\left[\begin{array}{c}5\\ 19\\ 9\end{array}\right]\end{array}\end{array}$$

, and another price and consumption bundle

$$\begin{array}{}\text{(2)}& \begin{array}{c}{p}^1=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],x^1=\left[\begin{array}{c}12\\ 12\\ 12\end{array}\right]\end{array}\end{array}$$

, and the third price and consumption bundle,

$$\begin{array}{}\text{(3)}& \begin{array}{c}{p}^2=\left[\begin{array}{c}1\\ 2\\ 1\end{array}\right],x^2=\left[\begin{array}{c}27\\ 11\\ 1\end{array}\right]\end{array}\end{array}$$

Check for the WARP

WARP implies that for any $x$$x$ and $y$$y$, if $p_{x}x\ge p_{y}y$$p_x x \geq p_y y$, then $p_{y}y<p_{y}x$$p_y y< p_y x$.

, which implies that $x^0$$x^0$ is affordable under price level $p^1$$p^1$, bundle 1 is WARP to bundle 0.

SARP and WARP

with budget balance and WARP, for 2 good cases WARP is equivalent to SARP.

Mean variance representation

(SS Page 406-408)

$$\begin{array}{}\text{(5)}& r\sim f(r)\end{array}$$

$$\begin{array}{}\text{(6)}& E(u(r))=\int u(r)f(r)dr\end{array}$$

When the utility function is CARA, i.e. $u=-\exp (-ax)$$u = -\exp\left(-a x\right)$, and

$$\begin{array}{}\text{(7)}& f(r)=\frac{1}{\sqrt{2\pi }\sigma }\exp (-\frac{(r-\mu {)}^2}{2{\sigma }^2})\end{array}$$

Then $E(u(r))$$E(u(r))$ only depends on $\mu$$\mu$ and ${\sigma }^2$$\sigma^2$.

Proof:

Therefore, the expected utility for a CARA function is a linear combination of mean and variance (after a monotonic transformation).

Variance 11.6?

General utility function $U(R)$$U(R)$

$R$$R$ is a random variable.

$$\begin{array}{}\text{(9)}& R/r\sim N(\mu ,{\sigma }^2)=f(r)\end{array}$$

(1) Show that $E(U(R))$$E(U(R))$ only depends on $\mu$$\mu$ and ${\sigma }^2$$\sigma^2$

Proof:

(2) Show that $\frac{\partial \mathrm{\Phi }}{\partial \mu }>0$$\frac{\partial \Phi}{\partial \mu} >0$, and $\frac{\partial \mathrm{\Phi }}{\partial \mu }<0$$\frac{\partial \Phi }{ \partial \mu } <0$.

Since we can do normalization (transformation) first to make $f(r)$$f(r)$ Simpler by utilizing $z=\frac{R-\mu }{\sigma }$$z = \frac{R - \mu}{\sigma}$, then $R=\sigma Z+\mu$$R = \sigma Z + \mu$.

Hence,

Therefore,

and, because we cannot determine the sign of $z$$z$, so it is not straightforward to determine the sign of $\frac{\partial \mathrm{\Phi }}{\partial {\sigma }^2}$$\frac{\partial \Phi}{\partial \sigma^2}$.

Combine them together,

Another question

Assume we have two $R$$R$, $R_1$$R_1$ and $R_2$$R_2$, they are i.i.d.

If a person is risk averse, she will devide her wealth between both assets.

If a persone is risk loving, she will invest all the walth on one asset.

Suppose this agent will devide share of $\alpha$$\alpha$ into $R_1$$R_1$ and $(1-\alpha )$$(1-\alpha)$ into $R_2$$R_2$.

The maximization of expected utility function is,

FOCs:

, which is equivalent to

Difference - firm theory and consumer theory

Elasticity of substitution

Elasticity measures changes in quantity in response to some other determinants.

SS 1.2

For a CES production function

$$\begin{array}{}\text{(19)}& y={({\alpha }_1{x}_1^{\rho }+{\alpha }_2{x}_2^{\rho })}^{\frac{1}{\rho }}\end{array}$$

Calculate the elasciticity of substitution.

By defination, a elasticity of substitution is

Output Elasticity

${ϵ}_i(x_i)=\frac{\partial f(x)}{\partial x_i}\frac{x_i}{f(x)}=\frac{d\log f(x)}{d\log x}$$\epsilon_i(x_i) = \frac{\partial f(x)}{\partial x_i} \frac{x_i}{f(x)} = \frac{d \log f(x)}{d\log x}$.

Elasticity of scale

$ϵ(x)=\frac{d\log f(tx)}{d\log t}{\mid }_{t=1}$$\epsilon(x) = \frac{d \log f(tx)}{d \log t} \mid_{t=1}$

And we can show that $ϵ(x)=\sum {ϵ}_i(x)$$\epsilon(x) = \sum \epsilon_i(x)$.

Specifically, for production function $f(x)$$f(x)$ that is homogeneous degree of k, we have $ϵ=k$$\epsilon = k$.

This is because: by Eular theorem

Therefore, for $f(x)$$f(x)$ is h.d.k, if $k>1$$k>1$ then it is IRS, $k=1\iff$$k = 1 \Leftrightarrow$ CRS, and $k<1\iff$$k<1 \Leftrightarrow$ DRS.

SS 1.8

For homothetic $f(x_1,x_2)$$f(x_1, x_2)$,

$MRTS(x_1,x_2)=MRTS(t{x}_1,t{x}_2)$$MRTS(x_1,x_2) = MRTS(tx_1,tx_2)$ for any $t\ne 0$$t \neq 0$. it is equivalent to MRTS is H.D.0.

As function $f(x_1,x_2)$$f(x_1,x_2)$ is homothetic, by defination of homothetic functions, there exists another function $g(x_1,x_2)$$g(x_1,x_2)$ such that $g$$g$ is H.D.1, $f=U(g)$$f = U(g)$, where $U$$U$ is a monotonic transformation.

And, $MRTS(tx)$$MRTS(tx)$ is

Since $g(x)$$g(x)$ is homogeneous degree of 1, then $g_1(tx)=g_1(x)$$g_1(tx) = g_1(x)$, then these two MRTS are equivalent. which means that MRTS is H.D.0.

HW7

JR3.28

A firm's technology possesses all the usual properties. It produces output using three inputs, with conditional input demands $x_i(w_1,w_2,w_3,y),i=1,2,3$$x_i\left(w_1, w_2, w_3, y\right), i=1,2,3$. Some of the following observations are consistent with cost minimisation and some are not. If an observation is inconsistent, explain why. If it is consistent, give an example of a cost or production function that would produce such behaviour. (a) $\partial x_2/\partial w_1>0$$\partial x_2 / \partial w_1>0$ and $\partial x_3/\partial w_1>0$$\partial x_3 / \partial w_1>0$. (b) $\partial x_2/\partial w_1>0$$\partial x_2 / \partial w_1>0$ and $\partial x_3/\partial w_1<0$$\partial x_3 / \partial w_1<0$. (c) $\partial x_1/\partial y<0$$\partial x_1 / \partial y<0$ and $\partial x_2/\partial y<0$$\partial x_2 / \partial y<0$ and $\partial x_3/\partial y<0$$\partial x_3 / \partial y<0$. (d) $\partial x_1/\partial y=0$$\partial x_1 / \partial y=0$. (e) $\partial \left(x_1/x_2\right)/\partial w_3=0$$\partial\left(x_1 / x_2\right) / \partial w_3=0$.

For $i\ne j$$i\neq j$, $\frac{\partial x_i}{{\partial }_j}>0$$\frac{\partial x_i}{\partial_j}>0$ implies net substitutes (which is represented by the hicksian demand, ) and $i\ne j,\frac{\partial x_i}{\partial w_j}<0$$i \neq j, \frac{\partial x_i}{\partial w_j}<0$ indicates net complements.

For (a), A CES/ CD production function could be one nice example.

For (b), $\partial x_2/\partial w_1>0$$\partial x_2 / \partial w_1>0$ and $\partial x_3/\partial w_1<0$$\partial x_3 / \partial w_1<0$, which means $x_1$$x_1$ and $x_2$$x_2$ are substitutes, and $x_1$$x_1$ and $x_3$$x_3$ are (net) complements. One example could be

In this case,

There is another good example for this statement is,

For (c), inconsistent, as we assume the marginal production to be positive.

For (d), if we require $\frac{\partial x_1}{\partial y}=0$$\frac{\partial x_1}{\partial y} = 0$ for all $w,y$$w,y$ , then the firm will not use input $x_1$$x_1$, if for some $w,y$$w,y$, Then we have one fixed input $x_1$$x_1$.

For(e), A perfect complement satisfies this statement.

Another example is cobb-douglus.

One conclusion that is important

Statement: The determinant of substitution matrix must be zero

(This result holds for both consumer theory and producer theory)

$$\begin{array}{}\text{(30)}& \begin{array}{c}\sigma =\left(\begin{array}{ccc}{e}_{1,1}& \dots & e_{1,n}\\ ⋮& & ⋮\\ e_{n,1}& \dots & e_{n,n}\end{array}\right)\end{array}\end{array}$$

This hold for both cost function and expenditure function.

Since we have hicksian third law, i.e. $\sum w_j{e}_{ij}=0$$\sum w_j e_{ij} = 0$, for $\mathrm{\forall }i$$\forall i$

This means that

This implies that in this substitution matrix, at least one element could be expressed as linear combination of others, then it is nonsingular, so the determinant of $\sigma$$\sigma$ is 0.

JR 3.42 Recovering production function.

Recall how do we derive the utility function from a expenditure function?

Expenditure function -> Indirect utility function -> Utility function

BUT we do not have a indirect production function.

Question

We have seen that every Cobb-Douglas production function, $y=A{x}_1^{\alpha }{x}_2^{1-\alpha }$$y=A x_1^\alpha x_2^{1-\alpha}$, gives rise to a CobbDouglas cost function, $c(\mathbf{w},y)=yA{w}_1^{\alpha }{w}_2^{1-\alpha }$$c(\mathbf{w}, y)=y A w_1^\alpha w_2^{1-\alpha}$, and every CES production function, $y=A(x_1^{\rho }+$$y=A\left(x_1^\rho+\right.$ ${x_2^{\rho })}^{1/\rho }$$\left.x_2^\rho\right)^{1 / \rho}$, gives rise to a CES cost function, $c(w,y)=yA{(w_1^r+w_2^r)}^{1/r}$$c(w, y)=y A\left(w_1^r+w_2^r\right)^{1 / r}$. For each pair of functions, show that the converse is also true. That is, starting with the respective cost functions, 'work backward' to the underlying production function and show that it is of the indicated form. Justify your approach.

Step1: We can use Sheppard's lemma.

Step2: Try to cancel out $w$$w$ and use $x$$x$ to replace $y$$y$,

Since

By sheppard lemma, $x_1=\frac{\partial c}{\partial w_1}$$x_1 = \frac{\partial c}{\partial w_1}$, and same for $x_2$$x_2$.

We can use this two to derive $w_1$$w_1$ and $w_2$$w_2$ as functions of $x_1$$x_1$ and $x_2$$x_2$. Then we can recover the production function.

Another example we can try to recover the production function

$c(\mathbf{w},y)=yA{(w_1^r+w_2^r)}^{1/r}$$c(\mathbf{w}, y)= y A \left(w_1^r+w_2^r\right)^{1/r}$.

We are able to recover a CES production function like this: $y=\frac{1}{A}{(x_1^{\rho }+x_2^{\rho })}^{1/\rho }$$y = \frac{1}{A} \left(x_1^\rho + x_2^\rho \right)^{1/\rho}$

Midterm 2021 Q2

[25] Suppose that a production function for a good is given by $q=f(k,l)=k+l+2\sqrt{kl}$$q=f(k, l)=k+l+2 \sqrt{k l}$. This function belongs to a class of functions known as generalized Leontief (fixed proportions) production functions. (a) What type of returns to scale does this function exhibit? (b) Calculate marginal productivities and show that they are positive and diminishing. (c) Calculate marginal rate of technical substitution and show that isoquants have the usual convex shape. (d) Calculate the elasticity of substitution for this production function.

Here in this question, $q$$q$ is the output,$v$$v$ is the price of capital $k$$k$ and $w$$w$ is the price of labor $l$$l$.

For (a),

Homothetic and homogeneous production function

Theorem 3.4 in J&R

1. For homothetic production function, the cost function could be writen down in this way:

   $$\begin{array}{}\text{(34)}& c(w,y)=h(y)\underset{\text{cost when }y=1}{\underbrace{c(w,1)}}\end{array}$$

   And the input demand function could be written as,

   $$\begin{array}{}\text{(35)}& x(w,y)=h(y)x(w,1)\end{array}$$

   This actually refers to separability.

2. For homogeneous production function with degree of $k$$k$, the cost function could be written as,

   $$\begin{array}{}\text{(36)}& \begin{array}{c}c(w,y)=y^{1/k}c(w,1)\\ x(w,y)=y^{1/k}x(w,1)\end{array}\end{array}$$

How do we understand Envelope theorem

1. For unconstraint problem (Profit maximization)

   Consider $f(x_1,x_2,\dots ,x_n;a)$$f(x_1,x_2,\dots, x_n; a)$ as a production function. $x$$x$ is something that we choose (choice variable), and $a$$a$ is an expgeneous variable.

   then we have a profit function like this,

   $$\begin{array}{}\text{(37)}& \pi =f(x_1,x_2,\dots ,x_n;a)-\sum _i{x}_i{w}_i\end{array}$$

   To solve this maximization problem,

   FOC: $\frac{\partial \pi }{\partial x_i}=0$$\frac{\partial \pi }{\partial x_i} = 0$, and we can obtain the optimal input demand $x^{\ast }(a)=x^{\ast }(p,\mathbf{w})$$x^*(a) = x^*(p,\mathbf w)$

   If we substitute back this imput demand back to our production function, we can obtain the obtimal production function, which we regard as a value function.

   $$\begin{array}{}\text{(38)}& y^{\ast }=f({\mathbf{x}}^{\ast }(a))\end{array}$$

   The envelope theorem says that

   $$\begin{array}{}\text{(39)}& \frac{\partial y^{\ast }}{\partial a}=\frac{\partial f({\mathbf{x}}^{\mathbf{\ast }},a)}{\partial a}\end{array}$$

   This only make sense at the optimal condition.

   In this profit maxmization case, we can also apply it to the optimal profit function (It becomes our value function in this case).

   $$\begin{array}{}\text{(40)}& \pi (p,\mathbf{w})=pf({\mathbf{x}}^{\ast })-\sum w_i{x}_i\end{array}$$

   Then, we see that

   $$\begin{array}{}\text{(41)}& \begin{array}{c}\frac{\partial \pi }{\partial p}=f({\mathbf{x}}^{\ast })=y^{\ast }\\ \frac{\partial \pi }{\partial w_i}=-x_i^{\ast }\end{array}\end{array}$$

2. For unconstraint case, if we are doing cost minimization, what we actually do is like this,

   $$\begin{array}{}\text{(42)}& \begin{array}{c}max/minf(x_1,\dots ,x_n;a)\\ s.t.g(x_1,\dots ,x_n;a)=0\end{array}\end{array}$$

   FOC: $\frac{\partial \mathcal{L}}{\partial x_i}=f_i+\lambda g_i=0$$\frac{\partial \mathcal L}{\partial x_i} = f_i + \lambda g_i = 0$

   Then,

   $$\begin{array}{}\text{(43)}& \frac{\partial y^{\ast }}{\partial a}=\frac{\partial \mathcal{L}}{\partial a}\end{array}$$

Comparative static analysis

step 1. Set up the optimization problem and write down the first order conditions.

Step 2. Differentiate with respect to the parameters that we are interested in, and we write the system of equations into a matrix form.

Step 3. Solve for the partial derivitives by using the Cramer's rule.