• Preface

• Chapter 1

  • General features
  • Theorey of value
  • Microeconomics

• Chapter 2

  • Derivates
    • first order derivatives
    • Second order derivatives
  • Function with several variable
    • Utility function
    • Partial derivatives
    • Total Differential
  • Constrained Maximisation, Lagrangian
    • $\lambda$ - Economic interpretation
  • Implicit function theorem: $g(x, y) = 0$, how $x,y$ are related
  • The Envelope theorem
  • Homogeneous Functions
  • Euler’s Theorem

Preface

• Midterms 10/25, 11/17 in class, 1hr 15 mins

  • 1st midterm chapter 1-4
  • 10 Mul Choice + 2 long question
  • Scantron 882-E

• Final 12/8 11:30-2:30

• Problems set will be graded due in one week

• HW 10%

• Exam 90%

  • 30/30/30, when final is lower than midterm
  • 50% higher midterm, 40% final, when final is higher than midterm.
    
    

Chapter 1

General features

• the ceteris paribus assumption, meaning other things stay same.
• Optimization assumptions, maximize the profits.
• positive - normative distinction
  • positive = real world;
  • normative = how things shoule be like.

Theorey of value

• Determine Equilibrium price,
  set quantity demanded = quantity supplied, $q_D=q_S$

• Production possibility frontier,

  • No free luch, oppotunity cost;
  • Oppotunity cost depends on how much each goods is produced.
    
    

Microeconomics

Behavior of individual decision makers (consumers and firms)

• Maximizing behaviours.
  e.g. stay bed/go to class, bacon/cereals,

• Price elasticity / Income effect.
  e.g. bus/car to school

• Price is not only the money you pay. Psychological cost (do not want to risk/ waste time).
  e.g. bank/parent loan/ investments (bonds, stocks)

• Marginal rate of substation. (cost of behavior), maximizing behavior.
  e.g. listen to the lecture/doze off, history/econ/job
  
  

Chapter 2

• Mathematics of Optimization.

  • Consumers attempt to maximize their welfare/utility when making decision.
  • Firms attempt to maximizing their profit when choosing input and outputs.

• Profits $\pi$ = total revenue - total cost

• Total revenue = $p*q$, $p$ is the price and $q$ is the quantity.

Derivates

first order derivatives

$f(x) = Q, ~ Q ~ is ~ a ~ constant, ~ f'(x) = 0$

$f(x) = Q*x, ~ f'(x) = a$

$f(x) = a*g(x),~ f'(x) = a ~ d ~ of ~ g(x)$

$f(x) = x^b, ~ b ~ is ~ constant, ~ f'(x) = b x^{(b-1)}$

$f(x) = a x^b = a g(x) ~ f'(x) = a*b*x^{(b-1)}$

$f(x) = ln(x) ~ f'(x) = \frac {1}{x}$

$e = ruler's~number~ \approx ~ 2.72$
$ln(e^{x}) = x$
$ln(e) =1$
$ln(a*b) = ln(a) + ln(b)$
$ln(\frac {a}{b}) = ln(a) - ln(b)$
$ln(b^a) = a*ln(b)$

$f(x) = a^x ~ f'(x) = ln(a) * a^x$

$f(x) = e^x ~ f'(x) = e^x$

$f(x) = g(x) + h(x) ~ f'(x) = g'(x) + h'(x)$

$f(x) = g(x) * h(x) ~ f'(x) = g'(x)h(x) + h'(x)g(x)$

$f(x) = \frac {g(x)}{h(x)} ~ f'(x) = \frac {[g'(x)h(x) - h'(x)g(x)]} {g^{2}(x)}$

$f(x) = g[h(x)] ~ f'(x) = \frac {dg}{dh} * \frac {dh}{dx} ~~~ chain ~ rule$

Second order derivatives

• first order condition max of a function with one variable, the derivative at that point is 0;
• second order condition to make sure the critical point is the max ( second derivative is negative). critical point is min when second derivative is positive.
• MAX: “mountain shape”, concave
• MIN: “smile face”, convex
  
  

Example:
$\pi = f(q) = 11q - 8 - q - q^2, ~ find ~ max:$

1. write down
   $f.o.c ~ \Rightarrow ~ f'(q) = 0$
   $f'(q) = 11 - 0 - 1 -2q = 10 - 2q$
   $~~~~~~~~~and ~ set$ $10 - 2q = 0, ~ \rightarrow ~ q* = 5$
2. check s.o.c $f''(q) \lt 0$
   $f''(q) = - 2 \lt 0$
3. ∴ $f(q)$ is the maximum when $q = 5$.
   
   

Example:
Max of $ln(15x) - 2x^2$,

$\displaystyle derivative \frac {1}{15x} * 15 - 4x = 0$

$\displaystyle \Rightarrow \frac {1}{x} = 4x$

$\displaystyle \Rightarrow x^* =\frac{1}{2}, (the \, critical \, point)$

Function with several variable

Utility function

x₁ = Food consumption, x₂ = Transportation, x₃ = Housing , … , x_n = leisure $U =f (x_1 , x_2 , x_3 , \cdots , x_n )$
and real problem is to solve the max of U. because of Budget constraint ( limited sources and unlimited wants).

set p₁ = price of x₁, p₂ x₂, p₃ x₃ … p_n = Price Leisure I = income $I = p_1x_1 + p_2x_2 + p_3x_3 +, \cdots, p_nx_n;$
$\rightarrow max ~ f(x_1, x_2, x_3, \ldots, x_n)$

Partial derivatives

• $f(x_1, x_2) = ax_1^2 + bx_1x_2 + cx_2^2$
  $f'(x_1) = \frac {\partial f}{\partial x_1} = 2ax_1 + bx_2$
  $f'(x_2) = \frac {\partial f}{\partial x_2} = bx_1 + 2cx_2$

• $f(x_1, x_2) = e^{ax_1 +bx_2} = g(h(x_1, x_2))$
  $g(h) = e^h, ~ h(x_1, x_2) = ax_1 +bx_2$
  $f'(x_1) = (\frac{\partial f}{\partial h}) * (\frac{\partial h}{\partial x_1}) = a2^{ax_1+bx_2}$
  $f'(x_2) = (\frac{\partial f}{\partial h}) * (\frac{\partial h}{\partial x_2}) = b2^{ax_1+bx_2}$

• $\frac{[\partial(\cfrac{\partial f}{\partial x_1})]}{\partial x_2} = f_{ij}$
  
  

Total Differential

• Consider $U=f(x_1,x_2,x_3, \cdots, x_n)$, How much f changes if all variable by a small amount ($dx_1, dx_2, dx_3, \ldots, dx_n$) ?

• Set $du = df = (\frac{\partial f}{\partial x_1}) * dx_1 + (\frac{\partial f}{\partial x_2}) * dx_2 + (\frac{\partial f}{\partial x_3}) * dx_3 + \ldots + (\frac{\partial f}{\partial x_n}) * dx_n$

• First order conditions, a necessary condition for a maximum (or a minimum ) of the function $f(x_1, x_2, x_3, \ldots, x_n)$, is that $du = 0$, for any combination of small changes in the $x's(dx_1, dx_2, dx_3, \ldots, dx_n)$ the only way for this to be true is if $f_1 = f_2 = f_3 = \ldots = f_n = 0;$

Constrained Maximisation, Lagrangian

• $L(x)$ is the function we want to maximise, $G(x)$ is the conditions
• $max = L(x_1, x_2) + \lambda G(x_1, x_2)$
• λ is called a Lagrangian multiplier, often called Lagrangian
• Find the first order conditions of the new objective function L;
  assume the answer is max/min, what the question asks for, no need to check second order derivative.

Example 1:
$c_1$ = consumption of food, $c_2$ = consumption of other goods;
$p_1$ = the price of $c_1$, $p_2$ = the price of $c_2$;
$U = f(c_1, c_2) = 2ln(c_1) + (1-\alpha)ln(c_2)$

Example 2, Find the max

$\alpha ln(c_1) + (1-\alpha)ln(c_2)$ under condition of $I = p_1c_1 + p_2c_2$
therefore, $I - p_1 * c_1- p_2 * c_2 = 0$
$L = 2ln(c_1) + (1-\alpha)ln(c_2) + \lambda(I - p_1 * c_1 + p_2 * c_2)$;

$\frac{\partial L}{\partial c_1} = \frac {\alpha}{c_1} - \lambda * p_1 = 0$

$\frac{\partial L}{\partial c_2} = \frac {1- \alpha}{c_2} - \lambda * p_2 = 0$

$\frac{\partial L}{\partial \lambda} = I - p_1 * c_1 + p_2 * c_2 = 0$

$\frac {2}{c_1} = \lambda * p_2$

$\frac {1-\alpha}{c_2} = \lambda * p_2$

$I - p_1 * c_1 + p_2 * c_2 = 0$, Divided both side $\frac {c_1}{c_2} = \frac {1-\alpha}{\alpha} * \frac {p_1}{p_2}$;
$c_1^{ * } = \alpha \frac {I}{p_1}$
$c_2^{ * } = (1-\alpha) \frac {I}{p_2}$
$\lambda^{ * } = \frac {1}{I}$
if $p_1$ increases, $\alpha \frac {I}{p_1} >0$.

Example 3, Profit, Cost, and Revenue, Second Order Condition:

$\displaystyle R = 6x_1 + \frac{99}{4}x_2$

$\displaystyle C = \frac{x_1x_2}{4} + ln(\frac{1}{x_{1}^{1/8}}) + \frac{x_2^2}{2}$

$\displaystyle max(x_1, x_2) = R - C$

$\displaystyle \max_{x_1, x_2} \left.(6x_1 + \frac{99}{4}x_2 - \frac{x_1x_2}{4} + \frac{1}{8}\ln(x_1) - \frac{x_2^2}{2})\right.$

→ FIRST ORDER
$\displaystyle \frac{\partial \pi}{\partial x_1} = 6 - \frac{x_2}{4} + \frac{1}{8x_1} = 0$

$\displaystyle \frac{\partial \pi}{\partial x_2} = \frac{99}{4} - \frac{x_1}{4} = x_2$

→ REPLACE $x_2$

$$\begin{align} \displaystyle 6 - \frac{1}{4}(\frac{99}{4} - \frac{x_1}{4}) + \frac{1}{8x_1} & = 0 \\ \frac{1}{16x_1}(x_1^2 - 3x_1 + 2) & = 0 \\ x_1 & = \frac{3 \pm \sqrt{9-8}}{2} \end{align}$$

→Find the critical points $\displaystyle x_1 = 2 ~ or ~ x_1 = 1, (x_1, x_2) = (2, \frac{97}{4}) ~ or ~ (1, \frac{49}{2})$

→SECOND ORDER FOR MAX $\displaystyle \frac{\partial^2 F}{\partial x_1^2} \lt 0, \frac{\partial^2 F}{\partial x_2^2} \lt 0$

$\displaystyle \frac{\partial^2 F}{\partial x_1^2} * \frac{\partial^2 F}{\partial x_2^2} \gt (\frac{\partial^2 F}{\partial x_1 \partial x_2})^2$

$\displaystyle \frac{\partial^2 \pi}{\partial x_1^2} = -\frac{1}{8x_1^2}$

$\displaystyle \frac{\partial^2 \pi}{\partial x_2^2} = -\frac{1}{}$

$\displaystyle \frac{\partial^2 \pi}{\partial x_1 \partial x_2} = -\frac{1}{4}$

→FOR $\displaystyle x_1 = 2, x_2 = \frac{97}{4}$ $\displaystyle \frac{\partial^2\pi}{\partial x_1^2} < 0$
$\displaystyle \frac{1}{32} < \frac{1}{16}$
This is not a MAX.

→ FOR $\displaystyle x_1 = 1, x_2 = \frac{49}{2}$
$\displaystyle \frac{1}{8} \gt \frac{1}{16}$ This is the only MAX.

$\lambda$ - Economic interpretation

meaning How much we can increase the objective function $f$ if the constraint is relaxed slightly.

• a HIGH value of $\lambda$ indicates that the utility could be increased substantially by relaxing the constraint (poor people );
• a LOW value of $\lambda$ indicates that there is not much to be gained by relaxing the constraint (rich people);
• $\lambda = 0$ implies that the constraint is not binding;

Implicit function theorem: $g(x, y) = 0$, how $x,y$ are related

• total differential is $dg = 0$, how much the function changes

• we solve for $dy$ and divide by $dx$, we get the implicit derivate;

• conditions are always satisfied in this class.

The Envelope theorem

Example of consumer problem, Envelope method
To find max of $\displaystyle x_1^{2/5} x_2^{3/5}$, such that

$\displaystyle p_1x_1 + p_2x_2 = I$

$\displaystyle max(x_1, x_2, \lambda):x_1^{2/5}x_2^{3/5} + \lambda(I - p_1x_1 + p_2x_2);$

$\displaystyle \frac{2x_2}{3x_1} = \frac{p_1}{p_2} \rightarrow x_2 = \frac{3x_1p_1}{2p_2}$

→ Replace in constraint: $\displaystyle x_1^* = \frac{2I}{5p_1}, x_2^* = \frac{3I}{5p_2}$

$\displaystyle \frac{\partial x_1^* }{\partial p_2} = 0$

→ NO CHANGE WHEN $P_2$ CHANGES

$\displaystyle \frac{\partial x_1^* }{\partial I} = \frac{2}{5p_1} \gt 0$

$\displaystyle \frac{\partial x_1^* }{\partial p_1} = -\frac{2I}{5p_1^2} \lt 0$

→ The value of the function:
$\displaystyle V(x_1^* ,x_2^* ) = (\frac{2I}{5p_1})^{\frac{2}{5}} * (\frac{3I}{5p_2})^{\frac{3}{5}}$

→ Change in well-being when I changes, $\displaystyle \frac{\partial V(x_1^* ,x_2^* )}{\partial I}$

• The definition of envelope theorem:
  Derivative of the value of the function =

Then plug in the value of $x_1^* ,x_2^* ~ and ~ \lambda$.

• a formula to compute , at the optimum , the derivative of $U$ with respect to parameter in the maximisation problem: $\frac {dU}{d\alpha} = f_{\alpha}$
  
  

Homogeneous Functions

called Homogeneous of degree K, if $f(tx_1,tx_2,\cdots, tx_n) = t^{k} f(x_1, x_2, \cdots, x_n)$

Euler’s Theorem

Special relationship between value of function and the partial derivatives of the function.