Game Theory

Math Review
1
1.1

Function
Definition

If we write down the relation of x and y as follows, y = f (x) this means y is related to x under the rule f . Also we say that the value of y depends on the value of x. This relation y = f (x) is called as a function if 1) the rule f assigns a single x value to single y value or 2) assigns multiple x values to single y value.

2
2.1

Shape of function
When f (x) = ax + b

Suppose that the function is given as follows. y = f (x) = ax + b ´ 1) Slope: f (x) = a and Y −intercept: b Y − axis is b.. b −a

2) Intersection with Y − axis: This is the case when x = 0. So from f (0) = b, the Intersection with 3) Intersection with X − axis: This is the case when y = 0. So the intersection with X − axis is from ax + b = 0.

Example Suppose y = f (x) = ax + b. Draw this function in following each case. 1) When a > 0, b > 0 2) When a > 0, b < 0 3) When a < 0, b > 0 4) When a < 0, b < 0

Now suppose you want to find the linear function that passes through following two points,. x = (a, b) and y = (c, d). Then the linear function is defined as follows. ¶ µ b−d (x − a) + b f (x) = a−c µ ¶ b−d = (x − c) + d a−c ´ ³ ´ ³ b−d b−d So from f (x) = a−c (x − a) + b or f (x) = a−c (x − c) + d, ¶ (bc − ad) b−d x+ f (x) = a−c (c − a) | {z } | {z } µ 1

Here, the first term is the slope and the second term is Y-intercept. Example Find the linear function that passes through following two points. A = (2, 4) and B = (−4, −2)

2.2

When f (x) = ax2 + bx + c

Suppose that the function is given as follows. y = f (x) = ax2 + bx + c 1) Suppose a > 0. Then the shape of function looks as follows. Example : y = 2x2 + 5x + 1 y 60

40

20

0 -5 -2.5 0 2.5 x 5

2) Suppose a < 0. Then the shape of function looks as follows. Example : y = −2x2 + 5x + 1
-5 -2.5 y 0 0 2.5 x 5

-20

-40

-60

2.3

Logarithmic functions: y = ln x

If function is give as y = ln x, ”x > 0” should be satisfied always. If x ≤ 0, function y = ln x can’t be defined. Example : y = ln x

2

y

20

10

0 0 5 10 15 x -10 20

-20

Also the basic rules of lug function are as follows.

ln 1 = 0 ln(ab) a ln( ) b ln(ab ) 1 ln( ) a = ln a + ln b = ln a − ln b = b ln a = ln 1 − ln a = − ln a = ln a−1

2.4

Function y = x

The basic rules are as follows.

x0 = 1 xm xn = xm+n xm = xm−n xn 1 x−m = xm m n (x ) = xmn xm ym = (xy)m

3

Differentiation (One variable)
Let f (x) be a continuous function. Then the first derivative of a function at x = x0 is

Definition

the function defined by following equation. f (x0 ) = lim .
0

h→0

f (x0 + h) − f (x0 ) f (x0 + h) − f (x0 ) = lim h→0 (x0 + h) − x0 h

3

The first derivative of a function at point x = x0 can be interpreted as a slope at that point. So f (x0 ) > 0 =⇒ Value of function is increasing at x =⇒ Positive slope at x f (x0 ) = 0 =⇒ Value of function is constant at x =⇒ Constant slope at x f (x0 ) < 0 =⇒ Value of function is decreasing at x =⇒ Negative slope at x Also, for every point x, f (x) > 0 =⇒ Value of function is increasing for all x f (x) = 0 =⇒ Constant for all x f (x) < 0 =⇒ Value of function is decreasing for all x Definition (Second derivative) f ” (x) = h 0 i ∂ f (x) ∂x
0 0 0 0 0 0

The second derivative of a function can be interpreted as a change of slope. That is, it gives the information about the curvature of function.

f ” (x) = 0 ⇒ the slope of function has no change =⇒ linear function f ” (x) < 0 ⇒ the slope of function is decreasing Example ´ 1) What is the shape of function, f (x) if f (x) > 0 and f ” (x) > 0 for all x? ´ 2) What is the shape of function, f (x) if f (x) > 0 and f ” (x) < 0 for all x? ´ 3) What is the shape of function, f (x) if f (x) < 0 and f ” (x) = 0 for all x?

f ” (x) > 0 ⇒ the slope of function is increasing

3.1

General rules

Let f and g are both differential function of x.

y = f (x) ± g(x) =⇒ y = f (x) ± g (x)
0 0

0

0

0

y = f (x) · g(x) =⇒ y = f (x) · g(x) + f (x) · g (x) y = f (x) · g(x) − f (x) · g (x) f (x) 0 =⇒ y = g(x) [g (x)]2
0 0 0 0 0

0

y = f [g (x)] =⇒ y = f [g(x)] · g (x)

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3.2

Special rules y = c =⇒ y = 0 (if c is constant) y = xa =⇒ y = a · x(a−1) (if a is constant) √ 1 1 1 1 0 x = x 2 =⇒ y = · x(− 2 ) = √ y = 2 2 x y = eax (exponential function) =⇒ y = a · eax (if a is constant) 1 0 (for x > 0 always) y = ln(x) =⇒ y = x
0 0 0

4

Partial Derivatives (Multivariable)

Definition 1 (Two variable case) if f = f (x1 , x2 ) ∂f ∂x1 ∂f ∂x2 For example,
∂f ∂x1

f (x1 + h, x2 ) − f (x1 , x2 ) h f (x1 , x2 + h) − f (x1 , x2 ) = lim h→0 h = h→0 lim

means you want to check what will happen to f if there is a change in x1 . Here, the
∂f ∂x2

point is that you just give a change only to x1 and regard x2 as a constant. Also only to x2 and regard x1 as a constant. Suppose that f (x, y) = 3x2 + 2xy2 + 4y 2 . Then ∂f ∂f = 6x + 2y 2 and = 4xy + 8y ∂x ∂y

means you want

to check what will happen to f if there is a change in x2 . Here, the point is that you just give a change

5
5.1

Unconstrained Optimization
One variable case

Assume that your objective function f (x) is continuous and differentiable (i.e., smooth and continuous) at x0 . Now suppose f (x) attains its local maximum value OR local minimum value at x = x0 . This means, f (x0 ) = 0 This is called as the First order condition for Maximum or Minimum. However, this does not give the information whether f (x) attains its MAX value or MIN value at x = x0 . For getting this information, we need the second derivatives. Check following: If f (x0 ) < 0 =⇒ f (x) attains its local max value at x0 If f (x0 ) > 0 =⇒ f (x) attains its local min value at x0 If f (x0 ) = 0 =⇒ x0 is a inflection point 5
00 00 00 0

These are called as the Second order condition. Theorem 1) Suppose that f (x) attains its local max value at x0 . If f (x) < 0 for all x, then f (x) attains its global max value at x0 . 2) Suppose that f (x) attains its local min value at x0 . If f (x) > 0 for all x, then f (x) attains its global min value at x0 . Example Draw the graph of each function in detail. Also check at which point each function attains the Max or Min value. 1) f (x) = x2 + 4x + 7. 3) h(x) = x2 − 3x + 2 and x ∈ [ 1 , 2]. Check the minimum value and maximum value of this function. 2 2) g(x) = −2x2 + 2x + 1.
00 00

5.2

Multivariable case

Suppose that y = f (x1 , x2 , x3 , · · · · ·, xn ) or minimize the value of y, what should we do? Step 1) Derive First order conditions. ∂y ∂y ∂y = 0, = 0, · · · ··, =0 ∂x1 ∂x2 ∂xn Step 2) For given n equations, solve simultaneously to find the values x1 , x2 , · · ·, xn that maximize and x1 , x2 , x3 , · · · · ·, xn are control variables. If we want to find the values x1 , x2 , · · ·, xn that maximize

or minimize the value of y. Remark

to check the second order condition as the one variable case. However, we need the knowledge of Matrix and it will not be covered in this course. Example Suppose z = f (x, y) = (10 − 2(x + y))(x + y) − x − ¡1 ¢

For checking whether the solved values x1 , x2 , · · ·, xn maximize or minimize the value of y, we need

2y

2

. Find x, y that maximize the value of z.

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