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Mathematical Expectation Spiegel et al (2000) - Chapter 3. Examples by Mansoor Al-Harthy Maria Sanchez Sara Russell DP Kar Alex Lombardia Presented by Professor Carol Dahl. Introduction. Green Power Co. investment

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mathematical expectation spiegel et al 2000 chapter 3

Mathematical ExpectationSpiegel et al (2000) - Chapter 3

Examples by Mansoor Al-Harthy

Maria Sanchez

Sara Russell

DP Kar

Alex Lombardia

Presented by Professor Carol Dahl

introduction
Introduction
  • Green Power Co. investment
  • solar insolation (X)
  • wind speed (W) Electricity
  • hybrid (X,W)
  • uncertainty
  • characterize
  • discrete random variable (X)
  • probability function p(x)
  • continuous random variable (W)
  • probability density function f(w)
mathematical expectation
Mathematical Expectation
  • Expected Value
  • Functions of Random Variables
  • Some Theorems on Expectation
  • The Variance and Standard Deviation
  • Some Theorems on Variance
  • Standardized Random Variables
  • Moments
  • Variances and Covariance for Joint Distributions
  • Correlation Coefficient
slide4

Mathematical Expectation

Conditional Expectation & Variance

Chebyshev\'s Inequality

Law of Large Numbers

Other Measures of Central Tendency

Percentiles

Other Measures of Dispersion

Skewness and Kurtosis

random variables
Random Variables
  • value with a probability attached
  • value never predicted with certainty
  • not deterministic
  • probabilistic
mathematical expectation discrete case
Mathematical ExpectationDiscrete Case
  • X is solar insolation in W/ft2
  • Want to know averages
mathematical expectation discrete case1
Mathematical ExpectationDiscrete Case
  • 3,4,5,6,7 with equal probability
  • from expectation theory:
  • n events with equal probability P(X)= 1/n,
  • Discrete Random Variable X
  • x1, x2, . . ., xn
  • x = E(X) = xj*(1/n) = xj/n
  • = 3*(1/5)+4*(1/5)+5*(1/5)+?*1/5 + ?*? = 5
mathematical expectation discrete case2
Mathematical ExpectationDiscrete Case
  • Don\'t have to be equal probability X P(X)
  • 3 1/6
  • 4 1/6
  • 5 1/3
  • 6 1/6
  • 7 1/6

= 3*(1/6) + 4*1/6+5*(1/3) + ?*1/6 + ?*? = 5

mathematical expectation discrete case3
Mathematical ExpectationDiscrete Case
  • don\'t have to be symmetric
  • P(X) may be a function
  • P(Xi) = i/15

= 3*(1/15) + 4*(2/15) + 5*3/15 + 6*? + ?*? = 85/15 = 5.67

mathematical expectation continuous case
Mathematical ExpectationContinuous Case
  • W represents wind a continuous variable
  • want to know average speed
  • from expectation theory:
  • continuous random variable W ~ f(w)
mathematical expectation continuous case1
Mathematical Expectation Continuous Case
  • Meteorologist has given us pdf
  • f(w)=w/50 () <w <10 m/s 

= w = 20/3 m/s = 6.67

functions of random variables
Functions of Random Variables
  • Electricity from solar X ~ P(X)

photovoltaics - 15% efficient

Y = 0.15X

E(Y)=?

functions of random variables1
Functions of Random Variables
  • X= {3, 4, 5, 6, 7}
  • P(xi) = i/15

=0.15* 3*(1/15) + 0.15*4*(2/15) + 0.15*5*3/15 +0.15*6*? + ?*?*? = 0.85

units W/ft2

linear functions of random variables
Linear Functions of Random Variables

E(g(X)) = 0.15* 3*(1/15) + 0.15*4*(2/15) +

0.15*5*(3/15) + 0.15*6*(4/15) + 0.15*7*(5/15)

= 0.15*(3*1/15 + 4*2/15 + 5*3/15 + 6*4/15 + 7*5/15)

mean of functions of random variables continuous case electricity from wind
Mean of Functions of Random Variables Continuous Case - Electricity from wind

y = -800 + 200w w>2 with y measured in Watts

fix diagram

mean of functions of random variables continuous case
Mean of Functions of Random Variables Continuous Case
  • w continuous random variable ~ f(w)
  • f(w)=w/50 0<w <10 m/s 
  • g(w) = -800 + 200w w>2
mean of functions of random variables continuous case1
Mean of Functions of Random Variables Continuous Case
  • w continuous random variable ~ f(w)
  • f(w)=w/50 () <w <10 m/s 
  • y = g(w) = -800 + 200w w>2
mean of functions of random variables continuous case2
Mean of Functions of Random Variables Continuous Case

= -800*10 + 200*102/2 - (-800*2+200*22)

= 533.33

functions of random variables2
Functions of Random Variables

Hybrid Electricity generation from wind and solar

W, S ~ ws/(1600) 0<W<10, 0<S<8

Check it’s a valid pdf

functions of random variables3
Functions of Random Variables

Hybrid Electricity generation from wind and solar

W, S ~ ws/(1600) 0<W<10, 0<S<8

Electricity generated E = g(w,s) = w2/2 + s2/4

E(E) = E(g(w,s)) = 08 010 g(w,s)f(w,s)dwds

functions of random variables4
Functions of Random Variables

work out this integral

some theorems on expectation
Some Theorems on Expectation
  • 1. If c is any constant, then
  • E(cX) = c*E(X)
  • Example:
  • f (x,y) = {

xy/96 0<x<4 , 1<y<5

0 otherwise

4 5

4 5

E(x) = ∫ ∫ x f(x,y) dx dy = ∫ ∫ x (xy/96) dx dy = 8/3

x=0 y=1

x=0 y=1

4 5

E(2x) = ∫ ∫ 2x f(x,y) dx dy = 16/3 = 2 E(x)

x=0 y=1

some theorems on expectation1
Some Theorems on Expectation
  • 2. X and Y any random variables, then 
  • E(X+Y) = E(X) + E(Y)
  • Example:
  • E(y) = ∫ ∫ y f(x,y) dx dy = 31/9
  • E(2x+3y) = ∫ ∫ (2x+3y) f(x,y) dx dy
  • = ∫ ∫ (2x+3y) (xy/96) dx dy = 47/3
  • E(2x+3y) = 2 E(x) + 3 E(y) = 2*(8/3)+ 3*(31/9) = 47/3

4 5

x=0 y=1

equivalent

some theorems on expectation2
Some Theorems on Expectation
  • Generalize
  • E(c1*X+ c2*Y) = c1*E(X) + c2* E(Y)
  • added after check
  • add simple numerical mineral economic example here
  • 3. If X & Y are independent variables, then
  • E(X*Y) = E(X) * E(Y)
  • add simple numerical mineral economic example here
some theorems on expectation slide 3 292
Some Theorems on Expectation slide 3-29
  • 2E(X) + 3E(Y) = 2*18 + 3*12 = 36 + 36 = 72
  • So,
  • E(2X+3Y) = 2E(X) + 3E(Y) = 72
some theorems on expectation slide 3 293
Some Theorems on Expectation slide 3-29
  • If X & Y are independent variables, then:
  • E(X*Y) = E(X) * E(Y)
some theorems on expectation slide 3 294
Some Theorems on Expectation slide 3-29
  • E(X) * E(Y) = 18 * 12 = 216
  • so, E(X*Y) = E(X) * E(Y) = 216
variance and standard deviation
Variance and Standard Deviation
  • variance measures dispersion or risk of X distributed f(x)
  • Defn: Var(X)= 2 = E[(X-)2]
  • Where  is the mean of the random variable X
  •  X discrete
  • X continuous
discrete example expected value
Discrete Example Expected Value
  • An example: Net pay thickness of a reservoir
  • X1 = 120 ft with Probability of 5%
  • X2 = 200 ft with Probability of 92%
  • X3 = 100 ft with Probability of 3%
  • Expected Value = E(X)=xP(x)
  • = 120*0.05 + 200*0.92 + 100*0.03 = 193.
discrete example of variance and standard deviation
Discrete Example of Variance and Standard Deviation
  • An example: Net pay thickness of a reservoir
  • X1 = 120 ft with Probability of 5%
  • X2 = 200 ft with Probability of 92%
  • X3 = 100 ft with Probability of 3%
  • Variance = E[(X-)2] =
variance and standard deviation1
Variance and Standard Deviation
  •  Standard deviation
  • = (571)0.5 = 23.896
continuous example variance and standard deviation
Continuous Example Variance and Standard Deviation
  • add continuous mineral econ example of mean and variance
variance and standard deviation theorems
Variance and Standard Deviation Theorems
  • $1,000,000 investment fund available,
  • Risk of Solar Plant Var(X)=69,
  • Risk of Wind Plant Var(Y)=61,
  • X and Y independent
  • Cov(X, Y)= XY = E[(X-X)(Y-Y)]=0
  • Expected Risk, if 50% fund in plant 1?
  • 1. Var(cX) = c2Var(X)
  • So, as c= 0.5
  • Var(cX) = 0. 52Var(x) = 0.52*69 = 17.25
variance and standard deviation theorems1
Variance and Standard Deviation Theorems

2. Var(X+Y) = Var(X) + Var(Y)

An example:

Var(X+Y)= 61+ 69= 130

3. Var(0.5X+0.5*Y)= 0.52*Var(X) +0.52*Var(Y) = 0.25* 69 + 0.25*61 = 32.5

Later we will generalize to non-independent4. Var(c1X+ c2Y) = c12Var(X) + c22Var(Y)

+2 c1c2Cov(X,Y)

summary theory expectations variances
Summary Theory Expectations & Variances

Expectations

Variances

E(cX) =c E(X)

Var(cX) = c2Var(X)

Independent

Var(X+Y) = Var(X) + Var(Y)

E(X+Y)= E(X)+ E(Y)

Independent

Var(X-Y) = Var(X) +Var(Y)

E(X-Y)= E(X)- E(Y)

Var(c1X+ c2Y) = c12Var(X) + c22Var(Y)+2c1c2Cov(X,Y)

E(c1*X+ c2*Y) =

c1*E(X) + c2*E(Y)

standardized random variables
Standardized Random Variables
  • X random variable
  • mean  standard deviation .
  • Transform X to Z
  • standardized random variable
standardized random variables1
Standardized Random Variables
  • Z is dimensionless random variable with
standardized random variables3
Standardized Random Variables
  • X and Z often same distribution
  • shifted by 
  • scaled down by 
  • mean 0, variance 1
standardized random variables4
Standardized Random Variables
  • Example:
  • Suppose wind speed W ~ (5, 22)
  • Normalized wind speed
measures of variation
Measures of Variation

WINDSOLAR

Discrete r.v. Continuous r.v.

X1 ~ p(x1) X2 ~ f(x2)

Variation is a function of Xi

1st measure of variation

  • (Xi – µ)2
  • g(x1) = (x1 - µ)2 g(x2) = (x2 - µ)2

E(g(x1)) = Σ g(x1)p(x1)

variance examples
Variance Examples
      • X1 ~ x/6 X2 ~ 0.0107x22+0.01x2
  • X1 = 1, .. 6 2 < X2< 6
  • squared deviation from mean
  • E(Xi – u)2
  • Σ(x1 – )2P(x1)
variance example discrete
Variance Example: discrete
      • X1 ~ x1/6 x1 = 1, 2, 3
  • squared deviation from mean
  • 2 = E(X1 –)2 = Σ(x1 – )2P(x1)
  •  = 1*(1/6) + 2 *(2/6) + 3*(3/6) = 2.5
  • 2 = (1-2.5)2(1/6) + (2-2.5)2 *(2/6)
  • + (3-2.5)2* (3/6)
    • = 0.58
    •  = 0.76
variance example continuous
Variance Example: continuous
      • X2 ~ 0.0107x2+0.01x
  • 2 < X < 7
  • squared deviation from mean
  • E(X2 – u)2 =
variance examples1

3-54

Variance Examples
  • X1 ~ x/6 X2 ~0.0107x22+0.01x2
  • X1 = 1 ... 6 2 < X2< 6.213277
  • squared deviation from mean
  • E(Xi – u)2
  • Σ(x1 – )2P(x1)
variance example discrete1

3-55

Variance Example: discrete
      • X1 ~ x1/6 x1 = 1, 2, 3
  • squared deviation from mean
  • 2 = E(X1 –)2 = Σ(x1 – )2P(x1)
  •  = 1*(1/6) + 2 *(2/6) + 3*(3/6) = 2.5
  • 2 = (1-2.5)2(1/6) + (2-2.5)2 *(2/6)
  • + (3-2.5)2* (3/6)
    • = 0.58 and  = 0.76
variance example continuous1

3-56

Variance Example: continuous
      • X2 ~ 0.0107x2 + 0.01x
  • 2 < X < 6.213277
  • find: E(X2 – u)2 =squared deviation from mean
  • 1. Find E(X2) =

= 4.786178 - .069467 = 4.716711 = μ

variance example continuous2

3-57

Variance Example: continuous
  • 2. Find Var(X2) = E[(X2 – μ)2] =
moments
Moments
  • Uncertainty measured by pdf
  • Mean and Variance characterize pdf
  • Moments other measure of pdf
  • rth moment of r.v. X = E(Xr)=xrf(x)dx
  • moments
  • zero = x0f(x)dx = f(x)dx
  • first = xf(x)dx
  • second = x2f(x)dx
  • third = x3f(x)dx
  • fourth = x4f(x)dx
moments about origin and mean
Moments about origin and mean
  • rth moment of r.v. X
  • r\' = E(Xr)=xrf(x)dx
  • sometimes called rth moment about origin
  • rth moment about the mean
  • E(X - )r =(x-)rf(x)dx
  • zero = (x-)0f(x)dx = ?
  • first = (x-)1f(x)dx =xf(x)dx- f(x)dx = ?
  • second = (x-)2f(x)dx = ?
relation between moments
Relation between moments
  • r\' = moment about the origin
  • r = moment about the mean
  • 1\' =  = mean
  • 0\' = 1
  • 2 =2\' - 2 = variance = E(X2) – E(X)2
  • In the above example:
  • 2\' = 12*(1/6) + 22*(2/6) + 32*(3/6) = 36/6
  • 2\' = 6
  • 2= 6 – (7/3)2 = 0.55
slide61

Show 0,1,2,3,4th moment and moment about mean for continuous variable X2 above

Verify 2=2\' - 2 = variance = E(X2) – E(X)2

slide62

Moment Example

X ~ oil well , +1 (positive result) , probability = 1/2

–1 (negative result) , probability = 1/2

First, find the moment generating function:

Second, moments about the origin:

slide64

Moment Example slide 3-52

Substituting in the moment generation function:

(1)

(2)

slide65

Moment Example slide 3-52

  • comparing equations 1 and 2 above
          • Odd moments are all zero
          • Even moments are all ones

See Schaum’s P. 93-94

slide66

Moment Generating Function

Another way to get moments

moment generating function

Mx(t) = E(etX)

X ~ oil well – P(1) = 1/2 , P(-1) = 1/2

E (etX) = et(1)(1/2) + et(-1)(1/2) = (1/2)(et +e-t)

moment generating function
Moment Generating Function

E (etX) = et(1)(1/2) + et(-1)(1/2) = (1/2)(et +e-t)

rth moment E(xr)= Mr(0)/rt

1st moment

(1/2)(et +e-t))/t = (1/2)(et - e-t)

Evaluated at zero

(1/2)(e0 - e-0)= 0

2st moment = 0

2(1/2)(et +e-t))/ t2 = (1/2)(et + e-t)

Evaluated at zero

(1/2)(e0 + e-0)= 2

moment generating function1
Moment Generating Function

3rd moment

3(1/2 (et - e-t))/t3 = 1/2(et - e-t)

Evaluated at zero

(1/2)(e0 - e-0)= 0

Variance = E(X2) – (E(X))2 = 2 – 02 = 2

slide69

Moment-Generating Function Theorems

X and Y same probability distribution iff

Mx(t) = My(t)

X and Y independent then

MX+Y = MxMY

Don’t always exist

slide70

Characteristic function

always exist

() = E(ei X)

can get the density function from it

so can get moments

joint distribution covariance
Joint Distribution Covariance…
  • Covariance is
  • XY = Cov(X,Y) = E[(X-X)(Y-Y)]
  • = E(X,Y) – E(X)E(Y)
  • Using the joint density function f(x,y)
joint distribution
Joint Distribution…
  • Discrete r.v.
  • X ~ world oil prices 1< X<2
  • Y ~ oil production of Venezuela 0<Y<1
  • f(x,y)=0.05x2+0.1xy+0.25y2
joint distribution1
Joint Distribution…
  • Discrete r.v.
  • f(x,y)=0.05x2+0.1xy+0.25y2 1< X<2 0<Y<1
  • E(X)=(1*0.4)+(2*0.6)=1.6
  • E(Y)= (0*0.5)+(1*0.5)=0.5
joint distribution covariance example
Joint Distribution Covariance-Example

E(XY) = (1*0.1*0)+(1*0.3*1)+(2*0.4*0)+(2*0.2*1)

= 0.7

Covariance is

XY = Cov(X,Y) = E[(X-X)(Y-Y)]

= E(XY) – E(X)E(Y)

Cov(X,Y) = 0.7 – (1.5)*(0.4) = 0.1

joint distribution covariance continuous example
Joint Distribution CovarianceContinuous Example

x ~ oil price 0 < x <1

y ~ oil specific gravity API 0 < y <2

and

theorems on covariance
Theorems on Covariance
  • 1. XY = E(XY) - E(X)E(Y) = E(XY) - XY
  • 2. If X and Y are two independent variables
  • XY = Cov(X,Y) = 0
  • (the converse is not necessarily true)
  • 3. Var(X  Y) = Var(X) + Var(Y)  2Cov(X,Y)
  • 4. XYXY
correlation coefficient
Correlation Coefficient
  • correlation coefficient
  • measure dependence X and Y
  • dimensionless
  • X and Y independent XY = 0  = 0
correlation coefficient discrete case
Correlation Coefficient: Discrete Case
  • Given the discrete r.v. on slide 60 where
    • E[X] = 1.5
    • E[Y] = 0.4
    • cov(X,Y) = -0.1
  • To find the correlation coefficient, we need to find σX and σY.
correlation coefficient discrete case1
Correlation Coefficient: Discrete Case

Start by finding var(X) and var(Y) where

var(X) = E[X2] – (E[X])2 (similar for var(Y))

We know E[X], so just need to find E[X2].

correlation coefficient discrete case3
Correlation Coefficient: Discrete Case

Using a similar method we find E[Y2] = 0.5

We can now figure out var(X) and var(Y)

var(X) = 2.8 – (1.6)2 = 2.8 – 2.56 = 0.24 and

var(Y) = 0.5 – (0.5)2 = 0.25

correlation coefficient discrete case4
Correlation Coefficient: Discrete Case

We can now figure out the correlation coefficient

correlation coefficient continuous case
Correlation Coefficient: Continuous Case
  • Ref. slide 3-62 for scenario, we know the following
    • E[X] = 0.6
    • E[Y] = 1.0
    • cov(X,Y) = 0.1
  • To find the correlation coefficient, we need to find σX and σY.
correlation coefficient continuous case2
Correlation Coefficient: Continuous Case
  • Using a similar technique, we find that var(Y) = 1.44
  • Giving us the following
    • σX = 0.671
    • σY = 1.2
correlation coefficient continuous case3
Correlation Coefficient: Continuous Case

Now we can find the correlation coefficient

correlation coefficient theorems
Correlation Coefficient Theorems
  • X and Y completely linearly dependent
  • (X=a + bY) XY = XY = 1
  • (X=a - bY) XY = -XY = -1
  • -1  1
  •  = 0  X and Y are uncorrelated

but not necessarily independent

applications of correlation
Applications of Correlation
  • Correlation important for risk management
  • energy and mineral companies
  • stock investors etc.
  • Negative correlated portfolio of assets
  • reduce business and market risk of companies
conditional probabilities means and variances
Conditional Probabilities, Means, and Variances
  • PDQ Oil & Gas acquires new oil &/or natural gas
    • X ~ random variable of oil production
    • Y ~ random variable of gas production
  • Discrete joint probability: P(X,Y)
  • Conditional probability: P(X|Y)
  • Conditional mean: E(X|Y)
  • Conditional variance: 2(X|Y)
discrete conditional probabilities
Discrete Conditional Probabilities

P[X|Y] = P(X,Y) = joint probability of X & Y

P(Y) marginal probability of Y

discrete conditional probabilities1
Discrete Conditional Probabilities

P[X=Xi| Y=Yi] = P(X= Xi,Y= Yi)

P (Y= Yi)

P[X=1000|Y=750] = 0.25/0.55 = 0.455

discrete conditional expectation
Discrete Conditional Expectation

E[Xi|Y=500]

E[Xi|Y=750]

Conditional Expectation:

E[X|Y=Yi] = ∑ Xi * P[X=Xi | Y=Yi]

E[X|Y=500] = 2000(.10/.45)+1000(.10/.45)+500(.25/.45)

= 944.44 bbl

E[X|Y=750] = 2000(0.5/.55)+1000(.25/.55)+500(.25/.55)

= 863.64 bbl

discrete conditional variance
Discrete Conditional Variance

Var[X|Y=750]

Var[X|Y=500]

Conditional Variance:

Var [X|Y=Yi] = ∑(X- E[X|Y=Yi])2 * P[X=Xi | Y=Yi]

Var [X|Y=500] = (2000-944.44)2 (.10/.45) + (1000-944.44)2 *(.10/.45) + (500-944.44)2 (.25/.45)

= 358,024 bbl

Cond. St. Dev. [X | Y=500] = (358,024)0.5 = 598.3 bbl

independence check
Independence Check

To check for independence:

P(X=x)P(Y=y) = P(X,Y)

P(X=2000)P(Y=500) = P(X,Y)

0.15*0.45 =? 0.10

0.6750.10

Gas production and oil production

not independent events!

uniform continuous distribution example
Uniform Continuous Distribution Example
  • Sunflower Inc. produces coal from open pit mine
  • probabilities and expectations of coal
  • carbon content (X)
  • ash (Y)
  • Suppose X,Y~ 25 (uniform)
    • With 0< X < 0.8 and 0 < Y < 0.05
conditional expectation of continuous function
Conditional Expectation of Continuous Function

If X and Y have joint density function f(x,y), then

Properties:

If X and Y are independent then

E(YX=x)=E(Y)

numerical example of a continuous conditional probability
Numerical Example of a Continuous Conditional Probability
  • f(X) = (00.05 25dy)
  • = 25y|00.05 = 25*0.5 – 25*0 = 1.25
  • f(Y) = (00.825dx) = 25x|00.5 = 25*0.8 – 25*0 = 20
  • E(X) = x=  00.8 (x*1.25)dx = 0.4
  • E(Y) = y =  00.05 (y*20)dy = 0.025
numerical example of a continuous conditional probability1
Numerical Example of a Continuous Conditional Probability
  • Var(X) =  00.8 (X-E(X))2f(X)dx
  • =  00.8 (X-0.4)21.25dx = 0.043
  • f(X|Y) = f(X,Y)/f(X) = 25/ (1.25) = 1.25
  • E(X|Y) = 00.8Xf(X|Y)dx = 00.8(X*1.25)dx
  • = X2*1.25/2| 00.8
  • = 0.82*1.25/2 - 02*1.25/2 = 0.4

0

numerical example of a continuous conditional probability2
Numerical Example of a Continuous Conditional Probability
  • Independence? f(X)f(Y)=? f(X,Y)
  • 1.25*20 = 25
  • 25=25
  • Independence between X & Y
chebyshev s inequality
Chebyshev\'s Inequality

X random variable (discrete or continuous)

finite mean , variance 2, and  >0

probability that X differs from its mean by more than is< variance divided by 2

chebyshev s inequality1
Chebyshev\'s Inequality

Take above example mean 0.4 and variance 0.043

Let  = 0.413

P(|X-0.4|>0.413) < 0.043/0.413²

P(|X-0.4|>0.413) < 0.25

probability that X differs from 0.4

by more than 0.413 is < 25%

law of large numbers
Law of Large Numbers

Theorem: x1, x2, . . xn

mutually independent random variables

finite mean  and variance 2.

If Sn = x1 + x2 + . . . +xn , (n=1,2, . . .), then

  • as sample size increases
    • sample mean converges to true mean
other measures of central tendency
Other Measures of Central Tendency

Sample mean = Average = S Obs. / # of Obs.

Example:

Sunflower Inc.’s expected profits last five years:

$1,509,600; $5,061,060; $250,800; $250,800, $752,500.

Mean =($1,509,600+$5,061,060+2($250,800)+$752,500)

5

= $1,564,952

other measures of central tendency1
Other Measures of Central Tendency

MEDIAN: Middle of a distribution

May not exist for discrete variables

Less sensitive to extreme values than mean:

=> a better measure than the mean for highly skewed distributions: e.g. income

Example: Profit =

{$250,800;$250,800; $752,500; $1,509,600; $5,061,060}

Median of profit = $752,500

other measures of central tendency2
Other Measures of Central Tendency
  • MODE:
  • - Value that occurs most frequently
  • - may not be at the middle
  • Can be “multimodal distributions”
  • Maximum of P.D.F

Example: profit =

{$250,800;$250,800; $752,500; $1,509,600; $5,061,060}

Mode of profit = $250,800

modes
Modes

UNIMODAL Distribution BIMODAL Distribution

Amount of mineral

Amount of mineral

Grade

Grade

Skinner‘s Thesis (1976)

other measures of dispersion
Other Measures of Dispersion

1. Range: difference largest and smallest values

Example:

profit = $5,061,060-$250,800= $4,810,260.

2. Interquartile range (IQR): difference x0.75 – x0.25

x0.25 and x0.75 are 25th and 75th percentile values.

Example:

IQR of profit = $1,509,600 – $250,800

=$1,258,800.

other measures of dispersion1
Other Measures of Dispersion

3. Mean Deviation (MD): E(X-).

Example:

Add a mineral economic example of mean deviation for a continuous and discrete r.v.

percentiles
Percentiles

Divide area under density curve

 percent to left

X = percentile

X10 = 10th percentile = decile

X50 = 50th percentile = median

P(x)

Area

x

x

example of a percentile
Example of a Percentile
  • Suppose a wind farm produces X megawatts of power
  • X ~ 3x3 0<X<1.075
  • Find the 70th percentile

P(x)

Area

=70%

x

x70

example of a percentile1
Example of a Percentile

0X70(3x3)dx=0.7

Find X70

Show computations to get x70

 x70=0.759

P(x)

Area

=70%

x

x70

skewness
Skewness

Coefficients of Skewness

describes symmetry of distribution

3: Dimensionless quantity

>0  distribution skewed to the right

<0  distribution skewed to the left

=0  symmetric distribution

Other measures of skewness possible

skewness1
Skewness

SKEWNESS – symmetry of a distribution

< 0 big tail to the left

--negatively skewed

> 0 big tail to the right

--positively skewed

= 0 symmetric

If

example of skewness
Example of Skewness

Wind farm p.d.f f(w) = w/50 0 ≤ w ≤ 10 m/s

The equation for skewness is as follows:

example of skewness1
Example of Skewness

Recall from slide 13, E[w] = 6.67 m/s

We need to find E[(W – E[W])2].

example of skewness2
Example of Skewness

Now we need to find σ3, which means we need var(w)

Since σ2 = 5.8, then σ = 2.41 and σ3 = 13.99

example of skewness3
Example of Skewness

And we finally can compute skewness

The distribution is slightly skewed to the left

kurtosis
Kurtosis

Coefficients of Kurtosis

describes distribution’s degree of peakedness

4: Dimensionless quantity

<3  flatter than normal

>3 taller than normal curve

=3  normal curve

Other measures of kurtosis possible

kurtosis1
Kurtosis

KURTOSIS = tallness or flatness

Which curve has kurtosis >3?

example of kurtosis
Example of Kurtosis
  • Using the wind farm data compute Kurtosis
skewness and kurtosis using eviews
Skewness and Kurtosis Using Eviews

8

Series: Residuals

Skewness=1.78>0

Kurtosis=8>3

Skewed to right

Taller than normal

Sample 1962:1 1967:4

Observations 24

6

Mean

8.96E-17

Median

0.013538

Maximum

0.447731

4

Minimum

-0.152261

Std. Dev.

0.124629

Skewness

1.784240

Kurtosis

7.990528

2

37.63941

Jarque-Bera

0.000000

Probability

0

-0.2

0.0

0.2

0.4

skewness and kurtosis using eviews1
Skewness and Kurtosis Using Eviews

Skewness = -.59<0  skewed to left

Kurtosis = 2.45  flatter than normal

Series: Residuals

Sample 1930 1960

Observations 25

8

Mean

2.19E-16

6

Median

0.000816

Max

0.013372

Min

-0.022015

4

Std. Dev.

0.010091

Skew.

-0.592352

Kurt.

2.446336

2

Jarque-Bera

1.781322

Probability

0.410384

0

-0.02

-0.01

0.00

0.01

chapter 3 sum up
Mathematical Expectations

Functions of Random Variables

Theorems on Expectations, Variance, & Standard D.

Variance and Standard Deviation

Standardized Variables

Moments/Theorems on Moments

Characteristic Functions

Variance & Covariance for Joint Distribution

Chapter 3 Sum Up
chapter 3 sum up1
Correlation coefficient

Conditional expectations and probabilities

Chebyshev’s inequality

Law of large numbers

Other Measures of central tendencies

Percentiles

Other Measure of Dispersion

Skewness & kurtosis

Chapter 3 Sum Up
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