Mathematical Expectation Spiegel et al (2000) - Chapter 3

1 / 128

# Mathematical Expectation Spiegel et al (2000) - Chapter 3 - PowerPoint PPT Presentation

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

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Mathematical Expectation Spiegel et al (2000) - Chapter 3' - chas

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### 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
• 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
• 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

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
• value with a probability attached
• value never predicted with certainty
• not deterministic
• probabilistic
Mathematical ExpectationDiscrete Case
• X is solar insolation in W/ft2
• Want to know averages
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 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 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 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 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
• Electricity from solar X ~ P(X)

photovoltaics - 15% efficient

Y = 0.15X

E(Y)=?

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

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

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

fix diagram

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 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 Case

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

= 533.33

g(w) = a + bw

= ?

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 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 Variables

work out this integral

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 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 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 3-29
• 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 3-29
• If X & Y are independent variables, then:
• E(X*Y) = E(X) * E(Y)
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 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
• 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
• 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 Deviation
•  Standard deviation
• = (571)0.5 = 23.896
Continuous Example Variance and Standard Deviation
• add continuous mineral econ example of mean and variance
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 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

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
• X random variable
• mean  standard deviation .
• Transform X to Z
• standardized random variable
Standardized Random Variables
• Z is dimensionless random variable with
Standardized Random Variables
• X and Z often same distribution
• shifted by 
• scaled down by 
• mean 0, variance 1
Standardized Random Variables
• Example:
• Suppose wind speed W ~ (5, 22)
• Normalized wind speed
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
• 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
• 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
• X2 ~ 0.0107x2+0.01x
• 2 < X < 7
• squared deviation from mean
• E(X2 – u)2 =

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)

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

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

3-57

Variance Example: continuous
• 2. Find Var(X2) = E[(X2 – μ)2] =
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
• 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
• 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

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

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:

Moment Example slide 3-52

Substituting in the moment generation function:

(1)

(2)

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

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

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 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

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

Characteristic function

always exist

() = E(ei X)

can get the density function from it

so can get moments

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…
• 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 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

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 CovarianceContinuous Example

x ~ oil price 0 < x <1

y ~ oil specific gravity API 0 < y <2

and

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
• measure dependence X and Y
• dimensionless
• X and Y independent XY = 0  = 0
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 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 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 Case

We can now figure out the correlation coefficient

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 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 Case

Now we can find the correlation coefficient

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
• 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
• 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

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

P(Y) marginal probability of Y

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

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

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

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
• 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

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
• 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 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 Probability
• Independence? f(X)f(Y)=? f(X,Y)
• 1.25*20 = 25
• 25=25
• Independence between X & Y
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 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

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

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 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 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

UNIMODAL Distribution BIMODAL Distribution

Amount of mineral

Amount of mineral

Skinner‘s Thesis (1976)

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 Dispersion

3. Mean Deviation (MD): E(X-).

Example:

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

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
• 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 Percentile

0X70(3x3)dx=0.7

Find X70

Show computations to get x70

 x70=0.759

P(x)

Area

=70%

x

x70

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

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

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

The equation for skewness is as follows:

Example of Skewness

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

We need to find E[(W – E[W])2].

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 Skewness

And we finally can compute skewness

The distribution is slightly skewed to the left

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

Kurtosis

KURTOSIS = tallness or flatness

Which curve has kurtosis >3?

Example of Kurtosis
• Using the wind farm data compute Kurtosis
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 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

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
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