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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


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


Wind continuous random variable purple red orange green blue m s
Wind Continuous Random Variablepurple>red>orange>green>blue (m/s)


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 Case

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 Case

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 Case

work out this integral


Summary of definitions expected value
SUMMARY OF DEFINITIONS CaseExpected Value


Some theorems on expectation
Some Theorems on Expectation Case

  • 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 Case

  • 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 Case

  • 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 Caseslide 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 Caseslide 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 Caseslide 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 Case

  • 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


The variance and standard deviation
The Variance and Standard Deviation Case

  • Standard Deviation


Discrete example expected value
Discrete Example Expected Value Case

  • 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 CaseVariance 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 Case

  •  Standard deviation

  • = (571)0.5 = 23.896


Continuous example variance and standard deviation
Continuous Example CaseVariance and Standard Deviation

  • add continuous mineral econ example of mean and variance


Variance and standard deviation theorems
Variance and Standard Deviation Theorems Case

  • $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 Case

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 CaseExpectations & 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 Case

  • X random variable

  • mean  standard deviation .

  • Transform X to Z

  • standardized random variable


Standardized random variables1
Standardized Random Variables Case

  • Z is dimensionless random variable with



Standardized random variables3
Standardized Random Variables Case

  • X and Z often same distribution

  • shifted by 

  • scaled down by 

  • mean 0, variance 1


Standardized random variables4
Standardized Random Variables Case

  • Example:

  • Suppose wind speed W ~ (5, 22)

  • Normalized wind speed


Measures of variation
Measures of Variation Case

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 Case

  • 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 Case

    • 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 Case

    • X2 ~ 0.0107x2+0.01x

  • 2 < X < 7

  • squared deviation from mean

  • E(X2 – u)2 =


  • Variance examples1

    3- Case54

    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- Case55

    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- Case56

    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- Case57

    Variance Example: continuous

    • 2. Find Var(X2) = E[(X2 – μ)2] =


    Moments
    Moments Case

    • 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 Case

    • 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 Case

    • 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 0,1,2,3,4 Caseth moment and moment about mean for continuous variable X2 above

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


    Moment Example Case

    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 Caseslide 3-52

    We have:


    Moment Example Caseslide 3-52

    Substituting in the moment generation function:

    (1)

    (2)


    Moment Example Caseslide 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 Case

    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 Case

    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 Case

    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 Case

    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 Case

    always exist

    () = E(ei X)

    can get the density function from it

    so can get moments


    Joint distribution covariance
    Joint Distribution Covariance… Case

    • 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… Case

    • 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… Case

    • 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 Case

    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 Covariance CaseContinuous Example

    x ~ oil price 0 < x <1

    y ~ oil specific gravity API 0 < y <2

    and


    Joint distribution covariance continuous example1
    Joint Distribution Covariance CaseContinuous Example


    Joint distribution covariance continuous example2
    Joint Distribution Covariance CaseContinuous Example


    Joint distribution covariance continuous example3
    Joint Distribution Covariance CaseContinuous Example


    Joint distribution covariance continuous example4
    Joint Distribution Covariance CaseContinuous Example


    Theorems on covariance
    Theorems on Covariance Case

    • 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 Case

    • correlation coefficient

    • measure dependence X and Y

    • dimensionless

    • X and Y independent XY = 0  = 0


    Correlation coefficient discrete case
    Correlation Coefficient: Discrete Case 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 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 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 Case

    We can now figure out the correlation coefficient


    Correlation coefficient continuous case
    Correlation Coefficient: Continuous Case 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 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 Case

    Now we can find the correlation coefficient


    Correlation coefficient theorems
    Correlation Coefficient Theorems Case

    • 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 Case

    • 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 CaseProbabilities, 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 Case

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

    P(Y) marginal probability of Y


    Discrete conditional probabilities1
    Discrete Conditional Probabilities Case

    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 Case

    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 Case

    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 Case

    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 Case

    • 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 CaseContinuous 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 CaseConditional 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 Case

    • 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 CaseConditional 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 Case

    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 Case

    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 Case

    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 Case

    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 Case

    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 Case

    • 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 Case

    UNIMODAL Distribution BIMODAL Distribution

    Amount of mineral

    Amount of mineral

    Grade

    Grade

    Skinner‘s Thesis (1976)


    Other measures of dispersion
    Other Measures of Dispersion Case

    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 Case

    3. Mean Deviation (MD): E(X-).

    Example:

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


    Percentiles
    Percentiles Case

    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 Case

    • 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 Case

    0X70(3x3)dx=0.7

    Find X70

    Show computations to get x70

     x70=0.759

    P(x)

    Area

    =70%

    x

    x70


    Skewness
    Skewness Case

    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 Case

    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 Case

    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 Case

    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 Case

    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 Case

    And we finally can compute skewness

    The distribution is slightly skewed to the left


    Kurtosis
    Kurtosis Case

    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 Case

    KURTOSIS = tallness or flatness

    Which curve has kurtosis >3?


    Example of kurtosis
    Example of Kurtosis Case

    • Using the wind farm data compute Kurtosis


    Skewness and kurtosis using eviews
    Skewness and Kurtosis Using Eviews Case

    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 Case

    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 Case

    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 Case

    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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