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# 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

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### 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

• 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)

• 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

Conditional Expectation & Variance

Chebyshev's Inequality

Law of Large Numbers

Other Measures of Central Tendency

Percentiles

Other Measures of Dispersion

Skewness and Kurtosis

• 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

Wind Continuous Random Variablepurple>red>orange>green>blue (m/s)

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

• Electricity from solar X ~ P(X)

photovoltaics - 15% efficient

Y = 0.15X

E(Y)=?

• 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

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

= ?

Hybrid Electricity generation from wind and solar

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

Check it’s a valid pdf

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

work out this integral

SUMMARY OF DEFINITIONS CaseExpected Value

• 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

• 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

• Generalize

• E(c1*X+ c2*Y) = c1*E(X) + c2* E(Y)

• 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 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 Caseslide 3-29

• If X & Y are independent variables, then:

• E(X*Y) = E(X) * E(Y)

Some Theorems on Expectation Caseslide 3-29

• E(X) * E(Y) = 18 * 12 = 216

• so, E(X*Y) = E(X) * E(Y) = 216

• 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

• 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%

• Expected Value = E(X)=xP(x)

• = 120*0.05 + 200*0.92 + 100*0.03 = 193.

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] =

•  Standard deviation

• = (571)0.5 = 23.896

Continuous Example CaseVariance and Standard Deviation

• add continuous mineral econ example of mean and variance

• \$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

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 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)

• X random variable

• mean  standard deviation .

• Transform X to Z

• standardized random variable

• Z is dimensionless random variable with

• X and Z often same distribution

• shifted by 

• scaled down by 

• mean 0, variance 1

• Example:

• Suppose wind speed W ~ (5, 22)

• Normalized wind speed

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)

• 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)

• 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

• X2 ~ 0.0107x2+0.01x

• 2 < X < 7

• squared deviation from mean

• E(X2 – u)2 =

• 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)

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

• 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

3- Case57

Variance Example: continuous

• 2. Find Var(X2) = E[(X2 – μ)2] =

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

• 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 = ?

• 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:

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

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)

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

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

X and Y same probability distribution iff

Mx(t) = My(t)

X and Y independent then

MX+Y = MxMY

Don’t always exist

always exist

() = E(ei X)

can get the density function from it

so can get moments

• 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)

• 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

• 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

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 CaseContinuous Example

x ~ oil price 0 < x <1

y ~ oil specific gravity API 0 < y <2

and

Joint Distribution Covariance CaseContinuous Example

Joint Distribution Covariance CaseContinuous Example

Joint Distribution Covariance CaseContinuous Example

Joint Distribution Covariance CaseContinuous Example

• 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

• measure dependence X and Y

• dimensionless

• X and Y independent XY = 0  = 0

• 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.

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].

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

We can now figure out the correlation coefficient

• 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.

Finding var(X)…

• Using a similar technique, we find that var(Y) = 1.44

• Giving us the following

• σX = 0.671

• σY = 1.2

Now we can find the correlation coefficient

• 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

• 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 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)

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

P(Y) marginal probability of Y

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

P (Y= Yi)

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

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

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

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!

• 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 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 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

• 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 CaseConditional Probability

• Independence? f(X)f(Y)=? f(X,Y)

• 1.25*20 = 25

• 25=25

• Independence between X & Y

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

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%

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

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

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

• 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 Case

UNIMODAL Distribution BIMODAL Distribution

Amount of mineral

Amount of mineral

Skinner‘s Thesis (1976)

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.

3. Mean Deviation (MD): E(X-).

Example:

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

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

• 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

0X70(3x3)dx=0.7

Find X70

Show computations to get x70

 x70=0.759

P(x)

Area

=70%

x

x70

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

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

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

The equation for skewness is as follows:

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

We need to find E[(W – E[W])2].

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

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

And we finally can compute skewness

The distribution is slightly skewed to the left

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

Kurtosis Case

KURTOSIS = tallness or flatness

Which curve has kurtosis >3?

• Using the wind farm data compute Kurtosis

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 = -.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

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

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