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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

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

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

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

The Variance and Standard Deviation

- Standard Deviation

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 =

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: 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: 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

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

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:

We have:

- 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)

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

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) - XY
- 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. XYXY

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

Finding var(X)…

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 = XY = 1
- (X=a - bY) XY = -XY = -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

0.6750.10

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(YX=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

Grade

Grade

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

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

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 UpCorrelation 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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