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
3-2
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)
3-3
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
3-4
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
3-5
Random Variables
value with a probability attached
value never predicted with certainty
not deterministic
probabilistic
3-6
Solar Insolation (m/s)
3-7
Mathematical Expectation
Discrete Case
X is solar insolation in W/ft2
Want to know averages
n
E(X)   x j P(X  x j )
j1

3-8
Mathematical Expectation
Discrete 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
3-9
Mathematical Expectation
Discrete Case
Don't have to be equal probability
n
E(X)   x j P(X  x j )
X P(X)
3 1/6
4 1/6
5 1/3
6 1/6
7 1/6
j1
= 3*(1/6) + 4*1/6+5*(1/3) + ?*1/6 + ?*? = 5

3-10
Mathematical Expectation
Discrete Case
don't have to be symmetric
P(X) may be a function
P(Xi) = i/15
n
E(X)   x j P(X  x j )
j1
= 3*(1/15) + 4*(2/15) + 5*3/15 + 6*? + ?*?
= 85/15 = 5.67
3-11
Wind Continuous Random Variable
purple>red>orange>green>blue (m/s)
3-12
Mathematical Expectation
Continuous Case
W represents wind a continuous variable
want to know average speed
from expectation theory:
continuous random variable W ~ f(w)

E( X ) 
 xf ( x)dx

3-13
Mathematical Expectation
Continuous Case
Meteorologist has given us pdf
f(w)=w/50 () <w <10 m/s

E ( w)   wf ( w)dw


 w / 150
3

10
  w * w / 50 * dw
0
10
0
= w = 20/3 m/s = 6.67
3-14
Functions of Random Variables
Electricity from solar X ~ P(X)
n
E(X)   x j P(X  x j )
j1

photovoltaics - 15% efficient
Y = 0.15X
E(Y)=?
n
E g(X)   g(x j ) p(x j )
j1
3-15
Functions of Random Variables
X = {3, 4, 5, 6, 7}
P(xi) = i/15
n
E g(X)   g(x j ) p(x j )
j1
=0.15* 3*(1/15) + 0.15*4*(2/15) +
0.15*5*3/15 +0.15*6*? + ?*?*? = 0.85
units W/ft2
3-16
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)
n
n
j1
j1
E g(X)   0.15x j p(x j )  0.15 x j p(x j )  0.15E(x)
n
n
j1
j1
E g(X)   ax j p(x j )  a x j p(x j )  aE(x)
3-17
Mean of Functions of Random Variables
Continuous Case - Electricity from wind
Electricity from Wind
Electricity (W/s)
2000
1500
1000
500
0
0
5
10
15
wind speed (m/s)
y = -800 + 200w
fix diagram
w>2 with y measured in Watts
3-18
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
E g ( X ) 

 g ( x) f ( x)dx

3-19
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
E g(w) 

 g(w) f (w)dw

E g(w) 

10
 (800  200w) f (w)dw
2

3-20
Mean of Functions of Random Variables
Continuous Case
2
w
E g(w)  800w  200
10
2
|
2
= -800*10 + 200*102/2 - (-800*2+200*22)
= 533.33
3-21
Mean of Linear Functions of Random
Variables - Continuous Case
E g(w) 
10
 (800  200w) f (w)dw
2

10
 (800 f (w)  200wf (w))dw
2



10
10
 800 f (w)dw  
2
2
200wf (w)dw
3-22
Mean of Linear Functions of Random
Variables - Continuous Case
10
10
2
2
E g(w)  800  f (w)dw  200  wf (w)dw
 800  200E(w)

3-23
Mean of Linear Functions of Random
Variables - Continuous Case
g(w) = a + bw
E g(w) 
10
 (a  bw) f (w)dw
2

10
 (af (w)  bwf (w))dw
2


10
10
 af (w)dw  
2
= ?
2
bwf (w)dw
3-24
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
10 8
P( w, s) 
  f (w, s) dw ds  1
0 0
10
ws 2
P( w, s)   (
)
2 *1600
0
10

8
0
10
64w
dw  
dw
3200
8
w 64 
100 * 64
P ( w, s ) 
1
 
2 * 3200  0
6400
2
3-25
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
8 10
2
2
w s
ws
0 0 ( 2  4 ) * 1600dwds 
3-26
Functions of Random Variables
work out this integral
8 10
2
2
w s
ws
0 0 ( 2  4 ) * 1600dwds 
3-27
SUMMARY OF DEFINITIONS
Expected Value
Discrete Random
E(X)=xP(x)
Continuous Random

E(X)   xf ( x )dx

E(X2)=x2P(x)
E g ( X ) 

 g ( x) f ( x)dx

E(g(X,Y)=
g(x,y)P(x,y))
Eg ( X , Y ) 
 
  g ( x, y ) f ( x, y )dxdy

3-28
Some Theorems on Expectation
1. If c is any constant, then
E(cX) = c*E(X)
Example:
f (x,y) = { xy/96
0
4 5
0<x<4 , 1<y<5
otherwise
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
3-29
Some Theorems on Expectation
2. X and Y any random variables, then
E(X+Y) = E(X) + E(Y)
Example:
4 5
E(y) =x=0∫ y=1
∫ 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
equivalent
E(2x+3y) = 2 E(x) + 3 E(y) = 2*(8/3)+ 3*(31/9) = 47/3
3-30
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
3-31
Some Theorems on Expectation slide 3-29
3
2
E (Y )    yf ( x , y )dxdy
0 0
3 2
3
1 3 2
2
   ( xy )dxdy   ( xy ) 0 dx
0 0
03
3
3
8
8 2
  xdx  x  12
6 0
0 3
3-32
Some Theorems on Expectation slide 3-29
3
2
0
0
E (2 X  3Y )    (2 x  3 y )( x , y )dxdy
3
2
   (2 x y  3xy )dxdy
2
0
3
0
  x y  xy
2
0
2
2
3
2
0
3
dx   (4 x  8 x )dx
2
0
3-33
Some Theorems on Expectation slide 3-29
4 3
2
 x  4x
3
3
0
 72
2E(X) + 3E(Y) = 2*18 + 3*12 = 36 + 36 = 72
So,
E(2X+3Y) = 2E(X) + 3E(Y) = 72
3-34
Some Theorems on Expectation slide 3-29
If X & Y are independent variables, then:
E(X*Y) = E(X) * E(Y)
3
2
0
0
E ( X * Y )    ( xy ) f ( x , y )dxdy
3
2
3
1 2 3 2
   ( x y )dxdy   ( x y ) 0 dx
0 0
03
2
2
3-35
Some Theorems on Expectation slide 3-29
3
8 2
3 3
  x dx  8 x 0  216
03
E(X) * E(Y) = 18 * 12 = 216
so,
E(X*Y) = E(X) * E(Y) = 216
3-36
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
n

2
X
  (x j  ) f (x j )
2
j 1
X continuous

 X2   ( x   ) 2 f ( x)dx

3-37
The Variance and Standard Deviation
Standard Deviation
 X  Var (X)  E[(X  ) 2 ]
3-38
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.
3-39
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]
=
n
2
X
  (x j  ) f (x j )
2
j 1
 y  (120  100) 0.05  ( 200  193) (0.92) 
2
2
(100  193) (0.03)  571
2
2
3-40
Variance and Standard Deviation
Standard deviation
X  Var (X)  E[(X  ) ]
2
= (571)0.5 = 23.896
3-41
Continuous Example
Variance and Standard Deviation
add continuous mineral econ example of mean and
variance
3-42
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
3-43
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)
3-44
Summary Theory
Expectations & Variances
Expectations
E(cX) =c E(X)
Variances
Var(cX) = c2Var(X)
Independent
E(X+Y)= E(X)+ E(Y)
Var(X+Y) = Var(X) + Var(Y)
Independent
E(X-Y)= E(X)- E(Y)
E(c1*X+ c2*Y) =
c1*E(X) + c2*E(Y)
Var(X-Y) = Var(X) +Var(Y)
Var(c1X+ c2Y) = c12Var(X) +
c22Var(Y)+2c1c2Cov(X,Y)
3-45
Standardized Random Variables
X random variable
mean  standard deviation .
Transform X to Z
standardized random variable
Z 
X 

3-46
Standardized Random Variables
X 
Z

Z is dimensionless
random variable with
X 
X

E( Z)  E
  E   E 
  


3-47
Standardized Random Variables
X 
X




Var ( Z)  Var 
  Var    Var  
  


2

 
 2 0
 
3-48
Standardized Random Variables
Z 
X 

X and Z often same distribution
shifted by 
scaled down by 
mean 0, variance 1
3-49
Standardized Random Variables
Example:
Suppose wind speed W ~ (5, 22)
Normalized wind speed
W  5
Z 
2
3-50
Measures of Variation
WIND
Discrete r.v.
SOLAR
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)
 g( x 2 )f ( x 2 )dx 2

3-51
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)

 ( x 2   ) f ( x 2 )dx 2
2

3-52
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
3-53
Variance Example: continuous
X2 ~ 0.0107x2+0.01x
2<X<7
squared deviation from mean

E(X2 –
u)2 =
 ( x 2   ) f ( x 2 )dx 2
2

3-54
Variance Examples
X1 ~ x/6
X1 = 1 ... 6
X2 ~0.0107x22+0.01x2
2 < X2 < 6.213277
squared deviation from mean
E(Xi – u)2
Σ(x1 – )2P(x1)

 ( x 2   ) f ( x 2 )dx 2
2

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

 ( x 2   ) f ( x 2 )dx 2
2

1. Find E(X2) =




6.213
2
x2 f ( x2 )dx2  
6.213
2
x2 (.0107 x22  .01x2 )dx2
(.0107 x  .01x )dx2  .002675 x  .003333x
3
2
2
2
4
2
= 4.786178 - .069467 = 4.716711 = μ
3
2
6.213
2
3-57
Variance Example: continuous

2. Find Var(X2) = E[(X2 –

6.213
2
μ)2]
=
2
(
x
2


)
f ( x 2 )dx 2


( x2  4.716711) 2 (.0107 x22  .01x2 )dx2
 .00214 x  .022734 x  .047904 x  .111237 x
5
2
4
2
3
2
 1.719518  .532916  1.186602  Var( X 2 )
2
2
6.213
2
3-58
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
3-59
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 = ?
3-60
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
3-61
Show
moment and moment about
mean for continuous variable X2 above
Verify 2 = 2' - 2 = variance = E(X2) – E(X)2
0,1,2,3,4th
3-62
Moment Example
X ~ oil well ,
+1 (positive result) , probability = 1/2
–1 (negative result) , probability = 1/2
First, find the moment generating function:
E (e )  e (1 / 2)  e
t
t
 1 / 2( e  e )
tX
t ( 1)
t (  1)
Second, moments about the origin:
(1 / 2)
3-63
Moment Example slide 3-52
We have:
2
3
4
t
t
t
e  1 t 


 ....
2 ! 3! 4 !
t
2
e
t
3
4
t
t
t
 1 t 


 ....
2 ! 3! 4 !
3-64
Moment Example slide 3-52
Substituting in the moment generation function:
2
4
t
t
1 / 2( e  e )  1 
  ....
2! 4!
t
t
2
3
(1)
t
t
M X (t )  1  t  2  . 3  .... (2)
2!
3!
3-65
Moment Example slide 3-52
comparing equations 1 and 2 above
0
2  1
3  0
Odd moments are all zero
Even moments are all ones
See Schaum’s P. 93-94
3-66
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)
3-67
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
3-68
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
3-69
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
3-70
Characteristic function
always exist
() = E(ei X)
can get the density function from it
so can get moments
3-71
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)
 
 XY    ( x   X )( y   Y )f ( x, y)dxdy
 
 XY    ( x   X )( y   Y )f ( x, y)
x y
3-72
Joint Distribution…
Discrete r.v.
X ~ world oil prices
Y ~ oil production of Venezuela
f(x,y)=0.05x2+0.1xy+0.25y2
1< X<2
0<Y<1
3-73
Joint Distribution…
Discrete r.v.
f(x,y)=0.05x2+0.1xy+0.25y2
f(x,y)
1
2
f(y)
0
0.1
0.4
0.5
1
0.3
0.2
0.5
f(x)
0.4
0.6
1
E(X)=(1*0.4)+(2*0.6)=1.6
E(Y)= (0*0.5)+(1*0.5)=0.5
1< X<2 0<Y<1
3-74
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
3-75
Joint Distribution Covariance
Continuous Example
x ~ oil price
y ~ oil specific gravity API
and f ( x, y)  0.5( x 2 
0 < x <1
0 < y <2
xy)
1
2
0
0
 x  E ( X )    xf ( x , y )dxdy
1
2
   (0.5x  0.5x y )dxdy
3
0
0
2
3-76
Joint Distribution Covariance
Continuous Example
1
  0.5x y  0.25x y dx
3
2
0
1
2 2
0
1 4 1 31
  x  x dx  x  x 0
4
3
0
1 1
   0.583
4 3
3
2
3-77
Joint Distribution Covariance
Continuous Example
1
2
0
0
 y  E (Y )    yf ( x , y )dxdy
1
2
   (0.5x y  0.5xy )dxdy
2
0
0
2
3-78
Joint Distribution Covariance
Continuous Example
1
1 2 2 1 32
  x y  xy 0 dx
6
04
1
8
1 3 8 21
2
  ( x  x )dx  x  x 0
6
3
12
0
1 8
  1
3 12
3-79
Joint Distribution Covariance
Continuous Example
1
2
0
0
 xy    ( x   x )( y   y ) f ( x , y )dxdy
1
2
   ( x  0.58)( y  1)(0.5x  0.5xy )dxdy
2
0
1
0
  (0.33x  019
. x )dx  01
.
2
0
3-80
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
3-81
Correlation Coefficient
 XY

XY
correlation coefficient
measure dependence X and Y
dimensionless
X and Y independent  XY = 0   = 0
3-82
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.
3-83
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].
3-84
Correlation Coefficient: Discrete Case
E[X 2 ]    x 2 f ( x, y )
x
y
 1 * [ f (1,0)  f (1,1)]  2 * [ f (2,0)  f (2,1)]
2
2
 1 * f x (1)  2 * f x (2)
2
2
 0.4  (4 * 0.6)  2.8
3-85
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
3-86
Correlation Coefficient: Discrete Case
We can now figure out the correlation coefficient

cov(x, y)
 X Y
 0.1

 0.408
0.24 0.25
3-87
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.
3-88
Correlation Coefficient: Continuous Case
Finding var(X)…
var( X )  E[X ]  ( E[X]) 
2
2
1 2
   x 2 f ( x, y )dydx  (0.6) 2
0 0
1 2
   x (0.5 x  0.5 xy ) dydx  0.45
2
0 0
2
3-89
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
3-90
Correlation Coefficient: Continuous Case
Now we can find the correlation coefficient

cov(x, y)
 X Y
0.1

 0.124
(0.671)(1.2)
3-91
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
3-92
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
3-93
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)
3-94
Discrete Conditional Probabilities
P[X|Y]
=
P(X, Y) = joint probability of X & Y
P(Y) marginal probability of Y
(Y)
(X)
X1
X2
X3
P(Y)
Y1
Y2
P(X=X1,Y=Y1) P(X=X1,Y=Y2)
P(X=X2,Y=Y1) P(X=X2,Y=Y2)
P(X=X3,Y=Y1) P(X=X3,Y=Y2)
P(Y=Y1)
P(Y=Y2)
P(X)
P(X=X1)
P(X=X2)
P(X=X3)
1.00
3-95
Discrete Conditional Probabilities
Gas (Y) ft3
Oil (X) Bbl 500
750
2,000
0.10
0.05
1,000
0.10
0.25
500
0.25 0.25
P(Y)
0.45 0.55
P(X)
0.15
0.35
0.50
1.00
P[X=Xi| Y=Yi] = P(X= Xi,Y= Yi)
P (Y= Yi)
P[X=1000|Y=750] = 0.25/0.55 = 0.455
3-96
Discrete Conditional Expectation
Conditional Expectation:
E[Xi|Y=500]
E[Xi|Y=750]
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
3-97
Discrete Conditional Variance
Conditional Variance:
Var[X|Y=750]
Var[X|Y=500]
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
3-98
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!
3-99
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
3-100
Conditional Expectation of
Continuous Function
If X and Y have joint density function f(x,y), then

E (Y X  x)   yf ( y x)dy

Properties:
If X and Y are independent then
E(YX=x)=E(Y)

E( Y )   E( Y X  x )f ( x )dx
1

3-101
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
3-102
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
0
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
3-103
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
3-104
Chebyshev's Inequality
X random variable (discrete or continuous)
finite mean , variance 2, and  >0
P ( X    ) 


2
2
probability that X differs from its mean by
more than  is < variance divided by 2
3-105
Chebyshev's Inequality
P ( X    ) 
2
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%
3-106
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
 Sn

lim P
 
0


n 
 n

as sample size increases
 sample mean converges to true mean
3-107
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
3-108
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
3-109
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
3-110
Modes
BIMODAL Distribution
Amount of mineral
Amount of mineral
UNIMODAL Distribution
Grade
Grade
Skinner‘s Thesis (1976)
3-111
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.
3-112
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.
3-113
Percentiles
Divide area under density curve
 percent to left
X =  percentile
X10 = 10th percentile = decile
X50 = 50th percentile = median
P(x)
Area

x
x
3-114
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
3-115
Example of a Percentile
0X (3x3)dx=0.7
70
Find X70
Show computations to get x70
 x70=0.759
P(x)
Area
=70%
x
x70
3-116
Skewness
Coefficients of Skewness
describes symmetry of distribution
3 

E (X  )

3

3
3: Dimensionless quantity
>0  distribution skewed to the right
<0  distribution skewed to the left
=0  symmetric distribution
Other measures of skewness possible
3-117
Skewness
SKEWNESS – symmetry of a distribution
< 0 big tail to the left
--negatively skewed
If
> 0 big tail to the right
--positively skewed
= 0 symmetric
3-118
Example of Skewness
Wind farm p.d.f
0 ≤ w ≤ 10 m/s
f(w) = w/50
The equation for skewness is as follows:
3 

E ( X  )

3
3
 

3
3
3-119
Example of Skewness
Recall from slide 13, E[w] = 6.67 m/s
We need to find E[(W – E[W])2].
10
w
E[( W   ) ]   3   ( w  6.67) ( ) dw
50
0
3
3
10
  w  19.97 w  133.11w  296.148w dw
4
1
50
0
 - 7.248
3
2
3-120
Example of Skewness
Now we need to find σ3, which means we need var(w)
10
10
w
  var( w )   ( w   ) f ( w ) dw   ( w  6.67 ) ( ) dw
50
0
0
2
2
2
10
  w  13.3w  44.4 w dw  5.8
3
2
0
Since σ2 = 5.8, then σ = 2.41 and σ3 = 13.99
3-121
Example of Skewness
And we finally can compute skewness
 3  7.248
3  3 
 0.518

13.99
The distribution is slightly skewed to the left
3-122
Kurtosis
Coefficients of Kurtosis
describes distribution’s degree of peakedness
4 

E (X  )
4
4

4: Dimensionless quantity
<3  flatter than normal
>3 taller than normal curve
=3  normal curve
Other measures of kurtosis possible
3-123
Kurtosis
KURTOSIS = tallness or flatness
Which
curve has
kurtosis
>3?
3-124
Example of Kurtosis
Using the wind farm data compute Kurtosis
4 

E ( X   )4

4

3-125
Skewness and Kurtosis Using Eviews
8
Skewness=1.78>0
Kurtosis=8>3
Skewed to right
Taller than normal
6
4
2
0
Series: Residuals
Sample 1962:1 1967:4
Observations 24
Mean
Median
Maximum
Minimum
Std. Dev.
Skewness
Kurtosis
8.96E-17
0.013538
0.447731
-0.152261
0.124629
1.784240
7.990528
Jarque-Bera 37.63941
Probability 0.000000
-0.2
0.0
0.2
0.4
3-126
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
Median
Max
Min
Std. Dev.
Skew.
Kurt.
6
4
2
0
2.19E-16
0.000816
0.013372
-0.022015
0.010091
-0.592352
2.446336
Jarque-Bera1.781322
Probability 0.410384
-0.02
-0.01
0.00
0.01
3-127
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
3-128
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