FIN 30210:
Managerial Economics
Statistical
Analysis
Part I: Probability
The Cubs have
a 12% chance
of winning the
world series
this year
Here are the odds for
Blackjack….remember, what
happens in Vegas stays in
Vegas
Probability is about
having the truth
“Patriots have no need for probability, win coin
flip at impossible clip” (Nov. 4th, 2015)
“Belichick has also been extremely lucky. The Pats have won
the coin toss 19 of the last 25 times, according to the Boston
Globe's Jim McBride.”
So, what are the odds that the Patriots can win at least
19 out of 25 flips?
To do this, we need a probability distribution…for a coin toss,
we have the following.
Probability
Side Note: 50 Super Bowl Coin Tosses
1/2
Heads: 24 (48%)
Tails: 26 (52%)
Head
Tail
Outcome
So, suppose that we wanted the odds that the Patriots got 19 wins in a
row….
Probability ( A and B) = Probability(A) * Probability (B)
Probability ( A and B) = Probability(A) * Probability (B)
Probability
1/2
Head
Tail
Outcome
So, we want the probability of
19 Wins
P W and W and ... and W   .5 .5  ... .5 
 .5 
19
 0.00000191
The odds of dying from an asteroid collision
with earth in the next 100 years is 1 in 500,000
(.000191% - 1 in 523,560)
This isn’t really what we want though…getting 19 wins in a row is one of many ways to
get 19 out of 25
What are the odds that the Patriots get 24 out of 25 wins
Probability ( A and B) = Probability(A) * Probability (B)
Probability ( A or B) = Probability(A) + Probability (B)
There are LOTS of ways to get
exactly 24 out of 25 wins
One way would be
P  .50   0.0000000298
25
LWWWWWWWWWWWWWWWWWWWWWWWW
One way would be
(.00000298%)
24 Wins
WLWWWWWWWWWWWWWWWWWWWWWWW
23 Wins
P  .50   0.0000000298
25
(.00000298%)
What are the odds that the Patriots get 24 out of 25 wins
Probability ( A and B) = Probability(A) * Probability (B)
Probability ( A or B) = Probability(A) + Probability (B)
In Fact, there are 25 ways to get 24 out of 25 wins, so the answer
would be
P  25 .50   0.00000075
25
(.000075% - 1 in 1.3 million)
The odds of becoming a movie star are 1 in 1.5
million
The probability for a number of wins out of a certain number of tries is given by
a binomial distribution:
k successes in n tries. Probability of success is p
Note: 24 out of 25 wins
n!
nk
k
P  k , n, p  
p 1  p 
k ! n  k  !
P  24, 25,.5   25 .5 
So, the probability that the patriots get EXACTLY 19 out of 25 wins would be
25!
25 19
19
P 19, 25;.5  
.5 .5 
 .0052
19! 25  19  !
(.52% - 1 in 192)
So, the probability that the patriots get AT LEAST 19 out of 25 wins would be
The odds of Notre Dame winning the national
title in football this year are 1 in 40
25
25!
P
.5i *.525i  .0073
i 19 i ! 25  i  !
(.73% - 1 in 137)
25
Here’s the binomial distribution for 25 tosses
18
Odds of 12 or Less = 50%
Odds of 12 or Less = 50%
16
14
Probability (%)
12
10
8
6
4
2
0
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
31%
.73%
On the other side of the proverbial coin is losing the toss
a lot. In 2011, the Cleveland Browns lost 11 in a row.
P  .50   0.00049
11
(.049% 1 in 2040)
The odds of fatally slipping in the shower are 1
in 2500
In 2012, the Carolina Panthers lost 12 in a row.
P  .50   0.00024
12
(.024% - 1 in 4,166)
The odds of getting a hole in 1 in golf are 1 in
5,000
What are your odds of winning at
craps?
Easiest Bet – Playing the Pass Line
The Game of Craps – Playing the Pass Line
•
•
•
•
•
If you roll a 2,3,or 12, you lose “crap out”
If you roll a 7 or 11, you “win”
If you roll a 4,5,6,8,9,10 the rolled number becomes the “point”
If you roll the point, you win, if you roll a seven before rolling the
“point” you lose.
The Pass Line Pays Even Odds
Probability
1/6
Number
1
2
3
4
5
Total on 2 dice
Combinations
Probability
Percentage
2
1+1
1/36
3%
3
1+2, 2+1
2/36
6%
4
1+3, 2+2, 3+1
3/36
8%
5
1+4, 2+3, 3+2, 4+1
4/36
11%
6
1+5, 2+4, 3+3, 4+2, 5+1
5/36
14%
7
1+6, 2+5, 3+4, 4+3, 5+2, 6+1
6/36
17%
8
2+6, 3+5, 4+4, 5+3, 6+2
5/36
14%
9
3+6, 4+5, 5+4, 6+3
4/36
11%
10
4+6, 5+5, 6+4
3/36
8%
11
5+6, 6+5
2/36
6%
12
6+6
1/36
3%
6
Come Out
• Win = 23%
• Lose = 12%
• Roll Again = 65%
• 4 or 10 = 16%
• 5 or 9 = 22%
• 6 or 8 = 27%
18
16
14
Probability (%)
12
10
8
6
4
2
0
2
3
“Craps” (9%)
4
5
“Point” (33%)
6
7
8
“Win” (17%)
9
10
11
12
“Craps” (3%)
“Point” (33%)
“Win” (6%)
65 Percent of the time, you have a “point” to make
Come Out = 4,10 Next Roll
• Win = 8%
• Lose = 17%
• Roll Again = 75%
Come Out = 6,8 Next Roll
• Win =14%
• Lose = 17%
• Roll Again = 69%
17%
16
Probability (%)
Come Out = 5,9 Next Roll
• Win =11%
• Lose = 17%
• Roll Again = 72%
18
14%
14
12
14%
11%
11%
10
8%
8%
8
6%
16%
6
4
3%
3%
2
0
2
3
4
5
6
7
8
9
10
11
12
So, what's the probability that you win with a pass bet?
Total on 2 dice
Probability
Roll a Seven
6/36
Roll a 11
2/36
Roll a 4 and then roll another 4 before rolling a 7
(3/36)*PR(4 before a 7)
Roll a 5 and then roll another 5 before rolling a 7
(4/36)*PR(5 before a 7)
Roll a 6 and then roll another 6 before rolling a 7
(5/36)*PR(6 before a 7)
Roll a 8 and then roll another 8 before rolling a 7
(5/36)*PR(8 before a 7)
Roll a 9 and then roll another 9 before rolling a 7
(4/36)*PR(9 before a 7)
Roll a 10 and then roll another 10 before rolling a 7
(3/36)*PR(10 before a 7)
These are a bit tricky…..
What’s the probability the you roll a 4 before you roll a7?
Total on 2 dice
Probability
4
3/36
7
6/36
All Other #s
27/36
2
Roll a 4
Roll something
other than a 4 or
7, then roll a 4
A useful bit of math
1  x  x 2  x3  ... 
1
1 x
3
 3   27  3   27   3   27   3 
P  4 before 7                    ......
 36   36  36   36   36   36   36 
Roll something
other than a 4
or 7 twice,
then roll a 4
Roll something
other than a 4
or 7 three
times, then roll
a4
2
3


 3 
 27   27   27 
   1           ......
 36    36   36   36 

 3   36  3
    
 36   9  9
1
36

27
1  36 9
So, what's the probability that you win with a pass bet?
Event
Probability
Total on 2 dice
Probability
Percentage (approx.)
4 before a 7
3/9
Roll a Seven
6/36
16.67%
5 before a 7
4/10
Roll a 11
2/36
5.56%
6 before a 7
5/11
Roll a 4 and then roll another 4 before rolling a 7
(3/36)*(3/9) = 9/324
2.78%
8 before a 7
5/11
Roll a 5 and then roll another 5 before rolling a 7
(4/36)*(4/10) = 16/360
4.44%
9 before a 7
4/10
Roll a 6 and then roll another 6 before rolling a 7
(5/36)*(5/11) = 25/396
6.31%
10 before a 7
3/9
Roll a 8 and then roll another 4 before rolling a 8
(5/36)*(5/11) = 25/396
6.31%
Roll a 9 and then roll another 9 before rolling a 7
(4/36)*(4/10) = 16/360
4.44%
Roll a 10 and then roll another 10 before rolling a 7
(3/36)*(3/9) = 9/324
2.78%
Total
244/495
49.3%
The Game of Craps – Playing the Pass Line
•
•
•
•
•
If you roll a 2,3,or 12, you lose “crap out”
If you roll a 7 or eleven, you win “win”
If you roll a 4,5,6,8,9,10 the rolled number becomes the “point”
If you roll the point, you win, if you roll a seven before rolling the “point” you lose.
The Pass Line Pays even odds
Playing the Pass Line
This is known as
the “House
Edge”
Win = 49.3%
- Loss = 50.7%
-1.4%
“The Gambler’s Ruin”
A gambler playing a negative expected value game will
eventually go broke with probability one!!
Event
Probability
(approx.)
Pass Line Win
22.2%
Pass Line Loss
11.1%
4,or 10 Win
5.6%
4 or 10 Loss
11.1%
5 or 9 Win
8.9%
5 or 9 Loss
13.3%
6 or 8 Win
12.6%
6 of 8 Loss
15.2%
The Game of Craps – Playing the Pass Line
•
•
•
•
If you roll a 2,3,or 12, you lose “crap out”
If you roll a 7 or eleven, you win “win”
If you roll a 4,5,6,8,9,10 the rolled number becomes the “point”
If you roll the point, you win, if you roll a seven before rolling the
“point” you lose.
• The Pass Line Pays even odds
Playing the Pass Line
Expected Value measures the average outcome over a large
number of attempts, given the probabilities of each outcome.
EV   Pr  xi xi
Event
Probability
(approx.)
Pass Line Win
22.2%
Pass Line Loss
11.1%
4,or 10 Win
5.6%
4 or 10 Loss
11.1%
5 or 9 Win
8.9%
5 or 9 Loss
13.3%
6 or 8 Win
12.6%
6 of 8 Loss
15.2%
Playing the Pass Line
Expected Percentage loss (House Edge)
 $.0141 

 *100  1.41%
 $1 
For a $1 Pass Bet
Event
Probability Total Bet Payout
Expected
Payout
Expected Total
Bet
Pass Line Win
22.22%
$1
$1
.222
.222
Pass Line Loss
11.11%
$1
-$1
-.111
.111
4,or 10 Win
5.56%
$1
$1
.392
.0556
4 or 10 Loss
11.11%
$1
-$1
-.444
.1111
5 or 9 Win
8.89%
$1
$1
.623
.0889
5 or 9 Loss
13.33%
$1
- $1
-.665
.1333
6 or 8 Win
12.63%
$1
$1
.882
.1263
6 of 8 Loss
15.15%
$1
-$1
-.912
.1515
-.0141
1.00
Total
100%
Suppose that the first roll is a 4. I can now make an additional bet. I can
make a bet that a 4 is rolled before a 7. This is called “Playing the odds”
Event
Probability
4 before a 7
3/9
5 before a 7
4/10
6 before a 7
5/11
8 before a 7
5/11
9 before a 7
4/10
10 before a 7
3/9
Bet
Payout
4 or 10
2 to 1
5 or 9
3 to 2
6 or 8
6 to 5
The house pays odds equal
to the to the true odds, so
the house edge on this
additional bet are
ZERO!!!!!!!
This is the only fair bet in Vegas!!!
Suppose that you can bet twice your initial bet on the odds
Whatever your initial Pass/Don’t Pass Wager, you can up your bet on a point as follows
• You can bet 2X your initial bet if your point is 4 or 10 (Pays 2 to 1)
• You can bet 2X your initial bet if your point is 5 or 9 (Pays 3 to 2)
• You can bet 2X your initial bet if your point is 6 or 8 (Pays 6 to 5)
For a $1 Initial Bet – Playing Pass/w 2x odds
Event
Probability
Total Bet
Payout
Expected
Payout
Expected
Bet
Pass Line Win
22.22%
$1
$1
.222
.222
Pass Line Loss
11.11%
$1
-$1
-.111
.111
4,or 10 Win (Pays 2-1)
5.56%
$3
$5
.392
.167
4 or 10 Loss
11.11%
$3
-$3
-.444
.333
5 or 9 Win (Pays 3-2)
8.89%
$3
$4
.623
.267
5 or 9 Loss
13.33%
$3
- $3
-.665
.400
6 or 8 Win (Pays 6-5)
12.63%
$3
$3.40
.882
.379
6 of 8 Loss
15.15%
$3
-$3
-.912
.455
-.0141
2.33
Total
100%
Expected Percentage loss (House Edge)
 $.0141 

 *100  .605%
$2.33


The expected loss is the
same, but your overall bet
is bigger, so the percentage
loss is smaller!!
A Common Casino Betting System for Casino Craps is the “3-4-5” System
Whatever your initial Pass/Don’t Pass Wager, you can up your bet on a point as follows
• You can bet 3X your initial bet if your point is 4 or 10 (Pays 2 to 1)
• You can bet 4X your initial bet if your point is 5 or 9 (Pays 3 to 2)
• You can bet 5X your initial bet if your point is 6 or 8 (Pays 6 to 5)
For a $1 Initial Bet – Playing Pass/w 3-4-5 odds
Event
Probability
Total Bet
Payout
Expected
Payout
Expected
Bet
Pass Line Win
22.22%
$1
$1
.222
.222
Pass Line Loss
11.11%
$1
-$1
-.111
.111
4,or 10 Win (Pays 2-1)
5.56%
$4
$7
.392
.224
4 or 10 Loss
11.11%
$4
-$4
-.444
.444
5 or 9 Win (Pays 3-2)
8.89%
$5
$7
.623
.444
5 or 9 Loss
13.33%
$5
- $5
-.665
.666
6 or 8 Win (Pays 6-5)
12.63%
$6
$7
.882
.7578
6 of 8 Loss
15.15%
$6
-$6
-.912
.909
-.0141
3.77
Total
100%
Expected Percentage loss (House Edge)
 $.0141 

 *100  .374%
 $3.77 
The bigger the multiple
allowed, the smaller the
house edge!!
House Edge for other craps bets
Bet
House Edge
Pass/Come
1.41%
Don’t Pass/Don’t Come
1.36%
Pass/Come(2X odds)
.606%
Don’t Pass/Don’t Come (2X odds)
.466%
Place 6 and 8
1.52%
Place 5 and 9
4.00%
Place 4 and 10
6.67%
Buy 6 and 8
4.76%
Buy 5 and 9
4.76%
Buy 4 and 10
4.76%
Lay 6 or 8
4.00%
Lay 5 or 9
3.23%
Lay 4 or 10
2.44%
Field Bet
5.56%
Any Craps
11.11%
6 or 8 Hard way
9.09%
4 or 10 Hard way
11.10%
11 or 3
11.10%
2 or 12
13.90%
Any 7
16.70%
Here’s a comparison of casino edges on other games…
Craps
Other Games
House Edge when you Take the Odds
Game
House Edge (w/ proper play)
Table Odds Taken
Pass Line
Don’t Pass
Blackjack
0.5%
0x
1.41%
1.36%
Video Poker
0.5% - 5%
1x
0.848%
0.682%
Baccarat
1.06%
2x
0.606%
0.455%
Roulette
5.5%
3x
0.471%
0.341%
Slot Machines
0% – 17%
3-4-5x
0.374%
0.273%
Progressive Slots
5% -17%
5x
0.326%
0.227%
Keno
25%+
10x
0.184%
0.124%
Typical State Lottery
50%+
20x
0.099%
0.065%
100x
0.021%
0.014%
What are the odds that it
will be 80 degrees
tomorrow in South Bend?
As with the first two examples, this
involves a probability distribution
Just as with the coin flip or the dice roll, we can imagine a “truth” out there governing South Bend
temperatures. This “truth”, again, is in the form of a probability distribution.
Probability
x = temperature
 = mean
 = standard deviation
  3
  2
 

 
Temperature
  2
  3
Probability
We can use the normal distribution to get the probability that the temperature lies within
various ranges
 = mean
 = standard deviation
0.2%
  3
34%
2.3%
  2
34%
2.3%
0.2%
13.5%
13.5%
 

68%
95%
99.6%
 
  2
Temperature
  3
So, for example……
 = 60
 = 15
Probability
34%
2.3%
0.2%
34%
Temperature
Range
Probability
<15
0.2%
15 – 30
2.3%
30 – 45
13.5%
45 – 60
34%
60 – 75
34%
75 – 90
13.5%
90 – 105
2.3%
>105
.2%
2.3%
13.5%
0.2%
13.5%
Temperature
15
30
45
60
68%
95%
99.6%
75
90
105
Conditional distributions give us probabilities conditional on some observable information
What is the probability that the Temperature in
south bend is greater than 15 degrees
Probability
Probability
Unconditional
 = 10
 =5
 = 60
 = 15
Temp
15
99.8%
30
45
Conditional on February
60
75
90
105
5
0
5 10 15 20 25
16%
Temp
Part II: Statistics
Statistics is about
finding the truth
Law of large numbers: In statistics, as the number of identically distributed, randomly
generated variables increases, their sample mean (average) approaches their theoretical
mean. The law of large numbers was first proved by the Swiss mathematician Jakob
Bernoulli.
Number of data points increases
Jakob Bernoulli
1655 - 1705
# of Heads
Total Flips
1 N
x     xi
 N  i 1
1 N
x     xi
 N  i 1
2
 1  N
s 
x

x


 i
N

1

 i 1
2
1
Pr  Heads 
2
 1 2  3  4  5  6 

  3.5
6



(Population Mean)
2
(Population Variance)
Average Monthly Temperatures in Indiana from 1894 - 2016
Sample Statistics
• Average = 50.7
• Std. Dev. = 16.3
• High = 78.1
• Low = 22.1
We have Average Monthly temperatures (1894 – 2016)
for 36 locations across Indiana. This is what we would
call a “cross sectional” dataset (multiple observations at
a single point in time)
4.5
Sample Statistics
• Average = 50.7
• Std. Dev. = 16.3
• High = 78.1
• Low = 22.1
4
3.5
Frequency (%)
3
2.5
2
1.5
1
0.5
0
20
25
30
35
40
45
50
55
60
65
70
75
80
85
Suppose that we condition on “Northern Indiana” or “Southern Indiana”
Sample Statistics
• Average = 47.9
• Std. Dev. = 16.8
• High = 72.6
• Low = 22.1
We have Average Monthly temperatures (1894 – 2016)
for 36 locations across Indiana. This is what we would
call a “cross sectional” dataset (multiple observations at
a single point in time)
“Northern Indiana”
7
Frequency (%)
6
Sample Statistics
• Average = 47.9
• Std. Dev. = 16.8
• High = 72.6
• Low = 22.1
5
4
3
2
1
0
20
25
30
35
40
45
50
55
60
65
70
75
80
85
“Southern Indiana”
7
Frequency (%)
6
Sample Statistics
• Average = 53.1
• Std. Dev. = 15.7
• High = 78.1
• Low = 24.2
5
4
3
2
1
0
20
25
30
35
40
45
50
55
60
65
70
75
80
85
January - March
14
Frequency (%)
12
Sample Statistics
• Average = 32.7
• Std. Dev. = 6.5
• High = 47.5
• Low = 22.1
10
8
6
4
2
0
20
25
30
35
40
18
45
50
55
60
65
70
75
80
I could also condition on
Month(s) of the year
85 Temperature
June - August
Frequency (%)
16
14
Sample Statistics
• Average = 70.8
• Std. Dev. = 2.6
• High = 78.8
• Low = 65.1
12
10
8
6
4
2
0
20
25
30
35
40
45
50
55
60
65
70
75
80
85
Temperature
February
Frequency (%)
25
20
Sample Statistics
• Average = 32.7
• Std. Dev. = 3.5
• High = 37.7
• Low = 23.6
15
10
5
0
20
25
30
35
40
20
45
50
55
60
65
70
75
80
Or individual months of the
year
85 Temperature
July
Frequency (%)
18
16
Sample Statistics
• Average = 72.6
• Std. Dev. = 1.9
• High = 78.1
• Low = 69.0
14
12
10
8
6
4
2
0
20
25
30
35
40
45
50
55
60
65
70
75
80
85
Temperature
Northern Indiana in February
35
30
Sample Statistics
• Average = 25.9
• Std. Dev. = 1.4
• High = 27.4
• Low = 23.6
25
20
15
10
5
Or individual months of the
year and locations
0
20
25
30
35
40
45
50
55
60
65
70
75
80
85
35
Northern Indiana in July
30
Sample Statistics
• Average = 70.8
• Std. Dev. = 1.1
• High = 72.6
• Low = 69.0
25
20
15
10
5
0
20
25
30
35
40
45
50
55
60
65
70
75
80
85
For Indiana
x = 50.7
Probability
s = 16.3
“I’m 95% sure that the
temperature for September will
be between 18 and 83 degrees”
34%
2.3%
0.2%
34%
2.3%
13.5%
0.2%
13.5%
Temperature
1.8
18.1
34.4
50.7
68%
95%
99.6%
67
83.3
99.6
So, for example……for Northern Indiana in September
Sample Statistics
x = 61.9
Probability
s = 1.1
0.2%
“I’m 95% sure that the
temperature for September will
be between 60 and 64 degrees”
34%
2.3%
34%
2.3%
13.5%
0.2%
13.5%
Temperature
58.6
59.7
60.8
61.9
68%
95%
99.6%
63
64.1
65.2
Regressions are about estimating conditional
distributions
Linear Regressions make several key assumptions
•
•
•
•
•
Linear Relationship
Multivariate Normality
No or Little Multicollinearity
No Auto-correlation
Homoscedasticity
Independent Variable
x is N   x ,  x2 
yi     xi   i
Explained Variable
Parameters to
be estimated
Error Term
 is N  0,  2 
Cov   , X   0
yi     xi   i
Frequency
Conditional Distribution of Y
 / 2

   xi
xi
x
yi     xi   i
y
E  y / xi      xi
Var  y / xi    2
The OLS (Ordinary Least Squares) method
estimates the parameters alpha and beta by
minimizing the sum of squared errors.
y
min  i 1  yˆ  y 
N
2
ˆ , ˆ
yˆi  ˆ  ˆ xi
Estimated Coefficients
N
ˆ 
 ŷ  y 
  x  x  y  y 
i 1
i
N
  xi  x 
i 1
̂
ˆ  y  ˆ x
x
i
2
We also have a set of error terms

 i  yi  ˆ  ˆ xi

Frequency
E    0
Var      2

0
x
 ŷ  y 
These errors are a sampling of the
population of errors

Each regression gives us a sample of the distribution of errors (not the entire population of
errors). Therefore, the estimated coefficients are not the true coefficients, but rather, they
are samples drawn from a distribution of possible true parameter values
E ˆ   
 
x2 
2 1
Var ˆ     
2 
N
N

x 

2

Var ˆ 
N x2
 
E ˆ  
Frequency
Frequency




A few important things regarding these parameter estimates…
The estimated parameters are
drawn from a distribution with a
mean equal to the true parameter
value – we are not making biased
prediction!
 
E ˆ  
E ˆ   
These parameters are unknown, so we
need to estimate them from the data
2

Var ˆ 
N x2
 
1
x2 
Var ˆ     
2 
N
N

x 

2
1) The variance of the parameters is smaller
(the estimates are more precise)when the
variance of x is large
2) As the number of observations gets large,
the variance approaches zero – we learn the
truth!
Law of large numbers: In statistics, as the number of identically distributed, randomly
generated variables increases, their sample mean (average) approaches their theoretical
mean. The law of large numbers was first proved by the Swiss mathematician Jakob
Bernoulli.
Sample Estimates
Population Parameters
N
ˆ 
2

i 1
2
i
Number of observations gets big
N 2
1

x2
ˆ
ˆ
Var      

N
N

1
Var
x






2
 
Var ˆ 
ˆ
Var  x    x2
Number of observations gets big
x  x
Var ˆ   Var    0
Var  x    x2
2
 N  1Var  x 
2
Number of observations gets big
x  x
 
Var ˆ  Var     0
We also have some additional “diagnostics” to check the performance of the regression
Total Sum of Squares
N
TSS    yi  y 
2
i 1
=
Total Variation in the Data we
are trying to explain
Regression Sum of Squares
N
RSS  
i 1

yˆi  yˆ

2
+
Total Variation in the data we
have actually
Residual Sum of Squared
Residuals
N
SSE    i 2
i 1
Total Variation in the Data left
unexplained
Standard Error of the Regression
N
R Squared of the Regression
R2 
RSS
TSS
The percentage of the
variation of Y explained in the
regression
SE 
  yi  yˆ 
2
i 1
N 2
The is the average error of
our estimates
If we would like to make a forecast using our regression
data, we need to calculate the conditional distribution
E  y0 / x0   ˆ  ˆ x0
Frequency
2


x

x


1
0
2
Var  y0 / x0   ˆ 1  

 N  N  1Var  x  
1) Note that since our estimates are unbiased, our
forecasts will also be unbiased!
2) As our sample size gets bigger, the variance of
our forecasts goes down (our forecasts get more
precise)
3) If the variance of X is big, we get better forecasts
   x0
ŷ
A forecast is only as good as
the error attached to it!!!
ŷ
yˆi  2
yˆi  ˆ  ˆ xi
95% Confidence
Interval
yˆi  2
̂
E  y0 / x0   ˆ  ˆ x0
x
We always get the best forecast at
the sample average
x
2


x

x


1
0
2
Var  y0 / x0   ˆ 1  

N
N

1
Var
x






Example: Does the striped ground cricket
chirp differently at different temperatures?
Temperature (F)
20.0
88.6
16.0
71.6
19.8
93.3
18.4
84.3
17.1
80.6
15.5
75.2
14.7
69.7
17.1
82.0
15.4
69.4
16.2
83.3
15.0
79.6
17.2
82.6
16.0
80.6
17.0
83.5
14.4
76.3
22.0
20.0
Chirps/Sec.
Chirps Per Sec
18.0
16.0
14.0
12.0
10.0
65.0
70.0
75.0
80.0
85.0
Temperature (F)
90.0
95.0
Example: Does the striped ground cricket
chirp differently at different temperatures?
Chirps     Temp.  
Temperature (F)
20.0
88.6
16.0
71.6
19.8
93.3
18.4
84.3
17.1
80.6
15.5
75.2
14.7
69.7
17.1
82.0
15.4
69.4
16.2
83.3
15.0
79.6
17.2
82.6
16.0
80.6
17.0
83.5
14.4
76.3
22.0
20.0
Chirps/Sec.
Chirps Per Sec
18.0
16.0
14.0
12.0
10.0
65.0
70.0
75.0
80.0
85.0
Temperature (F)
90.0
95.0
y
x
Chirps Per Sec Temperature (F)
y  y
x  x
 y  y  x  x 
x  x
2
20.0
88.6
3.3
8.6
28.6
73.3
16.0
71.6
-0.7
-8.4
5.5
71.2
19.8
93.3
3.1
13.3
41.7
175.8
18.4
84.3
1.7
4.3
7.4
18.1
17.1
80.6
0.4
0.6
0.3
0.3
15.5
75.2
-1.2
-4.8
5.6
23.4
14.7
69.7
-2.0
-10.3
20.2
106.9
17.1
82.0
0.4
2.0
0.9
3.8
15.4
69.4
-1.3
-10.6
13.3
113.2
16.2
83.3
-0.5
3.3
-1.5
10.6
15.0
79.6
-1.7
-0.4
0.7
0.2
17.2
82.6
0.5
2.6
1.4
6.6
16.0
80.6
-0.7
0.6
-0.4
0.3
17.0
83.5
0.3
3.5
1.2
12.0
14.4
76.3
-2.3
-3.7
8.4
14.0
Average = 16.7
Average = 80.0
Sum=133.5
Sum =629.8
133.5
ˆ

 .21
629.8
N
ˆ 
  x  x  y  y 
i 1
i
N
i
  xi  x 
2
i 1
ˆ  y  ˆ x
ˆ  16.7  .21 80   .309
Chirps Per
Sec
Temperature (F) Predicted
Error Squared Error
20.0
88.6
18.5
-1.5
2.3
16.0
71.6
14.9
-1.1
1.3
19.8
93.3
19.5
-0.3
0.1
18.4
84.3
17.6
-0.8
0.7
17.1
80.6
16.8
-0.3
0.1
15.5
75.2
15.6
0.1
0.0
14.7
69.7
14.5
-0.2
0.1
17.1
82.0
17.1
0.0
0.0
15.4
69.4
14.4
-1.0
1.0
16.2
83.3
17.3
1.1
1.3
15.0
79.6
16.6
1.6
2.4
17.2
82.6
17.2
0.0
0.0
16.0
80.6
16.8
0.8
0.6
17.0
14.4
83.5
76.3
17.4
15.9
0.4
1.5
0.1
2.1
Average = 16.7
Average = 80.0
Variance = 2.89
Variance = 44.98
1

x2
Var ˆ   ˆ  

N
N

1
Var
x






2
N
ˆ 2 
2

i
i 1
N 2
 
Var ˆ 
ˆ 2 
ˆ 2
 N  1Var  x 
12.3
 .946
13
 
Var ˆ 
 
.946
 .00149
14  44.98 
SE ˆ  .00149  .0387
Sum = 12.3
1

802
Var ˆ   .946  
  9.66
15
14
44.98



SE ˆ   9.66  3.109
Actual
y
ŷ
Predicted
y  y
2

yˆ  yˆ

2
 y  yˆ 
20.0
18.5
11.2
3.3
2.3
16.0
14.9
0.4
3.2
1.3
19.8
19.5
9.9
7.9
0.1
18.4
17.6
3.1
0.8
0.7
17.1
16.8
0.2
0.0
0.1
15.5
15.6
1.3
1.1
0.0
14.7
14.5
3.8
4.8
0.1
17.1
17.1
0.2
0.2
0.0
15.4
14.4
1.6
5.1
1.0
16.2
17.3
0.2
0.5
1.3
15.0
16.6
2.7
0.0
2.4
17.2
17.2
0.3
0.3
0.0
16.0
16.8
0.4
0.0
0.6
17.0
14.4
17.4
15.9
0.1
5.1
0.5
0.6
0.1
2.1
Sum = 28.3
Sum = 12.3
Average = 16.7 Average = 16.7 Sum = 40.6
R2 
28.3
 .70
40.6
Regression Sum of Squares
Total Sum of Squares
2
N
TSS    yi  y 
N
RSS  
2
i 1
i 1

yˆi  yˆ
R Squared of the Regression
RSS
R 
TSS
2
Standard Error of the Regression
N
SE 
SE 
12.3
 .97
13
  yi  yˆ 
i 1
N 2
2

2
Chirps  .309  .211 85   17.7
E  y0 / x0   ˆ  ˆ x0
 1  85  80 2 
Var  y / 85   .946 1  
  3.43
15
14
44.98





2


x

x


1
0
2
Var  y0 / x0   ˆ 1  

N
N

1
Var
x
    

SD  y / 85   3.43  1.85
Chirps  .309  .211Temp.
22.0
So, lets calculate a prediction
for the number of chirps at a
temperature of 85 degrees
20.0
17.7
18.0
95% Confidence Interval
17.7+/-2(1.85) = [21.4, 14]
Chirps/Sec.
16.0
14.0
12.0
10.0
65.0
70.0
75.0
80.0
85.0
Temperature (F)
90.0
95.0
Mult R  R 2  .70
Here are the results of the
regression done in Excel
1  R   N  1

1  .70 15  1

 1
 1
2
SUMMARY OUTPUT
Adj R 2
N  p 1
15  1  1
Regression Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
Observations
0.84
0.70
0.67
0.97
15
FStat 
RSS 28.29

 29.97
MSE
.94
ANOVA
df
Regression
Residual
Total
SS
1
13
14
28.29
12.27
40.56
-0.31
0.21
Standard Error
3.11
0.04
Coefficients
Intercept
Temperature (F)
MS
F
28.29
0.94
t Stat
29.97
P-value
-0.10
5.47
0.92
0.00
Significance F
0.00
Lower 95%
-7.02
0.13
Upper 95%
6.41
0.30
How does taking LSD affect your
performance on a math test?
Score     Conc.  
90
80
70
Test Score
60
50
40
30
20
“Correlation of performance test scores with “tissue
concentration” of Lysergic Acid Diethylamide in human
subjects”
John Wagner, George Aghajanian, and Oscar Bing
March 22, 1968
10
0
1
2
3
4
Tissue Concentration
5
6
7
How does taking LSD affect your
performance on a math test?
SUMMARY OUTPUT
Regression Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
0.963
0.927
0.912
5.710
Observations
7.000
Score  90.3  9.6  Conc.  
ANOVA
df
Regression
Residual
1
5
SS
2056.44358
163.0445059
Total
6
2219.488086
Coefficients
90.29465244
Standard Error
5.647752669
-9.565531392
1.204533226
Intercept
Tissue Conc (x)
MS
F
2056.44358 63.06387
32.60890117
Significanc
eF
0.00051
Lower
Upper
t Stat
P-value Lower 95% Upper 95% 95.0%
95.0%
15.98771365 1.74E-05 75.77664 104.8127 75.77664 104.8127
-7.941276492
0.00051
-12.6619
-6.46918
-12.6619
-6.46918
What would your predicted score be with
a concentration of 4?
Score  90.3  9.6Conc.  
90
Mean x = 4.3
Var(x) = 3.7
80
E  y0 / x0   90.3  9.6  4   51.9
70
 1  4  4.32 
Var  y / 4   32.6 1  
  37.4
7
6
3.7
  

50
SD  y / 4   37.4  6.1
Test Score
60
40
30
20
10
0
95% Confidence Interval
51.9 +/- 12.2
64.1, 39.7
1
2
3
4
Tissue Concentration
5
6
7
What about the possibility of a non-linear
relationship between LSD usage and math
performance?
Score     Conc.  

Score
Conc
Beta measures the unit change in
test score per unit change in LSD
concentration
VS
Score  e Conc 

%Score
Conc
Beta measures the percentage
change in test score per unit
change in LSD concentration
Both functional forms indicate a negative relationship, but one is linear
while the other is non-linear
120
100
80
60
40
Score   e  Conc.
Score     Conc.
20
0
0
1
2
3
4
5
6
7
I can estimate this nonlinear relationship
through a transformation of variable?
Score  e Conc 
Take the natural log of both sides…

ln Score   ln e
Conc 

A little math here…
ln Score   ln   Conc  
Define a new constant…
ln Score   Constant  Conc  
How does taking LSD affect your
performance on a math test?
ln  Score   4.64  .19  Conc.  
SUMMARY OUTPUT
ln    4.64
Regression Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
Observations
0.961663013
0.924795751
0.909754901
0.114937569
7
  e4.64  103.5
Score  103.5.e.19Conc 
ANOVA
df
Regression
Residual
Total
Intercept
Tissue Conc (x)
1
5
6
Coefficients
4.648346245
-0.19010758
SS
0.812264483
0.066053224
0.878317707
Standard Error
0.113676212
0.024244471
MS
0.812264483
0.013210645
Significan
F
ce F
61.4856 0.000541
Lower
Upper
Lower
Upper
t Stat
P-value
95%
95%
95.0%
95.0%
40.89110775 1.65E-07 4.356132 4.94056 4.356132 4.94056
-7.841275714 0.000541 -0.25243 -0.12779 -0.25243 -0.12779
What would your predicted score be with
a concentration of 4?
ln  Score   4.64  .19  Conc.  
90
Mean x = 4.3
Var(x) = 3.7
80
E ln y / 4   4.64  .194   3.88
e 3.88  48.4
 1 4  4.32 
Var ln y / 4   .0131  
  .015


7
6
3
.
7


SDln y / 4   .015  .122
Test Score
70
60
50
40
30
20
12.2%
10
0
95% Confidence Interval
48.4 +/- 24.4%
1
2
3
4
Tissue Concentration
60.2, 36.6
5
6
7
A linear regression can capture several
different non-linear relationships by
transforming the variables!
Functional Form
y   x
y  e   x 
y  ln  x  e 
y   x  e
Regression Equation
y   x
ln y     x  
y  ln    ln x  
ln y  ln    ln x  
Interpretation
A one unit change in X causes a
Beta units change in Y
A one unit change in X causes a
Beta percent change in Y
A one percent change in X causes
a Beta units change in Y
A one percent change in X causes
a Beta percent change in Y
“Northern Indiana”
7
Sample Statistics
• Average = 47.9
• Std. Dev. = 16.8
6
Frequency (%)
5
4
3
2
1
0
20
25
30
35
40
45
50
55
60
65
70
75
80
85
Indiana
4.5
Sample Statistics
• Average = 50.7
• Std. Dev. = 16.3
4
Frequency (%)
3.5
3
2.5
2
1.5
1
0.5
0
20
25
30
35
40
45
50
55
60
65
70
75
80
85
I could
accomplish the
same thing with a
“temperature
dummy”
Distribution for Northern Indiana
Temperature    DN  
E Temp / 1  51.8  3.9  47.9
 1, if North
DN  
0, if Not
2


1
1  .28 
Var Temp / 1  263.11 

  265.3


432
431
.
20


SDTemp / 1  265.3  16.3
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.107015815
R Square
0.011452385
Adjusted R
Square
0.009153437
Standard Error
16.22029062
Observations
Average Temperature for “Not Northern
Indiana”
Mean x = .28
Var(x) = .20
Temperature  51.8  3.9 DN  
432
ANOVA
df
Regression
Residual
1
430
SS
1310.641907
113132.0659
Total
431
114442.7079
Coefficients
51.79651334
Standard Error
0.918293128
t Stat
56.40520632
P-value
8.0168E-201
Lower 95%
49.99161169
Upper 95%
53.60141498
-3.888803317
1.742338706
-2.231944514
0.026133096
-7.313363408
-0.464243226
Intercept
North
MS
1310.641907
263.0978278
F
4.981576314
Significance F
0.026133096
“Southern Indiana”
7
Frequency (%)
6
5
Sample Statistics
• Average = 53.1
• Std. Dev. = 15.7
4
3
2
1
0
20
25
30
35
40
45
50
55
60
65
70
75
80
Indiana
4.5
4
Sample Statistics
• Average = 50.7
• Std. Dev. = 16.3
3.5
Frequency (%)
85
3
2.5
2
1.5
1
0.5
0
20
25
30
35
40
45
50
55
60
65
70
75
80
85
Suppose that I
repeat the
process for
Southern Indiana
Distribution for Northern Indiana
Temperature     DS  
E Temp /1  49.3  3.8  53.1
 1, if South
DS  
0, if Not
2

1  .39  

1
Var Temp /1  262.7 1 

  264.2
432
431
.24




SD Temp /1  264.2  16.3
Mean x = .39
Var(x) = .24
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.114475818
R Square
0.013104713
Adjusted R Square
0.010809608
Standard Error
16.20672908
Observations
432
Average Temperature for “Not Southern
Indiana”
Temperature  49.3  3.8 DS  
ANOVA
df
Regression
Residual
Total
Intercept
South
1
430
431
SS
1499.738841
112942.969
114442.7079
MS
1499.738841
262.6580675
F
5.709852566
Significance F
0.017299906
Coefficients
49.22994863
3.822021173
Standard Error
0.997455223
1.59948673
t Stat
49.35554748
2.389529779
P-value
3.1797E-179
0.017299906
Lower 95%
47.26945418
0.678236097
Upper 95%
51.19044307
6.96580625
Temperature      N DN   S DS  
 1, if North
DN  
0, if Not
 1, if South
DS  
0, if Not
Northern Indiana
E Temp / DN  1  50.3  2.7  47.6
Southern Indiana
E Temp / DS  1  50.3  2.9  53.2
SUMMARY OUTPUT
Average Temperature for Central Indiana
Regression Statistics
Multiple R
0.134093809
R Square
0.017981149
Adjusted R Square
0.013402973
Standard Error
16.18547051
Observations
432
Temperature  50.3  2.7 DN  2.9 DS  
ANOVA
df
Regression
Residual
Total
Intercept
North
South
2
429
431
SS
2057.811439
112384.8964
114442.7079
MS
1028.905719
261.9694555
F
3.927578951
Significance F
0.020403391
Coefficients
50.34616241
-2.728522571
2.900701864
Standard Error
1.255855211
1.869422778
1.717587401
t Stat
40.0891456
-1.459553507
1.68882344
P-value
3.6261E-147
0.145144669
0.09197992
Lower 95%
47.87776753
-6.402890097
-0.475231843
Upper 95%
52.81455728
0.945844954
6.27663557
Be mindful of what hypothesis you are testing!
Northern Dummy Only (T-stats in parentheses)
Temperature  51.8  3.9 DN  
(-2.2)
Northern Indiana has a different
average temperature than the rest
of the state
Significant
Southern Dummy Only (T-stats in parentheses)
Temperature  49.3  3.8 DS  
(2.4)
Northern Indiana has a different
average temperature than the rest
of the state
Significant
Northern and Southern Dummy (T-Stats in parentheses)
Temperature  50.3  2.7 DN  2.9 DS  
(-1.5)
(-1.7)
Not Significant!!!
Northern and Southern Indiana
have a different average
temperature than central Indiana
Suppose I put in Dummies for all three regions…
Temperature     N DN   C DC   S DS  
1, if Central
DC  
 0, if Not
There is no other region, so we know that
DN  DC  DS  1
One of our assumptions is violated!
•
•
•
•
•
Linear Relationship
Multivariate Normality
No or Little Multicollinearity
No Auto-correlation
Homoscedasticity
Example: The Famous 2000 Election
Al Gore
Democrat
Pat Buchanan
Reform Party
George W. Bush
Republican
Ralph Nader
Green Party
The Case of Palm
Beach County
Overall State Results
Candidate
Vote Total
Percentage
George W. Bush
2,909,815
49.039
Al Gore
2,909,578
49.035
Ralph Nader
96,844
1.633
Pat Buchanan
17,358
.293
Total
5,933,595
100.000
Palm Beach County Results
Candidate
Vote Total
Percentage
George W. Bush
152,954
35.44
Al Gore
269,696
62.48
Ralph Nader
5,564
1.29
Pat Buchanan
3,407
.79
Total
431,621
100.00
Did Pat Buchanan REALLY get 3,407 votes in Palm Beach
County
The Strategy: Use available data on demographics
from the counties in Florida (omitting Palm Beach
County) to estimate a relationship between
demographics and Pat Buchanan's vote total
“Are a function of”
B  F D
Pat
Buchanan’s
Votes
Demographic
Statewide Average
Palm Beach
% Black
15.9%
14.4%
% Hispanic
6.3%
9.8%
% Over 65 yrs.
16.9%
23.7%
% College Degree
13.9%
22.1%
Income (in thousand)
26.188
33.518
Observable
Demographics
Using Palm Beach demographics, forecast Pat
Buchanan’s vote total for Palm Beach
BPB  F DPB 
Turns out, the best fitting regression was as follows
LN  P      2 B   2 A65  3 H   4C   5 I  
Buchanan Votes
Total Votes
*100
Variable
Coefficient
Standard Error
t - statistic
Intercept
2.146
.396
5.48
Black (%)
-.0132
.0057
-2.88
Age 65 (%)
-.0415
.0057
-5.93
Hispanic (%)
-.0349
.0050
-6.08
College (%)
-.0193
.0068
-1.99
Income (000s)
-.0658
.00113
-4.58
R Squared = .73
Demographic
Palm Beach
% Black
14.4%
% Hispanic
9.8%
% Over 65 yrs.
23.7%
% College Degree
22.1%
Income (in thousand)
33.518
LN P   2.146  .0132B   .0415 A65   .0349H   .0193C   .0658I 
E  LnP / D   2.004
P  e 2.004  .134%
.00134431,621  578
This would be our prediction for Pat
Buchanan’s vote total!
+/- 2 Standard Deviation Confidence Interval
Var  LnP / D   .065
SD  LnP / D   .065  .2556
(25.56%)
.134(1.5112)  .2025%
.002025 431, 621  874
.134  0.4888   .065%
.00065 431, 621  280
Demographic
Palm Beach
% Black
14.4%
% Hispanic
9.8%
% Over 65 yrs.
23.7%
% College Degree
22.1%
Income (in thousand)
33.518
LN P   2.146  .0132B   .0415 A65   .0349H   .0193C   .0658I 
Frequency
Event
Odds
Win the Powerball
1 in 292,000,000
Struck by Lightning
1 in 960,000
Crushed by a Vending Machine
1 in 112,000,000
Becoming a Movie Star
1 in 1,505,000
Having Identical Quadruplets
1 in 15,000,000
7 Standard Deviations from the mean
• 1 in 390,882,215,445
280
578
+/1 2 Standard Deviations
874
3, 407
Votes
Speaking of Election, who
will win this year’s
election?
VS
Let’s ask Ray Fair….he should have a pretty good idea!
V    1  G * I    2  P * I   3  Z * I    4  DPER   
Democratic Share of Two
Party Presidential Vote
Variable
 DUR    6 I  
The Fair Presidential Election Model
Description
Coefficient Value
(T-Statistic)
Constant
47.75 (79.15)
Average Annual Growth in Real Per Capita GDP (First three
quarters of election year)
.667 (5.79)
P
Average Annual Growth in GDP Deflator (for first 15 quarters
of the current administration)
-.690 (-2.34)
Z
# of Quarters of the current administration with annual real
GDP per capita growth exceeds 3.2%
.968 (4.03)
DPER
1 if Democratic incumbent is running again, -1 if Republican
incumbent is running again, otherwise, 0
3.01 (2.14)
DUR
1 (-1) if Democrat (Republican) has been in office for 2 terms.
0 if ether party in for 1 term
-3.80 (-3.10)
1 If a democrat is the incumbent, -1 if a Republican is the
incumbent
-1.56 (-0.71)
Const
G
Ray Fair
Yale University
5
I
R Squared = .912
Democratic Share of Two Party Presidential Vote
V    1  G * I    2  P * I   3  Z * I    4  DPER   
5
 DUR    6 I  
Predictors for the 2016 Presidential Election
Description
Value
Average Annual Growth in Real Per Capita GDP (First three quarters of election year)
.87% (Estimated)
Average Annual Growth in GDP Deflator (for first 15 quarters of the current administration)
1.28%
# of Quarters of the current administration with annual real GDP per capita growth exceeds
3.2%
3 Quarters out of 15
1 if Democratic incumbent is running again, -1 if Republican incumbent is running again,
otherwise 0
0 (No)
1 (-1) if Democrat (Republican) has been in office for 2 terms. 0 if ether party in for 1 term
1 (2 Terms)
Democrat Incumbent
1 (Yes)
Since 1908, the Fair Model has correctly predicted 23 out of 27
elections (85% Success Rate)
He predicted every
election between
1908 and 1960
correctly!!
Ray Fair
Election
Year
Candidates
Predicted
Democrat
Predicted
Republican
Actual
Democrat
Actual
Republican
1960
Kennedy (D) vs. Nixon (R)
51.3
48.7
50.1
49.9
1964
Johnson (D) vs. Goldwater (R)
55.3
44.7
61.3
38.7
1968
Humphrey (D) vs. Nixon (R)
49.0
51
49.6
50.4
1972
McGovern (D) vs. Nixon (R)
39.9
60.1
38.2
61.8
1976
Carter (D) vs. Ford (R)
49.2
50.8
51.1
48.9
1980
Carter (D) vs. Reagan (R)
46.6
53.4
44.7
55.3
1984
Mondale (D) vs. Reagan (R)
42.8
57.2
40.1
59.9
1988
Dukakis (D) vs. Bush (R)
45.0
55
46.0
54
1992
Clinton (D) vs. Bush (R)
48.8
51.2
53.6
46.4
1996
Clinton (D) vs. Dole (R)
53.2
46.8
54.7
45.3
2000
Gore (D) vs. Bush (R)
49.3
50.7
50.3
49.7
2004
Kerry (D) vs. Bush (R)
45.5
54.5
48.8
51.2
2008
Obama (D) vs. McCain (R)
55.2
44.8
53.7
46.3
2012
Obama (D) vs. Romney (R)
49.0
51.0
51.3
48.7
Democratic Share of Two Party Presidential Vote
V    1  G * I    2  P * I   3  Z * I    4  DPER   
5
 DUR    6 I  
And the winner is…..
VS
45.0%
55.0%
prediction error  3%
Congratulations to President Elect Trump!
Cross Sectional Regressions vs. Time Series Regressions
y
y
0
1
2
3
A cross sectional regression focusses on
variations across locations (or other
factors) at a single point in time
t
0
1
2
3
t
A time series regression focusses entirely
on variation across time (ignoring variation
across location (or other factors)
Luckily, all the tools from the cross-sectional analysis carries over to
time series analysis
yt     t   t
ŷ
ŷ  ˆ  ˆ t
Time indicator
(t = 0,1,2,3,4)
All the properties of the
estimates are the same!
 
E ˆ  
E ˆ   
̂
1
t 2 
Var ˆ     
2 
 N N t 
2
2

Var ˆ 
N t2
 
0
t
We can forecast just as we did before as well
ŷ
In Sample
yˆi  2
ŷ  ˆ  ˆ t
The further out into the
future you try and predict,
the bigger your errors get!
yˆi  2
2


t

t


1
0
2
Var  y0 / x0   ˆ 1  

N
N

1
Var
t
    

̂
t
t
The longer your sample
period is, the better you
do!!
Example: South Bend Daily High Temperature (2013 – 2014)
16
14
Frequency (%)
12
10
8
Sample Statistics
• Average = 57.8
• Std. Dev. = 21.9
• Median = 60.1
• Mode = 84
• High = 97
• Low = 1.2
6
4
2
0
0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100 105
Temperature
Here’s the time series representation for Temperature in South Bend
120
100
80
60
40
20
0
1/2013
4/2013
7/2013
10/2013
1/2014
4/2014
7/2014
10/2014
Temp    1t  
Daily Observations
t = 0 is 1/1/2013
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.09
R Square
0.01
Adjusted R
Square
0.01
Standard Error
21.82
Observations
730.00
Temperature rises by .01
degrees per day (3.65
degrees per year)!!!
Temp  54.3  .01t  
ANOVA
df
Regression
Residual
Total
Intercept
Time
1.00
728.00
729.00
SS
2934.51
346550.24
349484.76
Coefficients
Standard Error
54.36
1.61
0.01
0.00
MS
2934.51
476.03
F
Significance F
6.16
0.01
t Stat
P-value
33.69
0.00
2.48
0.01
Lower 95%
51.20
0.00
Global Warming!
Somebody call Al Gore!!!
Upper 95%
57.53
0.02
120
Obviously, we have work to do!
100
80
Temp  54.4  .01t
60
40
20
0
1/2013
4/2013
7/2013
10/2013
1/2014
4/2014
7/2014
10/2014
My Sample is from
1/1/20013 to 12/31/2014.
Suppose that I want to
predict the temperature
for my birthday this year
Temp.
Date
Time
1/1/2013
0
12/31/2014
729
9/28/2016
1366
In Sample
E Temp / t  1,366   54.36  .011366   68.0
2

1 1,366  365  
Var Temp / t  1,366   476 1 

  491.3
 730  729  44, 469  
SD Temp / t  1,366   22.1
Out of Sample
120
112.2
100
80
68
60
40
23.8
20
0
1/1/2013
t=0
12/31/2013
t = 729
09/28/2016
t = 1,366
Time
120
Q1
Q4
Q3
Q2
There is obviously a regular
pattern here!!!
100
80
60
40
20
0
1/2013
4/2013
7/2013
10/2013
1/2014
4/2014
7/2014
10/2014
Lets Use some quarterly dummies
Dummies for quarters 1,2,3
Temp    1t   2 D1  3 D2   4 D3  
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.83
R Square
0.69
Adjusted R Square
0.69
Standard Error
12.28
Observations
730.00
Temp. in 4th
Quarter
Daily Observations
T = 0 is 1/1/2013
Temp  49.2  .004t  14.8D1  22.8D2  31.4 D3  
ANOVA
df
Regression
Residual
Total
Intercept
Time
Q1
Q2
Q3
4.00
725.00
729.00
SS
240117.82
109366.94
349484.76
Coefficients Standard Error
49.21
1.53
-0.00368
0.00
-14.75
1.45
22.76
1.36
31.43
1.30
MS
60029.45
150.85
F
Significance F
397.94
0.00
t Stat
P-value
32.13
0.00
-1.49
0.14
-10.14
0.00
16.72
0.00
24.17
0.00
Lower 95%
46.20
-0.01
-17.60
20.09
28.88
Upper 95%
52.22
0.00
-11.89
25.44
33.99
Looks like global
warming is just a
myth after all!
120
This look a lot better!
Temp  49.2  .004t  14.8D1  22.8D2  31.4 D3
100
80
60
40
20
0
1/2013
4/2013
7/2013
10/2013
1/2014
4/2014
7/2014
10/2014
My Sample is from
1/1/20013 to 12/31/2014.
Suppose that I want to
predict the temperature
for my birthday this year
Date
Time
1/1/2013
0
12/31/2014
729
9/28/2016
1366
E Temp / t  1,366   49.21  .003 1366   31.43  76.5
2

1 1,366  365  
Var Temp / t  1,366   150.8 1 

  155.7
730
729
44,
469
 
 

SD Temp / t  1,366   12.4
Temp.
Out of Sample
In Sample
101.3
Average (Q3,Q4)
Fourth Quarter
85.6
69.9
76.5
60.8
45
51.7
36
20.27
1/1/2013
t=0
12/31/2013
t = 729
12/31/2015
t = 729
09/28/2016
t = 1,366
Time
Just as with cross sectional analysis, I can capture non-linear
relationships by a transformation of the data
Linear Growth (unit change per unit time)
y    1t  
Linear Growth (percentage change per unit time)
ln y  ln  e   t  
y  e   t 
Take logs on
both sides
ln y     t  
Take logs on
both sides
Ln Temp     1t  
Daily Observations
t = 0 is 1/1/2013
SUMMARY
OUTPUT
Temperature rises by .021 percent per
day (7.6% per year)!!!
Regression Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
Observations
ln Temp   3.89  .00021t  
0.09
0.01
0.01
0.49
730.00
Global Warming!
Somebody call Al Gore!!!
ANOVA
df
Regression
Residual
Total
Intercept
Time
1.00
728.00
729.00
SS
1.37
172.58
173.95
Standard
Coefficients
Error
3.89
0.04
0.00021
0.00
MS
F
1.37
0.24
5.76
t Stat
107.95
2.40
P-value
0.00
0.02
Significance F
0.02
Lower 95%
Upper 95%
3.82
0.00
3.96
0.00
Lets Use some quarterly dummies
Dummies for quarters 1,2,3
ln Temp     1t   2 D1  3 D2   4 D3  
Temp. in 4th
Quarter
SUMMARY OUTPUT
Daily Observations
T = 0 is 1/1/2013
Regression Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
Observations
0.76
0.58
0.57
0.32
730.00
ln Temp   3.87  .00014t  .41D1  .42 D2  .55D3  
ANOVA
df
Regression
Residual
Total
Intercept
Time
Q1
Q2
Q3
4.00
725.00
729.00
SS
100.34
73.61
173.95
Standard
Coefficients
Error
3.87
0.04
-0.0001359
0.00
-0.41
0.04
0.42
0.04
0.55
0.03
MS
F
25.09
0.10
247.08
t Stat
97.37
-2.13
-10.79
11.75
16.42
P-value
0.00
0.03
0.00
0.00
0.00
Significance F
0.00
Lower 95%
3.79
0.00
-0.48
0.35
0.49
The second quarter
(Apr-June) is 42%
warmer than the 4th
quarter (Oct-Dec)
Let’s Compare the forecasts For September 28, 2016 (t=1,366)
No Dummies
Quarterly Dummies
ln Temp   3.89  .00021t  
ln Temp   3.87  .00014t  .41D1  .42 D2  .55D3  
E  ln Temp / t  1,366   3.89  .000211366   4.18
e4.18  65.4
Best Guess
2

1 1,366  365  
Var Temp / t  1,366   .24 1 

  .25
730
729
44,
469
 
 

SD Temp / t  1,366   .50 (50%)
Prediction: I’m 95% sure the
temperature will be between
32.7 and 130.8 degrees
E  ln Temp / t  1,366   3.87  .00014 1366   .55  4.61
e4.61  100.5
Best Guess
2

1 1,366  365  
Var Temp / t  1,366   .10 1 

  .103
 730  729  44, 469  
SD Temp / t  1,366   .32 (32%)
Prediction: I’m 95% sure the
temperature will be between
36.1 and 166.2 degrees
The Moral of the Story…
The exponential growth model is much more sensitive
to parameter changes than the liner model!!
12
  .30
  .25
  .20
10
8
600
  .30
500
400
6
300
4
2
0
0
3
6
9
12
15
18
200
  .25
100
  .20
0
0
Linear
3
6
9
Exponential
12
15
18
Gas Price: US Regular all Formulations
4.500
4.000
Exponential Trend
Dollars per Gallon
3.500
3.000
2.500
$2.37
2.000
1.500
1.000
0.500
0.000
1990
1992
1994
1996
1998
2000
2002
2004
2006
2008
2010
2012
2014
Source: US. Energy Information Administration
Let’s assume
exponential
growth
Price  e  1t  2 D1  3 D2  4 D3 
ln  Price     1t   2 D1  3 D2   4 D3  
SUMMARY OUTPUT
Gas prices increase (on average)
.45% per month (5.4% per year)
Regression Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
Observations
0.8929
0.7973
0.7946
0.2073
310
ln  Price   .14  .0045t  .0314 D1  .0568D2  .0692 D3  
ANOVA
df
Regression
Residual
Total
Intercept
Time
Q1
Q2
Q3
SS
MS
4
305
309
51.5481
13.1081
64.6562
12.8870
0.0430
Coefficients
-0.1378
0.0045
-0.0314
0.0568
0.0692
Standard Error
0.0308
0.0001
0.0332
0.0332
0.0334
t Stat
-4.4674
34.4261
-0.9452
1.7103
2.0711
Gas prices are (on average) 9.6%
higher in the 3rd quarter (July-Sept)
than they are in the 4th Quarter
(Oct-Dec)
Using the regression to seasonally adjust the data
ln  Price     2 D1  3 D2   4 D3     1t  
ln  P 
e
ln  P 
P
Seasonally
Adjusted
Price
ln  P     1t   2 D1  3 D2   4 D3  
Just to check, suppose that I run a
regression with my seasonally adjusted
price…
Intercept
Time
Q1
Q2
Q3
Coefficients
-0.1378
0.0045
0.0000
0.0000
0.0000
Standard Error
0.0308
0.0001
0.0332
0.0332
0.0334
t Stat
-4.4674
34.4261
0.0000
0.0000
0.0000
Let’s look at the residuals for a moment…
ln  Price     1t   2 D1  3 D2   4 D3   
Percentage difference between
predicted price and actual price
 
ln Pˆ
0.60
0.40
0.20
0.00
1990
1992
1994
1996
1998
2000
2002
2004
2006
2008
2010
2012
-0.20
-0.40
-0.60
What could cause a deviation of gas prices from trend?
-0.80
2014
Let’s look at the residuals for a moment…
Recession
Recession
0.60
Recession
0.40
0.20
0.00
1990
1992
1994
1996
1998
2000
2002
2004
2006
2008
2010
2012
2014
-0.20
-0.40
-0.60
-0.80
It could be because of changes in demand (i.e. the business cycle)
Let’s look at the residuals for a moment…
0.60
Recession
Recession
Recession
140.00
Oil
Residual
120.00
0.20
100.00
0.00
80.00
1990
1992
1994
1996
1998
2000
2002
2004
2006
2008
2010
2012
2014
-0.20
60.00
-0.40
40.00
-0.60
20.00
-0.80
0.00
Or, It could be because of changes in supply (i.e. oil)
Price of Oil
0.40