Download Quantitative Business Analysis

Quantitative Business Analysis:
Ken Choie
http://dasan.sejong.ac.kr/~kchoie/
Q:
A:
Part I:
How to make decisions under uncertainty?
Maximize the expected value!
Probability:
1. Random variables
Possible Outcomes
Probabilities
2. The Expected Value and Variance of a Random Variable
Common Probability Distributions
3. Summary of Observations/ Data
Central Tendency
Dispersion
Skewness
Kurtosis
Part II:
Statistics (Sample Data and Inferences):
1. Estimation of Population parameters
The Central Limit Theorem
Point Estimation
Interval estimation (Confidence Intervals)
2. Hypothesis Testing (Testing hypothesis on population parameters)
Concerning the Mean
Concerning the Variance
3. Regression
An Overview:
Possibility:
Winning a lottery
Probability:
Possible outcomes (events);
Attached likelihood (probability)
Random variable:
e.g., the # of heads in 10 coin-tosses
A random variable has many possible outcomes;
Frequency distribution
Each possible outcome has an associated probability
The probability distribution (outcome + probability combination)
Probability Theory:
Concerned with the parameters of the probability distribution of a random variable
The expected value,
The variance.
The probability distribution of a random variable;
Discrete random variables,
Continuous random variables,
The parameters of the probability distribution:
Central Tendency
Dispersion
Skewness
Kurtosis
Statistics:
Population with its parameters (the underlying probability distribution);
Sample (with its statistics) taken from the population
1. How do you estimate the unknown parameter values of a random variable?
Infer the population parameter values from samples (sample statistics).
Estimation of a parameter value:
Point estimate
Interval estimate (Confidence interval)
2. How do you verify/test hypotheses on the unknown parameter values of random
variable?
Use the sample statistics to evaluate the hypotheses on the population
parameters
Null hypothesis
Alternative hypothesis
3. Can you specify/quantify the relationship between random variables?
Do the regression analysis
Part I: Probability Theory:
Possible outcomes (events) of a random variable:
Methods of enumeration:
Multiplication principle: n * m
Permutation:
(n objects, r positions),
P(n,r) =
Combination:
n!
(n−r)!
(a distinguishable permutation),
n!
C(n,r) =
(n−r)!r!
The odds in favor of an event are 2 to 1 => the probability of the event is 1/3.
Algebra of sets:
Union:
Intersection:
∪
∩
Venn diagram:
P(A) = 1 – P(A)
P(A ∪ B) = P(A) + P(B) − P (A ∩ B)
; if A, B are not mutually exclusive
Sequential events:
Independent events:
Events A and B are independent if and only if,
P(A∩ B) = P(A)P(B)
Conditional probability:
P(A|B) =
P(A∩B)
P(B)
P(A∩ B) = P(B)P(A|B)
Bayes’ formula: a posterior probability conditional on the prior probability
P(Bj|A) =
P(Bj)P(A|Bj)
∑i P(Bi)P(A|Bi)
The mean and variance of a random variable:
Let X denote a random variable with a certain outcome set, R.
The probability distribution function of the random variable X is denoted by f(x).
The expected value of X is a weighted mean of X where the weights are the probabilities f(x);
the (weighted) mean of X is denoted as μ.
The expected value of X = the mean of X, μ
E [X]
= μ
= the mean
If X is a discrete random variable:
E [x] = ∑R x f(x)
In general,
E [g(x)] = ∑R g(x)f(x)
If c is a constant number, E(c) = c
If c is a constant and g is a function, E [cg(x)] = c E[g(x)]
In particular,
If g(x) = (X − μ)2
E [(X − μ)2 ]
= ∑R(X − μ)2 f(x)
= σ2
= the variance of X
σ = the standard deviation of X
Expected Value and Variance of a Random Variable:
E (X)
=
∑R f (X) ∗ X
: prob. weighted average
= ∑ P (Xi) ∗ Xi
Var (X) =
E [ Xi – E(Xi)]2
=
∑R f (X) ∗ [ Xi – E(Xi)]2
=
∑ P (Xi) ∗ [ Xi – E(Xi)]2
: prob. weighted dispersion
Example 1:
Probability:
.10
.60
.30
rate of return in %:
-10
12
15
Let the rate of return be X, then
E(X)
= .10 * (-10) + .60 *12 + .30*15
= 10.7;
Var(X)
= .10 * (10 − 10.7)2 + .60 * (12 − 10.7)2 + .30 ∗ (15 − 10.7)2
= 49.41
Discrete probability distribution function
Bernoulli trial:
Let X be a random variable, X(success) =1, X(failure) =o,
f(x) = P x (1 − P)1−x , x = 0,1
E [x]
σ2
=
∑R x f(x)
=
∑R x P x (1 − P)1−x
=
0*(1-P) +1*P
=
P
= ∑R(X − μ)2 f(x)
= ∑R(X − P)2 P x (1 − P)1−x
= P (1 - P)
The binomial distribution:
Let Y be a random variable whose value is the number of successes in n
Bernoulli trials.
f(x) = C(n, y) P y (1 − P)1−y , y = 0,1, … , n
μ = nP
σ2 = nP(1 – P)
The continuous probability distribution functions:
The uniform distribution
The exponential distribution
The normal distribution:
The standard normal distribution: N (0,1)
If X is N(μ, σ2 ) then Z =
(X− μ)
σ
is N (0,1)
Summary of Probability Distribution functions
Central Tendency
Dispersion:
Skewness
Kurtosis
The differences among mean, median, mode:
The connection between the mean and dispersion;
The coefficient of variance, the Sharpe ratio,
The relationship between the central tendency and skewness;
Positively skewed => median < mean
The normal distribution vs other distribution
Kurtosis = 3,
it is a normal distribution
Leptokurtic (Kurtosis >3)
fatter at the extremes and higher peak in the middle
Platykurtic (Kurtosis < 3)
Pressed down in the middle
The relationship between two random variables, X and Y:
The mean:
= ∑R X f(X)
E (X)
= μx
The variance:
= E [ (X − μx )2 ]
Var (X)
= ∑R(X − μx )2 f(X)
= σ2 x
The mean:
= ∑R Y f(Y)
E (Y)
= μy
The variance:
Var (Y)
= E
[ (Y − μy )2 ]
= ∑R(Y − μy )2 f(Y)
= σ2 y
The joint probability distribution function, f (X,Y):
The graphic representation of the relationship;
The covariance: Cov (X,Y)
= E [ (X – μx ) (Y − μy )]
= ∑R [(X − μx ) (Y - μy )]
= σxy
The correlation coefficient:
ρxy =
Cov (X,Y)
σx σy
f (X,Y)
Example 2:
The joint probability of returns for securities A, B.
Return on B = 30%
Return on A = 20%
Return on A = 10%
return on B = 15%
.6
0
0
.4
Let the rate of return on A be X, and the rate of return on B be Y.
E (X)
= 0.6 * 20% + 0.4 * 10%
= 16%
= Expected rate of return on A
E (Y)
= 0.6 * 30% + 0.4 * 15%
= 24%
= Expected rate of return on B
Var (X)
= 0.6 * (20 − 16)2
= 24
+ 0.4 * (10 − 16) 2
Var (Y)
= 0.6 * (30 − 24) 2
= 54
+ 0.4 * (15 − 24) 2
Cov (X,Y)
= 0.6 * (20 - 16)*(30 - 24) + 0.4 * (10 - 16)*(15 - 24)
= 0.6* 24 + 0.4 * 54
= 36
The correlation coefficient:
ρxy =
Cov (X,Y)
σx σy
=
36
√24∗54
 Positively correlated.
Example 3:
The joint probability of returns for securities A, B.
Return on B = 30%
Return on A = 20%
Return on A = 10%
return on B = 15%
.4
.1
.2
.3
Let the rate of return on A be X, and the rate of return on B be Y.
E (X)
= 0.6 * 20% + 0.4 * 10%
= 16%
= Expected rate of return on A
E (Y)
= 0.5 * 30% + 0.5 * 15%
= 22.5%
= Expected rate of return on B
Var (X)
= 0.6 * (20 − 16)2
= 24
Var (Y)
= 0.5 * (30 − 22.5) 2 + 0.5 * (15 − 22.5) 2
= 56.25
Cov (X,Y)
= 0.4 * (20 - 16)*(30 – 22.5) + 0.1 * (10 – 16) * (30-22.5)
+ 0.2 * (20 - 16) * (15 – 22.5) + .3 * (10 - 16)*(15 – 22.5)
= 16.5
The correlation coefficient:
+ 0.4 * (10 − 16) 2
ρxy =
Cov (X,Y)
σx σy
=
16
√24∗56.25
 Less positively correlated.
Example 4.
The expected return and variance of a portfolio
Portfolio P consists of the assets A and B whose weights in the portfolio are 55% and 45%,
respectively.
The return on the portfolio, K, is a linear combination of two random variables X and Y.
K = 0.55X + 0.45Y
Hence,
E(K)
=
=
=
=
E (0.55X + 0.45Y)
0.55 E(X) + 0.45 E(Y)
0.55 * 16%+ 0.45 * 24%
19.6 %
Var (K) = E [K – E(K)] 2
= E [(0.55X + 0.45Y) – E (0.55X + 0.45Y)]2
= E [ {0.55X – E(0.55X)} + {0.45Y − E(0.45Y)}] 2
2
= E [ {0.55X – E(0.55X)}
+ 2 ∗ {0.55X – E(0.55X)} ∗ {0.45Y − E(0.45Y)}
+ {0.45Y − E(0.45Y)2 ]
= 0.552 * Var (X)
+ 2 * 0.55 * 0.45 * Cov (X, Y)
+ 0.452 * Var (Y)
= 0.552 ∗ 24 + 2 * 0.55 * 0.45 *36 + 0.452 ∗ 54
Let Z be a linear combination of two random variables X and Y.
E (Z)
=
Var (Z) =
E(aX + bY)
= a E(X) + b E(Y)
Var(aX + bY)
=a2 Var(X) + 2ab Cov(X,Y) + b2 Var(Y)
Recapitulation of the Theory of Probability:
Probability =
a certain outcome
the number of possible outcome
 To compute the number of possible outcome, study combination and permutation;
 Independent probability
 Conditional probability
A random variable:
 A certain outcome may take on a particular value within a range of outcomes;
 Which particular value the random variable may take on has an associated
probability;
 The relationship between the value of the random variable and its associated
probability is called the probability distribution (or the probability distribution
function)
The probability distribution of a random variable X:
A measure of its central tendency:
E (X)
=
∑R f (X) ∗ X
: prob. weighted average
= ∑ P (Xi) ∗ Xi
A measure of the dispersion of outcomes:
Var (X) =
E [ Xi – E(Xi)]2
=
∑R f (X) ∗ [ Xi – E(Xi)]2
=
∑ P (Xi) ∗ [ Xi – E(Xi)]2
: prob. weighted dispersion
A measure of the relationship between the mean and the variance of X:
σ
The coefficient of variance:
The Sharpe’s ratio:
μ
E(X)− Rf
σ
Two or more random variables:
The relationship between two random variables, X and Y:
The mean:
= ∑R X f(X)
E (X)
= μx
The variance:
= E [ (X − μx )2 ]
Var (X)
= ∑R(X − μx )2 f(X)
= σ2 x
The mean:
= ∑R Y f(Y)
E (Y)
= μy
The variance:
Var (Y)
= E
[ (Y − μy )2 ]
= ∑R(Y − μy )2 f(Y)
= σ2 y
The joint probability distribution function, f (X,Y):
The graphic representation of the relationship;
The covariance: Cov (X,Y)
= E [ (X – μx ) (Y − μy )]
= ∑R [(X − μx ) (Y - μy )]
= σxy
f (X,Y)
The correlation coefficient:
ρxy =
Cov (X,Y)
σx σy
Let Z be a linear combination of two random variables X and Y.
E (Z)
=
Var (Z) =
E(aX + bY)
= a E(X) + b E(Y)
Var(aX + bY)
=a2 Var(X) + 2ab Cov(X,Y) + b2 Var(Y)
Part II. Statistics
The Probability theory:
For a random variable X with the probability distribution, f(x), we can determine the
central tendency, E(X), and the dispersion, Var(X). (These parameters of f(x) are
called “population parameters”.)
The issue:
In reality, the probability distribution of X, f(x), is unknown. Hence, the mean and the
variance of X are also unknown.
If we need to know the central tendency and the dispersion of the random variable X,
What can we do?
The solution:
Use the sample data to guess the parameter values of f(x).
Statistics:
If you take many samples, these samples themselves would have a certain probability
distribution with its own central tendency and dispersion. These parameters of sample
distribution are called “sample parameters”.
Using the characteristics of a sample (called the “sample statistics”) to infer
information on the “population parameters” of f(x) is the subject of “Statistics”.
Sampling: to save time and money
Minimize sampling error:
simple random sampling vs. stratified random sampling
Time-series data and cross-sectional data
Sampling biases:
selection (survivorship, time-period)
Data mining: search for patterns
the pattern found may be specific to the period, lacking an economic story.
The central limit theorem:
If ̅
X is the mean of a random sample with n observations from an unknown distribution of
the random variable X ~ --- (μ , σ2 ),
̅ ~N( μ ,
then X
σ2
n
)
Where,
σ2
the standard deviation (or error) of the sample mean, σ X̅ is √ n
let
W=
̅− μ
X
σ
⁄ n
√
W ~ N (0 , 1)
then,
as n
-->
∞
W is known as the standard normal distribution.
The Chebyshev’s Inequality:
If
Y ~ --- (μ , σ2 )
P[|Y– μ | > k*
then,
σ ]
<
1
k2
The Point Estimation of the population mean:
The sample mean is the BLUE :
unbiased,
efficient (smallest variance),
consistent (as n gets larger)
the MLE (the parameter of the population pdf that could have produced X)
The Confidence Interval (i.e., an interval estimation) for the population mean:
An interval estimation centered around the sample mean:
One can say with the
lies within the interval.
If
X ~ --- ( μ ,
σ2 ) , then
̅− μ
X
σ
⁄ n
√
W=
___ % degree of confidence that the population mean, μ ,
~ N (0 , 1)
as
n
-->
∞
Hence,
CI
If
=>
̅
X ±
z*
σ
√n
X ~ --- ( μ , unknown ) then,
Q =
̅− μ
X
s
⁄
√n−1
~ t distribution, when the sample size, n, is small;
~ approximately N(0,1), when n is large
CI
=>
̅
X ± t*
s
√n−1

The greater is the sample size (or the degrees of freedom), the narrower is the
confidence interval.

The greater is the reliability factor (i.e., the degree of confidence), the wider is the
confidence interval.

Given the range of the confidence interval and the degree of confidence, one can
compute the sample standard deviation of the test static.
Example 5.
stock.
Ten analysts have given the following fiscal year earnings forecasts for a
Forecast (X)
Number of Analysts(n)
1.40
1.43
1.44
1.45
1.47
1.48
1.50
1
1
3
2
1
1
1
What is the mean forecast and standard deviation of forecasts?
Provide a 95% confidence interval for the population mean of the forecast.
X i ni
(Xi − X̅ )
1
1
3
2
1
1
1
10
1.40
1.43
4.32
2.90
1.47
1.48
1.50
14.50
-0.05
-0.02
-0.01
0.00
0.02
0.03
0.05
̅
X
= 14.50/10
= 1.45
s2
=
s
= √0.0007778
Forecast (Xi )
analysts (ni )
1.40
1.43
1.44
1.45
1.47
1.48
1.50
̅ )2
∑(Xi − X
= 0.0007778
(n−1)
= 0.02789
the confidence interval,
̅ ± t*
X
s
√n
= 1.45 ± 2.262 ∗
0.02789
√10
(Xi − X̅ )2
0.0025
0.0004
0.0001
0.0000
0.0004
0.0009
0.0025
(Xi − X̅ )2 ni
0.0025
0.0004
0.0003
0.0000
0.0004
0.0009
0.0025
0.0070
̅.
The variance of X vs. the variance of 𝐗
X and ̅
X each has its own probability distrubution function
X ~ --- (μ , σ2 ),
σ2
=
∑(Xi − μ)2
n
We estimate σ2 , using the sample data:
s2
̅
X ~N( μ ,
σ2
n
=
̅ )2
∑(Xi − X
n−1
),
σ2
n
σ2
We estimate
σ2
n
n
, substituting s2 for σ2 :
s2
=>
n
The confidence interval for a variable over time:
If return, r (0,t), has a normal pdf and it is i,i,d, then
E [r (o,t)] = μ
*
t
; Var [r (0,t)] =
σ2 *
t
= > an envelope-shaped confidence interval.
= > can be used to calculate the probability of exhausting the (futures) margin
Hypothesis Testing :
How to use sample information to test a hypothesis about the population parameters?
=> How to infer conclusions about the population parameters from sample statistics?
Hypothesis:
Null hypothesis (H0 ) vs. alternative hypothesis (HA )
One-tailed test vs. two-tailed test
Test statistic:
Something based on a sample
Type I and type II errors:
True state of situation
Decision
H0 true
H0 false
Do not reject H0 :
Reject H0 :
correct
type I error
type II error
correct
The null hypothesis must be constructed in such a manner that
The cost (i.e., penalty) of not reject H0 is low.
The power of a test:
The probability of correctly rejecting the null hypothesis
The p – value:
The smallest level of significance at which the null hypothesis can be rejected.
The process:
Determine the hypotheses,
-> determine the distribution of the test statistic,
-> determine the decision rule,
One-tailed test vs. two-tailed test (depends on the alternative hypothesis);
the level of significance;
the type I (reject the null in error) & the type II (don’t reject the null in error)
-> take a sample and compute the value of the test statistic.
Reject the null hypothesis:

If the sample statistic is greater than the critical value;

if the p-value of the sample statistic (the area outside of the test statistic; the
smallest level of significance at which one can reject a null hypothesis) is smaller
than the significance level;

the confidence interval for the test statistic does not contain the null hypothesis
value on the test statistic
Concerning the mean: (the test statistic has the t- distribution.)
1. Tests concerning a single mean:
The test statistic is a mirror image of the confidence interval for the population mean.
Test statistic
=
̅ − μ0
X
s
√n
if the t statistic value is too large, reject the null hypothesis.
Example 6.
Investment analysts often use earnings per share (EPS) forecasts.
Performance in forecasting quarterly EPS
# of forecasts
mean forecast error
stan. dev. of erro
Analyst A
Analyst B
61
121
$0.05
$0.02
$0.10
$0.12
Question:
Are the analysts’ forecasting qualities good?
Answer:
the null hypothesis, H0 , is that each analyst ′ s mean forecasting error (=
predicted − actual) , μ0 , equal zero.
H0 : μ 0 = 0
Ha : μ0 ≠ 0
The test statistic:
Test statistic
=
̅ − μ0
X
s
√n
For analyst A, 60 = 61-1 degrees of freedom: (use the t-distribution)
at the 0.05 significance level, reject H0 if t > 2.000
test statistic =
(0.05−0)
0.10⁄
√61
=
0.05
0.0129
= 3.873
reject the null hypothesis:
(i.e., analyst A’s forecasts tend to be too high)
For analyst B, 120 = 121-1 degrees of freedom: (use the z-distribution)
at the 0.05 significance level, reject H0 if t > 1.96
test statistic
=
(0.02−0)
0.12⁄
√121
=
0.02
0.0109
= 1.8349
do not reject the null hypothesis:
(i.e., analyst B’s forecasts tend to be close to zero)
the p-value of the sample statistic = 0.5 - 0.4671 =0.0329
at the 0.05 significance level, each tail is 0.025
=> do not reject the null hypothesis
2. tests concerning differences between means:
(if the samples are independent, population variances are equal ):
Test statistic
=
̅1− X
̅ 2 )–(μ1 − μ2 )
(X
S2
S2
√( p + p )
n1
Example 7.
n2
Investment analysts often use earnings per share (EPS) forecasts.
Performance in forecasting quarterly EPS
# of forecasts
mean forecast error
stan. dev. of erro
Analyst A
Analyst B
61
121
$0.05
$0.02
$0.10
$0.12
Question:
Is analyst A’s forecasting quality worse than B’s?
Answer:
the null hypothesis is that analyst A’s forecasting error is no worse than B’s.
H0 : μ A − μ B ≤ 0
Ha : μ A − μ B > 0
The test statistic:
assuming that the forecast errors of both analysts are normally distributed
and that the samples are independent and from population with equal
variance,
test statistic
=
̅A− X
̅ B )–(μA − μB )
(X
S2
S2
√( p + p )
n1
n2
the degrees of freedom is 180 = 61-1+121-1: (use the z-distribution)
at the 0.05 significance level, reject H0 if z > 1.645
the pooled estimate of variance is:
Sp2
=
(n1 −1)
S2
(n1 + n2 −2) 1
=
=
+
(61−1)∗0.102
(61+121−2)
0.6+1.728
(61+121−2)
̅A− X
̅ B )–(μA − μB )
(X
S2
S2
√( p + p )
n1
=
(121−1)∗0.122
= 0.0129
180
Test statistic =
+
(n2 −1)
S2
(n1 + n2 −2) 2
n2
(0.05−0.02)−0
0.0129 0.0129
+
61
121
√
=
0.03
√0.0002+0.0001
= 1.7133
reject the null hypothesis:
(i.e., the forecast error of analyst A is greater than that of
analyst B)
the p-value of the sample statistic = 0.5 - 0.4564=0.0436
at the 0.05 significance level, one tail test
=> reject the null hypothesis
3. Tests concerning mean differences:
(the data consisting of paired observations; the random variable is the difference):
Test statistic
Example 8.
=
̅− μd
d
Sd
̅
The monthly returns on the S&P 500 and small-cap stocks.
January 1960 – December 1999, 480 months
Mean
Stand dev.
S&P 500 returns (%)
small-cap stock (%)
differences (%)
1.0542
4.2185
1.3117
5.9570
-0.258
3.752
Question:
Is there any difference between the mean returns on the S&P 500 and smallcap stocks?
Answer:
hypothesis: H0 : μd = 0
Ha : μ d ≠ 0
The test statistic:
The degrees of freedom = 480-1; use the z-distribution
Test statistic
=
̅− μd
d
Sd
̅
=
−0.258−0
3.752⁄
√480
= -1.5065
at the 0.05 significance level, we do not reject the null hypothesis.
(i.e., the mean difference between the two indexes in the period was 0.)
the p-value of the sample statistic = 0.5 - 0.4345= 0.0655
Concerning variance:
1. Tests concerning a single variance (normally distributed population):
The test statistic
(n−1)S2
~ 𝑋2
σ2o
The test statistic has the chi-square distribution with (n-1) degrees of freedom.
Example 9. Suppose that the variance of annual returns on your portfolio during
last ten years is 225% and the variance of annual returns on the benchmark in the same
period is 400%.
Question:
Is your underlying variance of return of your portfolio less than that of the
benchmark?
Answer:
hypothesis: H0 : σ2 ≥ 400
Ha : σ2 < 400
The test:
Use the 𝑋 2 -distribution; the degrees of freedom = 10-1;
at the 0.05 significance level, the 𝑋 2 -distribution is 3.33
i.e., we will reject the null hypothesis if the test statistic is less than
3.33
Test statistic
=
(n−1)S2
σ2o
We do not reject the null hypothesis.
=
9∗225
400
= 5.06
2. Tests concerning the equality of two variances (normally distributed population):
The test statistic
S21
S22
~ F
The test statistic has the F distribution with (n1 -1) and (n2 -1) degrees of
freedom.
Example 10. Determine if the variance of returns on the S&P 500 has changed.
Time period
n
monthly return(%)
variance
Before Oct 1987
After Oct 1987
120
120
1.416
1.436
22.367
15.795
Question:
Did the variance change subsequent to the October 1987 market crash?
Answer:
hypothesis: H0 : σ2b = σ2a
Ha : σ2b ≠ σ2a
The test:
Use the F-distribution;
the degrees of freedom = 120-1 both the numerator and the
denominator;
a two-tailed test;
120
at the 0.02 significance level, the F120
-distribution is 1.53
i.e., we will not reject the null hypothesis if the test statistic is less
than 1.53
Test statistic
=
S2b
S2a
=
22.367
15.795
= 1.416
We do not reject the null hypothesis.
Question:
Did the variance become smaller subsequent to the October 1987 market
crash?
Answer:
hypothesis: H0 : σ2b ≤ σ2a
Ha : σ2b > σ2a
The test:
Use the F-distribution;
the degrees of freedom = 120-1 both the numerator and the
denominator;
a one-tailed test;
120
at the 0.01 significance level, the F120
-distribution is 1.53
i.e., we will not reject the null hypothesis if the test statistic is less
than 1.35
Test statistic
=
S2b
S2a
=
We do not reject the null hypothesis.
22.367
15.795
= 1.416
A summary of Statistics:
(estimation methods & hypothesis tests on population parameters)
Estimation of population parameter:
Sample size:
small (less than 120), use the Student t-distribution
large (greater than 120), use the standard normal z-distribution
Hypothesis tests on population parameter:
H0 : parameter value = something
Ha : parameter value ≠ something
A two-tailed test
H0 : parameter value ≥ something
Ha : parameter value < 𝑠𝑜𝑚𝑒𝑡ℎ𝑖𝑛𝑔
A one-tailed test
Tests on μ:
Either the t-distribution or the z-distribution
Tests on σ2 :
Either the 𝑋 2 -distribution or the F-distribution
Probability theory vs. Statistics:
Both are concerned with a random variable X its probability distribution function.
Probability theory:
 Determine the probability distribution of X.
 Given the probability distribution function, f(X), determine the parameters of the
probability distribution such as the central tendency (e.g., mean) and the
dispersion (e.g., variance) of the distribution.
Statistics:
 Infer the parameters (such as the mean and the variance) of the probability
distribution function from a sample.
Non-parametric inference:
1.
A test that is not concerned with a parameter;
e.g., is the sample random or not?
runs test
2. Aa test that makes minimal assumptions on distribution of the population
Convert observation values into ranks or signs (+ or -)
The linear relationship between two variables (the correlation between them):
Corr. Coefficient
=
Cov [R(x),R(y)]
[std dev R(x) ∗ std dev R(y)]
The Spearman Rank Correlation Coefficient:
=1-
6 ∑ d2i
n(n2 −1)
Regression:
A linear relationship between two variables
= > a linear relationship between two economic variables.
Yi
Suppose
=
a
+
b * Xi
E (Yi ) =
a
+
b * Xi
+ ϵi
where, Yi ~ N [ E (Yi ) , σ2 ]
Now consider,
(Yi
̅ ) =
- Y
̅) + θi
β ( Xi - X
Visualize a graphic relationship: e.g.,
the expected weight given a height
The estimator for β that minimizes the variance of [ Yi − E (Yi )] is:
β̂ =
Cov (Xi , Yi )
Var (Xi )
̂i
Hence, the estimate of Yi is Y
Where
̂ = (Y
̅ - β̂ ∗ X
̅) + β̂ * Xi
Y
And the estimate of σ2 is ̂
σ2
Where
σ2
̂
=
1
n
̂i ] 2
∑ [ Yi − Y
Example 11. The following table shows the average price of gasoline per liter for the
year and the annual sales (in $mm).
Year (i)
Average Price (Xi )_
1
2
3
4
5
6
1.
2.
3.
4.
5.
$ 0.40
0.36
0.42
0.31
0.33
0.34
Sales (Yi )
$20
25
16
30
35
30
Calculate the (sample) mean and standard deviation for X and Y.
Calculate the (sample) covariance between X and Y.
Calculate the (sample) correlation coefficient between X and Y.
What is the linear relationship between X and Y?
If the average price is raised to $0.45, what is your the expected annual sale?
Answer:
Year(i)
Xi
Yi
(Xi − ̅
X)2
1
2
3
4
5
6
0.40
0.36
0.42
0.31
0.33
0.34
20
25
16
30
35
30
0.0016
0.0000
0.0036
0.0025
0.0009
0.0004
36
1
100
16
81
16
-0.24
0.00
-0.60
-0.20
-0.27
-0.08
Sum
2.16
156
0.0090
250
-1.39
1. For X,
̅ =
X
∑ Xi
n
=
2.16
6
̅ )2
∑(Xi − X
Sx = √
(n−1)
= 0.36
0.009
= √
5
= 0.042426
(Yi − ̅
Y)2
(Xi − ̅
X)*(Yi − ̅
Y)
For Y,
̅ =
Y
∑ Yi
n
=
156
6
̅ )2
∑(Yi − Y
Sy = √
(n−1)
2. Cov (X, Y) =
= 26
250
= √
5
= 7.0711
̅ )∗(Yi − Y
̅)
∑(Xi − X
(n−1)
3. Correlation coefficient =
=
−1.39
Cov (X,Y)
Sx ∗ Sy
= -0.278
5
=
−0.278
0.042426∗7.0711
4.
β̂
=
Cov (Xi , Yi )
Var (Xi )
=
−0.278
(0.042426)2
= -154.44
̅ - β̂ ∗ X
̅) = 26 + (-154.44) * 0.36 = 81.6
(Y
Hence,
̂
Yi = 81.6 − 154.44 ∗ Xi
5.
From
̂
̅ - β̂ ∗ ̅
Y = (Y
X) + β̂ * Xi
̂
Y
= 81.6 –
= 12.10
154.44 * (0.45)