Download - Statistics Review
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Review of Mathematics
&
Statistics Essentials
BSP 4513 Econometrics 1
Econometrics
Lecture 2
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References
2
(1) Principles of Econometrics :: Appendix A and Appendix B
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1) Summation and Operators
2) Logarithm – always natural
3) Linear Relationships
4) Nonlinear Relationships
Slide A-3
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Notation • Summation ∑i
n Xi = X1 + X2 + … + Xn ; ∑jn Rj = R + R2 + R3 + … + Rn
∑ikXi = k.∑iXi ; 0.8 ∑j
n Rj = 0.8R + 0.8R2 + 0.8R3 + … + 0.8Rn
∑i(aXi + bYi) = a∑iXi + b∑iYi • Lag Operator LXt = Xt-1 ; LnXt = Xt-n ; Yt = a0Xt +a1Xt-1 + a2Xt-2
=(a0+a1L+a2L2)Xt = A(L)Xt. • Difference Operator ∆Xt = Xt – Xt-1
∆Xt = Xt – LXt = (1-L)Xt ∆ = 1 – L ; ∆(aXt) = a∆Xt
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Slide A-5
1 1 1 1
1 21 1 1
1 1 1 1
( , ) ( )
( , ) [ ( , ) ( , ) ( , )]
( , ) ( , )
m n m n
i j i ji j i j
m n m
i j i i i ni j i
m n n m
i j i ji j j i
f x y x y
f x y f x y f x y f x y
f x y f x y
BSP 4513 Econometrics
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dln(Xt) = dXt/Xt growth of X
Discrete approximation
ln(Xt) = ln(Xt) – ln(Xt-1)
Slide A-6
ln ln
Rules:
ln( ) ln( ) ln( )
ln( / ) ln( ) ln( )
ln( ) ln( )
b
a
x e b
xy x y
x y x y
x a x
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Slide A-7
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The exponential function is the antilogarithm because we can recover
the value of x using it.
Slide A-8
lnexp ln
xx e x
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Slide A-9
(A.1)
(A.2)
1 2y x
2y x
1 2 1 2 10y x
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Figure A.1 A linear relationship Slide A-10
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Slide A-11
(A.3)
(A.4)
2
dy
dx
1 2 2 3 3y x x
(A.5)
2 3
2
3 2
3
given that is held constant
given that is held constant
yx
xy
xx
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Slide A-12
(A.6)
1 2 3Q L K
2 given that capital is held constant
, the marginal product of labor inputL
QK
L
MP
2 3
2 3
,y y
x x
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Slide A-13
(A.7)
∆y/y = the relative change in y, which is a decimal
%∆y = percentage change in y
(A.8a)
(A.8b)
(A.8c)
yx
y y y x xslope
x x x y y
% 100y
yy
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Figure A.2 A nonlinear relationship Slide A-14
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Slide A-15
(A.9)
(A.10)
2dy dx
yx
dy y dy x xslope
dx x dx y y
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Slide A-16
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Figure A.3 Alternative Functional Forms Slide A-17
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If β3 > 0, then the curve is U-shaped, and representative of average or
marginal cost functions, with increasing marginal effects. If β3 < 0,
then the curve is an inverted-U shape, useful for total product curves,
total revenue curves, and curves that exhibit diminishing marginal
effects.
Slide A-18
2
1 2 3y x x
2 3 2 32 0, or / (2 )dy dx x x
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• Example: the Phillips Curve
Slide A-19
1
1 2 1 2
1y x
x
1
1
1 2
% 100
1%
t tt
t
t
t
w ww
w
wu
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• In order to use this model all values of y and x must be positive. The
slopes of these curves change at every point, but the elasticity is
constant and equal to β2.
Slide A-20
1 2ln( ) ln( )y x
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• Both its slope and elasticity change at each point and are the same
sign as β2.
• The slope at any point is β2y, which for β2 > 0 means that the marginal
effect increases for larger values of y.
Slide A-21
1 2ln( )y x
1exp ln( ) exp( )y y x
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Random Variables
Probability Distributions
Joint, Marginal and Conditional Probability
Distributions
Properties of Probability Distributions
Slide B-22
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Random Variables
• A random variable is simply a variable whose values are determine by some chance mechanism.
• A random variable has a designated set of possible values and associated probabilities.
• A discrete random variable X consists of a set of possible values X1, X2…,Xk and associated non-negative fractions (probabilities) p1, p2…,pN such that
p1+ p2 + . . .+ pN = ∑pN = 1
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Probability Distribution
• A formula giving the probabilities for different values of the random variable X is called a probability distribution, and probability density function for continuous random variables.
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Probability Distribution Example: (a) Probability Function for outcome from
tossing 2 well-balanced coins:
Outcome (No of Heads) Probabilities X.P(X)
No Head X1 = 0 0.25 = P(X1) 0 x 0.25 = 0.0
1 Head X2 = 1 0.50 = P(X2) 1 x 0.50 = 0.5
2 Heads X3 = 2 0.25 = P(X3) 2 x 0.25 = 0.5
Total 1.00 ∑X.P(X) = 1.00
● H
T
H
H
T
T
2
1
1
0
½
½
½
½
½
½
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Probability Distribution
Example: (b) Probability Function for outcome from tossing
a die with the numbers shown on its faces.
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Probability Density Functions…
• Unlike a discrete random variable, a continuous random variable is one that can assume an uncountable number of values.
• We cannot list the possible values because there is an infinite number of them.
• Because there is an infinite number of values, the probability of each individual value is virtually 0.
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Point Probabilities are Zero
Because there is an infinite number of values, the probability of each individual value is virtually 0.
Thus, we can determine the probability of a range of values only.
• E.g. with a discrete random variable like tossing a die, it is meaningful to talk about P(X=5), say.
• In a continuous setting (e.g. with time as a random variable), the probability the random variable of interest, say task length, takes exactly 5 minutes is infinitesimally small, hence P(X=5) = 0.
• It is meaningful to talk about P(X ≤ 5).
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Probability Density Function…
• A function f(x) is called a probability density function (over the range a ≤ x ≤ b if it meets the following requirements:
1) f(x) ≥ 0 for all x between a and b, and
2) The total area under the curve between a and b is 1.0
f(x)
x b a
area=1
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Uniform Distribution… • Consider the uniform probability distribution (sometimes
called the rectangular probability distribution).
• It is described by the function:
f(x)
x b a
area = width x height = (b – a) x = 1
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Example 2.1(a)…
• The amount of gasoline sold daily at a service station is uniformly distributed with a minimum of 2,000 gallons and a maximum of 5,000 gallons.
• Find the probability that daily sales will fall between 2,500 and 3,000 gallons.
• Algebraically: what is P(2,500 ≤ X ≤ 3,000) ?
f(x)
x 5,000 2,000
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Example 2.1(a)…
• P(2,500 ≤ X ≤ 3,000) = (3,000 – 2,500) x = .1667
• “there is about a 17% chance that between 2,500 and 3,000 gallons of gas will be sold on a given day”
f(x)
x 5,000 2,000
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Probability Density Function
• PDF of a Continuous Random Variable
Example: the Normal Distribution for values of a random variable
taking values in the interval ( - , + )
45 60 0
Weight in kilogram
Pro
bab
ilit
y D
en
sit
y f(x) Probability that weight lies
between 45 and 60 kg is
given by area shaded
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Normal Distribution
• A bell-shaped distribution • Symmetrical about its mean,
median and mode • The uni-variate normal
probability density function is:
with mean = and variance = 2
f(X) = __1__ exp{-½[(X-)/]2}
(22)
-∞ +∞
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Measure of Central Tendency • Mean (average); Median; Mode • Expectation Operator, E( . ) E(x) = x.f(x)dx = μx = mean of distribution = x.f(x)dx More generally, for any function of x, g(x): + E[g(x)] = g(x).f(x)dx. In particular, E[a+bx] = a+bE(x), when a and b are non-stochastic. E[x-E(x)]2 = variance of distribution = E[x2 + (E(x))2 -2xE(x)] = Ex2 - (E(x))2
& Note that E(XY) = E(X).E(Y), if X and Y are independent; E(X/Y) ≠ E(X)/E(Y)
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Measure of Dispersion
• Range (max – min); Variance (σ2) ; Standard deviation (σ); coefficient of variation (CV =σ/μ).
• Definition of variance:
Var(x) = (σ2)
= E[x-E(x)]2
= E[x2 + (E(x))2 -2xE(x)]
= Ex2 - (E(x))2
Var(X) = ∑(Xi - X)2P(Xi)
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Measure of Dispersion Example: The variance of a random variable X,
the number shown on a die.
σ2 = 2.9167; σ = 1.708; CV = 0.488
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Dispersion with same mean
Hypothetical PDFs of continuous random variables
all with the same expected value.
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Joint Distribution; Marginal and Conditional Distributions
Consider the 16 possible outcomes from tossing a coin four times.
X = number of heads obtained on the first three tosses;
Y = number of heads obtained on four tosses;
X→
Y
0 1 2 3 g(Y) E(X|Y)
0 1/16 0 0 0 1/16 0.0
1 1/16 3/16 0 0 ¼ 0.75
2 0 3/16 3/16 0 3/8 1.5
3 0 0 3/16 1/16 ¼ 2.25
4 0 0 0 1/16 1/16 3.0
f(X) 1/8 3/8 3/8 1/8
E(Y|X) 0.5 1.5 2.5 3.5
Marginal distribution of X = f(X); E(X) = 12/8 = 3/2
Marginal distribution of Y = g(Y); E(Y)= 32/16 = 2
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Conditional Probabilities and Conditional Distribution
• P(Y|X) = P(X,Y)/P(X)
Or P(X,Y) = P(Y|X).P(X) = P(X|Y).P(Y)
• E.g. P(Y=3|X=2) = P(Y=3, X=2)/P(X=2)
= (3/16)/(3/8) = 0.5
Similarly, P(Y=4|X=2) = (3/16)/(3/8) = 0.5
The conditional distribution of Y given X = 2 is:
Values of Y given X = 2 Probabilities Y.P(Y|X)
Y = 2 0.5 = P(Y=3|X=2) 1.0
Y = 3 0.5 = P(Y=4|X=2) 1.5
Total 1.00 2.5 = E(Y|X=2)
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Conditional Distribution
X=2 P(Y|X=2)
Y=0 0 0
Y=1 0 0
Y=2 3/16 0.5
Y=3 3/16 0.5
Y=4 0 0
3/8 1.0
E(Y|X=2) = 2.5
P(Y=3|X=2)
= P(X=2,Y=3)/P(X=2)
= (3/16)/(3/8)
= 0.5
•Conditional Mean:
E(Y|X=2)=0.5*2+0.5*3 = 2.5
•Conditional Variance:
V(Y|X=2)=0.5*(2-2.5)2
+0.5*(3-2.5)2 = 0.25
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Statistical Independence
X and Y are said to be statistically independent if
P(X,Y) = P(X).P(Y)
X g(Y)
1 2 3
Y 1 0.12 0.18 0.30 0.6
2 0.08 0.12 0.20 0.4
f(X) 0.2 0.3 0.5
Example:
P(X=1,Y=2)
= P(X=1).P(Y=2)
= 0.2*0.4 =0.08.
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Conditional Distribution
• Conditional density function: f(Y|x)
• Conditional Expectation = μY|X
=E(Y|X) =EX(Y)
= YiP(Yi|X)
E.g. μY|X=2 = E(Y|X=2) = 2 * 0.5 + 3 * 0.5
= 5/2 = 2.5
• Conditional Variance = EX{Y- μY|X}2
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Conditional Expectation & Regression Line
• Note that the expectation of Y conditional on X can be expressed as a function of X.
• For the coin tossing example:
E(Y|X) = 0.5 + 1.0X regression line
• Also notice that the variance of (Y|X) in the coin-tossing example is the same for different values of X:
Var(Y|X) = EX{Y- μY|X}2 = 0.25
homoscedastic
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Regression Line
X
Y|X
E(Y|X) = 0.5 + 1.0X
0
3
2
1
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Covariance
• Cov(X,Y) = E[(X- X)(Y- Y)] = E(XY) - X Y
= ∑x∑y[(X- X)(Y- Y)P(X,Y)
= ∑x∑yXYf(X,Y) - XY
• Cov(X,X) = Var (X)
• Cov(a+bX, c+dY) = bdCov(X,Y)
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Covariance as an average of area
(x2,y2)
(x1,y1)
(x3,y3)
y
x
Cov(X,Y)
= ∑∑[(X- X)(Y-Y)f(X,Y)
= (1/n)∑∑xy
Let x = (X- X)
y = (Y - Y)
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Correlation Between X and Y
• Correlation Coefficient:
ρ = Cov(X,Y)/ X y
Computation formula:
∑∑(X- X)(Y-Y)
√{∑(X- X)2. ∑(Y-Y)2}
In the coin-tossing example, ρ =0.5/√(0.75*1.0)
= 0.5774
ρ =
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Properties of Correlation Coefficient
• A measure of linear association between two variables.
• The correlation coefficients always lies between -1 and +1.
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Figure 3-3
Some typical patterns of the correlation coefficient, ρ.
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Skewness of Distribution
Measured y the third moment, S=E(X – ux)3/σ3
Skewness =0, when distribution is symmetric
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Kurtosis
•Measured by the normalized fourth moments, K = E(X-ux)4/σ4
•For normal distribution, K = 3.
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Example
• According to the Registry of Business in a small city, there are 1,200 establishments. The tabulation below shows the size of the establishment in terms of the number of employees.
No of Employees
No of
Establishments
Relative
Frequency
0 - 50 768 0.640
51 - 100 180 0.150
101 - 200 180 0.150
201 - 500 72 0.060
Total 1200 1.000
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Example: Distribution of Establishments
Size Ref Freq (p)
0 - 50 0.640
51 - 100 0.150
101 - 200 0.150
201 - 500 0.060
0 50 100 200 500
Area of Shaded = 1.0
Prob
Density
Number of Employees
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Quick Quiz
• Express the following using the ‘L’ operator:
(a) Yt = 2 + 3Xt + 4Xt-1 + 8 Xt-4
(b) Qt = 1 + 0.3Qt-1 + 0.4Nt + 0.2Kt-1
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Quick Quiz
X→
Y
0 1 2 3 g(Y) E(X|Y)
0 1/8 1/16 0 0 3/16 ?
1 1/8 1/8 0 0 1/4 ?
2 0 1/16 1/8 0 3/16 ?
3 0 0 1/16 1/16 1/8 ?
4 0 0 1/16 3/16 1/4 ?
f(X) 1/4 1/4 1/4 1/4
E(Y|X) ? ? ? ?
Fill in the blanks
E(X) = ____
E(Y) = ____