statistics review
DESCRIPTION
Correlation, statisticsTRANSCRIPT
Review of Mathematics
&
Statistics Essentials
BSP 4513 Econometrics 1
Econometrics
Lecture 2
References
2
(1) Principles of Econometrics :: Appendix A and Appendix B
1) Summation and Operators
2) Logarithm – always natural
3) Linear Relationships
4) Nonlinear Relationships
Slide A-3
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
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
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
Slide A-7
The exponential function is the antilogarithm because we can recover
the value of x using it.
Slide A-8
lnexp ln
xx e x
Slide A-9
(A.1)
(A.2)
1 2y x
2y x
1 2 1 2 10y x
Figure A.1 A linear relationship Slide A-10
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
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
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
Figure A.2 A nonlinear relationship Slide A-14
Slide A-15
(A.9)
(A.10)
2dy dx
yx
dy y dy x xslope
dx x dx y y
Slide A-16
Figure A.3 Alternative Functional Forms Slide A-17
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
• 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
• 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
• 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
Random Variables
Probability Distributions
Joint, Marginal and Conditional Probability
Distributions
Properties of Probability Distributions
Slide B-22
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
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.
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
Probability Distribution
Example: (b) Probability Function for outcome from tossing
a die with the numbers shown on its faces.
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.
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).
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
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
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
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
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
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)
-∞ +∞
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)
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)
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
Dispersion with same mean
Hypothetical PDFs of continuous random variables
all with the same expected value.
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
●
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)
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
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.
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
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
Regression Line
X
Y|X
E(Y|X) = 0.5 + 1.0X
0
3
2
1
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)
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)
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 =
Properties of Correlation Coefficient
• A measure of linear association between two variables.
• The correlation coefficients always lies between -1 and +1.
Figure 3-3
Some typical patterns of the correlation coefficient, ρ.
Skewness of Distribution
Measured y the third moment, S=E(X – ux)3/σ3
Skewness =0, when distribution is symmetric
Kurtosis
•Measured by the normalized fourth moments, K = E(X-ux)4/σ4
•For normal distribution, K = 3.
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
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
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
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) = ____