ECMT1020-Business and Economic Statistics B

ECMT1020-Business and Economic Statistics B

Nektarios Aslanidis

Semester 2, 2014Lecture 1: Review of statistics 1

Aslanidis (University of Sydney) Review of statistics 1Semester 2, 2014 Lecture 1: Review of statistics 1 1

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Some notation

Summation operator

n

∑i=1Xi �

i=n

∑i=1Xi � ∑Xi � ∑

xXi = X1 + X2 + ...+ Xn

where Xi is a random variable.

Properties of summation operator

1.n

∑i=1k = (k + ...+ k

n�k) = nk

where k is a constant2.

n

∑i=1kXi = (kX1 + ....+ kXn) = k(X1 + ....+ Xn) = k

n

∑i=1Xi


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Some notation

3.n

∑i=1(Xi + Yi ) =

n

∑i=1Xi +

n

∑i=1Yi

where Yi is a random variable.4.

n

∑i=1(a+ bXi ) =

n

∑i=1a+

n

∑i=1bXi = na+ b

n

∑i=1Xi

where a and b are constants.


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Probability: Classical or a priori de�nition

De�nitionP(A) =

mn

where n ! total No of outcomes (mutually exlusive and equally likelyoutcomes) and m ! No of these outcomes favourable to event A.

ExampleA.6. Throw a die: Mutually exlusive and equally likely

P(A) =16...

A.6 Toss a coin: Mutually exlusive and equally likely

P(H) =12

P(T ) =12


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ExampleA.6. A deck of 52 cards: Probalility of drawing a single card

P(A) =152

Probability of drawing a spade

P(A) =1352


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Problems with the classical de�nition: What happens if the outcomesof an experiment are not �nite or are not equally likely?

ExampleToss an irregular coin: Not equally likely

P(H) = π

P(T ) = 1� π

How do we estimate (infer) π?


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Problems with the classical de�nition: What happens if the outcomesof an experiment are not �nite or are not equally likely?

ExampleProbability that GDP growth next year is 0.03(3%): Not �nite

P(GDP = 0.03)

How do we estimate (infer) P(GDP = 0.03)?


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Probability: Empirical de�nition

ExampleA.7. Distribution of marks (Table A-1).

Provided the number of observation is relatively large, the relativefrequencies are treated as probabilities.


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Properties of probabilities

Properties1.

0 � P(A) � 12. If A,B,C , ... are mutually exclusive events

P(A+ B + C + ...) = P(A) + P(B) + P(C ) + ...


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Properties3. If A,B,C , ... are mutually exclusive events and collectivelyexhaustive sets of events

P(A+ B + C + ...) = P(A) + P(B) + P(C ) + ... = 1

ExampleA.8 Throw a die

P(1+ ...+ 6) = P(1) + ...+ P(6) =16+ ...+

16= 1


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Properties1. If A,B,C , ... are statistically independent events

P(ABC ...) = P(A)P(B)P(C )...

where P(ABC ...)! joint probability and P(A),P(B),P(C ), ...!marginal, unconditional or individual probabilities.

ExampleA.9 Throw two (regular) coins simultaneously

P(HH) = P(H)P(H) =1212=14

P(TT ) = P(T )P(T ) =1212=14

P(HT ) = P(H)P(T ) =1212=14

P(TH) = P(T )P(H) =1212=14


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2. If A,B,C , ... are not mutually exclusive events

P(A+ B) = P(A) + P(B)� P(AB)

ExampleA.10. A deck of 52 cards: Probability of either a heart or a queen

P(heart or queen) = P(heart) + P(queen)� P(heart and queen)

Heart and queen are not mutually exclusive


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Conditional probabilities

Conditional probability of event A, conditional on event B

P(A/B) =P(AB)P(B)

, P(B) > 0

Conditional probability of event B, conditional on event A

P(B/A) =P(AB)P(A)

, P(A) > 0


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ExampleIn an macroeconomics class there are 1000 students, of which 450 aremales and 550 are females. Of these, 360 males and 150 females plan tomajor economics. Given that a randomly selected student wants plans tomajor economics, what is the probability that the student is female.

P(A/B) =P(AB)P(B)

=150/1000510/1000

= 0.294

The unconditional probability of selecting a female student isP(A) = 550/1000 = 0.55.


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TheoremBayes

P(A/B) =P(AB)P(B)

=P(B/A)P(A)

P(B/A)P(A) + P(B/A0)P(A0)

P(A)! prior probabilityP(A/B)! posterior (revised) probability. Knowledge about theconditional event, changes our assessment of the probabilities.


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If events A and B are independent

P(A/B) =P(AB)P(B)

=P(A)P(B)P(B)

= P(A)

P(B/A) =P(AB)P(A)

=P(A)P(B)P(A)

= P(B)


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Random variables and probability distributions

Discrete random variable (r.v.) that takes on a �nite number ofvalues. The function

f (X = xi ) � f (xi ) � P(X = xi )

is called probability mass function (PMF ) or probability function.

Properties

1.

0 � f (xi ) � 12.

∑xf (xi ) = f (x1) + f (x2) + ...+ f (xn) = 1


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ExampleA:13. X is number of heads obtained in two tosses of a coin (Figure A-2).Possibilities: (T\T), (T\H), (H\T), (T\T)

Number of heads : x1 = 0, x2 = 1, x3 = 2.3

∑i=1f (xi ) = f (x1) + f (x2) + f (x3) = 0.25+ 0.5+ 0.25 = 1


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Continuous random variable (r.v.) that takes on a certain range orinterval of values. The function

f (X ) � P(x1 < X < x2) =Z x2

x1f (x)dx

is called probability density function (PDF) or probability function.De�ned for all x1 < x2. Note that P(X = xi ) = 0!

Properties

1.

0 � P(x1 < X < x2) � 12. Z +∞

�∞f (x)dx = 1


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ExampleX is the height measured in inches (Figure A-3).

Example

A.14. Find P(0 < x < 1)

f (x) =x2

9, 0 � x � 3

Solution Z 1

0

x2

9dx =

19

Z 1

0x2dx =

19

�x3

3

�10=1913=127

Note:�x 33

�0= x2,

R 30x 29 dx =

19

hx 33

i30= 1,R 3

1x 29 dx =

19

hx 33

i31= 1

9

h333 �

133

i= 1

9

� 263

�= 26

27


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Cumulative distribution function (CDF) means the probability Xtakes a values less than or equal to xi .

F (X ) � P(X � xi )

ExampleDiscrete random variable

P(X � x5)where X takes values x1, x2, x3, x4, x5, ...

5

∑i=1f (xi ) = f (x1) + ...+ f (x5)


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Cumulative distribution function (CDF) means the probability Xtakes a values less than xi .

F (x) � P(X � xi ) = P(X < xi )

ExampleContinuous random variable

P(X � x5) = P(X < x5)

where X takes values, say, from -∞ to +∞.

F (x) =Z x5

�∞f (x)dx , f (x) = F 0(x)


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Properties of CDFs (assume X takes values in the range (-∞, +∞)).

1. As x ! �∞, F (x)! 0, and as x ! ∞, F (x)! 1.2. If x1 < x2, then F (x1) � F (x2) (nondecreasing function)3. P(X � k) = 1� P(X < k) � 1� F (k)4. P(x1 < X < x2) = P(X < x2)� P(X < x1) � F (x2)� F (x1)

ExamplesA.15-A.16 (Figures A-4, A-5)- - In class - -F (X = k) � F (k) � P(X � k)F (4)� F (3) = 1� 15/16 = 1/16F (3)� F (1) = 15/16� 5/16 = 10/16


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Multivariate probability distributions

Two-variable (bivariate) case (discrete random variables)De�ne bivariate or joint probability mass function (PMF )

f (X ,Y ) = P(X = xi ,Y = yi )

Note: Extend de�nition to multivariate probability mass function(PMF )

f (X ,Y , ...,Z ) = P(X = xi ,Y = yi , ...,Z = zi )


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Two-variable (bivariate) case (discrete random variables). Properties:

1.

f (X ,Y ) � 02.

∑x

∑yf (X ,Y ) = 1

Note: In case of the three random variables: Properties:1.

f (X ,Y ,Z ) � 02.

∑x

∑y

∑zf (X ,Y ,Z ) = 1


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ExampleA.17 (Tables A-2, A-3)- - In class - -


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For continuous random variables the joint PDF can be de�nedanalogously, though the expresions are more involved.

f (X ,Y ) = P(a < X < b, c < Y < d) =Z b

a

Z d

cf (X ,Y )dXdY

Note that continuous random variables cannot take a single value!

Properties1.

f (X ,Y ) � 02. Z

x

Zyf (X ,Y )dXdY = 1


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In relation to f (X ,Y ), f (X ) and f (Y ) are the univariate,unconditional or the marginal PMFs

f (X ) = ∑yf (X ,Y )

f (Y ) = ∑xf (X ,Y )

For continuous random variables the the marginal PDFs

f (X ) =Zyf (X ,S)ds

f (Y ) =Zxf (S ,Y )ds


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De�ne conditional probability mass function (PMF )

f (X/Y ) = P(X = xi/Y = yi )f (Y /X ) = P(Y = yi/X = xi )

Di¤erently,

f (X/Y ) =f (X ,Y )f (Y )

f (Y /X ) =f (X ,Y )f (X )


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ExampleA.17 (Table A-3)

f (Y = 4/X = 4) =f (Y = 4,X = 4)

f (X = 4)=0.150.32

� 0.47(vs. f (Y = 4) = 0.23)

f (Y = 4/X = 0) =f (Y = 4,X = 0)

f (X = 0)=0.010.08

= 0.125

f (Y = 1/X = 4) =f (Y = 1,X = 4)

f (X = 4)=0.010.32

= 0.03125

...


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Statistical independence

It followsf (X ,Y ) = f (X/Y )f (Y )

If two variables are statistical independent

f (X ,Y ) = f (X )f (Y )


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Statistical independence

ExampleTable A-3

f (X = 1,Y = 1) = 0.05

f (X = 1)f (Y = 1) = (0.12)(0.16) = 0.0192 6= 0.05f (X = 2,Y = 2) = 0.10

f (X = 2)f (Y = 2) = (0.24)(0.23) = 0.0552 6= 0.10...

X and Y are not independent!


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Bibliography

Appendix A from Gujarati Damodar N. and Dawn C. Porter (2010),Essentials of Econometrics, 4th ed, McGraw-Hill.

No tutorials this week.


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ECMT1020-Business and Economic Statistics B

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