Lecture Eleven

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Lecture Eleven

Probability Models

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Outline

• Bayesian Probability

• Duration Models

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Bayesian Probability

• Facts

• Incidence of the disease in the population is one in a thousand

• The probability of testing positive if you have the disease is 99 out of 100

• The probability of testing positive if you do not have the disease is 2 in a 100

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Joint and Marginal Probabilities

Sick: S Healthy: H

Test + Pr(+ S) Pr(+ H) Pr(+)

Test- Pr(- S) Pr(- H) Pr(-)

Pr(S) Pr(H)

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Filling In Our Facts

Sick: S Healthy: H

Test +

Test -

Pr(s) =0.001

Pr(H) =0.999

Using Conditional Probability

• Pr(+ H)= Pr(+/H)*Pr(H)= 0.02*0.999=.01998

• Pr(+ S) = Pr(+/S)*Pr(S) = 0.99*0.001=.00099

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Filling In Our Facts

Sick: S Healthy: H

Test + Pr(+ S)=0.00099

Pr(+ H)=0.01998

Test -

Pr(s) =0.001

Pr(H) =0.999

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By Sum and By Difference

Sick: S Healthy: H

Test + Pr(+ S)=0.00099

Pr(+ H)=0.00198

Pr(+)=0.02097

Test - Pr(- S)=0.00901

Pr(- H)=0.88802

Pr(s) =0.001

Pr(H) =0.999

False Positive Paradox

• Probability of Being Sick If You Test +

• Pr(S/+) ?

• From Conditional Probability:

• Pr(S/+) = Pr(S +)/Pr(+) = 0.00099/0.02097

• Pr(S/+) = 0.0472

Bayesian Probability By Formula

• Pr(S/+) = Pr(S +)/Pr(+) = PR(+/S)*Pr(S)/Pr(+)

• Where PR(+) = PR(+/S)*PR(S) + PR(+/H)*PR(H)

• And Using our facts;Pr(S/+) = 0.99*(0.001)/[0.99*.001 + 0.02*.999]

• Pr(S/+) = 0.00099/[0.00099+0.01998]

• Pr(S/+) = 0.00099/0.02097 = 0.0472

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Duration Models

• Exploratory (Graphical) Estimates– Kaplan-Meier

• Functional Form Estimates– Exponential Distribution

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Duration of Post-War Economic Expansions in Months

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Trough Peak DurationOct. 1945 Nov. 1948 37Oct. 1949 July 1953 45May 1954 August 1957 39April 1958 April 1960 24Feb. 1961 Dec. 1969 106Nov. 1970 Nov. 1973 36March 1975 January 1980 58July 1980 July 1981 12Nov. 1982 July 1990 92March 1991 March 2000 120

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Estimated Survivor Function for Ten Post-War Expansions

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Kaplan-Meyer Estimate of Survivor Function

• Survivor Function = (# at risk - # ending)/# at risk

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Duration # Ending # At Risk Survivor0 0 10 112 1 10 0.924 1 9 0.836 1 8 0.737 1 7 0.639 1 6 0.545 1 5 0.458 1 4 0.392 1 3 0.2106 1 2 0.1120 1 1 0

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Figure 2: Estimated Survivor Function for Post-War Expansions

0

0.2

0.4

0.6

0.8

1

1.2

0 20 40 60 80 100 120 140

Duration in Months

Su

rviv

or

Fu

nct

ion

Exponential Distribution

• Density: f(t) = exp[ - t], 0 t

• Cumulative Distribution Function F(t)

• F(t) =

• F(t) = - exp[- u]

• F(t) = -1 {exp[- t] - exp[0]}

• F(t) = 1 - exp[- t]

• Survivor Function, S(t) = 1- F(t) = exp[- t]

• Taking logarithms, lnS(t) = - t

f (u)du 0

t

exp[ u]du0

t

0t

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Postwar Expansions

y = -0.0217x + 0.1799

R2 = 0.9533

-2.5

-2

-1.5

-1

-0.5

0

0.5

0 20 40 60 80 100 120

Duration (Months)

Ln

Su

rviv

or

Fu

nct

ion

So

Exponential Distribution (Cont.)

• Mean = 1/ =

• Memoryless feature:

• Duration conditional on surviving until t = :

• DURC( ) = = + 1/

• Expected remaining duration = duration conditional on surviving until time , i.e DURC, minus

• Or 1/ , which is equal to the overall mean, so the distribution is memoryless

t * f (t)dt

t * f (t)dt / S( )

Exponential Distribution(Cont.)

• Hazard rate or function, h(t) is the probability of failure conditional on survival until that time, and is the ratio of the density function to the survivor function. It is a constant for the exponential.

• h(t) = f(t)/S(t) = exp[- t] /exp[- t] =

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Model Building

• Reference: Ch 20

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20.2 Polynomial Models

• There are models where the independent variables (xi) may appear as functions of a smaller number of predictor variables.

• Polynomial models are one such example.

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y = 0 + 1x1+ 2x2 +…+ pxp +

y = 0 + 1x + 2x2 + …+pxp +

Polynomial Models with One Predictor Variable

y01x

• First order model (p = 1)

y = 0 + 1x + 2x2 +

2 < 0 2 > 0

• Second order model (p=2)


y = 0 + 1x + 2x2 + 3x3 +

3 < 0 3 > 0

• Third order model (p = 3)


• First order modely = 0 + 1x1 +

Polynomial Models with Two Predictor Variables

x1

x2

y

2x2 + 1 < 0

1 > 0

x1

x2

y

2 > 0

2 <

0

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20.3 Nominal Independent Variables

• In many real-life situations one or more independent variables are nominal.

• Including nominal variables in a regression analysis model is done via indicator variables.

• An indicator variable (I) can assume one out of two values, “zero” or “one”.1 if a first condition out of two is met

0 if a second condition out of two is metI=1 if data were collected before 19800 if data were collected after 19801 if the temperature was below 50o

0 if the temperature was 50o or more1 if a degree earned is in Finance0 if a degree earned is not in Finance

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Nominal Independent Variables; Example: Auction Car Price (II)

• Example 18.2 - revised (Xm18-02a)– Recall: A car dealer wants to predict the auction

price of a car.– The dealer believes now that odometer reading

and the car color are variables that affect a car’s price.

– Three color categories are considered:• White

• Silver

• Other colors

Note: Color is a nominal variable.

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• Example 18.2 - revised (Xm18-02b)

I1 =1 if the color is white0 if the color is not white

I2 =1 if the color is silver0 if the color is not silver

The category “Other colors” is defined by:I1 = 0; I2 = 0

Nominal Independent Variables; Example: Auction Car Price (II)

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• Note: To represent the situation of three possible colors we need only two indicator variables.

• Conclusion: To represent a nominal variable with m possible categories, we must create m-1 indicator variables.

How Many Indicator Variables?

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• Solution– the proposed model is

y = 0 + 1(Odometer) + 2I1 + 3I2 + – The data

Price Odometer I-1 I-214636 37388 1 014122 44758 1 014016 45833 0 015590 30862 0 015568 31705 0 114718 34010 0 1

. . . .

. . . .

White car

Other color

Silver color

Nominal Independent Variables; Example: Auction Car Price

33Odometer

Price

Price = 16701 - .0555(Odometer) + 90.48(0) + 295.48(1)

Price = 16701 - .0555(Odometer) + 90.48(1) + 295.48(0)

Price = 16701 - .0555(Odometer) + 45.2(0) + 148(0)

16701 - .0555(Odometer)

16791.48 - .0555(Odometer)

16996.48 - .0555(Odometer)

The equation for an“other color” car.

The equation for awhite color car.

The equation for asilver color car.

From Excel (Xm18-02b) we get the regression equationPRICE = 16701-.0555(Odometer)+90.48(I-1)+295.48(I-2)

Example: Auction Car Price The Regression Equation

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From Excel we get the regression equationPRICE = 16701-.0555(Odometer)+90.48(I-1)+295.48(I-2)

A white car sells, on the average, for $90.48 more than a car of the “Other color” category

A silver color car sells, on the average, for $295.48 more than a car of the “Other color” category.

For one additional mile the auction price decreases by 5.55 cents.


SUMMARY OUTPUT

Regression StatisticsMultiple R 0.8355R Square 0.6980Adjusted R Square 0.6886Standard Error 284.5Observations 100

ANOVAdf SS MS F Significance F

Regression 3 17966997 5988999 73.97 0.0000Residual 96 7772564 80964Total 99 25739561

Coefficients Standard Error t Stat P-valueIntercept 16701 184.3330576 90.60 0.0000Odometer -0.0555 0.0047 -11.72 0.0000I-1 90.48 68.17 1.33 0.1876I-2 295.48 76.37 3.87 0.0002

There is insufficient evidenceto infer that a white color car anda car of “other color” sell for adifferent auction price.

There is sufficient evidenceto infer that a silver color carsells for a larger price than acar of the “other color” category.

Xm18-02b


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• Recall: The Dean wanted to evaluate applications for the MBA program by predicting future performance of the applicants.

• The following three predictors were suggested:– Undergraduate GPA– GMAT score– Years of work experience

• It is now believed that the type of undergraduate degree should be included in the model.

Nominal Independent Variables; Example: MBA Program Admission (

MBA II)

Note: The undergraduate degree is nominal data.

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Nominal Independent Variables; Example: MBA Program Admission

(II)

I1 =1 if B.A.0 otherwise

I2 =1 if B.B.A0 otherwise

The category “Other group” is defined by:I1 = 0; I2 = 0; I3 = 0

I3 =1 if B.Sc. or B.Eng.0 otherwise

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SUMMARY OUTPUT

Regression StatisticsMultiple R 0.7461R Square 0.5566Adjusted R Square 0.5242Standard Error 0.729Observations 89


Regression 6 54.75 9.13 17.16 0.0000Residual 82 43.62 0.53Total 88 98.37

Coefficients Standard Error t Stat P-valueIntercept 0.19 1.41 0.13 0.8930UnderGPA -0.0061 0.114 -0.05 0.9577GMAT 0.0128 0.0014 9.43 0.0000Work 0.098 0.030 3.24 0.0017I-1 -0.34 0.22 -1.54 0.1269I-2 0.71 0.24 2.93 0.0043I-3 0.03 0.21 0.17 0.8684

Nominal Independent Variables; Example: MBA Program Admission

(II)MBA-II

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20.4 Applications in Human Resources Management: Pay-Equity

• Pay-equity can be handled in two different forms:– Equal pay for equal work– Equal pay for work of equal value.

• Regression analysis is extensively employed in cases of equal pay for equal work.

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• Solution– Construct the following multiple regression model:

y = 0 + 1Education + 2Experience + 3Gender +

– Note the nature of the variables:• Education – Interval• Experience – Interval• Gender – Nominal (Gender = 1 if male; =0 otherwise).

Human Resources Management: Pay-Equity

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• Solution – Continued (Xm20-03)


SUMMARY OUTPUT

Regression StatisticsMultiple R 0.8326R Square 0.6932Adjusted R Square 0.6836Standard Error 16274Observations 100


Regression 3 57434095083 19144698361 72.29 0.0000Residual 96 25424794888 264841613.4Total 99 82858889971

CoefficientsStandard Error t Stat P-valueIntercept -5835.1 16082.8 -0.36 0.7175Education 2118.9 1018.5 2.08 0.0401Experience 4099.3 317.2 12.92 0.0000Gender 1851.0 3703.1 0.50 0.6183

Analysis and Interpretation• The model fits the data quite well.• The model is very useful.• Experience is a variable strongly related to salary.• There is no evidence of sex discrimination.

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• Solution – Continued (Xm20-03)


SUMMARY OUTPUT

Regression StatisticsMultiple R 0.8326R Square 0.6932Adjusted R Square 0.6836Standard Error 16274Observations 100


Regression 3 57434095083 19144698361 72.29 0.0000Residual 96 25424794888 264841613Total 99 82858889971

Coefficients Standard Error t Stat P-valueIntercept -5835.1 16082.8 -0.36 0.7175Education 2118.9 1018.5 2.08 0.0401Experience 4099.3 317.2 12.92 0.0000Gender 1851.0 3703.1 0.50 0.6183

Analysis and Interpretation • Further studying the data we find: Average experience (years) for women is 12. Average experience (years) for men is 17

• Average salary for female manager is $76,189 Average salary for male manager is $97,832

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Midterm Grade Distribution

• A: 68- 7

• A-: 65-67 7

• B+: 61-64 9

• B: -59 7

• total 30

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0

2

4

6

8

50 55 60 65 70 75

Series: MIDTERMSample 1 30Observations 30

Mean 63.40000Median 64.00000Maximum 73.00000Minimum 49.00000Std. Dev. 5.887450Skewness -0.666511Kurtosis 2.870864

Jarque-Bera 2.242033Probability 0.325948

Midterm Grade distribution: Normal Distribution

If you scored above the median, A- or Aotherwise B or B+

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Lecture Eleven