Transcript Logistic Regression

Logistic Regression
Logistic Regression

When ?

Just like multiple regression, but when the dependent variable is
dichotomous.


E.g. improved or not improved; successful or not successful.
Why ?

Logistic regression can be used for classification purpose (it includes c2).


Why not performed a discriminant analysis ?



Give probability of an effect (outcome) and evaluate the risk (odds).
Probability of success outside [0,1]
Normality
Why not performed a multiple regression ?



Probability of success outside [0,1]
Homoscedasticity
Normality
Logistic Regression

Example:

Suppose we want to predict whether someone has a coronary
disease (DV) using age in years (IV).

It is customary to code a binary DV either 0 or 1.
Logistic Regression

The logistic curve
u  b0  b1 x1 
yˆ 
1
1  eu
 bp x p
Linear part
Nonlinear part
Logistic Regression

The logistic curve
u  b0  b1 x1 
yˆ 
1
1  eu
 bp x p
Logistic Regression

Example:

Suppose we want to predict whether someone has a coronary
disease (DV) using age in years (IV).

It is customary to code a binary DV either 0 or 1.
Logistic Regression

The logistic curve
u  b0  b1 x1 
 bp x p
1
eu
yˆ 

1  e  u 1  eu

where ŷ is the probability of a 1, e is the base of the natural logarithm (about
2.718) and b is the parameters of the model. b adjusts how quickly the
probability changes when X increases by a single unit. Because the relationship
between X and ŷ is nonlinear, b does not have a straightforward interpretation in
this model; contrary to ordinary linear regression.
Logistic Regression

(Where did it came from)

Suppose we only know a person's age and we want to predict
whether that person has a coronary disease or not. We can talk
about the probability of having the disease, or we can talk about
the odds of having the disease. Let's say that the probability of
not having the disease for a given age is .95. Then the odds of
not having the disease is
yˆ
0.95
odds 

 19
1  yˆ 1  0.95

Now the odds of having the disease would be .05/.95 or 1/19 or
0.0526. This asymmetry is unappealing, because the odds of
having the disease should be the opposite of the odds of not
having the disease.
Logistic Regression

(Where did it came from)

We can take care of this asymmetry by using the natural logarithm, ln. The
natural log of 19 is 2.9444 (ln(0.95/0.05)=2.9444). The natural log of 1/19 is
- 2.9444 (ln(0.05/0.95)=-2.9444), so the log odds of having a coronary
disease is exactly the opposite of the log odds of not having a disease.
Log (odds )  Log (
Solving for
yˆ
)  b0  b1 x1 
ˆ
1 y
 bp x p  u
yˆ
)u
ˆ
1 y
yˆ

 eu
1  yˆ
ŷ Log (
eu
1
 yˆ 

1  eu 1  e  u
In term of probability
In term of odds
Logistic Regression

Finding the regression weights.

In multiple regression, we wanted to minimize the residual sum of squares.
b  ( X T X) 1 X T y


With the logistic curve, there is no mathematical solution that will produce
least squares estimates of the parameters. We will use instead the maximum
(log) likelihood.
A likelihood is a conditional probability: P( ŷ |X), the probability of ŷ given
X). The idea is to choose the regression weights that will give the maximum
(log) likelihood between the data and the logistic curve.
n
L(b)    yi yˆì  (1  yi )(1  yˆì ) , b  (b0 , b1 ,
i 1
, bp )
Maximum likelihood
n
LL(b)    yi ln( yˆì )  (1  yi ) ln(1  yˆì ) 
i 1
Maximum log likelihood
Logistic Regression

Finding the regression weights.

The maximum of this expression can then be found numerically using an
optimization algorithm
Logistic Regression

Finding the regression weights.

The maximum of this expression can then be found numerically using an
optimization algorithm
Logistic Regression

Finding the regression weights.

The maximum of this expression can then be found numerically using an
optimization algorithm
Logistic Regression

Hypothesis testing

The idea is to compare the full model with only the constant using chisquare.
c 2  2ln( LL(b)  LL(0))  -53.6765-(-68.3315) = 29.3099
c 2 (1)  29.3099, p <0.01
There is only 1
predictor

This indicates that age can reliably distinguished between people having a
coronary disease from those who do not.
Logistic Regression

Hypothesis testing

We can use the same idea to build a regression model.
c 2  2 ln( LL(bigger model)  LL(smaller model))

Also, the Wald statistic can be used (Z test).
zi 
bi
SEii
SE 
Fisher information
matrix
I (b) 1
I (b)  X T WX, where, W  diag  yˆi 1  yˆ i 
Logistic Regression

Hypothesis testing

Also, the Wald statistic can be used
Constant
SE 
I (b) 1 
I (b)  XT WX 
z const 
bconst
-5.30945

 -4.68348
SEconst 1.13365
z disease 
b disease
0.110921

 4.61022
SEdisease 0.0240598
IV (coronary disease)
Logistic Regression

Explained variability

There are three popular measures that approximate the variance
interpretation found in linear regression (R2).
1- McFadden's p 2  1 
2
CS
2- R
 1 e

LL(b)
53.6765
 1
 0.214468
LL(0)
68.3315
2
 LL ( b )  LL ( 0 ) 
n
 1 e

2
 53.6765  ( 68.3315) 
n
 0.254052
2
RCS
0.254052
3 R  2 =
=0.340993,
RMax 0.745035
2
N
2
Max
where, R
 1 e
2
 LL ( 0 )
n
 1 e
2
 ( 68.3315)
n
 0.745035
Logistic Regression

Odds Ratio (OR)




The odds ratio is the increase (or decrease) in odds of being in one outcome
category when the value of the predictor increases by on unit.
If the odds are the same across groups, then OR=1.
If the odds are greater than 1, then there is an increase probability of being
classify into the category.
If the odds are smaller than 1, then there is a decrease probability of being
classify into the given category.
ORi  ebi
ORDisease  ebdisease  e0.110921  1.11731

Thus, at each of my birthdays I increase my odds of having a coronary
disease by 1.12. In other words, each year I increase the risk of developing a
coronary disease by 12 percents.
Logistic Regression

Odds Ratio (OR)


For a 5 year age difference, say, the increase is exp(b)5 [= 1.117315] = 1.74,
or a 74% increase.
Classification table

Cutoff = 0.5
Constant only
Total correct percentage = 57
All predictors
Total correct percentage = 74
Logistic Regression

Prediction

If I have (x’=)50 years old, what is my probability of having a coronary
disease ?
yˆ 
yˆ 
1
1  e  (-5.30945 0.110921x1 )
1
1 e
 (-5.30945  0.110921*50)

1
1 e
0.236604
 0.558877
Logistic Regression

Confidence intervals

CI=0.95
SE (u (x))  x T I (b) 1 x  x TVar (b)x
 1 
SE (u    ) 
 50  
0.026677   1 
 1.28517
1
50

  0.026677 0.000578876  50 

 
 1 
SE (u    )  0.0646601  0.254284
 50  
u ( x)  Z / 2 SE ( yˆ ( x))
0.236604  1.96*0.254284
-0.261783, 0.73499
1
1  eu ( x )
1

1 e 
0.434925, 0.675899
 -0.261783, 0.73499
Logistic Regression

Confidence bands

CI=0.95
SE (u (x))  xT I(b) 1 x  xTVar (b)x
u (x)  Z / 2 SE ( yˆ (x))
1
1  eu ( x )
Logistic Regression

Recoding a continuous variable into a dichotomous
variable


Cutoff at 55
Contingency table
Logistic Regression

Recoding a continuous variable into a dichotomous
variable


Cutoff at 55
Regression weights
yˆ 

Wald test
1
1  e  (-0.840783 2.09355x1 )
Logistic Regression

Recoding a continuous variable into a dichotomous
variable


Cutoff at 55
Explained variability
1- McFadden's p 2  0.136861
2
2- RCS
 0.170588
3  RN2 = 0.228967
Logistic Regression

Recoding a continuous variable into a dichotomous
variable


Cutoff at 55
Classification table
Total correct percentage = 57
Total correct percentage = 72
Logistic Regression

Recoding a continuous variable into a dichotomous
variable


Cutoff at 55
Odds ratio
ORDisease  ebdisease  e2.09355  8.11364

If I am 55 years old and up, I have 8 times more chances to have a coronary
disease.
ORDisease 
21
22  8.11364
6
51
Logistic Regression

Recoding a continuous variable into a dichotomous
variable


Cutoff at 55
Confidence intervals
ebi  Z a / 2 SEbi
2.87956, 22.8615

The CI (0.95) is asymmetric. It suggests that coronary disease is 2.9 to 22.9
more likely to occur if I am 55 yrs and up.