CENTRAL LOGISTIC MODEL

CENTRAL BINARY LOGISTIC MODEL

Imaging  Standard  Charted  Bank  wants  to determine  which  customers  are  most  likely  to  repay  their  loan. Thus, they  want  to record  a number  of  independent  variables  that  describe  the  customer’s  reliability  and then determine whether these variables are related to the binary variables    x_2m =1 if the customer repays the  loan and x_2m=0 if the customer fails to repay the loan. This simply telling us that when the mean response variable x_2m is binary, the distribution of x_2m reduces to a single value, the probability  p=pr(x_2m=1).

In this central logistic model, the natural logarithm of odds ratio is the mean explanatory variables by a central logistic model. Here, we are to consider the situation where we have a single mean independent variable, but this model can be generalized by relating the binary mean response variables x_2m, x_3m,..,x_jm,..,x_rm to single mean predictor variable x_1m.

Considered p(x_1m)=x_2m be the probability that x_2m equals 1 when the mean independent variable equals x_1m. By modeling the log-odds ratio to a model in x_1m, a simple central logistic model is given as:

P(x_1m)=[℮^β0+β1X1m]/[1+℮^β0+β1X1m]

EXAMPLE

Assuming a doctor recorded the level of an enzyme, Creatinine  Kinase(CK), for patients who he suspected of having a heart attack. The Objective of the study was to asses whether measuring the amount of CK on admission to the hospital was a useful diagnostic indicator of whether patients admitted with a diagnosis of a heart attack had really had a heart attack. The enzyme CK was measured in 360 patients on admission to the hospital. After a period of time a doctor review the records of these patients to decide which of the 360 patients had actually had a heart attack. The data are given in the table below with CK values given as the 

midpoint of the range of values in each of 13 classes of values.          

CK Values	Number Of Patients With Heart Attack	Number Of Patients Without Heart Attack
20	2	88
60	13	26
100	30	8
140	30	5
180	21	0
220	19	1
260	18	1
300	13	1
340	19	0
380	15	0
420	7	0
460	8	0
500	35	0

1)      Use the formula to calculate the exact probability that a patient had a heart attack when the average CK level in the patient was 160.

SOLUTION

β₀=230/2_*13=8.45

β₁=230/2_*3380=0.034

P(160)=[℮^{8.45+0.034*160}]/[1+℮^{8.45+0.034*160}]=1

NOTE: This central logistic model does not account random error, since it includes the mean component only.




REFERENCE
Galton, Francis.(1886). 'Regression Towards Mediocrity in Hereditary Stature'. Volume 15.