Seeking help with LOGIT model

[Bogaso]

Dear all, I am fitting a LOGIT model on this Data...........

Data <- structure(c(1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1,
0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1,
0, 1, 1, 0, 1, 0, 47, 58, 82, 100, 222, 164, 161, 70, 219, 81,
209, 182, 185, 104, 126, 192, 95, 245, 97, 177, 125, 56, 85,
199, 298, 145, 78, 144, 178, 146, 132, 98, 120, 148, 123, 282,
79, 34, 104, 91, 199, 101, 109, 117, 1.1, 0.92, 1.72, 2.18, 1.75,
2.26, 2.07, 1.43, 1.92, 1.82, 2.34, 2.12, 1.81, 1.35, 1.26, 2.07,
2.04, 1.55, 1.89, 1.68, 0.76, 1.96, 1.29, 1.81, 1.72, 2.39, 1.68,
2.29, 2.34, 2.21, 1.42, 1.97, 2.12, 1.9, 1.15, 1.7, 1.24, 1.55,
2.04, 1.59, 2.07, 2, 1.84, 2.04, 51.2, 48.5, 50.8, 54.4, 52.4,
56.7, 54.6, 52.7, 52.3, 53, 55.4, 53.5, 51.6, 48.5, 49.3, 53.9,
55.7, 51.2, 54, 52.2, 51.1, 54, 55, 52.9, 53.7, 55.8, 50.4, 58.8,
54.5, 53.5, 48.8, 54.5, 52.1, 56, 56.2, 53.3, 50.9, 53.2, 51.7,
54.3, 53.7, 54.7, 47, 56.9, 0.321, 0.224, 0.127, 0.063, 0.021,
0.027, 0.139, 0.218, 0.008, 0.012, 0.076, 0.299, 0.04, 0.069,
0.33, 0.017, 0.166, 0.003, 0.01, 0.076, 0.454, 0.032, 0.266,
0.018, 0.038, 0.067, 0.075, 0.064, 0.065, 0.065, 0.09, 0.016,
0.061, 0.019, 0.389, 0.037, 0.161, 0.127, 0.017, 0.222, 0.026,
0.012, 0.057, 0.022, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1,
1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1,
0, 1, 1, 0, 1, 0, 0, 1, 0), .Dim = c(44L, 6L), .Dimnames = list(
    c("Obs 1", "Obs 2", "Obs 3", "Obs 4", "Obs 5", "Obs 6", "Obs 7",
    "Obs 8", "Obs 9", "Obs 10", "Obs 11", "Obs 12", "Obs 13",
    "Obs 14", "Obs 15", "Obs 16", "Obs 17", "Obs 18", "Obs 19",
    "Obs 20", "Obs 21", "Obs 22", "Obs 23", "Obs 24", "Obs 25",
    "Obs 26", "Obs 27", "Obs 28", "Obs 29", "Obs 30", "Obs 31",
    "Obs 32", "Obs 33", "Obs 34", "Obs 35", "Obs 36", "Obs 37",
    "Obs 38", "Obs 39", "Obs 40", "Obs 41", "Obs 42", "Obs 43",
    "Obs 44"), c("Y", "X 1", "X 2", "X 3", "X 4", "X 5")))


glm(Data[,1] ~ Data[,-1], binomial(link = logit))

Call:  glm(formula = Data[, 1] ~ Data[, -1], family = binomial(link = logit))

Coefficients:
  (Intercept)  Data[, -1]X 1  Data[, -1]X 2  Data[, -1]X 3  Data[,
-1]X 4  Data[, -1]X 5
     10.99326        0.01943       10.61013       -0.66763
70.98785       17.33126

Degrees of Freedom: 43 Total (i.e. Null);  38 Residual
Null Deviance:      44.58
Residual Deviance: 17.46        AIC: 29.46
Warning message:
glm.fit: fitted probabilities numerically 0 or 1 occurred


However I am getting a warning mesage as "fitted probabilities
numerically 0 or 1 occurred". Here my question is, have I made any
mistakes with my above implementation? Is it just because, I have too
less number of '0' in my response Variable?

Thanks for your help.

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[Ken Knoblauch]

Christofer Bogaso <bogaso.christofer <at> gmail.com> writes:
> Dear all, I am fitting a LOGIT model on this Data...........
---- << snip >>---

> glm(Data[,1] ~ Data[,-1], binomial(link = logit))
>
> Call:  glm(formula = Data[, 1] ~ Data[, -1], family = binomial(link = logit))
>
> Coefficients:
>   (Intercept)  Data[, -1]X 1  Data[, -1]X 2  Data[, -1]X 3  Data[,
> -1]X 4  Data[, -1]X 5
>      10.99326        0.01943       10.61013       -0.66763
> 70.98785       17.33126
>
> Degrees of Freedom: 43 Total (i.e. Null);  38 Residual
> Null Deviance:      44.58
> Residual Deviance: 17.46        AIC: 29.46
> Warning message:
> glm.fit: fitted probabilities numerically 0 or 1 occurred
>
> However I am getting a warning mesage as "fitted probabilities
> numerically 0 or 1 occurred". Here my question is,

have I made any
> mistakes with my above implementation? I
s it just because, I have too
> less number of '0' in my response Variable?
>
Look at the output of summary, especially the standard errors.  
You seem to be getting complete
separation on X5 and X4 doesn,'t look so hot either.

Ken

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[Bogaso]

Thanks Ken for your reply. No doubt your english is quite tough!! I
understand something is not normal with the 5th explanatory variable
(se:2872.17069!) However could not understand what you mean by "You
seem to be getting complete separation on X5 "?

Can you please be more elaborate?

Thanks,

On Thu, Apr 12, 2012 at 4:06 PM, ken knoblauch <[hidden email]> wrote:

> Christofer Bogaso <bogaso.christofer <at> gmail.com> writes:
>> Dear all, I am fitting a LOGIT model on this Data...........
> ---- << snip >>---
>> glm(Data[,1] ~ Data[,-1], binomial(link = logit))
>>
>> Call:  glm(formula = Data[, 1] ~ Data[, -1], family = binomial(link = logit))
>>
>> Coefficients:
>>   (Intercept)  Data[, -1]X 1  Data[, -1]X 2  Data[, -1]X 3  Data[,
>> -1]X 4  Data[, -1]X 5
>>      10.99326        0.01943       10.61013       -0.66763
>> 70.98785       17.33126
>>
>> Degrees of Freedom: 43 Total (i.e. Null);  38 Residual
>> Null Deviance:      44.58
>> Residual Deviance: 17.46        AIC: 29.46
>> Warning message:
>> glm.fit: fitted probabilities numerically 0 or 1 occurred
>>
>> However I am getting a warning mesage as "fitted probabilities
>> numerically 0 or 1 occurred". Here my question is,
> have I made any
>> mistakes with my above implementation? I
> s it just because, I have too
>> less number of '0' in my response Variable?
>>
> Look at the output of summary, especially the standard errors.
> You seem to be getting complete
> separation on X5 and X4 doesn,'t look so hot either.
>
> Ken
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[Ken Knoblauch]

You should look up the Hauck-Donne phenomenon, which shows that
with binomial GLMs, the standard error can grow faster than
the effect size.  Complete separation results, for example,
when one predictor (or a combination of several predictors)
perfectly predicts the response.  Something like this seems
to be happening for variables 4 and 5.  You could try the
brglm function from the package of the same name, which
uses bias correction.  Compare (after coercing your Data to
a data frame):

summary(glm(Y ~ ., binomial, Data))

Call:
glm(formula = Y ~ ., family = binomial, data = Data)

Deviance Residuals:
      Min        1Q    Median        3Q       Max
-2.00979   0.00000   0.00006   0.27987   1.82302

Coefficients:
               Estimate Std. Error z value Pr(>|z|)
(Intercept)   10.99326   20.77336   0.529   0.5967
`X 1`          0.01943    0.01040   1.868   0.0617 .
`X 2`         10.61013    5.65409   1.877   0.0606 .
`X 3`         -0.66763    0.47668  -1.401   0.1613
`X 4`         70.98785   36.41181   1.950   0.0512 .
`X 5`         17.33126 2872.17069   0.006   0.9952


summary(brglm(Y ~ ., binomial, Data))

Call:
brglm(formula = Y ~ ., family = binomial, data = Data)


Coefficients:
              Estimate Std. Error z value Pr(>|z|)
(Intercept) 12.017791  14.337183   0.838   0.4019
`X 1`        0.014898   0.008263   1.803   0.0714 .
`X 2`        8.307941   4.010792   2.071   0.0383 *
`X 3`       -0.576309   0.352097  -1.637   0.1017
`X 4`       35.627644  16.638766   2.141   0.0323 *
`X 5`        2.134544   2.570756   0.830   0.4064


Good luck.

Ken



Quoting Christofer Bogaso <[hidden email]>:

> Thanks Ken for your reply. No doubt your english is quite tough!! I
> understand something is not normal with the 5th explanatory variable
> (se:2872.17069!) However could not understand what you mean by "You
> seem to be getting complete separation on X5 "?
>
> Can you please be more elaborate?
>
> Thanks,
>
> On Thu, Apr 12, 2012 at 4:06 PM, ken knoblauch  
> <[hidden email]> wrote:
>> Christofer Bogaso <bogaso.christofer <at> gmail.com> writes:
>>> Dear all, I am fitting a LOGIT model on this Data...........
>> ---- << snip >>---
>>> glm(Data[,1] ~ Data[,-1], binomial(link = logit))
>>>
>>> Call:  glm(formula = Data[, 1] ~ Data[, -1], family =  
>>> binomial(link = logit))
>>>
>>> Coefficients:
>>>   (Intercept)  Data[, -1]X 1  Data[, -1]X 2  Data[, -1]X 3  Data[,
>>> -1]X 4  Data[, -1]X 5
>>>      10.99326        0.01943       10.61013       -0.66763
>>> 70.98785       17.33126
>>>
>>> Degrees of Freedom: 43 Total (i.e. Null);  38 Residual
>>> Null Deviance:      44.58
>>> Residual Deviance: 17.46        AIC: 29.46
>>> Warning message:
>>> glm.fit: fitted probabilities numerically 0 or 1 occurred
>>>
>>> However I am getting a warning mesage as "fitted probabilities
>>> numerically 0 or 1 occurred". Here my question is,
>> have I made any
>>> mistakes with my above implementation? I
>> s it just because, I have too
>>> less number of '0' in my response Variable?
>>>
>> Look at the output of summary, especially the standard errors.
>> You seem to be getting complete
>> separation on X5 and X4 doesn,'t look so hot either.
>>
>> Ken
>>
>> ______________________________________________
>> [hidden email] mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>

--
Ken Knoblauch
Inserm U846
Stem-cell and Brain Research Institute
Department of Integrative Neurosciences
18 avenue du Doyen Lépine
69500 Bron
France
tel: [phone]
fax: [phone]
portable: [phone]
http://www.sbri.fr/members/kenneth-knoblauch.html

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and provide commented, minimal, self-contained, reproducible code.