Proc logistic results with missing intercept

[YennyS]

I'm trying to run a logistic model but the below code generated output without intercept. The output is essentially the same as specifying " /noint" in the model line. Can someone please shed some light?

proc logistic data = out.analysis descending;
model answer =  duration treat;
strata respid;
run;

[PaigeMiller]

Please show us the output from your PROC LOGISTIC. Screen capture is fine, please click on the "Insert Photos" icon to include your screen capture in your reply. Please do NOT attach files.

[YennyS]

[StatDave]

The conditional model resulting from using the STRATA statement does not have an intercept. See "Details: Conditional Logistic Regression" in the LOGISTIC documentation and the conditional logistic example in the Examples section.

[YennyS]

Thanks! Have seen the example and details.

May I ask why there is no intercept?

[StatDave]

As explained in the Details section of the LOGISTIC documentation that I referred to, the intercept and the separate intercepts that would otherwise be for the individual strata, are conditioned out by the method. The point of the conditional model, much like with GEE models, is to better assess the effects of the predictors in the model by taking into account the correlations within the strata. The intercepts that would otherwise appear are considered nuisance parameters. You can find a lot more about this method including discussion about its usage and limitations in Allison's Logistic Regression book cited in the References section of the LOGISTIC documentation and his book on Fixed Effects Regression. You can find these at RedShelf.com or Amazon.com

[YennyS]

Thank you very much for the explanation.

With the estimation results, if I were to reconstruct the utility derived from a certain combination of predictors, can this still be achieved without intercepts reported?

I was conceptualizing this with a logistic regression without stratification, where the utility is derived as beta_0+beta_k*X_k, where beta_0 is the intercept and X_k represents the predictors.

[ballardw]

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@YennyS wrote:

Thank you very much for the explanation.

With the estimation results, if I were to reconstruct the utility derived from a certain combination of predictors, can this still be achieved without intercepts reported?

I was conceptualizing this with a logistic regression without stratification, where the utility is derived as beta_0+beta_k*X_k, where beta_0 is the intercept and X_k represents the predictors.

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First I will say I am not sure what you are looking for.

If you think an intercept might help you might try Surveylogistic with Strata. There the Strata is treated as part of the sample design and can provide intercepts for each level of the response variable.

[StatDave · Accepted solution]

That computation is the linear predictor as computed in an unconditional model (model not using the STRATA statement), and would be available using the XBETA= option in the OUTPUT statement of PROC LOGISTIC when fitting the unconditional model. Since the conditional model removes the intercept, the linear predictor is not available. The linear predictor estimates the log odds in a logistic model, but it is usually of more interest to then transform the linear predictor to obtain an estimate of the probability of the event level of the response. For a conditional model, you can estimate the conditional, stratum-specific probabilities of the event using the PRED= option in the OUTPUT statement. This again is described in the Conditional Logistic Regression section of the LOGISTIC documentation. Regarding SURVEYLOGISTIC, it does not fit the conditional model.

[YennyS]

Thanks! This is very helpful!