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I have a Logisitc Regression model with a McFadden Pseudo-$R^2=0.7113$. Based on the answers to this question: McFadden's Pseudo-R2 Interpretation, it seems my model has a good fit. However if I want to present this result in a report, what is the interpretation of good fit? According to this link: http://www.statisticalassociates.com/logistic10.htm, Pseudo-$R^2$ is NOT a measure of percent of variance explained in the dependent variable by the model. This is exactly what I thought it was but apparently it is wrong.

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    $\begingroup$ You'll find an excellent paper on this topic by Mittlbock and Schemper: onlinelibrary.wiley.com/doi/abs/10.1002/… In summary, they recommend the use of the squared Pearson correlation coefficient of observed outcome with the estimated probability and another $R^2$ measure like linear sum of squares. Other popular measures were found to be lacking on one of their 12 judgement criteria. $\endgroup$ – StatsStudent Jan 20 at 21:37
  • $\begingroup$ Hi StatsStudent, thanks for the link. I have glanced over the paper quickly but I will take some time soon to read and understand it fully. So just to clarify, are you suggesting it is inappropriate to report on Pseudo-$R^2$ as a metric in reports? $\endgroup$ – Suraj Iyer Jan 20 at 22:48
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    $\begingroup$ Well, "inappropriate" might be a bit strong and whether it's appropriate or not is somewhat subjective, depending on what criteria one chooses to judge the measure by. I like to use the squared Pearson correlation coefficient of observed outcomes with the estimated probabilities as described by Mittlbock and Schemper, because the it can attain a lower bound of zero and an upper bound of one, it has an easy interpretation, and it's not changed by linear transformations of the predictor variables. $\endgroup$ – StatsStudent Jan 20 at 23:00
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    $\begingroup$ Lastly, I wouldn't spent too much time wringing your hands on which measure to use. As long as it's well reasoned and you can explain why you chose to use the measure you chose. $\endgroup$ – StatsStudent Jan 20 at 23:01
  • $\begingroup$ Any particular reason you choose the squared pearson over the sum-of-squares measure? I can understand ofcourse, as you said if its only for simplicity, or are there some cases where you would pick the latter? $\endgroup$ – Suraj Iyer Jan 20 at 23:28

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