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I'm running a logistic regression model by using SAS PROC LOGISTIC.

In the Association Statistics table that SAS provides as analysis output there is the Somers' D stat.

Could you provide an interpretation of such statistics and a reference about it, please?

Thanks you all in advance.

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Somers’ $D$ is an index that you want to be closer to 1 and farther from $-1$.  First, you use your model to generate the predicted scores for your dependent variable, $\hat{y}_i$.  Then you think about every possible pairing of data points.  A pair of predicted scores are “in agreement” if the rank order of the predicted scores match the rank order of the observed scores.  That is, if $y_1 < y_2$ and $\hat{y}_1 < \hat{y}_2$, then the pair of points is considered to agree with the prediction.  (To make life easy for this explanation, we will assume that ties are not possible.)  If $\hat{y}_1 > \hat{y}_2$, then the predicted values are not in agreement with the observed values.  After considering all possible pairs of points in your data set, $n_c$ is the number of agreements and $n_d$ is the number of disagreements (I'm using the same mathematical notation as the SAS PROC page, but using slightly different vocabulary.)

Let $p=\frac{n(n-1)}{2}$, the total number of possible pairs for a data set with sample size $n$.  The formula for Somers” $D$ is $$D = \frac{n_c-n_d}{p}$$

Hope this explanation is clear.

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  • $\begingroup$ Thanks for the answer @Gregg! Could you provide the document/link where you found this explanation? For instance the SAS PROC page. Thanks again! $\endgroup$ – Quant.Pi Mar 29 '18 at 7:28
  • $\begingroup$ I believe a google search of "sas proc logistic somer's d" will bring up the sas manual page in the first few results $\endgroup$ – Gregg H Mar 29 '18 at 12:01
  • $\begingroup$ Why would this be a correct interpretation: "Somers’ D is an index that you want to be closer to 1 and farther from −1" ??? $\endgroup$ – Sal Mangiafico Mar 29 '18 at 13:55
  • $\begingroup$ You (ideally) want perfect agreement, $n_c=p$, as opposed to perfect disagreement, $n_d=p$. $\endgroup$ – Gregg H Mar 29 '18 at 13:58
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    $\begingroup$ @GreggH going through the correspondung wikipedia article, I believe your formula might be wrong. I think you are showing kendall's tau and sommers d is a ration of two different kendall's tau calculations. Please check again and correct me if I am wrong. $\endgroup$ – PalimPalim Jun 11 '18 at 9:15

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