I have done a logistic regression with two independent variables (x1 and x2) and a dependent binary variable (y). The AUC (roc curve) is 0.7915.

After making a heatmap, I noticed the logistic regression was not flexible enough since the range of y for x1=10 is 0.25 – 0.76 while in reality this range is 0.25 – 1.

enter image description here

For that reason I have tried to add splines with package mgcv. This results in an improvement. The AUC (roc curve) improves to 0.8069. The improvement is also visible in the graph below.

enter image description here

Red line: logistic regression Blue line: logistic regression with splines, setting sp=0.1 Green line: logistic regression with splines, setting sp=0.9 Black line: Raw data: for every (x1,x2): y=1 / total y

model_1 <- glm(y ~ x2+x1, data = mydata3, family=binomial())
model_2 <- gam(y ~ s(x2, bs="tp", sp=0.1) + s(x1, bs="tp", sp=0.1), data = mydata3, family=binomial())
model_3 <- gam(y ~ s(x2, bs="cr", sp=0.9) + s(x1, bs="cr", sp=0.9), data = mydata3,  family=binomial())
pivot10 <- mydata3 %>% group_by(x1, x2) %>% summarize(mean = mean(y)) %>% filter(x1==10)

It seems to me the logistic regression with splines is still not flexible enough since the black line has more curve than the blue and green lines. I have tried to modify the settings to make the splines more flexible. I have done this by altering the settings for parameter sp. I have also experimented with several types for parameter bs: “ds”, “cr”, “ps” and “te”. Although all these influences are negligible. Illustrative is the small difference between the green and the blue line.

Does someone has solution to make the splines more flexible?

Thank you!

  • $\begingroup$ To be clear, the effect of various parameters (in particular the basis dimension $k$ and wiggliness penalty $λ$, with the latter being something you should probably not mess with directly) on "flexibility" is with respect to the fitted response function, not necessarily the ROC curve. Also, shouldn't all ROC curves meet at (0,0) and (1,1)? I know if large increments are used for the threshold value they may not appear to intersect, but yours look smooth enough that I'd expect them to meet at the limits, and they don't appear to do so. $\endgroup$
    – Josh
    Jun 29, 2019 at 23:26
  • $\begingroup$ Thank you Josh. The AUC is the complete area under the ROC-curve so I suppose the thresholds are not relevant in this case. The chart above is not the ROC curve. It is a chart of y for every x2 while x1=10. $\endgroup$
    – Marcel
    Jun 30, 2019 at 18:26

1 Answer 1


I not sure what you are trying to achieve by setting the smoothness parameter of the smooth function directly. If you are simply trying to fix the wiggliness of the smooth at some value then you would be better off fixing the effective degrees of freedom (EDF) of the smooth by setting k to the required value and also using fx = TRUE in the definition of the smooth.

Note that the EDF of the smooth is equal to k - 1 because one basis function is removed from the smooth as it is a constant function and is unidentifiable in a model that also has an intercept.

gam(y ~ s(x2, bs="tp", k = 15, fx = TRUE) + 
      s(x1, bs="tp", k = 15, fx = TRUE),
    method = 'REML',
    data = mydata3, family=binomial())

In the above code block, we would have a model with two smooths, each of 14 effective degrees of freedom, with fitting done using REML.

Unless you have very good reason to do this however, I would be very careful how you interpret such a model. Normally, one would just set k to be large enough, which can be checked using gam.check() and let the wiggliness penalty decide how complex the estimated smooth functions should be. Otherwise you could be badly under or over-fitting the data and not know it.

As to the specific question, I think you'd need to vary the smoothness parameter over much larger ranges to get the desired effect. This isn't a value that should be in the range 0–1, instead it should be some value in the range 0–∞, with 0 being no wiggliness penalty (hence the full wiggliness implied by EDF == k - 1), and ∞ meaning a purely linear fit (on the link scale).

  • $\begingroup$ Thanks a lot Gavin! I have changed my settings and tried k = 15 and also k = 100 for both variables, although it only results in slightly more wiggliness. After setting k =15 the gam.check result is: k' edf k-index p-value s(x2) 99 99 1.01 0.96 s(x1) 99 99 0.98 0.44. For k = 100 the result is: s(x2) 14 14 0.97 0.095. s(x1) 14 14 0.99 0.435. $\endgroup$
    – Marcel
    Jun 30, 2019 at 13:37
  • $\begingroup$ So actually I am trying to overfit to see if it is possible to make the green line (gam) identical to the black line (pivot). Could it be the case that the amount of overfitting is limited to a certain amount imposed by the gam-method? Apart from this it seems that the gam is only slightly reacting to any modification of it's settings in my case (although it is a large improvement compared to the original log regression model (red line). $\endgroup$
    – Marcel
    Jun 30, 2019 at 13:51

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