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Currently I have a dataset and I am trying to predict whether someone will default on their bank loan.

The dataset is quite tricky. It covers those who have defaulted in the past, but is also tracking those with current loans. So we don’t know whether they will default in the future.

This means that the target variable (default, not default) has one unique value which is 1. There aren’t any observations that can describe the 0 class.

Assuming I can’t use external datasets, what’s the solution to this problem? How can I best understand those who have defaulted if I don’t have something to compare them to?

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Assuming I can’t use external datasets, what’s the solution to this problem? How can I best understand those who have defaulted if I don’t have something to compare them to?

If you wanna use this only this data as classification task, you can't perform this task. You could way this around, by generated fake data with label 0(you think about which value of parameters comes to default).

But you could try clustering algorithms(like k-means) and find the groups of specific clients. Perhaps one of these groups its group of default.

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To expand on fuwiak's answer, you can cluster the current loan group, declare clusters to be classes, and see whether a good fraction from your default set gets classified in one of the classes/clusters. If yes, this class is predictive of default.

Another take would be to do anomaly detection: use you default set to train the detector and apply it on the current loan data set. The "anomalous" ones are those not likely to default.

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  • $\begingroup$ Hi! you mentioned I should cluster the current loan group only? If I do that, how am I able to see whether the default set gets classified as one of a class/cluster? Can I not just perform k-means clustering on the entire dataset (current + defaulted)? $\endgroup$ – snoopdawg Apr 6 at 9:46
  • $\begingroup$ Clustering the entire data set can work, too. My approach only gives you some additional confidence: You cluster on one set and validate on the other. $\endgroup$ – Igor F. Apr 6 at 17:44

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