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Currently I'm working on a project involving loan data where I am trying to build a model that will predict a probability of an individual defaulting on a loan. I have binary dependent variable where defaulted loans are denoted with 1. I decided to use logistic regression for this task, but However, I have difficulty interpreting the results of the regression and would greatly appreciate some clarification.

Namely, I would like to know, whether the probabilities calculated by logistic regression should be interpreted as a probability of default? For example, if for some observation, the predicted value is 0.8, should I assume that according to the model, this individual has 80% chance of default?

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    $\begingroup$ You mean logistic regression, not logarithmic, right? And when you are talking about probabilities calculated by the regression, do you mean the predicted values? $\endgroup$
    – Misius
    Jul 8, 2021 at 15:37
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    $\begingroup$ Yes I mean logistic regression and by probabilities, I do mean predicted values. Sorry for the confusion $\endgroup$
    – Sandro Alavidze
    Jul 8, 2021 at 16:57

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Yes, as long as the input for the outcome is failure (0,1) where 0-No 1-Yes (default), then the predicted probability from your logistic regression model will be the probability of defaulting. For your use case of 0.8 or 80% that person does have a default prediction of 80%, so probably not approved. I would think a cutoff of 50% (0.5) is used, but this would depend on the model, the credit agency, and the input predictors.

Please do consider that a lot of times in finance several models could be used. It may work out that a model for predicting default with outcome coded 0-no default, 1-default will sometimes employ features which are different from optimal features used for a model for predicting no default with outcome coded 0-default, 1-no default.

Reason I am raising this is because a lot of people try to front-load everything into one model, when in fact if you "divide and conquer" to break up a complex problem into multiple smaller problems your prediction may be better.

In fact, I would never approach a bank or credit agency with results of one model!

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  • $\begingroup$ Thanks a lot for a detailed answer! Could you possibly name some of the models that could be used for such problem? Or refer me to some literature exploring this topic? My current model is not quite as accurate as I would like it to be so I'm looking for ways to improve it. $\endgroup$
    – Sandro Alavidze
    Jul 9, 2021 at 6:46
  • $\begingroup$ "...a model for predicting default...can fit differently from a model for predicting no default...". Really? Shouldn't the models be exact opposites (all coefficients opposite signs, and hence output probabilities complementary)? $\endgroup$
    – Ben Reiniger
    Jul 9, 2021 at 20:43
  • $\begingroup$ Of course the $P(\textrm{no default}) = 1-P(\rm{default})$, which if modeled separately after swapping 0,1 would be true if the same features are used for both models. However, it's quite common in financial engineering to also employ features selected a priori which are hinged to historical informativeness, which are typically different for success/failure models. Glad you know that swapping the outcome for 0,1 in logistic models results in 1 minus predicted outcome. What happens to the sign of coefficients? They swap $\pm$ sign so e.g. $\beta_1\rightarrow\ - \beta_1$. $\endgroup$
    – user318288
    Jul 9, 2021 at 23:54
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For the sake of notation, I am using the model with only one independent variable $X$. All of the results hold true with more independent variables as well.

In logistic regression, we use the logistic function

$$ p(X) = \frac{e^{\beta_0 + \beta_1 X}}{1 + e^{\beta_0 + \beta_1 X}}$$

to model the relationship between $p(X) = \mathbb{P}(Y = 1|X)$ and $X$. In your case, $p(X)$ denotes the probability of the default conditional on $X$.

For the predictions, for an individual with the value of the independent variable $x_0$, we predict the probability of default by just plugging it in the estimated function:

$$ \hat{p}(x_0) = \frac{e^{\hat{\beta}_0 + \hat{\beta}_1 x_0}}{1 + e^{\hat{\beta}_0 + \hat{\beta}_1 x_0}}.$$

If you get $\hat{p}(x_0) = 0.8$, then you predict the probability of default for this individual to be $0.8$.

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