This is a cross-posting from CrossValidated: In support vector regression (with linear loss), we minimise the objective function: \begin{align} \min_{\mathbf{w}, b, \mathbf{\xi}} \quad & \frac{...

If you are distinguishing between Statistics and Machine Learning, then you need to define boundary between the two, and that boundary is going to be opinion-based. It is a matter of definition which ...

As Kashra said, your "system" has an infinite number of valid solutions. However, there is one "canonical" solution, that might make more sense than others, depending what you are after. A matrix is ...

The point of regularization is to avoid overfitting, and overfitting happens when you have too many predictor variables (i.e. neurons) contributing to the outcome. So, by regularization you are ...

In addition to the ncasas' answer, which is good in my opinion, I'd like to point out that ReLU is computationally inexpensive, in contrast to sigmoid activation functions. They require only an if / ...

Your example shows that K-means (and clustering in general) is not a suitable tool to detect anomalies. Anomalies are, by definition, points (observations) deviating from normality, however that ...

I believe you can use a classification algorithm where you manually overrepresent the "anomalies" class. By how much, depends on the cost induced by the anomalies. Just to illustrate what I mean: ...

Let's start with k-means: If you add class information on top of it, you get the learning vector quantization (LVQ). On the other hand, if you impose a topology on the means (force them on an elastic ...

This question is similar to this one and this one, but seems to be ill-posed. Either because it implies an unknown (undefined) way how the neurons process the inputs, or because the provided solution (...

If you perform linear regression, encoding the categorical variables by dummy numerical variables, the p-value of the corresponding coefficients will show you whether they significantly affect the ...

You are partially right and partially wrong: $f'(\textbf{x})$ is a matrix, but $\text{diag}(f'(\textbf{x}))$ means taking the diagonal of that matrix and making a vector out of it.

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/...