Can a power of y (dependent variable) be used in neural network as input? I am using MLP (Multilayer perceptron) for regression and predicting a continuous variable y (which is the waiting time of passengers in an intercity station). Unfortunately, the fit of the model is bad and It does not predict the variable y well. The residuals of the model have ascending linear pattern. I want to know, can I insert a power of y (for example y^2) as an input to my neural network model (my transfer functions are linear and tansig)? I have seen many times that in situations of having an ascending linear pattern of residual in linear regression we can use a power of y as an input to manage the errors. Can we do the same in MLP?
Thanks in advance for your time and consideration.
 A: NO
You will run into trouble when it comes time to make new predictions. If you don't know the $\text{wait_time}$, you cannot calculate $\text{wait_time}^p$.
Nothing will stop you from inputting this into your model when you are writing your code, since (I assume) you have all of the $\text{wait_time}$ values, but it is illegitimate to use $\text{wait_time}$  to predict $\text{wait_time}$.
A: Well no, but with some modification, yes. You cannot use $y^2$ to predict $y$, because your model will make every coefficient $0$ and simply predict $y$ as $\sqrt{y^2}$. However, you can use other $y^2$ to predict $y$. Let me explain, if you have a time series you can use $y_{t-1}$ to predict $y_t$. In your case, if you know your variable depends on another one (which may not be time), you can use the target of another similar data point to the input to predict that target.
In order to make it more precise, I'll add a method for doing what I explained. Take your input, apply $K$-nearest neighbours based on the atributes and select $K$ neighbours. Then, add the targets of those $K$ neighbours as input to your model. When predicting test you just need to select the nearest training neighbours.
¿Will this solve your problem? Probably not. Ascending linear residuals in linear models means that the hypothesis for linear models to apply are wrong, and therefore the model doesn't work. Here, there are no hypothesis for a MLP to work, but it is an indicator that the architecture doesn't have the complexity to fit the data. I don't know the trick you mention, but I do know that modifying the input with some transformation may be enough for it to work properly. Since you are using a MLP this could mean changing the architecture (or adding more data which always helps).
