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How many zeros in an independent variable are too many for regression?
Depending upon a dependent variable, for some simple regressions, I have 13 observations, and there are only 3 non-zero values out of 13. For other simple regressions, I have 150 observations and only 12 are non-zeros out of 150.
How to fit a model to a variable with many zeros?
If the number of counts where Count ≠ 0 is small then you can just handle this as a classification problem. Otherwise you can firstly separate the data based on the target variable into two groups: You can use a classification method (for example a logistic regression) to model each of the above outcomes.
How to fit a model to a dependent variable?
I would like to fit a model to a dependent variable distributed like the one below (see picture). The distribution is a count of people (with specific characteristics) in various districts.
When do you add a lagged dependent variable to a differenced model?
The way I see it, once you add a lagged dependent variable to a differenced model, you are really doing lagged-dep-var control and not fixed effects. Steve may disagree (he’s generally less dogmatic than me). This is not always exactly true but it is a theorem for the simple example we use to contrast f.e. and lagged-dep-var control in Section 5.4
How to fit a regression to a variable?
Balancing the data: If you fit a regression model with least square it would be heavily biased towards 0, as most of your data is located at count==0. When you separate your data into two groups then all the data with Count ≠ 0 are put into one bin and they will have more weight against Count ≠ 0.
When to consider the meaning of the zero values?
The meaning of the zero values is an important consideration when choosing an analytic approach.
What does it mean to have zero values in log income?
At worst, zero really means missing rather than zero. There can’t be a single correct identifiable way to proceed without knowing whether the zeros really are correct. I am a great fan of cube roots, but I suspect the problem is that for whatever you are doing log income is a good version of income; it’s just that those zeros get in the way.