Take the 2-minute tour ×
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

I have created a multiple linear regression model to predict prices and about 17% of the predicted prices have come out to be negative. Is there a way to correct for this error, or does it mean that the independent variables are not good predictors? The R^2 coefficient is 0.865.

share|improve this question
    
A negative predicted price is not necessarily an error. We have no basis to expect perfect predictions. Although such a model should give one pause--clearly it does not fully or correctly capture the relationship between prices and the explanatory variables--it could still have great predictive value. (Nevertheless, I have to agree with @Glen_b's answer suggesting that you consider using a more realistic model.) –  whuber Jul 11 at 12:51

2 Answers 2

up vote 11 down vote accepted

It's not the predictors (IVs) that are the problem, but the analysis.

Indeed, the mere existence of an impossible part of the range of the response (you can't have negative prices) would be a hint to consider something other than multiple regression (at least on the untransformed variable) - since it clearly can have negative predictions.

There are a number of ways of dealing with non-negative variables, but two fairly simple approaches might be worth considering:

  • modelling log-price; this is a common strategy with price-like variables in economics

  • using generalized linear models (GLMs). A gamma-model with a log-link would be quite similar to modelling log-price, but the model would be for the expected price rather than expected log-price. This may have some advantages. If you need the relationship with the predictors to be linear in actual price, this can be done (identity link), but a log link for this sort of data would be more common.

As Bill mentions in comments, there's an issue with log-price when you transform back to original units - you no longer have a model for the expectation, but for the median. That said, if you assume normality on the log-scale you can easily compute an ML or MOM estimate of the mean on the original scale.

(And if you can't assume normality on the log-scale you can still approximate the expectation on the original scale via a Taylor expansion.)

Prediction intervals for a new observation, however, transform just fine.

share|improve this answer
1  
Also, since OP is interested in prediction, he is going to want to look at posts on the re-transformation problem. –  Bill Jul 10 at 17:17
    
@Bill yes, it's an important consideration –  Glen_b Jul 11 at 4:26
    
@Bill, Thanks I will look into that –  Anfänger Jul 11 at 13:45

One option would be to fit a generalized linear model with a different reference distribution that is bounded at zero (and probably more appropriate than the normal distribution, depending on what kind of prices you're modeling). For example, the negative binomial distribution suits discrete distributions of counts; this could work if your data are whole numbers representing counts of currency units. The gamma distribution is a continuous alternative (with an otherwise similar shape) that may be more suitable, as it's much more commonly used for financial data (see "Real-life examples of common distributions").

Negative predictions do not reflect badly on the utility of your predictors.

share|improve this answer
7  
It doesn't strike me as valid to say that \$12 is a count variable with 12 currency units as the realization. You probably want something like Gamma. –  gung Jul 9 at 22:47
1  
@gung: good to know. Is there a simple explanation for your reasoning, or somewhere I should look for a not-so-simple explanation? I see from Greg Snow's answer here that, 'Conceptually [the negative binomial distribution] is the number of "failures" before k "successes".' That sure doesn't seem right, but why does it matter? (Good material for a new question...?) –  Nick Stauner Jul 9 at 22:51
2  
I just don't think it's meaningful to call it a count in the sense that the negative binomial is a distribution of counts. Gamma is a continuous non-negative distribution and is often used for things like that. When you aren't right up against 0, it can look fairly normal, which can be a plus when people think the data ought to be normal. –  gung Jul 9 at 22:54
2  
All counts are discrete & non-negative, but not all non-negative discrete variables are counts. It may be that you can use the NB successfully in this situation, but it feels a bit like shoehorning to me. –  gung Jul 9 at 23:10
3  
12 surgical complications at a given hospital in May are 12 different discrete events; 12 accidents at a given intersection in May are 12 different discrete events; etc.; \$12 for a given purchase is 1 event. Yes, you are counting out dollars, but I don't see it as really being the same ontologically. –  gung Jul 9 at 23:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.