Tell me more ×
Cross Validated is a question and answer site for statisticians, data analysts, data miners and data visualization experts. It's 100% free, no registration required.
up vote 3 down vote favorite
1

I am trying to run an econometric panel data (random effects) model with about 950 observations (so not a small dataset). My data consists of different European public companies and of a few financial statistics of theirs during the last 10 years. However, I am having enormous amount of trouble with the following aspect: According to Jarque-Bera test (p value of 0) I am violating the assumption of normality. While my skewness seems small enough (under 3), the kurtosis has a value of 14. However, I am very puzzled as to what to do now. As one of the options I read that you could take logarithms, but one of my variables (market value added) is often negative and Eviews, obviously, doesn't let me opt for that option. Is there any other solutions to this problem or can anyone offer some advice to an econometric novice? Thank you in advance.

share|improve this question
1  
(1) Are you applying the JB test to (a) an dependent variable, (b) an independent variable, or (c) residuals in a regression? It probably is most useful in case (c); its interpretation and your response could be quite different in cases (a) or (b). Could you clarify? (2) "Market Value Added" sounds like a difference of two variables. If you do elect to re-express its values, you are best off re-expressing the original two variables and then recomputing MVA. For example, you might compute the log of the ratio of the original two variables. Do you have these variables? – whuber♦ Dec 7 '11 at 18:03
Thank you for all the answers. As to the first post then, yes, I meant the residuals (c) and I did happen to have the original values of my market-value-added variable. For some reason I couldn't come up to using those instead myself, so that was very helpful. However, after now trying taking logarithms or roots (tried 3rd, 5th and 7th) the problem remains the same. While the actual values of skewness and kurtosis are now much smaller, the kurtosis is still valued around 5 and, thus, my Jarque-Bera test fails. Are there any other options on what I could still try? – Marie Dec 13 '11 at 15:31
1  
Let's back up, Marie: could you please explain why you are worried about normality? In almost every situation (a) only approximate normality of (b) a test statistic (such as the mean) of the (c) residuals is required and (d) one can tolerate larger departures from normality as the number of observations goes up. Moreover, it's relatively rare to use formal tests to check normality: it's more insightful to use exploratory methods such as probability plots. You might therefore be concerned about nothing in this situation. – whuber♦ Dec 13 '11 at 17:07
Well, this is basically the second econometric project I have ever done in my life so I am not 100% certain this is very important. However, after reading about main requirements for a project/model from different econometrics books and other thesises a few things kept popping up. They always mention non-normality, heteroskedacity, multicollinearity etc as things that are always to be avoided. I, therefore, conducted that if my model would be suffering from any of the previous it would not be acceptable thesis material. – Marie Dec 15 '11 at 12:37
If I have misinterpreted the importance of the normality assumption it would, honestly, lift a weight off my shoulders, because at the moment I have been stuck with normality for 2 weeks and cannot move onwards with my work, because I thought without normality my model is close to worthless. – Marie Dec 15 '11 at 12:42

1 Answer

up vote 2 down vote

The cube root is the simplest transformation that pulls in tails and treats values around 0 symmetrically. A gated reference is Nicholas J. Cox's Stata Journal paper. You don't need to be a Stata user to appreciate the wisdom therein.

share|improve this answer
You can also try fifth or seventh roots. – Dimitriy V. Masterov Dec 7 '11 at 15:53

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.