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I am running multiple regression analysis and my dependent variable is not normally distributed (skewness=-1.794 and kurtosis: 4.643). In order to correct this, I applied log transformation but because it is negatively skewed I used this formula: $log(1+max value of DV - DV)$. The inversely and log transformed DV was closer to normally distributed (though inverse transformation was much closer, I still chose log transformation as there is literature on how to interpret log transformation, while none on how to interpret inverse transformation is available to me).

When I run the regression on the transformed DV, I got negative values for unstandardised b which implies that increase in one unit of the predictor would lead to beta*100% decrease in DV. But it is very well known in literature that the increase of that independent variable cannot cause decrease in the dependent variable. I also see reversal of the signs of beta coefficients for some of the other independent variables (as I would expect also to see the increase in both dependent and indepenent variables, not one increase and other decrease).

Does anyone know whether this formula applied to the negatively skewed variable can actually produce such a result (reversal of signs), or something else is wrong? Thank you.

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    $\begingroup$ A couple points: 1) OLS regression does NOT require that the DV be normally distributed, it requires that the error, as estimated by the residuals, be normally distributed. 2) Look into Box-Cox transformations (lots of info on the web). $\endgroup$
    – Peter Flom
    Commented Aug 29, 2012 at 11:44
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    $\begingroup$ Hi @Beka I don't know where you read that. It is not true. The assumptions of OLS regression are not about the DV, and you can have a non-normal DV with normal residuals (usually you don't, but it is possible). $\endgroup$
    – Peter Flom
    Commented Aug 29, 2012 at 12:02
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    $\begingroup$ Hi @Beka, your comment "if there is no normality, it would be detectable in both pre and post analysis" is a common belief but it is not true in general (+1, Peter). It's not difficult to give an example where the errors are normal but the outcome variable itself is not - just make your predictor(s) highly skewed! For an R example, take x <- rexp(200); y <- 1 + 2*x + rnorm(200) - the errors are normal but y is not. Note: non-normal predictors are not a violation of the model assumptions. $\endgroup$
    – Macro
    Commented Aug 29, 2012 at 12:19
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    $\begingroup$ @Michael, OK. All I'm saying (and I think Peter, as well) is that the distribution of $Y|X$ can be normal while the marginal distribution of $Y$ is not. Therefore, checking the response variable, $Y$, for normality, alone, doesn't check the "normality" assumption of the model - it actually tests a much stronger assumption (i.e. that the errors and all of the predictors are normal). Many novices misinterpret what they learn in stat classes and do exactly that. This misunderstanding comes up on this site (and in teaching undergrads) frequently so I think Peter's comment is valuable. $\endgroup$
    – Macro
    Commented Aug 29, 2012 at 14:46
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    $\begingroup$ Beka, @Peter has given you a good answer to that already: your particular transformation is an order-reversing one. As far as contradictory opinions go, I urge you to investigate the bases for those opinions: those that are bare assertions or appeals to authority ("so-and-so says..." or "I believe...") can be ignored, while those that provide references can be checked, but best of all are those that are accompanied by careful, clear reasons and do not shy away from mathematics where that can help: you can check those yourself and decide whether they apply in your case. $\endgroup$
    – whuber ♦
    Commented Aug 29, 2012 at 14:59

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Your original x and your transformed x are inversely related, so, naturally, any relationship that is positive on one will be negative on the other. One way to see this inverse relationship is 

x <- rnorm(100)
x2 <- log(1 + max(x) - x)
plot(x, x2)

(I used a normally distributed X here, but it does not matter for these purposes; you could substitute your variable and its transformation).

Further explanation after reading one of @Beka 's comments above.

x2 is a transformed version of x, using the transformation you used. Then I plotted x vs. x2. When x goes up, x2 goes down. So, any relationship between x and some other variable will be reversed between x2 and that variable.

In other words, your findings do not contradict earlier work.

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