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I have seen a question that claims transformation changes a distribution into a normal distribution. Is this a correct assertion?

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    $\begingroup$ We have many threads about transformations, including several that point out no transformation can possibly make a Binomial distribution into a Normal one. $\endgroup$
    – whuber
    Dec 28, 2016 at 16:25
  • $\begingroup$ @whuber Did you notice in my comment I said that the distribution has to be continuous and completely known to do this? Am I missing any other restrictions? $\endgroup$ Dec 28, 2016 at 16:29
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    $\begingroup$ @Michael An appeal to the Probability Integral Transformation establishes the sufficiency of your restrictions. That they are necessary follows from the fact that a transformation (whether it is continuous or even measurable, for that matter) of any non-continuous distribution will remain non-continuous. $\endgroup$
    – whuber
    Dec 28, 2016 at 16:34
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    $\begingroup$ "I have seen a question..." - Which question would that be? $\endgroup$
    – Firebug
    Dec 28, 2016 at 19:20

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No. There are a lot of cases where a particular transformation may make the transformed distribution closer to normal, like using taking the square root of data to address rightward skew. But usually when transforming data this way the goal is to get a result that is "normal enough" for various statistical assumptions to hold (or come close enough to holding).

Other transformations are done to better model true relationships among data elements. If two variables have a logarithmic relationship but the unit of measure does not reflect that, then a log transformation may make that relationship more clear, easier to work with, and more interpretable, whether the underlying distribution is normal or not (either before or after transformation).

It is definitely not correct to say that transformations, in general, change any non-normal distributions into normal distributions.

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It depends on the underlying starting distribution. Many transformations can make skewed data, into a more "normal looking" type of data. Often "Log" transformation is used to that purpose (as it can compress heavily skewed data). The Box-Cox transformations are also sometimes used.

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    $\begingroup$ Note that f.g. said normal looking. These transformation do not make the distribution exactly normal. If the starting distribution is a completely known distribution with cumulative distribution F then its inverse gives you ,a uniform [0,1] distribution. Then apply a normal cdf to that and you get a normal distribution. But this cannot be done in practice to data from an unknown distribution. This approach can be used to simulate data from a continuous distribution. $\endgroup$ Dec 28, 2016 at 16:23
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    $\begingroup$ box-cox transform, even if often presented as a normalizing transformation, more often is variance stabilizing. $\endgroup$ Dec 28, 2016 at 21:05
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In order to convince you that the question is correct suppose that $x\sim N(0,1)$ is standard normal value and $y=sign(x)*x^2$ is another random value. So now it is clear that $y$ is not normaly distributed. However taking inverse transformation gives $x=\sqrt{|y|}*sign(y)\sim N(0,1)$. That is, transformation of $y$ changes its distribution to normal.

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  • $\begingroup$ @whuber If x is negative y is negative since y= -x^2. |y|=x^2 is positive. However sqrt(|y|) can be -x or +x and sign(y) does not turn these two values into 1. So the transformation is not 1-1. Is that what you meant? $\endgroup$ Dec 28, 2016 at 16:38
  • $\begingroup$ @Michael The comment to which you are responding was quickly deleted when I realized it was based on a misreading of the transformation. $\endgroup$
    – whuber
    Dec 28, 2016 at 17:30
  • $\begingroup$ Wouldn't log and exp be easier to transformations to use to illustrate your point? $\endgroup$
    – Dason
    Dec 28, 2016 at 18:20
  • $\begingroup$ @whuber Even though the deleted comment misread the transformation. don't you think my comment still points to a problem with the transformation? $\endgroup$ Dec 28, 2016 at 19:39
  • $\begingroup$ @Michael By convention, unless we are working in the Complex domain, the square root is assumed to be nonnegative. Thus the transformations described here are well-defined and inverses of each other. (I don't see that this particular example really responds to the question or provides much insight, but then again the question is vague.) $\endgroup$
    – whuber
    Dec 28, 2016 at 19:43

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