Does Applying a Function to Gaussian Variables Preserve Normality?

[aaaa202]

Say I draw a set of numbers {x1,x2,x3,...} from a normal distribution and apply a function f(x) to these.
Will the new set of numbers {f(x1),f(x2),f(x3),...} be gaussianly distributed? I guess it depends on f(x), since for example f(x)=x would certainly mean that the new set is gaussianly distributed, whereas for general f(x) I am not sure. Where can I read about the general theory of this?

 

[Samy_A]

Say I draw a set of numbers {x1,x2,x3,...} from a normal distribution and apply a function f(x) to these.
Will the new set of numbers {f(x1),f(x2),f(x3),...} be gaussianly distributed? I guess it depends on f(x), since for example f(x)=x would certainly mean that the new set is gaussianly distributed, whereas for general f(x) I am not sure. Where can I read about the general theory of this?

If f is affine, the new set will be normally distributed. For general functions not necessarily. For example if f(x)=x², you get a χ2 distribution.
Look for "transformation of normal random variable".

For example: www.math.uiuc.edu/~r-ash/Stat/StatLec1-5.pdf

 

[zinq]

One way to see that the answer depends on the function f(x) is to pick a function that takes only values in an interval that is a proper subset of the real numbers.

Since a Gaussian can have any real number as its value, this shows that with positive probability a Gaussian random variable will take a value that f(x) does not take. Hence the function f applied to a Gaussian r.v. cannot have a Gaussian distribution in this case. For example,

f(x) = x²

as above, which of course takes only nonnegative values. Or alternatively

f(x) = 1 - 1/(x² + 1),

which takes values only in the half-open interval [0, 1).