Is a Cubed Normally Distributed Variable Still Normal?

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SUMMARY

A normally distributed variable raised to the power of three does not remain normally distributed. This conclusion is based on the principle that any non-linear transformation of a normally distributed variable results in a distribution that is not normal. The discussion also highlights the importance of correctly applying transformations to obtain the standard deviation of the new variable, which varies depending on the specific transformation applied.

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Jellis78
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Hi everybody,

I'm wrestling with the following problem:

Suppose a variable is normaly distributed, then is this same variable still normal distributed when raised to the power of 3? I know if the variable is raised to the power of 2 a chi-squared distribution is obtained, but what happens when raised to the power of 3?

I have a feeling that the variable is still normaly distributed but I can't prove it. Let me take this question one bit further; does anyone know what kind of transformation I have to apply to obtain the standard deviation of the transformed varaible?

Thank you and nice weekend

Jellis
 
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The answer to your first question is, in general no. Any non-linear transformation of a normally distributed variable gives one that is not linearly distributed. I'm not sure why you would accept that the square of a normally distributed variable is not normally distributed but think that the cube would be!
How you would obtain the standard deviation of a transformed variable depends strongly on the transformation.
 
Hi,

Thanks for the reply.

Of course, If the chi-square distribution is not normal then the cubic isn't normaly distributed either.

I made a excel sheet in which I made the mistake to raise the 3rd power of the frequency instead of the variable of the distribution... No wonder it still looked normally distributed

I still meant the transformation of raising the 3rd power of a normal distributed variable. So let me rephrase th equestion:

What kind of mathematical operation do I need to do to obtain the standard deviation of this ‘new’ variable?

Hope you can link me some sort of solution on the web...:rolleyes:

Jellis
 

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