Distribution of higher powers than 2 of a gaussian distribution

If $X \sim \mathcal{N}(0,1)$, then $X^2 \sim \chi^2(1)$. What about higher powers of $X$? I know that the Gamma Distribution is a generalization of the $\chi^2$ distribution, but I don't know how the Gamma Distribution parameters relate to the square part of $\chi^2$.

In particular I'm trying to calculate $X^4$, where $X \sim \mathcal{N} \left(0,\frac{1}{N} \right)$. How do you even take on such a problem?

Thanks for any tips!

• @YiorgosS.Smyrlis Yet again an incorrect edit approved by you... Please revise your behaviour in this domain.
– Did
Feb 20 '14 at 11:54

For every positive $a$, the distribution of $T=|X|^a$ has density $$\frac2{a\sqrt{2\pi}}t^{(1/a)-1}\exp\left(-\tfrac12t^{2/a}\right)\,\mathbf 1_{t>0}.$$ This follows from the usual change of variables method explained there.

For example, the distribution of $Z=X^4$ has density $$\frac{1}{2\sqrt{2\pi}z^{3/4}}\exp\left(-\tfrac12\sqrt{z}\right)\,\mathbf 1_{z>0}.$$

• According to the general formula, for a=4 the exponent for z should be 1/3, not -3/4 (unless i'm missing something)? Feb 20 '14 at 11:11
• Read (1/a)-1, not 1/(a-1) (which would be absurd, say for a=1).
– Did
Feb 20 '14 at 11:52
• Sorry for the edit, I don't know why I interpreted this exponent in such a silly way. It obviously is all correct right know. Feb 20 '14 at 12:10
• @Did this distribution seems not normalizable? I put it into mathematica and the integral over all $z$ is divergent. How do you explain that? Apr 10 '15 at 13:49
• @user1936752 The density in my answer has integral 1, as every distribution has.
– Did
Apr 10 '15 at 14:02