If $x$ is a normally distributed random variable, then what is the distribution of $x^4$ ? Does it follow a well-known distribution? I am interested in the PDF of $x^4$ if $x$ is a normal distributed variable with non-zero mean.
The question is related to this post that considers the cubic of a normal distributed variable. However there only an answer could be given for the zero-mean case.
 A: Let $Y=X^4$ where $X \sim N(u,\sigma^2)$ so:
$G_{Y}(y)=P(Y\leq y)$
$G_{Y}(y)=P(X^4\leq y)
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G_{Y}(y)=P(\sqrt X^4\leq \sqrt{y})
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G_{Y}(y)=P(|X^2|\leq \sqrt{y})
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G_{Y}(y)=P(X^2\leq \sqrt{y})
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G_{Y}(y)=P(|X|\leq y^{1/4})
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G_{Y}(y)=P(- y^{1/4}\leq X\leq y^{1/4})
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G_{Y}(y)=G_{X}(y^{1/4})-G_{X}(-y^{1/4})
$
Where $G_{X}$ is the cumulative of $X$, remember that $Y$ distribution can be get as $f_{Y}(y)=G_{Y}^{'}(y)$ then:
$f_{Y}(y)=4y^{-3/4}f_{X}(y^{1/4})+4y^{-3/4}f_{X}(-y^{1/4})$
where $f_X(.)$ is the $X$ distribution applied to $(.)$
Now we are seeking for its support:
As $x^4$ is a non negative function so its minimal is equal to $0$ and its maximum is $\infty$ so $Y$ support is given by:
$A_Y=(0,\infty)$
A: Although whuber means there is no simple expression for the probability density distribution of $x^4$ with $x\sim \mathcal{N}(\mu,\sigma^2)$
we can use the CDF from whuber's comments to give a simple expression for the PDF:
For the CDF we get
$${\rm Pr}\left(x \le t\right)=\Phi\left(\frac{\mu}{\sigma}+\frac{t^{1/4}}{\sigma}\right)-\Phi\left(\frac{\mu}{\sigma}-\frac{t^{1/4}}{\sigma}\right)\\
=\frac{1}{2}\left[1+{\rm erf}\left(\frac{1}{\sqrt{2}}\left(\frac{\mu}{\sigma}+\frac{t^{1/4}}{\sigma}\right)\right)\right]-\frac{1}{2}\left[1+{\rm erf}\left(\frac{1}{\sqrt{2}}\left(\frac{\mu}{\sigma}-\frac{t^{1/4}}{\sigma}\right)\right)\right]\tag{1}$$
and for the PDF
$${\rm Pr}\left(t\right){\rm d}t=\frac{{\rm d}}{{\rm d}t}{\rm Pr}\left(x \le t\right){\rm d}t=\frac{1}{4 \sqrt{2 \pi } \sigma  t^{3/4}}\left({\rm exp}\left(-\frac{\left(t^{1/4}-\mu\right)^2}{2 \sigma ^2}\right)+{\rm exp}\left(-\frac{\left(t^{1/4}+\mu\right)^2}{2 \sigma ^2}\right)\right){\rm d}t \tag{2}$$
The expected value is
$$\mu ^4+6 \mu ^2 \sigma ^2+3 \sigma ^4\tag{3}$$
and variance is
$$8 \left(2 \mu ^6 \sigma ^2+21 \mu ^4 \sigma ^4+48 \mu ^2 \sigma ^6+12 \sigma ^8\right)\tag{4}$$
Check by simulation
The histogram for 300 million samples with $\mu=4.5,\sigma=0.4$ agrees well with the theoretical PDF (black). The simulated (theoretical) values for the expectation are $429.591 (429.579)$ and for the variance $151.8317^2 (151.8326^2)$.

