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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.

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    $\begingroup$ The post you reference begins, "The general case of the cube of an normal random variable with any mean is quite complicated,..." That's correct--and the case of higher powers is even worse. Use numerical methods. $\endgroup$
    – whuber
    Jan 12 at 21:00
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    $\begingroup$ There is no "complicated" answer: the best one can do is express it as an integral, but that's perfectly straightforward. For instance, the characteristic function $\exp(itx^4)$ is an integral. So is the distribution function; it is given by $$\Pr(x\le t)=\Phi(\mu/\sigma+t^{1/4}/\sigma)-\Phi(\mu/\sigma -t^{1/4}/\sigma).$$ Differentiating w.r.t. $t$ yields the density function. But these are trivial applications of definitions: the problem is that there's no further simplification. $\endgroup$
    – whuber
    Jan 12 at 22:56

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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 $151.8317^2 (151.8326^2)$.

enter image description here

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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) \\ \\ G_{Y}(y)=P(\sqrt X^4\leq \sqrt{y}) \\ \\ G_{Y}(y)=P(|X^2|\leq \sqrt{y}) \\ \\ G_{Y}(y)=P(X^2\leq \sqrt{y}) \\ \\ G_{Y}(y)=P(|X|\leq y^{1/4}) \\ \\ G_{Y}(y)=P(- y^{1/4}\leq X\leq y^{1/4}) \\ \\ 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:

$A_Y=(0,\infty)$

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    $\begingroup$ cristalline clear ! $\endgroup$
    – G Cab
    Jan 16 at 20:04

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