# 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.

• 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.
– whuber
Jan 12 at 21:00
• 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.
– whuber
Jan 12 at 22:56

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 $$151.8317^2 (151.8326^2)$$.

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:

$$A_Y=(0,\infty)$$

• cristalline clear ! Jan 16 at 20:04