Probability distribution of $x^4$ for $x\sim \mathcal{N}(\mu,1)$

Question

I am interested in the distribution of $x^4$ for normal distributed $x\sim \mathcal{N}(\mu,1)$.

For $\mu=0$ I know the probability distribution (source): $$f(z)=\frac{1}{2\sqrt{2\pi} z^{3/4}}{\rm e}^{-\frac{\sqrt{z}}{2}} \,\,{\rm with }\,z\ge 0$$

Answer 1

If $Y=X^4\Rightarrow F_Y(y)=P(Y\leq y)=P(X^4\leq y)$

Thus $F_Y(y)=P(-\sqrt[4]{y}\leq X\leq\sqrt[4]{y})=F_X(\sqrt[4]{y})-F_X(-\sqrt[4]{y})$

The density: $f_Y(y)=\frac d{dy}F_Y(y)=\frac1{4y^{3/4}}(f_x(\sqrt[4]{y})+f_X(-\sqrt[4]{y}))$

When you plug-in and simplify you'll get your answer.