${\displaystyle X=(X_1,X_2)}$, ${\displaystyle X_1,X_2\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathcal{N}(0,1)}$, ${\displaystyle T(X)=X_1^2+X_2^2}$.

This transformation is not invertible, but preimage ${\displaystyle T^{-1}(A)}$ is well defined.

For such a case, ${\displaystyle \mathrm{\Theta }\sim \mathrm{Uniform}(-\frac{\pi }{2},\frac{\pi }{2})}$, ${\displaystyle Y=\tan \mathrm{\Theta }}$.
For ${\displaystyle y\in \mathbb{R}}$, $$\begin{array}{rl}& \mathbb{P}(y<\tan \mathrm{\Theta }<y+\mathrm{d}y)\\ =& \mathbb{P}({\tan }^{-1}(y)<\mathrm{\Theta }<{\tan }^{-1}(y+\mathrm{d}y))\\ \approx & \mathbb{P}({\tan }^{-1}(y)<\mathrm{\Theta }<{\tan }^{-1}(y)+\mathrm{d}y\frac{\mathrm{d}}{\mathrm{d}y}({\tan }^{-1}(y)))\\ \approx & f_{\mathrm{\Theta }}({\tan }^{-1}(y))\frac{\mathrm{d}y}{1+y^2}.\end{array}$$
Meanwhile $$\mathbb{P}(y<\tan \mathrm{\Theta }<y+\mathrm{d}y)\approx f_Y(y)\mathrm{d}y,$$ so $$f_Y(y)=\frac{1}{\pi (1+y^2)},y\in \mathbb{R}.$$ So ${\displaystyle Y\sim \mathrm{Cauchy}}$.

Commonly used in economics, stock market analysis, engineering, biology, etc.

${\displaystyle X\sim \mathcal{N}(\mu ,{\sigma }^2)}$, ${\displaystyle Y=T(X)={\mathrm{e}}^X}$. Since ${\displaystyle T^{-1}(y)=\log y}$, by the formula (1.2) $$f_Y(y)=f_X(\log y)\frac{\mathrm{d}(\log y)}{\mathrm{d}y}=\frac{1}{y\sqrt{2\pi {\sigma }^2}}{\mathrm{e}}^{-\frac{(\log y-\mu {)}^2}{2{\sigma }^2}},y>0.$$

A noticeable fact is that, all moments of ${\displaystyle Y}$ exist: $$\mathbb{E}[Y^n]={\int }_0^{+\mathrm{\infty }}{y}^n{f}_Y(y)\mathrm{d}y={\mathrm{e}}^{n\mu +\frac{n^2{\sigma }^2}{2}}.$$ However MGF doesn't exist, because $$\mathbb{E}[{\mathrm{e}}^{tY}]={\int }_0^{+\mathrm{\infty }}{\mathrm{e}}^{ty}{f}_Y(y)\mathrm{d}y=+\mathrm{\infty },\mathrm{\forall }t>0.$$ So we cannot find the interval ${\displaystyle (-\epsilon ,\epsilon )}$.

${\displaystyle X\sim \mathcal{N}(0,1)}$, ${\displaystyle Y=T(X^2)}$. For ${\displaystyle y>0}$, $$\begin{array}{rl}& \mathbb{P}(y<Y<y+\mathrm{d}y)\\ =& \mathbb{P}(\sqrt{y}<X<\sqrt{y+\mathrm{d}y})+\mathbb{P}(-\sqrt{y+\mathrm{d}y}<X<-\sqrt{y})\\ \approx & \mathbb{P}(\sqrt{y}<X<\sqrt{y}+\frac{\mathrm{d}y}{2\sqrt{y}})+\mathbb{P}(-\sqrt{y}-\frac{\mathrm{d}y}{2\sqrt{y}}<X<-\sqrt{y}),\end{array}$$
i.e. $$\begin{array}{rl}{f}_Y(y)\mathrm{d}y& \approx f_X(\sqrt{y})\frac{\mathrm{d}y}{2\sqrt{y}}+f_X(-\sqrt{y})\frac{\mathrm{d}y}{2\sqrt{y}}\\ \Rightarrow f_Y(y)& =\frac{f_X(\sqrt{y})}{\sqrt{y}}=\frac{1}{\sqrt{2\pi }}{y}^{-\frac{1}{2}}{\mathrm{e}}^{-\frac{y}{2}},y>0.\end{array}$$
This is ${\displaystyle {\chi }_1^2}$ (see here) (or ${\displaystyle \mathrm{Gamma}(\frac{1}{2},2)}$) distribution. We can calculate the MGF $$\begin{array}{rl}{M}_Y(t)& =\mathbb{E}[{\mathrm{e}}^{tY}]={\int }_0^{+\mathrm{\infty }}\frac{1}{\sqrt{2\pi y}}{\mathrm{e}}^{y(t-\frac{1}{2})}\mathrm{d}y\\ & =\frac{1}{\sqrt{1-2t}}{\int }_0^{+\mathrm{\infty }}\frac{1}{\sqrt{2\pi u}}{\mathrm{e}}^{-\frac{u}{2}}\mathrm{d}u=\frac{1}{\sqrt{1-2t}},t<\frac{1}{2}.\end{array}$$

Let ${\displaystyle V=[X_1,\cdots ,X_n{]}^{\mathrm{T}},X_1,\cdots ,X_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathcal{N}(0,1)}$. ${\displaystyle S_n=X_1^2+\cdots +X_n^2}$, then $$M_{S_n}=[M_{X_1}(t){]}^n=\frac{1}{(1-2t{)}^{\frac{n}{2}}}.$$
For ${\displaystyle Y\sim {\chi }_n^2}$ (or ${\displaystyle \mathrm{Gamma}(\frac{n}{2},2)}$) with p.d.f $$f_Y(x)=\frac{1}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{2^{\frac{n}{2}}}{x}^{\frac{n}{2}-1}{\mathrm{e}}^{-\frac{x}{2}},x\ge 0,$$ it also has ${\displaystyle M_Y(t)=\frac{1}{(1-2t{)}^{\frac{n}{2}}},t<\frac{1}{2}}$, so ${\displaystyle X_1^2+\cdots +X_n^2\sim {\chi }_n^2}$.

Now let ${\displaystyle R_n=\sqrt{X_1^2+\cdots +X_n^2}}$, then $$f_{R_n}(r)=f_{X_1^2+\cdots +X_n^2}(r^2)\left|\frac{\mathrm{d}{r}^2}{\mathrm{d}r}\right|=\{\begin{array}{rl}& f_Y(r),r\ge 0,\\ & 0,r<0,\end{array}$$ so $$\mathbb{E}[R_n]=\sqrt{2}\frac{\mathrm{\Gamma }(\frac{n}{2}+\frac{1}{2})}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}.$$

${\displaystyle F_X(x)=\mathbb{P}(X\le x)=\mathbb{P}(\omega \in \mathrm{\Omega }|X(\omega )\le x)}$.
Here ${\displaystyle X:\mathrm{\Omega }\to {\mathcal{R}}_X,F_X:{\mathcal{R}}_X\to [0,1]}$, ${\displaystyle F_X(X):\mathrm{\Omega }\to [0,1]}$.
So $$F_X(X(\omega ))=\mathbb{P}(\{{\omega }^{\prime }\in \mathrm{\Omega }|X({\omega }^{\prime })\le X(\omega )\})={\int }_{-\mathrm{\infty }}^{X(\omega )}{f}_X(x)\mathrm{d}x.$$
If ${\displaystyle F_X}$ continuous, inverse exists and ${\displaystyle q_X(u)=F_X^{-1}(u),\mathrm{\forall }u\in (0,1)}$.
For ${\displaystyle u\in (0,1)}$, since ${\displaystyle F_X^{-1}}$ is strictly increasing, $$\begin{array}{rl}\mathbb{P}(F_X(X)\le u)& =\mathbb{P}(F_X^{-1}(F_X(X))\le F_X^{-1}(u))\\ & =\mathbb{P}(X\le F_X^{-1}(u))=F_X(F_X^{-1}(u))=u.\end{array}$$