$X$,$Y$,$Z$,$T$ are independant random variables with probability density functions $f_{X}(x)$,$f_{Y}(y)$,$f_{Z}(z)$,$f_{T}(t)$.
What is the pdf of $U$ if $U=F(X,Y,Z,T)$ knowing the pdf of $X$,$Y$,$Z$,$T$ ? The answer should be a function of $U$: $f_{U}(u)$
Maybe this general question has no general answer so here is a practical example:
let $U=\sqrt{(X-Y)^2+(Z-T)^2}$ and $X$,$Y$,$Z$,$T$ are uniformly distributed function between $-a$ and $a$.
With Monte-Carlo simulation, I find that the histogram (the pdf) of U is some sort of bell function. How to have its analytical or numerical value without running a Monte-Carlo spraying?
You have to first find the cdf, differentiate it w.r.t $u$ to get the pdf. Let's start with the cdf computation. For $0\leq u\leq 2a$, \begin{align} F_U(u) &= \mathbb{P}(U\leq u)=\mathbb{P}(U^2\leq u^2)\\ &= \int_{ \left\{\substack{(x,y,z,t):-a\leq x,y,z,t \leq a\\ (x-y)^2+(z-t)^2\leq u^2}\right\}}f_X(x)f_Y(y)f_Z(z)f_T(t)dxdydzdt. \end{align} Here, we first integrate with respect to $y$. The range of $y$ is obtained by rearranging the constraints on $y$: \begin{align} (x-y)^2+(z-t)^2\leq u^2 & \implies (x-y)^2+\leq u^2-(z-t)^2\\ & \implies \begin{cases} x-\sqrt{u^2-(z-t)^2}\leq y\leq x+\sqrt{u^2-(z-t)^2} \\ (z-t)^2\leq u^2 \end{cases}\\ -a\leq y \leq a & \implies \begin{cases} \max\{x-\sqrt{u^2-(z-t)^2},-a\}\leq y\leq \min\{ x+\sqrt{u^2-(z-t)^2},a\}\\ (z-t)^2\leq u^2 \end{cases}\\ &\implies \begin{cases} \begin{cases} x-f(u,z,t)\leq y\leq x+f(u,z,t)\\ (z-t)^2\leq u^2 \end{cases} &\text{ if } \begin{cases} x-f(u,z,t)\geq -a\\ x+f(u,z,t)\leq a \end{cases}\\ \begin{cases} -a\leq y\leq x+f(u,z,t)\\ (z-t)^2\leq u^2 \end{cases} &\text{ if } \begin{cases} x-f(u,z,t)\leq -a\\ x+f(u,z,t)\leq a \end{cases}\\ \begin{cases} x-f(u,z,t)\leq y\leq a\\ (z-t)^2\leq u^2 \end{cases} &\text{ if } \begin{cases} x-f(u,z,t)\geq -a\\ x+f(u,z,t)\geq a \end{cases}\\ \begin{cases} -a\leq y\leq a\\ (z-t)^2\leq u^2 \end{cases} &\text{ if } \begin{cases} x-f(u,z,t)\leq -a\\ x+f(u,z,t)\geq a \end{cases} \end{cases}, \end{align} where $f(u,z,t)=\sqrt{u^2-(z-t)^2}$. Thus, we get \begin{align} F_U(u) &= \int_{ \left\{\substack{(x,y,z,t):-a\leq x,y,z,t \leq a\\ x-f(u,z,t)\leq y\leq x+f(u,z,t) \\ (z-t)^2\leq u^2 }\right\}}\frac{1}{16a^4}dxdydzdt\\ &=\frac{1}{16a^4}\underset{I_1}{\underbrace{\int_{ \left\{\substack{(x,z,t):-a\leq x,z,t \leq a\\ x-f(u,z,t)\geq -a\\x+f(u,z,t)\leq a \\ (z-t)^2\leq u^2}\right\}}2f(u,z,t)dxdzdt }}\\ &\hspace{1cm}+ \frac{1}{16a^4} \underset{I_2}{\underbrace{\int_{ \left\{\substack{(x,z,t):-a\leq x,z,t \leq a\\ x-f(u,z,t)\leq -a\\x+f(u,z,t)\leq a \\ (z-t)^2\leq u^2}\right\}}\left(x+f(u,z,t)+a \right)dxdzdt }} \\ &\hspace{1cm}+ \frac{1}{16a^4}\underset{I_3}{\underbrace{\int_{ \left\{\substack{(x,z,t):-a\leq x,z,t \leq a\\ x-f(u,z,t)\geq -a\\x+f(u,z,t)\geq a \\ (z-t)^2\leq u^2}\right\}}\left(a-x+f(u,z,t) \right)dxdzdt}} \\ &\hspace{1cm}+ \frac{1}{16a^4}\underset{I_4}{\underbrace{\int_{ \left\{\substack{(x,z,t):-a\leq x,z,t \leq a\\ x-f(u,z,t)\leq -a\\x+f(u,z,t)\geq a \\ (z-t)^2\leq u^2}\right\}}2a dxdzdt}} . \end{align} Simplifying each integral, \begin{align} I_1&=2\int_{ \left\{\substack{(x,z,t):-a\leq z,t \leq a\\ -a+\sqrt{u^2-(z-t)^2}\leq x\leq a-\sqrt{u^2-(z-t)^2}\\ (z-t)^2\leq u^2\leq a}\right\}}\sqrt{u^2-(z-t)^2}dxdzdt\\ &=2\int_{ \left\{\substack{(z,t):-a\leq z,t \leq a\\ (z-t)^2\leq u^2}\right\}}2\sqrt{u^2-(z-t)^2}(a-\sqrt{u^2-(z-t)^2})dzdt\\ &=4\int_{ \left\{\substack{(s,w):-a\leq \frac{w+s}{2},\frac{w-s}{2} \leq a\\ -u\leq s\leq u}\right\}}\sqrt{u^2-s^2}(a-\sqrt{u^2-s^2})\frac{1}{2}dsdw\\ &=2\int_{ \left\{\substack{(s,w):-2a+s\leq w \leq 2a-s\\ 0\leq s\leq u}\right\}}\sqrt{u^2-s^2}(a-\sqrt{u^2-s^2})dsdw \\ &\hspace{2cm}+ \frac{1}{8a^4}\int_{ \left\{\substack{(s,w):-2a-s\leq w\leq 2a+s\\ -u \leq s\leq 0}\right\}}\sqrt{u^2-s^2}(a-\sqrt{u^2-s^2})dsdw\\ &=2\int_0^u\sqrt{u^2-s^2}(a-\sqrt{u^2-s^2})2(2a-s)ds \\ &\hspace{1cm}+ \frac{1}{8a^4}\int_{-u}^0\sqrt{u^2-s^2}(a-\sqrt{u^2-s^2})2(2a+s)ds\\ &=4\int_0^u(2a-s)\sqrt{u^2-s^2}(a-\sqrt{u^2-s^2})ds\\ &=4\int_0^u 2a^2\sqrt{u^2-s^2} ds+4\int_0^u a \sqrt{u^2-s^2} \frac{1}{2}d(u^2-s^2) - 4\int_0^u (2a-s)(u^2-s^2)ds\\ &= 4\left(\frac{\pi}{2}a^2u^2-\frac{5}{3}au^3+\frac{1}{4}u^4\right). \end{align} \begin{align} I_2=\int_{ \left\{\substack{(x,z,t):-a\leq x,z,t \leq a\\ x-\sqrt{u^2-(z-t)^2}\leq -a\\x+\sqrt{u^2-(z-t)^2}\leq a \\ (z-t)^2\leq u^2}\right\}}\left(x+\sqrt{u^2-(z-t)^2}+a \right)dxdzdt \end{align} Similarly, simplify the other integrals!