For example, I have three independent random variables $x$,$y$,$z$, all of them are uniformly distributed between $0$ and $1$. Now I want to find the probability satisfying the two constraints simultaneously:
$$P\left[\frac{\log\left(x\right)}{\log\left(y\right)}<\frac{4}{3},\frac{\log\left(y\right)}{\log\left(z\right)}<\frac{3}{2}\right]$$ I know the basic method to find the volume by integration manually, and I can obtain the answer $4/15$. However it is too complicated and I could have more of these variables and more constraints.
My question is, for such problem, is there a systematic way or formula to obtain the probability directly? Or, can these be transformed into exponential distributions and simplify the calculation?
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ The answer is given by: \begin{align} &\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \bracks{{\ln\pars{x} \over \ln\pars{y}} < {4 \over 3}} \bracks{{\ln\pars{y} \over \ln\pars{z}} < {3 \over 2}}\dd x\,\dd y\,\dd z \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \bracks{x > y^{4/3}}\bracks{z < y^{2/3}}\dd x\,\dd z\,\dd y = \int_{0}^{1}\int_{0}^{y^{\large{2/3}}}\int_{y^{\large{4/3}}}^{1} \dd x\,\dd z\,\dd y \\[5mm] = &\ \int_{0}^{1}y^{2/3}\pars{1 - y^{4/3}}\,\dd y = \int_{0}^{1}\pars{y^{2/3} - y^{2}}\,\dd y = \left.\pars{{y^{5/3} \over 5/3} - {y^{3} \over 3}}\right\vert_{\ 0}^{\ 1} = {3 \over 5} - {1 \over 3} = \bbx{4 \over 15} \end{align}