finding the limits of integration for joint probability

Question

I have three variables $x_1$, $x_2$ and $x_3$. Their joint dist. is $f(x_1,x_2,x_3)= \exp(-x_1-x_3)$, where limits of $x_3 = 0$ to $\infty$, $x_2 = x_3$ to $\infty$ and $x_1 = x_2-x_3$ to $\infty$. How do I set the limits of integration so that I can find the probability that $P(x_2>x_1>x_3)$ ? Please explain.

Answer 1

Once again, an illustration of the principle that one ought to include indicator functions in the densities, first to get true density functions, and second, to be able to manipulate them...

Here, using Iverson brackets, we are told that the density is the function $f$ such that, for every $(x_1,x_2,x_3)$, $$ f(x_1,x_2,x_3)=\mathrm e^{-x_1-x_3}\cdot[x_3\gt0,\,x_2\gt x_3,\,x_1\gt x_2-x_3]. $$ On the other hand, for every $(X_1,X_2,X_3)$ with density $f$, the probability $P(X_2\gt X_1\gt X_3)$ is, by definition, $$ p=\int\!\!\!\iint_{\mathbb R^3}[x_2\gt x_1\gt x_3]\,f(x_1,x_2,x_3)\,\mathrm dx_1\mathrm dx_2\mathrm dx_3. $$ These are only the definitions, now the task is to evaluate the triple integral $p$. The product of indicator functions reads $$ [x_2\gt x_1\gt x_3]\cdot[x_3\gt0,\,x_2\gt x_3,\,x_1\gt x_2-x_3]=[x_1+x_3\gt x_2\gt x_1\gt x_3\gt0], $$ hence $$ p=\int_0^\infty\mathrm e^{-x_3}\int_{x_3}^\infty\mathrm e^{-x_1}\int_{x_1}^{x_1+x_3}\mathrm dx_2\mathrm dx_1\mathrm dx_3=\int_0^\infty x_3\mathrm e^{-x_3}\int_{x_3}^\infty\mathrm e^{-x_1}\mathrm dx_1\mathrm dx_3, $$ that is, finally, $$ P(X_2\gt X_1\gt X_3)=\int_0^\infty x_3\mathrm e^{-2x_3}\mathrm dx_3=\frac14\int_0^\infty x\mathrm e^{-x}\mathrm dx=\frac14. $$ Nota: The density of $(X_1,X_2,X_3)$ corresponds to some specific representation in terms of independent random variables, namely, $$ (X_1,X_2,X_3)=(U+V,V+W,W), $$ where $(U,V,W)$ are i.i.d. standard exponential random variables, hence $$ p=P(V\lt W\lt U+V), $$ or, still equivalently, $p$ is the volume of the domain $$ \{(r,s,t)\in[0,1]^3\mid rs\lt t\lt r\}. $$ The value $p=\frac14$ is then immediate.