# finding the limits of integration for joint probability

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.

• Got something from the answer below?
– Did
Commented Aug 14, 2014 at 8:06

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.
• Thanks @Did, I checked the pdf and it should be $exp(−x_1−x_3)$, So total probability is 1. when you integrate x1 from x2-x3 to inf, x2 from x3 to inf and x3 from 0 to inf. Can you please comment again with this updated pdf ? Commented Jun 23, 2014 at 18:03