It is a question 59 on page 87 from Ross's book (Introduction to Probability Models)

Let $X_1,X_2,X_3,X_4$ are independent continuous random variables with a common distribution function F and let

$p = P(X_1 < X_2 > X_3 < X_4)$

Just as the Title, what does it mean? Or similar questions with such an inequity?


Update 1:

The solution says: Use the fact that F(Xi) is a uniform (0,1) random variable to obtain. But where is this fact?

Update: A similar question

How can I compute an expression for $P(X_1>X_2>X_3>X_4)$ if $X_1,X_2,X_3,X_4$ are normal and mututally independent?


I am not a native-English speaker, and I am learning it by myself.

  • $\begingroup$ Usually, a "multiple" inequality is a conjucntion: $X_1 < X_2 > X_3$ is $X_1 < X_2$ and $X_2 > X_3$. $\endgroup$ Dec 27, 2017 at 13:41
  • $\begingroup$ Of course "$X_1 < X_2 > X_3 < X_4$" means "$X_1 < X_2$ and $X_2 > X_3$ and $X_3 < X_4$". $\endgroup$
    – GEdgar
    Dec 27, 2017 at 13:41
  • $\begingroup$ Or maybe a typo and is $\mathbb{P}(X_1<X_2<X_3<X_4)$ $\endgroup$ Dec 27, 2017 at 13:42
  • $\begingroup$ Are you sure that is the correct name of the book? Ross has a book called "Introduction to Probability Models" and there are many different editions. $\endgroup$
    – user9464
    Dec 27, 2017 at 13:45
  • $\begingroup$ @Jack Yes. 11th Edition $\endgroup$ Dec 27, 2017 at 13:49

2 Answers 2


It's the probability that $X_1$ is less than $X_2$ AND that $X_2$ is greater than $X_3$ AND that $X_3$ is less than $X_4$. That's all.


This seems to be a rather unconventional use of inequalities as GEdgar points out in his comment and there is no typo. The "chain" of inequalities $X_1<X_2>X_3<X_4$ means all of the following hold: $$ X_1<X_2,\ \ X_2>X_3,\ \ X_3<X_4. $$

Here is the original problem:

enter image description here

Here is the official solution by Ross:

enter image description here

Thanks to http://math2.uncc.edu/~imsonin/Ross_Probability10ed_Student_Solutions2010.pdf

  • $\begingroup$ Thanks. How do you know F(X_i) is a uniform distribution function? $\endgroup$ Dec 27, 2017 at 16:53
  • $\begingroup$ @evergreenhomeland: this is a very instructive exercise. Try it! $\endgroup$
    – user9464
    Dec 27, 2017 at 17:09

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