What does $p = P(X_1 < X_2 > X_3 < X_4)$ mean?

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?

Thanks

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?

BTW

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

• Usually, a "multiple" inequality is a conjucntion: $X_1 < X_2 > X_3$ is $X_1 < X_2$ and $X_2 > X_3$. – Mauro ALLEGRANZA Dec 27 '17 at 13:41
• Of course "$X_1 < X_2 > X_3 < X_4$" means "$X_1 < X_2$ and $X_2 > X_3$ and $X_3 < X_4$". – GEdgar Dec 27 '17 at 13:41
• Or maybe a typo and is $\mathbb{P}(X_1<X_2<X_3<X_4)$ – Martín Vacas Vignolo Dec 27 '17 at 13:42
• 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. – Jack Dec 27 '17 at 13:45
• @Jack Yes. 11th Edition – evergreenhomeland Dec 27 '17 at 13:49

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: Here is the official solution by Ross: Thanks to http://math2.uncc.edu/~imsonin/Ross_Probability10ed_Student_Solutions2010.pdf

• Thanks. How do you know F(X_i) is a uniform distribution function? – evergreenhomeland Dec 27 '17 at 16:53
• @evergreenhomeland: this is a very instructive exercise. Try it! – Jack Dec 27 '17 at 17:09

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.