# Find $P(X_1 < X_2 < X_3 < X_4)$

The annual rainfall figures in Bandrika are independent identically distributed continuous random variables $$\{ X_r : r \geq 1\}$$. Compute $$P(X_1

## Try

My books gives $$\frac{1}{24}$$ as answer with no explanation. I started by considering the case with two random varaibles say $$P(X < Y )$$. Since $$f_X(x) = f_Y(y) = f$$, then

$$P(X

since the domain $$(x,y) : x > \infty$$ is half plane and the domain of integration is the entire plane then we must get half since the integral of entire plane is $$1$$. Now, for three variables

$$P(X

So, it seems like a patter and I would say that for the case $$P(X_1. Why is the answer key different? What is my mistake here?

• If they are i.i.d, then any order is as probable as any other...there are $4!$ orders.
– lulu
Nov 14, 2018 at 11:40
• How can this give 1/4! Nov 14, 2018 at 11:50
• I gave you a complete argument. Note: your computation doesn't make sense...$\int_x^{\infty} f(y)dy$ is obviously a function of $x$. It can't be $\frac 12$.
– lulu
Nov 14, 2018 at 11:52
• Note, one technical detail: I am assuming that the distributions are continuous. I believe you need that assumption (or more information). Assuming continuity, we can ensure that, for instance $P(X_1=X_2)=0$. That wouldn't be true for discrete distributions.
– lulu
Nov 14, 2018 at 12:00

Since regardless of the outcome the random variables can always be put into ascending order and we can neglect the cases were random variables coincide (these events have probability $$0$$) we can easily deduct that:
$$1=\underbrace{P(X_1
since all terms have to have the same probability, it has to be $$\frac{1}{4!}$$.