The annual rainfall figures in Bandrika are independent identically distributed continuous random variables $\{ X_r : r \geq 1\}$. Compute $$P(X_1<X_2<X_3<X_4)$$


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

$$ P(X<Y) = \int\limits_{- \infty}^{\infty} \int\limits_{x}^{\infty} f^2 dy dx = \int\limits_x^{\infty} f(y) dy = \frac{1}{2}$$

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

$$ P(X<Y<Z) = \int\limits_{- \infty}^{\infty} \int\limits_{- \infty}^z \int\limits_x^{\infty} f^3 dydxdz = \int_{- \infty}^z \int_x^{\infty} f^2 dx dz = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} $$

So, it seems like a patter and I would say that for the case $P(X_1<X_2<X_3<X_4) = \frac{1}{2^3}= \frac{1}{8} $. Why is the answer key different? What is my mistake here?

  • 3
    $\begingroup$ If they are i.i.d, then any order is as probable as any other...there are $4!$ orders. $\endgroup$
    – lulu
    Nov 14, 2018 at 11:40
  • $\begingroup$ How can this give 1/4! $\endgroup$ Nov 14, 2018 at 11:50
  • $\begingroup$ 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 $\frac 12$. $\endgroup$
    – lulu
    Nov 14, 2018 at 11:52
  • 3
    $\begingroup$ 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. $\endgroup$
    – lulu
    Nov 14, 2018 at 12:00

1 Answer 1


As pointed out bu lulu in the comments you need an additional information on the distribution namely that it is continuous.

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:

$1=\underbrace{P(X_1<X_2<X_3<X_4)+P(X_1<X_2<X_4<X_3)+...}_{\text{all permutations of the four random variables } = 4! \text{ terms}}$

since all terms have to have the same probability, it has to be $\frac{1}{4!}$.

  • $\begingroup$ You should give a more detailed example how we can "deduct" that. i.e. what @lulul said in the other thread. $\endgroup$
    – Nearoo
    Nov 14, 2018 at 13:11

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