The annual rainfall figures in Bandrika are independent identically
distributed continuous random variables $\{ X_r : r \geq 1\}$. Compute
$$P(X_1<X_2<X_3<X_4)$$
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
$$ P(X<Y) = \int\limits_{- \infty}^{\infty} \int\limits_{x}^{\infty} f^2 dy dx = \int\limits_x^{\infty} f(y) dy = \frac{1}{2}$$
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?
Best Answer
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 $\frac{1}{4!}$.