Find $P(X_1<X_2<\cdots < X_n)$

$\displaystyle P(X_{1}<X_{2}<X_{3}<...<X_{8})$ is given by the integral:-

$\displaystyle \int_{0}^{\infty}\int_{X_{1}}^{\infty}\int_{X_{2}}^{\infty}.....\int_{X_{7}}^{\infty}\lambda^{n} e^{-\lambda\sum_{i=1}^{8}X_{i}}\,\,dX_{8}\,dX_{7}........dX_{3}\,dX_{2}\,dX_{1} = \frac{1}{8!}$

The integral is easy enough to calculate.

Otherwise we can approach it in a more general way.

Let $\displaystyle P(X_{1}<X_{2}<X_{3}<...<X_{n})= k$ .

The number of ways these random variables $X_{1},X_{2}...X_{n}$ can be arranged in ascending order is equivalent to the permutation of $n$ distinct objects = $n!$ . All of those cases will have equal probability $=k$ . Also these $n!$ events constitute the entire event space(except for the cases when two or more variables are equal which happens with a probability which in limit tends to 0 in p) . So summation of all the probabilities will be equal to $1$ .

Hence $\displaystyle \sum_{i=1}^{n!} k = k\cdot n! = 1 \implies k = \frac{1}{n!}$ . This is in general true for all distributions. For the case $n=8$ . The probability is $\displaystyle \frac{1}{8!}$

Lets consider the case when there is only two random variables.

$P(X_1<X_2)$ which can be easily shown to be $\frac{1}{2}$ . How about for the case when there is three random variables? $P(X_1<X_1<X_3) = \frac{1} {6}$

First, lets consider the complimentary case of $P(X_1 < X_2)$ The compliment would be $P(X_1> X_2)$ . This probability is also just $\frac{1}{2}$

This seems to suggest 2 things about our sample space.

1) Obviously when there are 2 Random variables : $P(X_1< X_2)$ and $P(X_1>X_2)$ span the entire sample space.

2) Though nontrivial, due to independence the $P(X_1 <X_2) = P(X_1>X_2)$ .

Could we generalize this to case with $n$ variables? Well, we could take the brute force approach and evaluate the general expression evaluated in the beginning of this post. However it is evident that some pattern exists among these random variables.

Let us now consider the case of $3$ random variables. Can we partition the sample space in such a way that makes the computation of $P(X_1<X_2<X_3)$ easier?

Well if we notice $P(X_1<X_2<X_3) = P(X_1<X_3<X_2) = P(X_2<X_3<X_1) = P(X_2<X_1<X_3) = P(X_3<X_2<X_1)= P(X_3<X_1<X_2)$ Then their sum is obviously equal to $1$ and therefore each individual event occurs with probability $\frac{1}{6}$

Finally, for the case where there are $n$ random variables there are $n!$ ways of choosing an ordering of $X_1..X_n$ thus the $P(X_1<X_2<..X_n) = \frac{1} {n!}$ .

The answer to the question is $\frac{1}{8!}$