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Since X1, ..., Xn are iid, any ordering among them are equally likely. Also, since they are continuous, the probability of having equal signs in X1<X2<...<Xn is zero. Therefore,
P(X1<...<Xn) = 1/(number of all permutations of X1,..,Xn) = 1/n!
If X's are not continuous, then it is possible to have equal signs in the ordering with non-zero probability. So the argument breaks down and the answer would be wrong. One simple counter example is that X=0 with probability p and X=1 with prob 1-p. Then for n > 2, it is impossible to have X1<...<Xn, so prob = 0. And for n = 2, it is obvious that P(X1<X2) = P(X1=0,X2=1) = p(1-p), not quite 1/2! = 0.5 in the continuous case in general. |
