All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
probability of X's (Posted on 2006-08-23) Difficulty: 2 of 5
Suppose X1, ..., Xn are independent, identically distributed, purely continuous random variables, find P(X1<...<Xn).

Is the result the same if the random variables are not purely continuous?

  Submitted by Bon    
Rating: 3.0000 (2 votes)
Solution: (Hide)
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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Puzzle Thoughts K Sengupta2023-07-21 00:06:26
Answers - No PeekingRichard2006-08-23 16:21:51
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (14)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information