Proof that sum of the probabilities of variables being minimum of a set is 1 I'm trying to prove that the sum of the probabilities of variables being minimum of a set is 1.
Say S is a set of random variables, $ S = \{ X_i \} $.
The probability of $X_i$ being the minimum of the set is $p_i = P(X_i = min(S)) = \prod_{j \neq i} P(X_i < X_j)$. So, the probability that $X_i$ is smaller than all the other variables.
There is obviously only one minimum in the set, which is either one of the $X_i$, so $\sum p_i = 1$. Right?
But how to prove this? How to prove that $\sum_i \prod_{j \neq i} P(X_i < X_j) = 1$?
Many thanks,
Misel
 A: It is not necessarily true that the probabilities sum to $1$. Whether they do depend on how the variables relate to each other.
In the most extreme case, suppose all of the $X_i$s are equal almost certainly. Then all of the $p_i$s are $1$, and their sum is the number of variables.
In the particular case that the $X_i$s are independent and continuously distributed, the probability that two of them are equal is $0$, and then your sum of probabilities is $1$. But you can't make those assumptions if all you know is that you have a set of random variables.

And even in the best of cases, you won't have $P(X_i=\min S)=\prod_{j\ne i}(X_i<X_j)$. This assumes that the events on the right-hand side are independent, but they cannot be (except in degenerate cases). For example, if we have $4$ independent normal distributed variables, then your formula would say that $p_i = 1/8$ for all $i$, and if that were true the probability that there's any minimum at all would be at most $4/8=1/2$, which is absurd.
A: Although not necessary for this particular question, a slightly more elaborate approach to show your desired statement is to use the pdf (assuming the random variables you refer to are continuous) of the minimum order statistic (pls see http://en.wikipedia.org/wiki/Order_statistic#Order_statistics_sampled_from_a_uniform_distribution).
