Hello,
I would to know how to calculate the probability that X > Y when Y is greater than a given value p (X and Y are random variable following normal distribution). I think that it is equivalent to caculate Y - X > 0 with X and Y are normal discutribution truncated at p.
Thanks,
Pascal
Both X and Y are truncated normal.
My problem is the follow : I have two prices (X and Y) that follow a normal distribution and I would like to compare this prices (X > Y) above a given minimum price. In a more formal way, I think it could be expressed as P(X > Y | Y > p) where p is the minimum price.
hi,
maybe 1 -P(X<=Y| Y> p) could do the trick? This would be 1- P(p<X<Y) I guess.
regards
rogojel
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