# Thread: comparison with normal distribution truncated

1. ## comparison with normal distribution truncated

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

2. ## Re: comparison with normal distribution truncated

Do you mean is truncated normal but is ordinary normal?

3. ## Re: comparison with normal distribution truncated

Both X and Y are truncated normal.

4. ## Re: comparison with normal distribution truncated

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.

5. ## Re: comparison with normal distribution truncated

Originally Posted by pascal34
Both X and Y are truncated normal.
Why is X truncated as well? I took your original statement to mean P(X > Y | Y > p)

6. ## Re: comparison with normal distribution truncated

Originally Posted by pascal34
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.
Just one of them being above the minimum price or both? Are the two random variables independent?

7. ## Re: comparison with normal distribution truncated

hi,
maybe 1 -P(X<=Y| Y> p) could do the trick? This would be 1- P(p<X<Y) I guess.

regards
rogojel

8. ## Re: comparison with normal distribution truncated

Originally Posted by Dason
Just one of them being above the minimum price or both? Are the two random variables independent?
Both are above the minimum price. Yes variables are independent.

Sorry for the lack of precision.

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