# comparison with normal distribution truncated

#### pascal34

##### New Member
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

#### BGM

##### TS Contributor
Do you mean $$Y$$ is truncated normal but $$X$$ is ordinary normal?

#### pascal34

##### New Member
Both X and Y are truncated normal.

#### pascal34

##### New Member
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.

#### Dason

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)

#### Dason

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?

#### rogojel

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

regards
rogojel

#### pascal34

##### New Member
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.