# probability of one random variable being greater than another

#### manzell

##### New Member
I took statistics many moons ago, and remember performing these calculations, but for the life of me I can't remember what this process is called, or how to do it.

I've got two means and standard devs; how do I determine what the probability that one of the variables will be greater than the other?

Thanks!

#### JohnM

##### TS Contributor
Compute the difference distribution and determine the probability that it has a value of < 0.

mean of difference distribution = mean1 - mean2

std dev of difference distribution = sqrt(var1 + var2)
note--> var = (std dev)^2

#### Paulo

##### New Member
Manzel,
I am sorry it´s been so long you asked. I believe you what you realy want to know is: Guiven 2 probab. disdtr. functions, what is the probability that one sample from the 1st will be greater than a sample from the other. If that is so, I would like to know it too. If someone read that, and know the answer or where to find it , please replay to both of us.

#### BGM

##### TS Contributor
Suppose you have two continuous random variables $$X, Y$$ with a joint pdf $$f_{X, Y}$$. Then

$$\Pr\{X > Y\} = \iint\limits_{\{(x,y):x > y\}} f_{X,Y}(x,y)dxdy$$

$$= \int_{-\infty}^{+\infty} \int_y^{+\infty} f_{X,Y}(x,y)dxdy$$

$$= \int_{-\infty}^{+\infty} \int_{-\infty}^x f_{X,Y}(x,y)dydx$$