Distribution function

1. Nov 6, 2009

Rane3

I've computed a distribution function f(m,v) by taking partials of P(X<m, Y<v) with respect to m, v. Suppose I wanted the distribution function for P(X-Y > a). Since I know f(m,v), can I use that to help me compute my new distribution function by taking partials? If so, how? I'm a little confused about this. Any good resources/references?

2. Nov 6, 2009

You mean you have a density function $$f(m,v)$$ - the (joint) distribution function
would be $$F(m,v) = P(X < m, Y < v)$$.

I'm not sure which of the following two items you want for your second question:

i) You want a specific calculation of $$P(X - Y > a)$$ for a given value of $$a$$. In this case you calculate this double integral

$$\iint_{\{X-Y > a\}} f(x,y) \, dx dy$$

ii) You want an expression for the distribution of the random variable $$Z = X - Y$$.
You can either work out it out as an integral:

$$P(Z \le z) = \iint_{\{X-Y \le z\}} f(x,y) \, dx dy$$

or you can do a transformation of variables approach.

3. Nov 6, 2009

Rane3

I am looking for the second description, although I just want the probability density. If I know that:
X>0
X-Y>Z
and I know f(x,y), how can I find the density for X-Y>Z by taking partial derivatives of the integral? I'm getting myself confused. Should it again be partials with respect to Y,X, like I used to find f(x,y) in the first place? It seems that when I setup my limits, there is no dependence on Y and that throws me off.
$$\int_{-\infty}^{X}\int_0^{X-z}f(x,y)dydx$$ Is this even the correct integral?