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Distribution function

  1. Nov 6, 2009 #1
    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. jcsd
  3. Nov 6, 2009 #2


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    Homework Helper

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

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

    i) You want a specific calculation of [tex] P(X - Y > a) [/tex] for a given value of [tex] a [/tex]. 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 [tex] Z = X - Y [/tex].
    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.
  4. Nov 6, 2009 #3
    I am looking for the second description, although I just want the probability density. If I know that:
    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.
    [tex]\int_{-\infty}^{X}\int_0^{X-z}f(x,y)dydx[/tex] Is this even the correct integral?
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