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Density function

  1. Nov 15, 2007 #1

    1. The problem statement, all variables and given/known data
    what's the density function for [tex]X-Y[/tex] if [tex]X[/tex] and [tex]Y[/tex] are independent and continously distributed on [tex][0,1][/tex]?
  2. jcsd
  3. Nov 15, 2007 #2


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    you must show some work before getting help; what have you tried?
  4. Nov 15, 2007 #3
    The answer of course depends on how X and Y are distributed. Are you given two specific distributions or do you want to calculate the answer in all its generality? See the Jacobi method to get a way of calculating the pdf of X-Y.
  5. Nov 15, 2007 #4
    ok... here's my try...

    [tex]X[/tex] has the uniform distribution [tex]f_X(x) = 1[/tex]. to get the distribution for [tex]Y[/tex]: [tex]F_{-Y}(y) = P(-Y \leq y) = P(Y \geq -y) = 1 - F_Y(-y) \Rightarrow f_Y(y) = f_Y(-y) = -1[/tex]

    the formula for convulsion in this case is [tex]f_Z(z) = \int_\infty^\infty f_X(z-y)f_Y(y)dy[/tex].

    combining this with [tex]f_Y(y)[/tex] leads to [tex]f_Z(z) = -\int_0^1 f_X(z-y)dy[/tex]

    the integrand is zero if the condition 0 <= z-y <= 1 (z-1 <= y <= z) isn't fulfilled.

    we get three cases:

    1. if 0 <= z <= 1: [tex]f_Z(z) = -\int^z_0 dy = -z[/tex]
    2. if 1 < z <= 2: [tex]f_Z(z) = -\int^1_{z-1} dy = z - 2[/tex]
    3. if z < 0 or z > 2: [tex]f_Z(z) = 0[/tex]

    it seems correct to me, but i'm not sure..
  6. Nov 16, 2007 #5
    here is my hint:

    1. first define X - Y as Z

    2. then graph X - Y <= Z on a graph

    3. find the limits of integration

    4. solve it and this gives the cumulative distribution function

    5. take the derivative of the CDF with respect to Z
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