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Transformation Of Probability Density Functions

  1. Jan 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Let X and Y be random variables. The pdfs are [itex]f_X(x)=2(1-x)[/itex] and [itex]f_Y(y) = 2(1-y)[/itex]. Both distributions are defined on [0,1].

    Let Z = X + Y. Find the pdf for Z, [itex]f_Z(z)[/itex].


    2. Relevant equations
    I'm using ideas, not equations.


    3. The attempt at a solution
    I'm dying of curiosity about where I'm going wrong. I'm so sure of each step, but my answer can't be correct because [itex]\int_0^2 f_Z(z)\,dz[/itex] is zero! Here's my logic.

    Consider the cdf (cumulative distribution function) for Z:

    [tex]
    F_Z(z) = P(Z\le z) = P(X+Y \le z)
    [/tex]

    Here, [itex]F_Z(z)[/itex] is the volume above the triangle shown in the image I attached to this message (in case something happens to the attachment, it's the triangle in quadrant 1 bounded by x=0, y=0 and x+y=z.)

    The volume above the shaded region represents [itex]F_Z(z)[/itex].

    [tex]
    F_Z(z) = \int_{x=0}^{x=z} 2(1-x)\int_{y=0}^{y=z-x} 2(1-y)\,dy\,dx
    [/tex]

    Performing the integrals gives [itex]F_Z(z) = \frac{1}{6}z^4 - \frac{4}{3}z^3 + 2z^2[/itex]. Then taking the derivative of the cdf gives the pdf:

    [tex]
    f_Z(z) = \partial_z F_Z(z) = \frac{2}{3}z^3 - 4z^2 +4z
    [/tex]

    Unfortunately, this can't be right because the integral of this function over [0,2] gives zero.

    I also would've expected that the maximum of [itex]f_Z(z)[/itex] would be at z=0 since individually, X and Y are most likely to be zero.

    I checked my algebra and calculus with Mathematica; it looked fine. I think there's a conceptual problem. There must be something I don't understand or some point I'm not clear about.

    What did I do wrong?
     

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    Last edited by a moderator: Jan 20, 2007
  2. jcsd
  3. Jan 21, 2007 #2

    quasar987

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    You have to split the problem into different cases.

    i) z<0
    ii) z<1
    iii) 1<z<2
    iv) z>2

    In each case the area of integration is different. But for each case, your model for the integral is

    [tex]F_Z(z)=\int\int_{\{(x,y)\in [0,1]\times[0,1]:\ y\leq z-x, \}}f_{XY}(x,y)dxdy[/tex]

    So you're integrating over the area that's the intersection of the square [0,1] x [0,1] with the area under the curve y=z-x.


    (Btw, you never said that X and Y are independant but I assume they are?)
     
    Last edited: Jan 21, 2007
  4. Jan 21, 2007 #3
    Ahhh.... yes. I *completely* understand.

    We're integrating that portion of the unit square (in QI with lower left corner at origin) that lies underneath x+y=1.

    So when 0<z<1, there's only one function that represents the "top": the line x+y=z.

    And when 1<z<2, there's two functions: y=1 and x+y=z, and therefore we need to break the region of integration into two portions.

    Got it. Thanks.

    And, yes. :-)
     
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