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Convolution integral

  1. Dec 6, 2006 #1
    Can someone explain convolution to me. I have read three different books and gone to office hours and am not getting the fundamentals.
     
  2. jcsd
  3. Dec 6, 2006 #2

    quasar987

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    In what context? Do you mean you don't understand some of the "applications"?
     
  4. Dec 6, 2006 #3
    I'm trying to understand in the context of probability distributions. What the convolution of the sum of two random variables represents.
     
  5. Dec 7, 2006 #4

    quasar987

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    Oh.

    I never really took time to ponder about this. The way it was presented to me was that the convolution appeared kind of coincidentally:

    We set out to find the density f of Z=X+Y by finding it's repartition function F and then differentiating it. So we proceed from definition

    [tex]F_{Z}(z)=P(X+Y<z)=\int_{-\infty}^{+\infty}\int_{-\infty}^{z-y}f_X(x)f)Y(y)dxdy=\int_{-\infty}^{+\infty}F_X(z-y)f_Y(y)dy[/tex]

    This is the convolution [itex]F_X[/itex] and [itex]f_Y[/itex]. The density of Z is found simply by differentiating [itex]F_Z[/itex] wrt z and it gives the convolution of [itex]f_X[/itex] and [itex]f_Y[/itex].


    There is probably a way to understand something from this and gain some insights about the relation btw the sum of two random variables.

    Let me know if you find something interesting.
     
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