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Transformation of two dimensional random var

  1. Feb 16, 2012 #1
    given the following Probability density function:

    f x,y(x,y) = { 0.5, ( 0<y<1, 2y-x<2, 2y+x<2 }
    0, else

    and i need to find f z(z) while z=y-x

    i got really confused while trying to calculate the borders of x and y for the integration.

    i would be really thankful for someone explaining me how to come to the borders slow and step by step.

    i got to the following equations: y < 2-z , y < (2+z)/3 , x < 2-z, x < (2-2z)/3

    but i don't know how to resume from here. it said on the solution that borders are 0 to (2-2z)/3 for y and y-z to 2-2y for x. why?

    thanks in advance for any help
     
  2. jcsd
  3. Feb 24, 2012 #2
    Hi dannee.
    The inequalities (0<y<1, 2y-x<2, 2y+x<2) define a region A in the (x,y)-plane.
    You can follow these steps, in order to graphically "visualize" the problem:

    1) Draw the region A in the (x,y)-plane

    2) Introduce the two dimensional transformation u=x , z=y-x

    3) Obtain the density function g(u,z) of the two dimensional random variable (U,Z)

    4) From the inequalities between (x,y) derive the corresponding inequalities between (u,z).

    5) From the inequalities between (u,z) draw the region A' in the (u,z)-plane. A' is the image of A under the transformation introduced in step 2).

    6) Finally, from g(u,z) derive the marginal density function g(z)
     
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