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Help on to find probability density function

  1. Jun 29, 2011 #1
    hey guys, i am really confused on something.here is the thing:
    i have;

    i=x+(x^2-y)^(1/2)

    and here x is uniform distribution on (a,b)
    y is uniform distribution on (c,d)
    x and y independent
    i need to find the probability density function of i but how??????
    actually i dont know how to start!!
     
  2. jcsd
  3. Jun 29, 2011 #2

    micromass

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    Hi musademirtas! :smile:

    First you will need to know the joint distribution of X and Y. This is easy because of independence:

    [tex]f_{X,Y}(x,y)=f_X(x)f_Y(y)[/tex]

    Now, to find the pdf, you will need to find the cdf first. That is, for each a, you will want to calculate

    [tex]P\{X+\sqrt{X^2-Y}\leq a\}=\iint_{\{(x,y)~\vert~x+\sqrt{x^2-y}\leq a\}}{f_{X,Y}(x,y)dxdy}[/tex]

    To evaluate this integral, you'll need to know the region [itex]\{(x,y)~\vert~x+\sqrt{x^2-y}\leq a\}[/itex] somewhat better.

    So I suggest you first find out for which tuples the equality holds. That is, for which x and y does it hold that

    [tex]x+\sqrt{x^2-y}=a[/tex]

    (hint: the answer will be a straight line!)
     
  4. Jul 2, 2011 #3
    hi micromass
    thanks for the help but can you solve it? because i used your help to solve it but i couldnt do it.
     
  5. Jul 3, 2011 #4

    micromass

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    Well, first you're going to need to write

    [tex]x+\sqrt{x^2-y}=a[/tex]

    in function of y. What do you get for that?
     
  6. Jul 3, 2011 #5
    ok.i got y=2ax-x^2 then what? how am i going to use this?
     
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