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Homework Help: Continuous random variable - transformation using sin

  1. Feb 5, 2010 #1
    1. The problem statement, all variables and given/known data

    There is a pin of length 4 which appear on a photograph, and the length of the image observed is y, an observation on the random variable Y. The pin is at an angle x, 0[tex]\leq[/tex]x[tex]\leq[/tex][tex]\pi[/tex]/2, to the normal to the film, this is an observation on the r.v. X.

    1. If all angles X are equally likely then derive the distribution of Y.

    2. What is E(Y)?

    2. Relevant equations

    3. The attempt at a solution

    X is uniform so fX(x) = 2/[tex]\pi[/tex] , 0[tex]\leq[/tex]x[tex]\leq[/tex][tex]\pi[/tex]/2 and fX (x) = 0 otherwise.

    So FX (x) = 0, x<0
    = 2x/[tex]\pi[/tex], 0[tex]\leq[/tex]x[tex]\leq[/tex][tex]\pi[/tex]/2
    =1, x>[tex]\pi[/tex]/2


    FY(y)= P(Y[tex]\leq[/tex]y)
    =P(4sinX [tex]\leq[/tex]y)
    =P([tex]\pi[/tex] - arcsin(y/4) [tex]\leq[/tex] X [tex]\leq[/tex] 2[tex]\pi[/tex] - arcsin(y/4)) (*)
    = FX (2[tex]\pi[/tex] - arcsin(y/4)) - FX ([tex]\pi[/tex] - arcsin(y/4)
    = 2 + (2/[tex]\pi[/tex])arcsin(y/4)

    So fY(y) = (2/[tex]\pi[/tex])(1/4)(1/[tex]\sqrt{1-y2 /4}[/tex]) for 0[tex]\leq[/tex]y[tex]\leq[/tex]4 and 0 otherwise.

    Is this ok so far? I'm very unsure of the stage marked by (*).
    Last edited: Feb 5, 2010
  2. jcsd
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