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Uniform Probability over Real Line?

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a two-ended laser spinner; that is a pen-like laser acting as the arrow mounted on a pin at the center of a spinner. Suppose the center of the disk is one meter away from a wall of infinite extent marked with a linear scale, with zero at the point closest the center of the spinner and negative numbers to left, positive to the right.

    The laser is spun and comes to a rest projecting for one of its ends at a point Y on the
    scale (with probability zero the laser will stop parallel to the wall and miss it; we ignore
    that possibility). Suppose that the angle X the laser makes to the perpendicular to the
    wall is uniformly distributed over –π/2 to π/2. Find the probability density of Y.

    2. Relevant equations



    3. The attempt at a solution

    The angle at which the laser lands must have PDF [itex]f_X(x)=1/{\pi}[/itex]

    That is, it is the uniform distribution over pi radians.

    Next, I set up a right triangle and find that the angle opposite the the 1 meter side is equal to [itex]{\pi}/2-X[/itex] and that the angle opposite the base (call it B) is simply X (given by the PDF above). Using the law of sines (and that the distance from the "infinite wall" to the spinner is 1 meter) to find:

    [itex]B=sinx/sin({\pi}/2-X)[/itex]

    It seems clear to me that Y is then the uniform distribution over [itex]lim B [/itex] as [itex] x->{\pi}/2 + - (lim B [/itex] as [itex] x->-{\pi}/2)[/itex]

    This is the entire real number line and so the PDF of Y does not exist.

    Is this right (even if my steps are less than rigorous)? Any help would be appreciated.
     
  2. jcsd
  3. Oct 16, 2013 #2

    LCKurtz

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    Let the origin of the Y axis along the wall be the foot of the 1 meter perpendicular side. Then for a point Y on that axis you have ##\frac Y 1 =\tan X##. Try starting with that.
     
  4. Oct 16, 2013 #3
    So much simpler! My trig is too rusty.

    Ok, so then I find that g(Y)=arctan(Y) and, after taking the derivative, I make the transformation assuming the uniform distribution described before for X.

    I get [itex]f_Y(y)=1/{\pi}(1+y^2)[/itex]

    Is it that simple? Thanks!
     
    Last edited: Oct 16, 2013
  5. Oct 17, 2013 #4
    Confirmed this elsewhere. It is indeed that simple. Appreciate your help!
     
  6. Oct 17, 2013 #5

    LCKurtz

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    Yes, it is. Sorry I couldn't get back to you yesterday. Too busy.
     
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