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Math Help - Find PDF

  1. #1
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    Find PDF

    Start at a point (0,1) in a rectangular coordinate system, choose an angle \displaystyle\theta\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right] and draw the line to cross the x-axis at X. The range for X is (-\infty,\infty). Find the pdf of X.

    I picked the angle \displaystyle\frac{\pi}{4} and now have no idea what to do.
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  2. #2
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    What an odd question. Anyway, you do not seem to have a useful idea. Let's see if I can make some sense of it...

    I believe the intent is to choose ANY angle, not a specific one, but ALL of them in the given range. Perhaps examples would be useful.

    P(X<0) = 1/2 and this corresponds to \theta = \frac{\pi}{2}

    P(X<-1) = 1/4 and this corresponds to \theta = \frac{\pi}{4}

    P(X<1) = 3/4 and this corresponds to \theta = -\frac{\pi}{4}

    Essentially, what I am trying to define is a mapping from a uniform distribution on [-pi/2,pi/2] to the entire x-axis. Sounds like it might be related to a tangent, perhaps.
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  3. #3
    Senior Member Sambit's Avatar
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    Quote Originally Posted by dwsmith View Post
    Start at a point (0,1) in a rectangular coordinate system, choose an angle \displaystyle\theta\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right] and draw the line to cross the x-axis at X. The range for X is (-\infty,\infty). Find the pdf of X.

    I picked the angle \displaystyle\frac{\pi}{4} and now have no idea what to do.


    It reduces to the pdf of a standard Cauchy distribution. Angle OQP is \theta in the figure and OP=x.

    The pdf of \theta can be written as: f(\theta)=\frac{1}{\pi}, when -\frac{\pi}{2}\leq x\leq \frac{\pi}{2}.

    Also, -\infty<x<\infty. Then the cdf of X can be given by:

    G(x)=P(X\leq x)

    =P(\frac{OP}{OQ}\leq x)

    =P(\tan\theta\leq x)

    =P(\theta\leq \tan^{-1}x)

    F(\tan^{-1}x) , -\infty<x<\infty, where F is the cdf of \theta

    So the pdf of X is given by:

    g(x)=\frac{d}{dx}G(x)

    =f(\tan^{-1}x).\frac{d}{dx}\tan^{-1}x, where f is the pdf of \theta

    =f(\tan^{-1}x).\frac{1}{1+x^2}

    = f(\theta)\frac{1}{1+x^2}

    = \frac{1}{\pi}\frac{1}{1+x^2} which is a Standard Cauchy pdf.
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