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Math Help - Cauchy distribuation and maths help

  1. #1
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    brighton
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    Cauchy distribuation and maths help

    Hello i have this question that i am struggling to answer regarding changing variables, i really dont know how to do it

    A gun fires at random in the angular range
    −π/2 < θ < π/2 towards a wall a distance 'l' away. If 'y' is the coordinate along the wall,


    show that

    g(y)dy=(1/π) (1/(1 +(y/l)2)) (dy/l)










    This is the Cauchy distribution. Assuming that the mean should be zero from symmetry considerations, try to find the standard deviation; what problem do you have? Truncate the distribution at a distance |y| = L either side of the peak. Calculate the new normalisation constant, and find the standard deviation.
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  2. #2
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    Re: Cauchy distribuation and maths help

    id appreciate if someone could even point me in the right direction, im abit stumped...
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  3. #3
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    Re: Cauchy distribuation and maths help

    Quote Originally Posted by bobby84 View Post
    Hello i have this question that i am struggling to answer regarding changing variables, i really dont know how to do it

    A gun fires at random in the angular range
    −π/2 < θ < π/2 towards a wall a distance 'l' away. If 'y' is the coordinate along the wall,



    show that

    g(y)dy=(1/π) (1/(1 +(y/l)2)) (dy/l)










    This is the Cauchy distribution. Assuming that the mean should be zero from symmetry considerations, try to find the standard deviation; what problem do you have? Truncate the distribution at a distance |y| = L either side of the peak. Calculate the new normalisation constant, and find the standard deviation.
    I am not 100% sure what you are asking, but I think this is what you mean....



    Cauchy distribuation and maths help-capture.png

    From the diagram we see that

    y=I\tan(\theta)

    If we take the derivative we get

    dy=I\sec^2(\theta)d\theta

    We can rewrite the secant function interms of tangent using the pythagorean identity

    dy=I(1+\tan^2(\theta) d \theta

    Now substitute tangent out for y to get

    dy=I\left(1+\left(\frac{y}{I}\right)^2\right) d \theta

    Solving for d\theta gives

    d\theta=\frac{1}{1+\left(\frac{y}{I}\right)^2} \frac{dy}{I}

    So for this to be a distribution the integral over the entire real line must be 1. You can use this to find the constant.
    Thanks from bobby84
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  4. #4
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    Re: Cauchy distribuation and maths help

    im not sure if you read this bit...

    This is the Cauchy distribution. Assuming that the mean should be zero from symmetry considerations, try to find the standard deviation; what problem do you have? Truncate the distribution at a distance |y| = L either side of the peak. Calculate the new normalisation constant, and find the standard deviation.
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  5. #5
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    Re: Cauchy distribuation and maths help

    Quote Originally Posted by bobby84 View Post
    im not sure if you read this bit...

    This is the Cauchy distribution. Assuming that the mean should be zero from symmetry considerations, try to find the standard deviation; what problem do you have? Truncate the distribution at a distance |y| = L either side of the peak. Calculate the new normalisation constant, and find the standard deviation.
    Well if \mu = 0

    Then we need to calculate the integral

    \sigma^2=\frac{1}{\pi}\int_{-\infty}^{\infty}y^2\frac{1}{1+\left( \frac{y}{I}\right)^2}\frac{dy}{I}

    So now try to calculate this integral. What do you get?
    Thanks from bobby84
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  6. #6
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    Re: Cauchy distribuation and maths help

    hey i got there in the end actually,

    my final solution was....

    (sigma) = sqrt(4-(pie)/2(pie)) l

    cheers for the help
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