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Math Help - Cauchy Distribution

  1. #1
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    Question Cauchy Distribution

    Can I get help on the following HW question:
    If X is a standard Cauchy Random Variable then 1/X is also a standard Cauchy Random Variable. How can it be proved?
    Is it correct to say that: P(1/X<x) = P(X>1/x) = 1 - P(X<1/x) for x>0?
    What about if x<0
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  2. #2
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    You seem to be on the right track. I would start with defining Y = \frac{1}{X}, and begin with F_{Y}(y) = P \left (Y \le y \right ) = P\left (\frac{1}{X} \le y \right ) = P \left (X \ge \frac{1}{y} \right ) = 1 - F_{X}(1 / y), then take the derivative and see if you recognize it.
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  3. #3
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    Question

    Thanks a lot. But what I am not clear about is that Cauchy Dist is from -infinity to +infinity. So is it correct that
    both when y is positive and when y is negative. I would think that if y<0
    then I should have P(Y<y) = P(1/X<y) = P(X<1/y) but then the derivative will
    be negative???
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  4. #4
    MHF Contributor matheagle's Avatar
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    Use f_Y(y)=f_X(x)\biggl|{dx\over dy}\biggr|

    ={1\over \pi(1+{1\over y^2})}\biggl|{-1\over y^2}\biggr|

    ={1\over \pi(y^2+1)}
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  5. #5
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    Quote Originally Posted by matheagle View Post
    Use f_Y(y)=f_X(x)\biggl|{dx\over dy}\biggr|

    ={1\over \pi(1+{1\over y^2})}\biggl|{-1\over y^2}\biggr|

    ={1\over \pi(y^2+1)}
    I'm not sure this works, since g(X) = 1/X is not monotone over the whole real line, and moreover is undefined for part of the support of X. The method DOES give the right answer, but I think if you are going to use it, you should handle to problem point X = 0. I think it's easier just to use work directly with the cdf.
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  6. #6
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    Quote Originally Posted by thomas_donald View Post
    Thanks a lot. But what I am not clear about is that Cauchy Dist is from -infinity to +infinity. So is it correct that
    both when y is positive and when y is negative. I would think that if y<0
    then I should have P(Y<y) = P(1/X<y) = P(X<1/y) but then the derivative will
    be negative???
    Taking the reciprocal reverses the ordering regardless of where you are on the real line; e.g. -4 < -3 and -1/4 > -1/3.
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  7. #7
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    Wink

    Thank you very much. Now I get it.
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  8. #8
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by theodds View Post
    I'm not sure this works, since g(X) = 1/X is not monotone over the whole real line, and moreover is undefined for part of the support of X. The method DOES give the right answer, but I think if you are going to use it, you should handle to problem point X = 0. I think it's easier just to use work directly with the cdf.

    I've never worried about a set of measure zero
    and 1/x is decreasing for both x<0 and x>o, so that part is fine
    This technique is just the derivative of the technique using F(x).
    If you worry about it, just do one case at a time, x<0 and then x>0.
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  9. #9
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    Smile

    Thank you very much for the help.
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