CDF and Inverse CDF of Wrapped Cauchy Distribution

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The standard wrapped-up Cauchy distribution has the following probability density function:



$$f(x,p)= frac1-p^22pi(1+p^2)-2pcos(x)$$



where x is from $0$ to $2pi$



Can anybody know, what is the CDF and InvCDF of this distribution?



Can you please also suggest any other wrapped distribution?










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    The standard wrapped-up Cauchy distribution has the following probability density function:



    $$f(x,p)= frac1-p^22pi(1+p^2)-2pcos(x)$$



    where x is from $0$ to $2pi$



    Can anybody know, what is the CDF and InvCDF of this distribution?



    Can you please also suggest any other wrapped distribution?










    share|cite|improve this question

























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      The standard wrapped-up Cauchy distribution has the following probability density function:



      $$f(x,p)= frac1-p^22pi(1+p^2)-2pcos(x)$$



      where x is from $0$ to $2pi$



      Can anybody know, what is the CDF and InvCDF of this distribution?



      Can you please also suggest any other wrapped distribution?










      share|cite|improve this question















      The standard wrapped-up Cauchy distribution has the following probability density function:



      $$f(x,p)= frac1-p^22pi(1+p^2)-2pcos(x)$$



      where x is from $0$ to $2pi$



      Can anybody know, what is the CDF and InvCDF of this distribution?



      Can you please also suggest any other wrapped distribution?







      probability-distributions






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      edited Sep 10 at 14:20









      gammatester

      16.1k21529




      16.1k21529










      asked Sep 10 at 10:42









      Wasim Shaikh

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          Wolfram Alpha nows the antiderivate of your PDF. Simply enter
          int((1-p^2)/(2*Pi*(1+p^2)-2*p*cos(x)), x) and get a complicated looking expession.
          $$frac(p^2 - 1) tanh^-1left(frac(π p^2 + p + pi) tan(x/2)sqrtp^2 - π^2 (p^2 + 1)^2right)sqrtp^2 - pi^2 (p^2 + 1)^2 + c$$



          But the dependence on $x$ is in a single $tan(x/2)$ term, so it can be easily inverted.






          share|cite|improve this answer




















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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

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            up vote
            0
            down vote













            Wolfram Alpha nows the antiderivate of your PDF. Simply enter
            int((1-p^2)/(2*Pi*(1+p^2)-2*p*cos(x)), x) and get a complicated looking expession.
            $$frac(p^2 - 1) tanh^-1left(frac(π p^2 + p + pi) tan(x/2)sqrtp^2 - π^2 (p^2 + 1)^2right)sqrtp^2 - pi^2 (p^2 + 1)^2 + c$$



            But the dependence on $x$ is in a single $tan(x/2)$ term, so it can be easily inverted.






            share|cite|improve this answer
























              up vote
              0
              down vote













              Wolfram Alpha nows the antiderivate of your PDF. Simply enter
              int((1-p^2)/(2*Pi*(1+p^2)-2*p*cos(x)), x) and get a complicated looking expession.
              $$frac(p^2 - 1) tanh^-1left(frac(π p^2 + p + pi) tan(x/2)sqrtp^2 - π^2 (p^2 + 1)^2right)sqrtp^2 - pi^2 (p^2 + 1)^2 + c$$



              But the dependence on $x$ is in a single $tan(x/2)$ term, so it can be easily inverted.






              share|cite|improve this answer






















                up vote
                0
                down vote










                up vote
                0
                down vote









                Wolfram Alpha nows the antiderivate of your PDF. Simply enter
                int((1-p^2)/(2*Pi*(1+p^2)-2*p*cos(x)), x) and get a complicated looking expession.
                $$frac(p^2 - 1) tanh^-1left(frac(π p^2 + p + pi) tan(x/2)sqrtp^2 - π^2 (p^2 + 1)^2right)sqrtp^2 - pi^2 (p^2 + 1)^2 + c$$



                But the dependence on $x$ is in a single $tan(x/2)$ term, so it can be easily inverted.






                share|cite|improve this answer












                Wolfram Alpha nows the antiderivate of your PDF. Simply enter
                int((1-p^2)/(2*Pi*(1+p^2)-2*p*cos(x)), x) and get a complicated looking expession.
                $$frac(p^2 - 1) tanh^-1left(frac(π p^2 + p + pi) tan(x/2)sqrtp^2 - π^2 (p^2 + 1)^2right)sqrtp^2 - pi^2 (p^2 + 1)^2 + c$$



                But the dependence on $x$ is in a single $tan(x/2)$ term, so it can be easily inverted.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 10 at 14:20









                gammatester

                16.1k21529




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