Determine the distribution function of a random variable












-1












$begingroup$


I am a bit lost with the following exercise, I would really appreciate your help:



$$
f_X,_Y(x,y)
begin{cases}
c & x^2 + y^2leq 4\
0 & otherwise
end{cases}
$$



What is the distribution function of :



$$
theta=
begin{cases}
arctan frac{X}{Y} & X ne 0\
0 & X=0
end{cases}
$$



Well, I don't even have a clue of how to start with this one.
Thanks










share|cite|improve this question









$endgroup$

















    -1












    $begingroup$


    I am a bit lost with the following exercise, I would really appreciate your help:



    $$
    f_X,_Y(x,y)
    begin{cases}
    c & x^2 + y^2leq 4\
    0 & otherwise
    end{cases}
    $$



    What is the distribution function of :



    $$
    theta=
    begin{cases}
    arctan frac{X}{Y} & X ne 0\
    0 & X=0
    end{cases}
    $$



    Well, I don't even have a clue of how to start with this one.
    Thanks










    share|cite|improve this question









    $endgroup$















      -1












      -1








      -1


      0



      $begingroup$


      I am a bit lost with the following exercise, I would really appreciate your help:



      $$
      f_X,_Y(x,y)
      begin{cases}
      c & x^2 + y^2leq 4\
      0 & otherwise
      end{cases}
      $$



      What is the distribution function of :



      $$
      theta=
      begin{cases}
      arctan frac{X}{Y} & X ne 0\
      0 & X=0
      end{cases}
      $$



      Well, I don't even have a clue of how to start with this one.
      Thanks










      share|cite|improve this question









      $endgroup$




      I am a bit lost with the following exercise, I would really appreciate your help:



      $$
      f_X,_Y(x,y)
      begin{cases}
      c & x^2 + y^2leq 4\
      0 & otherwise
      end{cases}
      $$



      What is the distribution function of :



      $$
      theta=
      begin{cases}
      arctan frac{X}{Y} & X ne 0\
      0 & X=0
      end{cases}
      $$



      Well, I don't even have a clue of how to start with this one.
      Thanks







      probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 15 at 14:54









      superuser123superuser123

      47628




      47628






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          First note that since $$int f_{XY}(x,y)dxdy=1$$we obtain $$c={1over 4pi}$$and since the PDF is a circle (more precisely a cylinder) and is maintained under rotation then the distribution of $theta$ is uniform, i.e.$$thetasim Uleft(-{piover 2},{piover 2}right)$$according to standard definition of $arctan$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
            $endgroup$
            – superuser123
            Jan 15 at 15:21












          • $begingroup$
            Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
            $endgroup$
            – Mostafa Ayaz
            Jan 16 at 12:02











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          1 Answer
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          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          First note that since $$int f_{XY}(x,y)dxdy=1$$we obtain $$c={1over 4pi}$$and since the PDF is a circle (more precisely a cylinder) and is maintained under rotation then the distribution of $theta$ is uniform, i.e.$$thetasim Uleft(-{piover 2},{piover 2}right)$$according to standard definition of $arctan$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
            $endgroup$
            – superuser123
            Jan 15 at 15:21












          • $begingroup$
            Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
            $endgroup$
            – Mostafa Ayaz
            Jan 16 at 12:02
















          1












          $begingroup$

          First note that since $$int f_{XY}(x,y)dxdy=1$$we obtain $$c={1over 4pi}$$and since the PDF is a circle (more precisely a cylinder) and is maintained under rotation then the distribution of $theta$ is uniform, i.e.$$thetasim Uleft(-{piover 2},{piover 2}right)$$according to standard definition of $arctan$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
            $endgroup$
            – superuser123
            Jan 15 at 15:21












          • $begingroup$
            Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
            $endgroup$
            – Mostafa Ayaz
            Jan 16 at 12:02














          1












          1








          1





          $begingroup$

          First note that since $$int f_{XY}(x,y)dxdy=1$$we obtain $$c={1over 4pi}$$and since the PDF is a circle (more precisely a cylinder) and is maintained under rotation then the distribution of $theta$ is uniform, i.e.$$thetasim Uleft(-{piover 2},{piover 2}right)$$according to standard definition of $arctan$.






          share|cite|improve this answer









          $endgroup$



          First note that since $$int f_{XY}(x,y)dxdy=1$$we obtain $$c={1over 4pi}$$and since the PDF is a circle (more precisely a cylinder) and is maintained under rotation then the distribution of $theta$ is uniform, i.e.$$thetasim Uleft(-{piover 2},{piover 2}right)$$according to standard definition of $arctan$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 15 at 15:14









          Mostafa AyazMostafa Ayaz

          15.6k3939




          15.6k3939












          • $begingroup$
            could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
            $endgroup$
            – superuser123
            Jan 15 at 15:21












          • $begingroup$
            Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
            $endgroup$
            – Mostafa Ayaz
            Jan 16 at 12:02


















          • $begingroup$
            could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
            $endgroup$
            – superuser123
            Jan 15 at 15:21












          • $begingroup$
            Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
            $endgroup$
            – Mostafa Ayaz
            Jan 16 at 12:02
















          $begingroup$
          could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
          $endgroup$
          – superuser123
          Jan 15 at 15:21






          $begingroup$
          could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
          $endgroup$
          – superuser123
          Jan 15 at 15:21














          $begingroup$
          Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
          $endgroup$
          – Mostafa Ayaz
          Jan 16 at 12:02




          $begingroup$
          Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
          $endgroup$
          – Mostafa Ayaz
          Jan 16 at 12:02


















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