Polar form of normal random vector , angle and length are independent ,and angle is spherical distribution











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Represent $g sim N(0,I_n)$in polar form as $g=r theta$ where $r = |g|_2$ is the length and $theta = frac{g}{|g|_2} $ is the direction



prove that $r$ and $theta$ are independent ?



prove that $theta$ is uniformly distributed on sphere $S^{n-1}$



for first one :
The only things I know how to do is to show the product pdf of both of $theta$ and $r$ is same as n-dimeinal pdf of of standard Gaussian vector ? but how to find pdf of $|g|_2$? I have some problem in trasformation of random variable in this case . since the transformationare not bijective , is any simple way to do that?










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    up vote
    1
    down vote

    favorite












    Represent $g sim N(0,I_n)$in polar form as $g=r theta$ where $r = |g|_2$ is the length and $theta = frac{g}{|g|_2} $ is the direction



    prove that $r$ and $theta$ are independent ?



    prove that $theta$ is uniformly distributed on sphere $S^{n-1}$



    for first one :
    The only things I know how to do is to show the product pdf of both of $theta$ and $r$ is same as n-dimeinal pdf of of standard Gaussian vector ? but how to find pdf of $|g|_2$? I have some problem in trasformation of random variable in this case . since the transformationare not bijective , is any simple way to do that?










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Represent $g sim N(0,I_n)$in polar form as $g=r theta$ where $r = |g|_2$ is the length and $theta = frac{g}{|g|_2} $ is the direction



      prove that $r$ and $theta$ are independent ?



      prove that $theta$ is uniformly distributed on sphere $S^{n-1}$



      for first one :
      The only things I know how to do is to show the product pdf of both of $theta$ and $r$ is same as n-dimeinal pdf of of standard Gaussian vector ? but how to find pdf of $|g|_2$? I have some problem in trasformation of random variable in this case . since the transformationare not bijective , is any simple way to do that?










      share|cite|improve this question















      Represent $g sim N(0,I_n)$in polar form as $g=r theta$ where $r = |g|_2$ is the length and $theta = frac{g}{|g|_2} $ is the direction



      prove that $r$ and $theta$ are independent ?



      prove that $theta$ is uniformly distributed on sphere $S^{n-1}$



      for first one :
      The only things I know how to do is to show the product pdf of both of $theta$ and $r$ is same as n-dimeinal pdf of of standard Gaussian vector ? but how to find pdf of $|g|_2$? I have some problem in trasformation of random variable in this case . since the transformationare not bijective , is any simple way to do that?







      probability probability-theory random-variables normal-distribution independence






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      edited Nov 20 at 2:33

























      asked Nov 20 at 2:06









      ShaoyuPei

      1307




      1307






















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          Big hint: The joint density of $G = (X, Y)$ is $f(x,y) , dx , dy = frac{1}{2pi} e^{-(x^2+y^2)/2} , dx , dy$. By performing a change of variables to polar coordinates $R^2 = X^2 + Y^2$ and $Theta = arctan(Y/X)$, we have $$f(r, theta) , dr , dtheta = frac{1}{2pi} e^{-r^2/2} r , dr , dtheta,$$
          where the extra factor of $r$ comes from the Jacobian when performing the change of variables.



          Hint: If you know the joint density can be written as $f(r, theta) = g(r) h(theta)$ for some densities $g$ and $h$, then $R$ and $Theta$ are independent.






          share|cite|improve this answer





















          • actually I have already seen that , but I think we still need to know the density of θ and r
            – ShaoyuPei
            Nov 20 at 2:38










          • @ShaoyuPei But the above argument already shows you that the two densities are $frac{1}{2 pi c}$ and $c e^{-r^2/2} r$ for some constant $c$. (Now you just need to find what $c$ is.)
            – angryavian
            2 days ago













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          1 Answer
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          active

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          up vote
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          down vote













          Big hint: The joint density of $G = (X, Y)$ is $f(x,y) , dx , dy = frac{1}{2pi} e^{-(x^2+y^2)/2} , dx , dy$. By performing a change of variables to polar coordinates $R^2 = X^2 + Y^2$ and $Theta = arctan(Y/X)$, we have $$f(r, theta) , dr , dtheta = frac{1}{2pi} e^{-r^2/2} r , dr , dtheta,$$
          where the extra factor of $r$ comes from the Jacobian when performing the change of variables.



          Hint: If you know the joint density can be written as $f(r, theta) = g(r) h(theta)$ for some densities $g$ and $h$, then $R$ and $Theta$ are independent.






          share|cite|improve this answer





















          • actually I have already seen that , but I think we still need to know the density of θ and r
            – ShaoyuPei
            Nov 20 at 2:38










          • @ShaoyuPei But the above argument already shows you that the two densities are $frac{1}{2 pi c}$ and $c e^{-r^2/2} r$ for some constant $c$. (Now you just need to find what $c$ is.)
            – angryavian
            2 days ago

















          up vote
          0
          down vote













          Big hint: The joint density of $G = (X, Y)$ is $f(x,y) , dx , dy = frac{1}{2pi} e^{-(x^2+y^2)/2} , dx , dy$. By performing a change of variables to polar coordinates $R^2 = X^2 + Y^2$ and $Theta = arctan(Y/X)$, we have $$f(r, theta) , dr , dtheta = frac{1}{2pi} e^{-r^2/2} r , dr , dtheta,$$
          where the extra factor of $r$ comes from the Jacobian when performing the change of variables.



          Hint: If you know the joint density can be written as $f(r, theta) = g(r) h(theta)$ for some densities $g$ and $h$, then $R$ and $Theta$ are independent.






          share|cite|improve this answer





















          • actually I have already seen that , but I think we still need to know the density of θ and r
            – ShaoyuPei
            Nov 20 at 2:38










          • @ShaoyuPei But the above argument already shows you that the two densities are $frac{1}{2 pi c}$ and $c e^{-r^2/2} r$ for some constant $c$. (Now you just need to find what $c$ is.)
            – angryavian
            2 days ago















          up vote
          0
          down vote










          up vote
          0
          down vote









          Big hint: The joint density of $G = (X, Y)$ is $f(x,y) , dx , dy = frac{1}{2pi} e^{-(x^2+y^2)/2} , dx , dy$. By performing a change of variables to polar coordinates $R^2 = X^2 + Y^2$ and $Theta = arctan(Y/X)$, we have $$f(r, theta) , dr , dtheta = frac{1}{2pi} e^{-r^2/2} r , dr , dtheta,$$
          where the extra factor of $r$ comes from the Jacobian when performing the change of variables.



          Hint: If you know the joint density can be written as $f(r, theta) = g(r) h(theta)$ for some densities $g$ and $h$, then $R$ and $Theta$ are independent.






          share|cite|improve this answer












          Big hint: The joint density of $G = (X, Y)$ is $f(x,y) , dx , dy = frac{1}{2pi} e^{-(x^2+y^2)/2} , dx , dy$. By performing a change of variables to polar coordinates $R^2 = X^2 + Y^2$ and $Theta = arctan(Y/X)$, we have $$f(r, theta) , dr , dtheta = frac{1}{2pi} e^{-r^2/2} r , dr , dtheta,$$
          where the extra factor of $r$ comes from the Jacobian when performing the change of variables.



          Hint: If you know the joint density can be written as $f(r, theta) = g(r) h(theta)$ for some densities $g$ and $h$, then $R$ and $Theta$ are independent.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 20 at 2:19









          angryavian

          37k13178




          37k13178












          • actually I have already seen that , but I think we still need to know the density of θ and r
            – ShaoyuPei
            Nov 20 at 2:38










          • @ShaoyuPei But the above argument already shows you that the two densities are $frac{1}{2 pi c}$ and $c e^{-r^2/2} r$ for some constant $c$. (Now you just need to find what $c$ is.)
            – angryavian
            2 days ago




















          • actually I have already seen that , but I think we still need to know the density of θ and r
            – ShaoyuPei
            Nov 20 at 2:38










          • @ShaoyuPei But the above argument already shows you that the two densities are $frac{1}{2 pi c}$ and $c e^{-r^2/2} r$ for some constant $c$. (Now you just need to find what $c$ is.)
            – angryavian
            2 days ago


















          actually I have already seen that , but I think we still need to know the density of θ and r
          – ShaoyuPei
          Nov 20 at 2:38




          actually I have already seen that , but I think we still need to know the density of θ and r
          – ShaoyuPei
          Nov 20 at 2:38












          @ShaoyuPei But the above argument already shows you that the two densities are $frac{1}{2 pi c}$ and $c e^{-r^2/2} r$ for some constant $c$. (Now you just need to find what $c$ is.)
          – angryavian
          2 days ago






          @ShaoyuPei But the above argument already shows you that the two densities are $frac{1}{2 pi c}$ and $c e^{-r^2/2} r$ for some constant $c$. (Now you just need to find what $c$ is.)
          – angryavian
          2 days ago




















           

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