Mixing
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Polar form of normal random vector , angle and length are independent ,and angle is spherical distribution
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
asked Nov 20 '18 at 2:06
ShaoyuPei
1698
add a comment |
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
asked Nov 20 '18 at 2:06
ShaoyuPei
1698
add a comment |
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
asked Nov 20 '18 at 2:06
ShaoyuPei
1698
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
probability probability-theory random-variables normal-distribution independence
asked Nov 20 '18 at 2:06
ShaoyuPei
1698
asked Nov 20 '18 at 2:06
ShaoyuPei
1698
edited Nov 20 '18 at 2:33
asked Nov 20 '18 at 2:06
ShaoyuPei
1698
asked Nov 20 '18 at 2:06
ShaoyuPei
1698
asked Nov 20 '18 at 2:06
ShaoyuPei
1698
1698
add a comment |
add a comment |
1 Answer
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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.
answered Nov 20 '18 at 2:19
angryavian
39k23180
actually I have already seen that , but I think we still need to know the density of θ and r
– ShaoyuPei
Nov 20 '18 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
Nov 20 '18 at 22:20
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
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.
answered Nov 20 '18 at 2:19
angryavian
39k23180
actually I have already seen that , but I think we still need to know the density of θ and r
– ShaoyuPei
Nov 20 '18 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
Nov 20 '18 at 22:20
add a comment |
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.
answered Nov 20 '18 at 2:19
angryavian
39k23180
actually I have already seen that , but I think we still need to know the density of θ and r
– ShaoyuPei
Nov 20 '18 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
Nov 20 '18 at 22:20
add a comment |
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.
answered Nov 20 '18 at 2:19
angryavian
39k23180
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.
answered Nov 20 '18 at 2:19
angryavian
39k23180
answered Nov 20 '18 at 2:19
angryavian
39k23180
answered Nov 20 '18 at 2:19
angryavian
39k23180
answered Nov 20 '18 at 2:19
angryavian
39k23180
39k23180
actually I have already seen that , but I think we still need to know the density of θ and r
– ShaoyuPei
Nov 20 '18 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
Nov 20 '18 at 22:20
add a comment |
actually I have already seen that , but I think we still need to know the density of θ and r
– ShaoyuPei
Nov 20 '18 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
Nov 20 '18 at 22:20
actually I have already seen that , but I think we still need to know the density of θ and r
– ShaoyuPei
Nov 20 '18 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 '18 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
Nov 20 '18 at 22:20
@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
Nov 20 '18 at 22:20
add a comment |
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